electronic devices

and circuit theory

ROBERT L. BOYLESTAD LOUIS NASHELSKY
 SIGNIFICANT EQUATIONS

1 Semiconductor Diodes W = QV, 1 eV = 1.6 X 10- 19 J, In = ls (eVv/nVr - 1), VT= kT/q, TK = Tc + 273°,
k = 1.38 X 10-23 J/K, VK =0.7 V (Si), VK = 0.3 V(Ge), VK = l.2 V (GaAs), Rn = Vn/In, rd = 26 rnV/In, rav = .<l Vd/.Mdlpt. topt.,
Pn = Vnln,Tc = (ilVz/Vz)/(T1 - To) X 100%/°C

2 Diode Applications Silicon: VK =0.7 V, germanium.: VK =0.3 V, GaAs: VK = l.2 V; half-wave: Vctc = 0.318Vm;
full-wave: Vdc = 0.636Vm

3 Bipolar Junction Transistors le = le + IB, le = Icmajority + Icominority' le = le, VBE = 0.7 V, adc = le/le, le = ale + ICBo,
aac = Mc/ME, ICEo = ICBo/(l - a), f3ctc = lc/IB, /3ac = Mc/MB, a = /3/(/3 + 1), /3 = a/(1 - a), le = f3/B, IE = (/3 + 1)/B,
Pcmax. = VcEic

4 DC Biasing-BJTs In general: VBE = 0.7 V, le = le, le = f3IB; fixed-bias: IB = (Vcc - VBE) /RB, VCE = V cc - IcRc,
lcsat = Vee/Re; emitter-stabilized: IB = <Vee - VBE)/(RB + (/3 + l)RE), R; = (/3 + l)RE, VcE = Vee - lc(Rc + RE),
Icsat = Vcc/(Rc + RE); voltage-divider: exact: RTh = R1 IIR2, ETh = R2Vcc/(R1 + R2),IB = (ETh - VBE)/(RTh + (/3 + l)RE),
VCE = Vee - Ic(Rc + RE), approximate: f3RE ~ lOR2, VB= R2Vcc/(R1 + R2), VE= VB - VBE,lc = le= VE/RE; voltage-feedback:
IB = <Vee - VBE)/(RB + {3(Rc + RE)); cornrnon-base: IB = (VEE - VBE)/RE; switching transistors: t00 = t, + td, t0 ff = ts + t1;
stability: S(Ico) = Mc/Mc0 ; fixed-bias: S(Ico) = /3 + 1; emitter-bias: S(Ico) = (/3 + 1)(1 + RB/RE)/(l + /3 + RB/RE);
voltage-divider: S(/c0 ) = (/3 + 1)(1 + RTh/RE)/(1 + /3 + RTh/RE); feedback-bias: S(/c0 ) = (/3 + 1)(1 + RB/Rc)/(l + /3 + RB/Re),
S(VBE) = Mc/.<l VBE; fixed-bias: S(VBE) = -{3/RB; emitter-bias: S(VBE) = -{3/(RB + (/3 + l)RE); voltage-divider: S(VBE) =
-{3/(RTh + (/3 + l)RE); feedback bias: S(VBE) = -{3/(RB + (/3 + l)Rc), S(/3) = Mc/ilf3; fixed-bias: S(/3) = lcJ/3 1;
emitter-bias: S(/3) = IcP + RB/RE)/(/31(1 + /32 + RB/RE)); voltage-divider: S(/3) = IcP + RTo/RE)/(/31(1 + /32 + RTh!RE));
feedback-bias: S(/3) = IcP + RB/Rc)/(/31(1 + /32 + RB/Re)), Mc = S(/co) Meo + S(VBE) .<l VBE + S(/3) .<l/3

5 BJTACAnalysis r, = 26rnV/le;CEfixed-bias:Z; ={3r,,Z =Rc,Av = -Rc/r,;voltage-dividerbias:Z; = R1IIR2ll/3r,,Z =Re,

0 0

Av = -Rc/r,; CE emitter-bias: Z; =RBllf3RE, Z =Re, Av =-Re/RE; emitter-follower: Z; =RB II f3RE, Z = r,, Av = l;

0 0

cornrnon-base:Z; = =Rc,Av =Rc/r,;collectorfeedback:Z; = r,/(1//3 + Rc/RF),Z =RcllRF,Av = -Rc/r,;collector

REllr,,Z0 0

dcfeedback:Z; = RF1 llf3r,,Z0 = RcllRF2,Av = -(RF2 II Rc)/r,;effectofloadirnpedance:Av = RLAvNL/(RL + R 0 ),A; = -AvZdRL;

effect of source impedance: V; = R;Vs/(R; + Rs), Avs = R;AvNL/(R; + Rs), ls = Vs/(Rs + R;); combined effect ofload and source
impedance: Av = RLAvNL/(RL + R 0 ), Avs = (R;/(R; + Rs))(Rd(RL + R 0 ))AVNL, A; = -AvR;/RL, A;s = -Av.{Rs + R;)/RL; cascade
connection: Av = Av 1Av2 ; Darlington connection: /3n = /31/32; emitter-follower configuration: IB = <Vee - VBE)/(RB + f3nRE),
le= le= f3nlB,Z; = RBll/31/32RE,A; = f3nRB/(RB + f3nRE),Av = l,Z0 = r,J/32 + r, 2;basicarnplifierconfiguration:Z; = R1IIR2IIZ/,
Z;' = /31(r, 1 + /32r, 2),A; = /3n(R1IIR2)/(R1IIR2 + Z(),Av = /3nRc/Z(,Z0 = Rcllr02 ;feedbackpair:/B1 = <Vee - VBE1)/(RB + /31f32Rc),
Z; = RB II Z/, z; = /31r,1 + /31/32Rc, A; = -f31/32RB/(RB + /31/32Rc) Av = f32Rc/(r, + f32Rc) = 1, Zo = r,J /32-

6 Field-Effect Transistors le= 0 A,ln = lnss(l - Vcs/Vp) 2, In= ls, Vas= Vp (l - ~ ) , I n = lnss/4 (if Vas= Vp/2),
In = lnss/2 (if Vas = 0.3 Vp), Pn = Vnsln, rd = r0 /(l - Vcs/Vp)2; MOSFET: In = k(Vcs - VT)2, k = In(onil<Vcs(on) - VT) 2

7 FETBiasing Fixed-bias: Vas= -Vee, Vns = Vnn - InRn; self-bias: Vas= -InRs, Vns = Vnn - In(Rs + Rn), Vs= lnRs;
voltage-divider: Ve= R 2Vnn/(R 1 + R2), Vas= Ve - InRs, Vns = Vnn - ln(Rn + Rs); cornrnon-gateconfiguration: Vas= Vss - lnRs,
Vns = Vnn + Vss - In(Rn + Rs); special case: VcsQ = 0 V: hQ = lnss, Vns = Vnn - InRn, Vn = Vns, Vs = 0 V. enhancement-type
MOSFET: In = k(Vcs - Vcs(To/, k = In(onil<Vcs(on) - Vcs(To/; feedback bias: Vns = Vas, Vas = Vnn - InRn; voltage-divider:
Ve= R 2Vnn/(R 1 + R2), Vas= Ve - lnRs; universal curve: m = IVpl/InssRs, M = m X Vc/lVpl,Vc = R 2Vnn/(R 1 + R2)

8 FET Amplifiers gm = YJs = illn/.<l Vas, gmo = 2lnss/lVPI, gm = gmo(l - Vcs/Vp), gm = gmo Vln/Inss, rd= l/y0 s =
.<l Vns/Mnlvas=constant; fixed-bias: Z; = Re, Z 0 = Rn, Av = -gmRn; self-bias (bypassed Rs): Z; = Re, Z 0 = Rn, Av = -gmRn; self-bias
(unbypassedRs): Z; = Re, Z 0 = Rn, Av = -gmRn/(l + gmRs); voltage-divider bias: Z; = R1 II R2, Z 0 = Rn,Av = -gmRn; source follower:
Z; = Re, Z 0 = Rs 111/gm, Av = gmRs/(l + gmRs); cornrnon-gate: Z; = Rs 111/gm, Z 0 = Rn, Av = gmRn; enhancement-type MOSFETs:
gm= 2k(VcsQ - Vas(To));drain-feedbackconfiguration:Z; = RF/(l + gmRn),Z0 = Rn,Av = -gmRn;voltage-dividerbias:Z; = Rill R2,
Z0 = Rn, Av = - gmRn.
9 BJT and JFET Frequency Response log,a = 2.3 log 10a, log 10 1 = 0, log 10 a/b = log 10 a - log 10b, log 10 1/b = -log 10b,
log10ab = log10a + log10b, GdB = l0log10P2/P1, GdBm = l0log10P2/l mWl6oon, GdB = 20log10 V2/V1,
GdBT = GdBi + GdB2 + · · · + GdBn P 0HPF = 0.5P0mid' BW = /1 - fz; low frequency (BIT): As = 1/27r(Rs + R;)Cs,
Ac= 1/2TT(R0 + RL)Cc,AE = 1/2TTR,CE,Re = REll(R;/13 + r,),R~ = RsllR1IIR2,FET:AG = 1/27r(Rsig + R;)Cc,
Ac = 1/2TT(R0 + RL)Cc, As = 1/2TTR,qCs, R,q = Rs 111/gm(rd "' 00 D); Miller effect: CM; = (1 - Av)CJ, CM0 = (1 - 1/Av)Cf;
high frequency (BIT): fH; = l/2'TTRTh;ci, RTh; = RsllR1 II R2II R;, C; = Cw; + Cbe + (1 - Av)Cbc, fHo = l/2'TTRThoco,
RTho = Re II RL II ro, Co = Cwo + Cc, + CMo' /13 "' 1/2'TT/3mictr,(Cbe + Cbc), fT = /3mictf{3; FET: fH; = 1/2'TTRThpi, RTh; = Rsig II Re,
C; = Cw;+ Cgs + CM;, CM;= (1 - Av)Cgd fHo = 1/2'TTRThoco, RTho = Rvll Rd rd, Co= Cwo + eds+ CMo; CMo = (1 - l/Av)Cgd;
multistage: f{ = fi/Y2 1/n - 1, f 2 = (Y2 1/n - l)fz; square-wave testing: fH; = 0.35/t,, % tilt = P% = ((V - V')/V) X 100%,
A 0 = (P/'TT)fs

10 Operational Amplifiers CMRR = Ad/Ac; CMRR(log) = 20 log10(Ad/Ac); constant-gain multiplier: V0/V1 = -R1/R1;
noninverting amplifier: V0/V1 = 1 + R1/R1; unity follower: V0 = V1; summing amplifier: V0 = -[(RJ/R1)V1 + (R1/R2W2 + (RJ/R3)V3];
integrator: vo(t) = -(l/R1C1)fv1dt

11 Op-Amp Applications Constant-gain multiplier: A = - RJ!R 1; noninverting: A = l + RJ/Ri= voltage summing:

V0 = -[(R1/R1W1 + (RJ/R2W2 + (R1/R3)V3]; high-pass active filter: foL = 1/2TTR1C1; low-pass active filter: foH = 1/2TTR1 C1

12 Power Amplifiers
Powerin: P; = VcclcQ
power out: P 0 = VCEic = lf;Rc = VtE/Rcrms
= VCEic/2 = (/f;/2)Rc = VtE/(2Rc) peak
= VCEic/8 = (/t/8)Rc = VtE/(8Rc) peak-to-peak
efficiency: %YJ = (P 0/ P;) X 100%; maximum efficiency: Class A, series-fed = 25%; Class A, transformer-coupled = 50%; Class B,
push-pull= 78.5%; transformer relations: V2/V1 = N2/N1 = Ji/]z, R2 = (N2/N1) 2R1; power output: P 0 = [(VcE max - VCE nun . )
(le max - le mm. )]/8; class B power amplifier: P; = Vcd (2/TT)lpeak]; P0 = V[(peak)/(2RL); %YJ = ('TT /4)[ VL(peak)/Vcc] X 100%;
PQ = P2Q/2 = (P; - P0)/2; maximumP0 = Vtc/2RL; maximum P; = 2Vtc/'TTRL; maximumP2Q = 2Vtc/TT 2RL; % total harmonic
distortion(% THD) = YD~ + D~ + D~ + ... X 100%; heat-sink: T1 = Pv01A + TA, 01A = 40°C/W (free air);
Pv = (T1 - TA)/(01c + 0cs + 0sA)

13 Linear-Digital ICs Ladder network: V0 = [(Do X 2° + D 1 X 2 1 + D 2 X 22 + · · · + Dn X 2n)/2nlVref;

555 oscillator: f = 1.44(RA + 2RB)C; 555 monostable: Thigh = 1.lRAC; VCO:/0 = (2/R1C1)[(V+ - Vc)/V+]; phase-
locked loop (PLL): /0 = 0.3/R1C1,A = ±8/0/V,Jc = ±(1/2TT)Y2TTA/(3.6 X 103)C2

14 Feedback and Oscillator Circuits A1 = A/(l + f3A); series feedback; Zif= Z;(l + f3A); shunt feedback: Zif= Z;/(1 + f3A);
voltage feedback: z01 = Z0 /(l + f3A); current feedback; z01 = Zo(l + f3A); gain stability: dA1/A1 = 1/(11 + /3Al)(dA/A); oscillator;
f3A = 1; phase shift:/= 1/2TTRCV6, /3 = 1/29, A > 29; FET phase shift: IA I = gmRL, RL = Rvrd/(Rv + rd); transistor phase shift:
f = (1/2TTRC)[l/Y6 + 4(Rc/R)], hf,> 23 + 29(Rc/R) + 4(R/Rc); Wien bridge: R3/R4 = Ri/R2 + C2/C1, f 0 = 1/2TTVR1C1R2C2;
tuned: f 0 = 1/2TT~, Ceq = C1C2/(C1 + C2), Hartley: Leg= L1 + Lz + 2M, f 0 = 1/2TT vr;;;c
15 Power Supplies (Voltage Regulators) Filters: r = V,(rms)/Vctc X 100%, V.R. = (VNL - VFL)/VFL X 100%, Vctc = Vm - V,(p-p)/2,
V,(rms) = V,(p-p)/2\/3, V,(rms) "' (lctc/4 V3)(Vctc/Vm); full-wave, light load V,(rms) = 2.4lctc/C, Vctc = Vm - 4.17/ctc/C, r =
(2.4lctcCVctc) X 100% = 2.4/RLC X 100%,/peak = T/T1 X lctc;RCfilter: Vi1c = RL Vctc/(R + RL),Xc = 2.653/C(half-wave),Xc =
1.326/C (full-wave), v;(rms) = (Xc/YR2 + X~); regulators: IR = (/NL - IFL)/IFL X 100%, VL = Vz(l + Ri/R2), V0 =
Vref(l + R2/R1) + lactjR2

16 Other Two-Terminal Devices Varactordiode: Cr= C(0)/(1 + IV,/Vrlt,TCc = (l1C/Co(T1 - T0)) X 100%;photodiode:
W = /if, A = v /f, 1 lm = 1.496 X 10- 10 W, 1 A = 10-10 m, 1 fc = 1 lm/ft2 = 1.609 X 10-9 W /m2

17 pnpn and Other Devices Diac: VBR, = VBR 2 ± 0.1 VBR 2 UIT: RBB = (RB 1 + RB2)lh=O, VRB, = YJVBBlh=O,
TJ = RB/(RB 1 + RB2)liE=o, Vp = YJVBB
+ Vv; phototransistor: le "' h1,1>._; PUT: T/ = RB/(RB 1 + RB2),Vp = YJVBB + Vv
Electronic
Devices and
Circuit Theory
•
Eleventh Edition

Robert L. Boylestad
Louis Nashelsky

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Library of Congress Cataloging-in-Publication Data

Boylestad, Robert L.
Electronic devices and circuit theory/ Robert L. Boylestad, Louis Nashelsky.-1 lth ed.
p.cm.
ISBN 978-0-13-262226-4
1. Electronic circuits. 2. Electronic apparatus and appliances. I. Nashelsky, Louis. II. Title.
TK7867.B66 2013
621.3815-dc23
2011052885

10 9 8 7 6 5 4 3 2 1

PEARSON ISBN 10: 0-13-262226-2

ISBN 13: 978-0-13-262226-4
 DEDICATION

To Else Marie, Alison and Mark, Eric and Rachel, Stacey and Jonathan,
and our eight granddaughters: Kelcy, Morgan, Codie, Samantha, Lindsey,
Britt, Skylar, and Aspen.

To Katrin, Kira and Thomas, Larren and Patricia, and our six grandsons:
Justin, Brendan, Owen, Tyler, Colin, and Dillon.
This page intentionally left blank
 PREFACE

The preparation of the preface for the 11th edition resulted in a bit of reflection on the 40
years since the first edition was published in 1972 by two young educators eager to test
their ability to improve on the available literature on electronic devices. Although one may
prefer the term semiconductor devices rather than electronic devices, the first edition was
almost exclusively a survey of vacuum-tube devices-a subject without a single section in
the new Table of Contents. The change from tubes to predominantly semiconductor devices
took almost five editions, but today it is simply referenced in some sections. It is interest-
ing, however, that when field-effect transistor (FET) devices surfaced in earnest, a number
of the analysis techniques used for tubes could be applied because of the similarities in the
ac equivalent models of each device.
We are often asked about the revision process and how the content of a new edition is
defined. In some cases, it is quite obvious that the computer software has been updated,
and the changes in application of the packages must be spelled out in detail. This text
was the first to emphasize the use of computer software packages and provided a level
of detail unavailable in other texts. With each new version of a software package, we
have found that the supporting literature may still be in production, or the manuals lack
the detail for new users of these packages. Sufficient detail in this text ensures that a
student can apply each of the software packages covered without additional instruc-
tional material.
The next requirement with any new edition is the need to update the content reflecting
changes in the available devices and in the characteristics of commercial devices. This
can require extensive research in each area, followed by decisions regarding depth of
coverage and whether the listed improvements in response are valid and deserve recog-
nition. The classroom experience is probably one of the most important resources for
defining areas that need expansion, deletion, or revision. The feedback from students
results in marked-up copies of our texts with inserts creating a mushrooming copy of the
material. Next, there is the input from our peers, faculty at other institutions using the
text, and, of course, reviewers chosen by Pearson Education to review the text. One
source of change that is less obvious is a simple rereading of the material following the
passing of the years since the last edition. Rereading often reveals material that can be
improved, deleted, or expanded.
For this revision, the number of changes far outweighs our original expectations. How-
ever, for someone who has used previous editions of the text, the changes will probably
be less obvious. However, major sections have been moved and expanded, some 100-plus
problems have been added, new devices have been introduced, the number of applications
has been increased, and new material on recent developments has been added through-
out the text. We believe that the current edition is a significant improvement over the
previous editions.
As instructors, we are all well aware of the importance of a high level of accuracy
required for a text of this kind. There is nothing more frustrating for a student than to
work a problem over from many different angles and still find that the answer differs
from the solution at the back of the text or that the problem seems undoable. We were
pleased to find that there were fewer than half a dozen errors or misprints reported since
vi PREFACE the last edition. When you consider the number of examples and problems in the text
along with the length of the text material, this statistic clearly suggests that the text is as
error-free as possible. Any contributions from users to this list were quickly acknowl-
edged, and the sources were thanked for taking the time to send the changes to the pub-
lisher and to us.
Although the current edition now reflects all the changes we feel it should have, we
expect that a revised edition will be required somewhere down the line. We invite you to
respond to this edition so that we can start developing a package of ideas and thoughts that
will help us improve the content for the next edition. We promise a quick response to your
comments, whether positive or negative.

NEW TO THIS EDITION

• Throughout the chapters, there are extensive changes in the problem sections. Over 100
new problems have been added, and a significant number of changes have been made to
the existing problems.
• A significant number of computer programs were all rerun and the descriptions updated
to include the effects of using OrCAD version 16.3 and Multisim version 11.1. In addi-
tion, the introductory chapters are now assuming a broader understanding of computer
methods, resulting in a revised introduction to the two programs.
• Throughout the text, photos and biographies of important contributors have been added.
Included among these are Sidney Darlington, Walter Schottky, Harry Nyquist, Edwin
Colpitts, and Ralph Hartley.
• New sections were added throughout the text. There is now a discussion on the impact
of combined de and ac sources on diode networks, of multiple BJT networks, VMOS
and UMOS power FETs, Early voltage, frequency impact on the basic elements,
effect of Rs on an amplifier's frequency response, gain-bandwidth product, and a
number of other topics.
• A number of sections were completely rewritten due to reviewers' comments or
changing priorities. Some of the areas revised include bias stabilization, current
sources, feedback in the de and ac modes, mobility factors in diode and transistor
response, transition and diffusion capacitive effects in diodes and transistor response
characteristics, reverse-saturation current, breakdown regions (cause and effect), and
the hybrid model.
• In addition to the revision of numerous sections described above, there are a number of
sections that have been expanded to respond to changes in priorities for a text of this
kind. The section on solar cells now includes a detailed examination of the materials
employed, additional response curves, and a number of new practical applications. The
coverage of the Darlington effect was totally rewritten and expanded to include detailed
examination of the emitter-follower and collector gain configurations. The coverage of
transistors now includes details on the cross-bar latch transistor and carbon nanotubes.
The discussion of LEDs includes an expanded discussion of the materials employed,
comparisons to today's other lighting options, and examples of the products defining
the future of this important semiconductor device. The data sheets commonly included
in a text of this type are now discussed in detail to ensure a well-established link when
the student enters the industrial community.
• Updated material appears throughout the text in the form of photos, artwork, data
sheets, and so forth, to ensure that the devices included reflect the components avail-
able today with the characteristics that have changed so rapidly in recent years. In
addition, the parameters associated with the content and all the example problems are
more in line with the device characteristics available today. Some devices, no longer
available or used very infrequently, were dropped to ensure proper emphasis on the
current trends.
• There are a number of important organizational changes throughout the text to ensure
the best sequence of coverage in the learning process. This is readily apparent in the
early de chapters on diodes and transistors, in the discussion of current gain in the ac
chapters for BJTs and JFETs, in the Darlington section, and in the frequency response
chapters. It is particularly obvious in Chapter 16, where topics were dropped and the
order of sections changed dramatically.
INSTRUCTOR SUPPLEMENTS PREFACE vii
To download the supplements listed below, please visit: http://www.pearsonhighered.
com/ire and enter "Electronic Devices and Circuit Theory" in the search bar. From there,
you will be able to register to receive an instructor's access code. Within 48 hours after
registering, you will receive a confirming email, including an instructor access code.
Once you have received your code, return to the site and log on for full instructions on
how to download the materials you wish to use.
PowerPoint Presentation-(ISBN 0132783746). This supplement contains all figures
from the text as well as a new set of lecture notes highlighting important concepts.
TestGen® Computerized Test Bank-(ISBN 013278372X). This electronic bank of test
questions can be used to develop customized quizzes, tests, and/or exams.
Instructor's Resource Manual-(ISBN 0132783738). This supplement contains the solu-
tions to the problems in the text and lab manual.

STUDENT SUPPLEMENTS
Laboratory Manual-(ISBN 0132622459) . This supplement contains over 35 class-tested
experiments for students to use to demonstrate their comprehension of course material.
Companion Website-Student study resources are available at www.pearsonhighered.
com/boylestad

ACKNOWLEDGMENTS
The following individuals supplied new photographs for this edition.
Sian Cummings International Rectifier Inc.
Michele Drake Agilent Technologies Inc.
Edward Eckert Alcatel-Lucent Inc.
Amy Flores Agilent Technologies Inc.
Ron Forbes B&K Precision Corporation
Christopher Frank Siemens AG
Amber Hall Hewlett-Packard Company
Jonelle Hester National Semiconductor Inc.
George Kapczak AT&T Inc.
Patti Olson Fairchild Semiconductor Inc.
Jordon Papanier LEDtronics Inc.
Andrew W. Post Vishay Inc.
Gilberto Ribeiro Hewlett-Packard Company
Paul Ross Alcatel-Lucent Inc.
Craig R. Schmidt Agilent Technologies, Inc.
Mitch Segal Hewlett-Packard Company
Jim Simon Agilent Technologies, Inc.
Debbie Van Velkinburgh Tektronix, Inc.
Steve West On Semiconductor Inc.
Marcella Wilhite Agilent Technologies, Inc.
Stan Williams Hewlett-Packard Company
J. Joshua Wang Hewlett-Packard Company
This page intentionally left blank
 BRIEF CONTENTS

Preface v

CHAPTER 1: Semicondudor Diodes 1

CHAPTER 2: Diode Applications 55

CHAPTER 3: Bipolar Jundion Transistors 129

CHAPTER 4: DC Biasing-BJTs 160

CHAPTER 5: BJT AC Analysis 253

CHAPTER 6: Field-Effed Transistors 378

CHAPTER 7: FET Biasing 422

CHAPTER 8: FET Amplifiers 481

CHAPTER 9: BJT and JFET Frequency Response 545

CHAPTER 10: Operational Amplifiers 607

CHAPTER 11 : Op-Amp Applications 653

CHAPTER 12: Power Amplifiers 683

CHAPTER 13: Linear-Digital ICs 722

CHAPTER 14: Feedback and Oscillator Circuits 751

CHAPTER 15: Power Supplies (Voltage Regulators) 783

CHAPTER 16: Other Two-Terminal Devices 811

CHAPTER 17: pnpn and Other Devices 841

Appendix A: Hybrid Parameters-Graphical

Determinations and Conversion Equations (Exad
and Approximate) 879
x BRIEF CONTENTS Appendix B: Ripple Fador and Voltage Calculations 885
Appendix C: Charts and Tables 891
Appendix D: Solutions to Seleded
Odd-Numbered Problems 893
Index 901
 CONTENTS

Preface V

CHAPTER 1: Semicondudor Diodes 1

.1 Introduction 1
.2 Semiconductor Materials: Ge, Si, and GaAs 2
.3 Covalent Bonding and Intrinsic Materials 3
.4 Energy Levels 5
.5 n-Type and p-Type Materials 7
.6 Semiconductor Diode 10
.7 Ideal Versus Practical 20
.8 Resistance Levels 21
.9 Diode Equivalent Circuits 27
.10 Transition and Diffusion Capacitance 30
.11 Reverse Recovery Time 31
.12 Diode Specification Sheets 32
.13 Semiconductor Diode Notation 35
.14 Diode Testing 36
.15 Zener Diodes 38
.16 Light-Emitting Diodes 41
.17 Summary 48
.18 Computer Analysis 49

CHAPTER 2: Diode Applications 55

2.1 Introduction 55
2.2 Load-Line Analysis 56
2.3 Series Diode Configurations 61
2.4 Parallel and Series-Parallel Configurations 67
2.5 AND/OR Gates 70
2.6 Sinusoidal Inputs; Half-Wave Rectification 72
2.7 Full-Wave Rectification 75
2.8 Clippers 78
2.9 Clampers 85
2.10 Networks with a de and ac Source 88
xii CONTENTS 2.11 Zener Diodes 91
2.12 Voltage-Multiplier Circuits 98
2.13 Practical Applications 101
2.14 Summary 111
2.15 Computer Analysis 112

CHAPTER 3: Bipolar Jundion Transistors 129

3.1 Introduction 129
3.2 Transistor Construction 130
3.3 Transistor Operation 130
3.4 Common-Base Configuration 131
3.5 Common-Emitter Configuration 136
3.6 Common-Collector Configuration 143
3.7 Limits of Operation 144
3.8 Transistor Specification Sheet 145
3.9 Transistor Testing 149
3.10 Transistor Casing and Terminal Identification 151
3.11 Transistor Development 152
3.12 Summary 154
3.13 Computer Analysis 155

CHAPTER 4: DC Biasing-BJTs 160

4.1 Introduction 160
4.2 Operating Point 161
4.3 Fixed-Bias Configuration 163
4.4 Emitter-Bias Configuration 169
4.5 Voltage-Divider Bias Configuration 175
4.6 Collector Feedback Configuration 181
4.7 Emitter-Follower Configuration 186
4.8 Common-Base Configuration 187
4.9 Miscellaneous Bias Configurations 189
4.10 Summary Table 192
4.11 Design Operations 194
4.12 Multiple BJT Networks 199
4.13 Current Mirrors 205
4.14 Current Source Circuits 208
4.15 pnp Transistors 210
4.16 Transistor Switching Networks 211
4.17 Troubleshooting Techniques 215
4.18 Bias Stabilization 217
4.19 Practical Applications 226
4.20 Summary 233
4.21 Computer Analysis 235
CHAPTER 5: BJT AC Analysis 253 CONTENTS xiii
5.1 Introduction 253
5.2 Amplification in the AC Domain 253
5.3 BJT Transistor Modeling 254
5.4 The re Transistor Model 257
5.5 Common-Emitter Fixed-Bias Configuration 262
5.6 Voltage-Divider Bias 265
5.7 CE Emitter-Bias Configuration 267
5.8 Emitter-Follower Configuration 273
5.9 Common-Base Configuration 277
5.10 Collector Feedback Configuration 279
5.11 Collector DC Feedback Configuration 284
5.12 Effect of RL and R5 286
5.13 Determining the Current Gain 291
5.14 Summary Tables 292
5.15 Two-Port Systems Approach 292
5.16 Cascaded Systems 300
5.17 Darlington Connection 305
5.18 Feedback Pair 314
5.19 The Hybrid Equivalent Model 319
5.20 Approximate Hybrid Equivalent Circuit 324
5.21 Complete Hybrid Equivalent Model 330
5.22 Hybrid 7T Model 337
5.23 Variations of Transistor Parameters 338
5.24 Troubleshooting 340
5.25 Practical Applications 342
5.26 Summary 349
5.27 Computer Analysis 352

CHAPTER 6: Field-Effed Transistors 378

6.1 Introduction 378
6.2 Construction and Characteristics of JFETs 379
6.3 Transfer Characteristics 386
6.4 Specification Sheets (JFETs) 390
6.5 Instrumentation 394
6.6 Important Relationships 395
6.7 Depletion-Type MOSFET 396
6.8 Enhancement-Type MOSFET 402
6.9 MOSFET Handling 409
6.10 VMOS and UMOS Power and MOSFETs 410
6.11 CMOS 411
6.12 MESFETs 412
6.13 Summary Table 414
xiv CONTENTS 6.14 Summary 414
6.15 Computer Analysis 416

CHAPTER 7: FEY Biasing 422

7.1 Introduction 422
7.2 Fixed-Bias Configuration 423
7.3 Self-Bias Configuration 427
7.4 Voltage-Divider Biasing 431
7.5 Common-Gate Configuration 436
7.6 Special Case Vc 50 = O V 439
7.7 Depletion-Type MOSFETs 439
7.8 Enhancement-Type MOSFETs 443
7.9 Summary Table 449
7.10 Combination Networks 449
7.11 Design 452
7.12 Troubleshooting 455
7.13 p-Channel FETs 455
7.14 Universal JFET Bias Curve 458
7.15 Practical Applications 461
7.16 Summary 470
7.17 Computer Analysis 471

CHAPTER 8: FEY Amplifiers 481

8.1 Introduction 481
8.2 JFET Small-Signal Model 482
8.3 Fixed-Bias Configuration 489
8.4 Self-Bias Configuration 492
8.5 Voltage-Divider Configuration 497
8.6 Common-Gate Configuration 498
8.7 Source-Follower (Common-Drain) Configuration 501
8.8 Depletion-Type MOSFETs 505
8.9 Enhancement-Type MOSFETs 506
8.10 E-MOSFET Drain-Feedback Configuration 507
8.11 E-MOSFET Voltage-Divider Configuration 510
8.12 Designing FET Amplifier Networks 511
8.13 Summary Table 513
8.14 Effect of RL and Rsig 516
8.15 Cascade Configuration 518
8.16 Troubleshooting 521
8.17 Practical Applications 522
8.18 Summary 530
8.19 Computer Analysis 531
CHAPTER 9: BJT and JFET Frequency Response 545 CONTENTS xv
9.1 Introduction 545
9.2 Logarithms 545
9.3 Decibels 550
9.4 General Frequency Considerations 554
9.5 Normalization Process 557
9.6 Low-Frequency Analysis-Bode Plot 559
9.7 Low-Frequency Response-BJT Amplifier with RL 564
9.8 Impact of R5 on the BJT Low-Frequency Response 568
9.9 Low-Frequency Response-FET Amplifier 571
9.10 Miller Effect Capacitance 574
9.11 High-Frequency Response-BJT Amplifier 576
9.12 High-Frequency Response-FET Amplifier 584
9.13 Multistage Frequency Effects 586
9.14 Square-Wave Testing 588
9.15 Summary 591
9.16 Computer Analysis 592

CHAPTER 1O: Operational Amplifiers 607

10.1 Introduction 607
10.2 Differential Amplifier Circuit 610
10.3 BiFET, BiMOS, and CMOS Differential Amplifier Circuits 617
10.4 Op-Amp Basics 620
10.5 Practical Op-Amp Circuits 623
10.6 Op-Amp Specifications-DC Offset Parameters 628
10.7 Op-Amp Specifications-Frequency Parameters 631
10.8 Op-Amp Unit Specifications 634
10.9 Differential and Common-Mode Operation 639
10.10 Summary 643
10.11 Computer Analysis 644

CHAPTER 11 : Op-Amp Applications 653

11.1 Constant-Gain Multiplier 653
11.2 Voltage Summing 657
11.3 Voltage Buffer 660
11.4 Controlled Sources 661
11.5 Instrumentation Circuits 663
11.6 Active Filters 667
11.7 Summary 670
11.8 Computer Analysis 671

CHAPTER 12: Power Amplifiers 683

12.1 Introduction-Definitions and Amplifier Types 683
12.2 Series-Fed Class A Amplifier 685
xvi CONTENTS 12.3 Transformer-Coupled Class A Amplifier 688
12.4 Class B Amplifier Operation 695
12.5 Class B Amplifier Circuits 699
12.6 Amplifier Distortion 705
12.7 Power Transistor Heat Sinking 709
12.8 Class C and Class D Amplifiers 712
12.9 Summary 714
12.10 Computer Analysis 715

CHAPTER 13: Linear-Digital ICs 722

13.1 Introduction 722
13.2 Comparator Unit Operation 722
13.3 Digital-Analog Converters 729
13.4 Timer IC Unit Operation 732
13.5 Voltage-Controlled Oscillator 736
13.6 Phase-Locked Loop 738
13.7 Interfacing Circuitry 742
13.8 Summary 745
13.9 Computer Analysis 745

CHAPTER 14: Feedback and Oscillator Circuits 751

4.1 Feedback Concepts 751
4.2 Feedback Connection Types 752
4.3 Practical Feedback Circuits 758
4.4 Feedback Amplifier-Phase and Frequency Considerations 763
4.5 Oscillator Operation 766
4.6 Phase-Shift Oscillator 767
4.7 Wien Bridge Oscillator 770
4.8 Tuned Oscillator Circuit 771
4.9 Crystal Oscillator 774
4.1 0 Unijunction Oscillator 777
4.11 Summary 778
4.12 Computer Analysis 779

CHAPTER 15: Power Supplies (Voltage Regulators) 783

15.1 Introduction 783
15.2 General Filter Considerations 784
15.3 Capacitor Filter 786
1U ~Fl~r H9
15.5 Discrete Transistor Voltage Regulation 791
15.6 IC Voltage Regulators 798
15.7 Practical Applications 803
15.8 Summary 805
15.9 Computer Analysis 806
CHAPTER 16: Other Two-Terminal Devices 811 CONTENTS xvii
16.1 Introduction 811
16.2 Schottky Barrier (Hot-Carrier) Diodes 811
16.3 Varactor (Vari cap) Diodes 815
16.4 Solar Cells 819
16.5 Photodiodes 824
16.6 Photoconductive Cells 826
16.7 IR Emitters 828
16.8 Liquid-Crystal Displays 829
16.9 Thermistors 831
16.10 Tunnel Diodes 833
16.11 Summary 837

CHAPTER 17: pnpn and Other Devices 841

7.1 Introduction 841
7.2 Silicon-Controlled Rectifier 841
7.3 Basic Silicon-Controlled Rectifier Operation 842
7.4 SCR Characteristics and Ratings 843
7.5 SCR Applications 845
7.6 Silicon-Controlled Switch 849
7.7 Gate Turn-Off Switch 851
7.8 Light-Activated SCR 852
7.9 Shockley Diode 854
7.10 Diac 854
7.11 Triac 856
7.12 Unijunction Transistor 857
7.13 Phototransistors 865
7.14 Opto-lsolators 867
7.15 Programmable Unijunction Transistor 869
17.16 Summary 874

Appendix A: Hybrid Parameters-Graphical Determinations

and Conversion Equations (Exad and Approximate) 879
A.1 Graphical Determination of the h-Parameters 879
A.2 Exact Conversion Equations 883
A.3 Approximate Conversion Equations 883

Appendix B: Ripple Factor and Voltage Calculations 885

8.1 Ripple Factor of Rectifier 885
8.2 Ripple Voltage of Capacitor Filter 886
8.3 Relation of Vdc and Vm to Ripple r 887
8.4 Relation of V,(rms) and Vm to Ripple r 888
8.5 Relation Connecting Conduction Angle, Percentage
Ripple, and lpeaklldc for Rectifier-Capacitor Filter Circuits 889
xviii CONTENTS Appendix C: Charts and Tables 891

Appendix D: Solutions to Selected

Odd-Numbered Problems 893

Index 901
CHAPTER OBJECTIVES
•
• Become aware of the general characteristics of three important semiconductor
materials: Si, Ge, GaAs.
• Understand conduction using electron and hole theory.
• Be able to describe the difference between n- and p-type materials.
• Develop a clear understanding of the basic operation and characteristics of a diode in
the no-bias, forward-bias, and reverse-bias regions.
• Be able to calculate the de, ac, and average ac resistance of a diode from the
characteristics.
• Understand the impact of an equivalent circuit whether it is ideal or practical.
• Become familiar with the operation and characteristics of a Zener diode and
light-emitting diode.

1.1 INTRODUCTION
•
One of the noteworthy things about this field, as in many other areas of technology, is how
little the fundamental principles change over time. Systems are incredibly smaller, current
speeds of operation are truly remarkable, and new gadgets surface every day, leaving us to
wonder where technology is taking us. However, if we take a moment to consider that the
majority of all the devices in use were invented decades ago and that design techniques
appearing in texts as far back as the 1930s are still in use, we realize that most of what we
see is primarily a steady improvement in construction techniques, general characteristics,
and application techniques rather than the development of new elements and fundamen-
tally new designs. The result is that most of the devices discussed in this text have been
around for some time, and that texts on the subject written a decade ago are still good ref-
erences with content that has not changed very much. The major changes have been in the
understanding of how these devices work and their full range of capabilities, and in
improved methods of teaching the fundamentals associated with them. The benefit of all
this to the new student of the subject is that the material in this text will, we hope, have
reached a level where it is relatively easy to grasp and the information will have applica-
tion for years to come.
The miniaturization that has occurred in recent years leaves us to wonder about its limits.
Complete systems now appear on wafers thousands of times smaller than the single element
of earlier networks. The first integrated circuit (IC) was developed by Jack Kilby while
working at Texas Instruments in 1958 (Fig. 1.1). Today, the Intel® Core™ i7 Extreme
2 SEMICONDUCTOR Edition Processor of Fig. 1.2 has 731 million transistors in a package that is only slightly
DIODES larger than a 1.67 sq. inches. fu 1965, Dr. Gordon E. Moore presented a paper predicting that
the transistor count in a single IC chip would double every two years. Now, more than
45 years, later we find that his prediction is amazingly accurate and expected to continue
for the next few decades. We have obviously reached a point where the primary purpose
of the container is simply to provide some means for handling the device or system and to
provide a mechanism for attachment to the remainder of the network. Further miniaturiza-
tion appears to be limited by four factors: the quality of the semiconductor material, the
network design technique, the limits of the manufacturing and processing equipment, and
the strength of the innovative spirit in the semiconductor industry.
The first device to be introduced here is the simplest of all electronic devices, yet has a
range of applications that seems endless. We devote two chapters to the device to introduce
the materials commonly used in solid-state devices and review some fundamental laws of
electric circuits.

Jack St. Clair Kilby, inventor of the

integrated circuit and co-inventor of
1.2 SEMICONDUCTOR MATERIALS: Ge, Si, AND GaAs
•
The construction of every discrete (individual) solid-state (hard crystal structure) electronic
device or integrated circuit begins with a semiconductor material of the highest quality.
the electronic handheld calculator. Semiconductors are a special class of elements having a conductivity between that of a
(Courtesy of Texas Instruments.) good conductor and that of an insulator.
Born: Jefferson City, Missouri,1923. In general, semiconductor materials fall into one of two classes: single-crystal and
MS, University of Wisconsin. compound. Single-crystal semiconductors such as germanium (Ge) and silicon (Si) have a
Director of Engineering and Tech- repetitive crystal structure, whereas compound semiconductors such as gallium arsenide
nology, Components Group, Texas (GaAs), cadmium sulfide (CdS), gallium nitride (GaN), and gallium arsenide phosphide
Instruments. Fellow of the IEEE. (GaAsP) are constructed of two or more semiconductor materials of different atomic
Holds more than 60 U.S. patents. structures.
The three semiconductors used most frequently in the construction of electronic
devices are Ge, Si, and GaAs.
In the first few decades following the discovery of the diode in 1939 and the transis-
tor in 1947 germanium was used almost exclusively because it was relatively easy to
find and was available in fairly large quantities. It was also relatively easy to refine to
obtain very high levels of purity, an important aspect in the fabrication process. How-
ever, it was discovered in the early years that diodes and transistors constructed using
germanium as the base material suffered from low levels of reliability due primarily to
its sensitivity to changes in temperature. At the time, scientists were aware that another
material, silicon, had improved temperature sensitivities, but the refining process for
manufacturing silicon of very high levels of purity was still in the development stages.
The first integrated circuit, a phase-
Finally, however, in 1954 the first silicon transistor was introduced, and silicon quickly
shift oscillator, invented by Jack S.
became the semiconductor material of choice. Not only is silicon less temperature sensi-
Kilby in 1958. (Courtesy of Texas
Instruments.) tive, but it is one of the most abundant materials on earth, removing any concerns about
availability. The flood gates now opened to this new material, and the manufacturing
FICi. 1.1 and design technology improved steadily through the following years to the current high
Jack St. Clair Kilby. level of sophistication.
As time moved on, however, the field of electronics became increasingly sensitive to
issues of speed. Computers were operating at higher and higher speeds, and communica-
tion systems were operating at higher levels of performance. A semiconductor material
capable of meeting these new needs had to be found. The result was the development of
the first GaAs transistor in the early 1970s. This new transistor had speeds of operation
up to five times that of Si. The problem, however, was that because of the years of intense
design efforts and manufacturing improvements using Si, Si transistor networks for most
applications were cheaper to manufacture and had the advantage of highly efficient design
strategies. GaAs was more difficult to manufacture at high levels of purity, was more ex-
pensive, and had little design support in the early years of development. However, in time
the demand for increased speed resulted in more funding for GaAs research, to the point that
today it is often used as the base material for new high-speed, very large scale integrated
(VLSI) circuit designs.
 This brief review of the history of semiconductor materials is not meant to imply that COVALENT BONDING 3
GaAs will soon be the only material appropriate for solid-state construction. Germanium AND INTRINSIC
MATERIALS
devices are still being manufactured, although for a limited range of applications. Even
though it is a temperature-sensitive semiconductor, it does have characteristics that find
application in a limited number of areas. Given its availability and low manufacturing costs,
it will continue to find its place in product catalogs. As noted earlier, Si has the benefit of
years of development, and is the leading semiconductor material for electronic components
and ICs. In fact, Si is still the fundamental building block for Intel's new line of processors.

1.3 COVALENT BONDING AND INTRINSIC MATERIALS •

To fully appreciate why Si, Ge, and GaAs are the semiconductors of choice for the elec-
tronics industry requires some understanding of the atomic structure of each and how the
atoms are bound together to form a crystalline structure. The fundamental components of
an atom are the electron, proton, and neutron. In the lattice structure, neutrons and protons
form the nucleus and electrons appear in fixed orbits around the nucleus. The Bohr model
for the three materials is provided in Fig. 1.3.

Valence shell (F/nr valenc7electrons) Valence electron

FICi. 1.2
Shell Intel® Core™ i7 Extreme Edition
Processor.
+ Orbiting
electrons
Nucleus

Silicon Germanium

(a) (b)

Three valence Five valence

/
electrons

Gallium Arsenic

(c)

FICi. 1.3
Atomic structure of (a) silicon; (b) germanium; and
(c) gallium and arsenic.

As indicated in Fig. 1.3, silicon has 14 orbiting electrons, germanium has 32 electrons,
gallium has 31 electrons, and arsenic has 33 orbiting electrons (the same arsenic that is
a very poisonous chemical agent). For germanium and silicon there are four electrons in
the outermost shell, which are referred to as valence electrons. Gallium has three valence
electrons and arsenic has five valence electrons. Atoms that have four valence electrons
are called tetravalent, those with three are called trivalent, and those with five are called
pentavalent. The term valence is used to indicate that the potential (ionization potential)
required to remove any one of these electrons from the atomic structure is significantly
lower than that required for any other electron in the structure.
4 SEMICONDUCTOR
DIODES

FIG. 1.4
Covalent bonding of the silicon atom.

In a pure silicon or germanium crystal the four valence electrons of one atom form a
bonding arrangement with four adjoining atoms, as shown in Fig. 1.4.
This bonding of atoms, strengthened by the sharing of electrons, is called covalent
bonding.
Because GaAs is a compound semiconductor, there is sharing between the two different
atoms, as shown in Fig. 1.5. Each atom, gallium or arsenic, is surrounded by atoms of the
complementary type. There is still a sharing of electrons similar in structure to that of Ge
and Si, but now five electrons are provided by the As atom and three by the Ga atom.

FIG. 1.5
Covalent bonding of the GaAs crystal.

Although the covalent bond will result in a stronger bond between the valence electrons
and their parent atom, it is still possible for the valence electrons to absorb sufficient kinetic
energy from external natural causes to break the covalent bond and assume the "free" state.
The term free is applied to any electron that has separated from the fixed lattice structure and
is very sensitive to any applied electric fields such as established by voltage sources or any
difference in potential. The external causes include effects such as light energy in the fonn
of photons and thermal energy (heat) from the surrounding medium. At room temperature
there are approximately 1.5 X 10 10 free carriers in 1 cm3 of intrinsic silicon material, that
is, 15,000,000,000 (15 billion) electrons in a space smaller than a small sugar cube-an
enormous number.
The term intrinsic is applied to any semiconductor material that has been carefully ENERGY LEVELS 5
refined to reduce the number of impurities to a very low level-essentially as pure as
can be made available through modern technology.
The free electrons in a material due only to external causes are referred to as intrinsic car-
riers. Table 1.1 compares the number of intrinsic carriers per cubic centimeter (abbreviated ni)
for Ge, Si, and GaAs. It is interesting to note that Ge has the highest number and GaAs the
lowest. In fact, Ge has more than twice the number as GaAs. The number of carriers in the
intrinsic form is important, but other characteristics of the material are more significant
in determining its use in the field. One such factor is the relative mobility (µ,n) of the free
carriers in the material, that is, the ability of the free carriers to move throughout the mate-
rial. Table 1.2 clearly reveals that the free carriers in GaAs have more than five times the
mobility of free carriers in Si, a factor that results in response times using GaAs electronic
devices that can be up to five times those of the same devices made from Si. Note also that
free carriers in Ge have more than twice the mobility of electrons in Si, a factor that results
in the continued use of Ge in high-speed radio frequency applications.

TABLE 1.1
Intrinsic Carriers n; TABLE 1.l
Relative Mobility Factor J.Ln
Intrinsic Carriers
Semiconductor (per cubic centimeter) Semiconductor

GaAs 1.7 X 106 Si 1500

Si 1.5 X 10 10 Ge 3900
Ge 2.5 X 10 13 GaAs 8500

One of the most important technological advances of recent decades has been the abil-
ity to produce semiconductor materials of very high purity. Recall that this was one of the
problems encountered in the early use of silicon-it was easier to produce germanium of
the required purity levels. Impurity levels of 1 part in 10 billion are common today, with
higher levels attainable for large-scale integrated circuits. One might ask whether these
extremely high levels of purity are necessary. They certainly are if one considers that the
addition of one part of impurity (of the proper type) per million in a wafer of silicon material
can change that material from a relatively poor conductor to a good conductor of electricity.
We obviously have to deal with a whole new level of comparison when we deal with the
semiconductor medium. The ability to change the characteristics of a material through this
process is called doping, something that germanium, silicon, and gallium arsenide readily
and easily accept. The doping process is discussed in detail in Sections 1.5 and 1.6.
One important and interesting difference between semiconductors and conductors is their
reaction to the application of heat. For conductors, the resistance increases with an increase
in heat. This is because the numbers of carriers in a conductor do not increase significantly
with temperature, but their vibration pattern about a relatively fixed location makes it in-
creasingly difficult for a sustained flow of carriers through the material. Materials that react
in this manner are said to have a positive temperature coefficient. Semiconductor materials,
however, exhibit an increased level of conductivity with the application of heat. As the tem-
perature rises, an increasing number of valence electrons absorb sufficient thermal energy to
break the covalent bond and to contribute to the number of free carriers. Therefore:
Semiconductor materials have a negative temperature coefficient.

1.4 ENERGY LEVELS

•
Within the atomic structure of each and every isolated atom there are specific energy levels
associated with each shell and orbiting electron, as shown in Fig. 1.6. The energy levels
associated with each shell will be different for every element. However, in general:
The farther an electron is from the nucleus, the higher is the energy state, and any
electron that has left its parent atom has a higher energy state than any electron in
the atomic structure.
Note in Fig. 1.6a that only specific energy levels can exist for the electrons in the atomic
structure of an isolated atom. The result is a series of gaps between allowed energy levels
6 SEMICONDUCTOR Energy
DIODES Valence level (outermost shell)
Energy gap t
Second level (next inner shell)
Energy gap t
Third level (etc.)
etc.

i Nucleus

(a)

Energy
Electrons Energy Energy
Conduction band "free" to
establish
Conduction band
conduction
The bands Conduction band
r
E >5 eV
8
overlap
Unable to reach Valence band
conduction level 8 8 Valence
electrons
Valence band Conductor
bound to
Valence band tbe atomic
stucture
Insulator
Eg = 0.67 eV (Ge)
Eg = l.l eV (Si)
Eg = 1.43 eV (GaAs)
Semiconductor
(b)

FIG. 1.6
Energy levels: (a) discrete levels in isolated atomic structures; (b) conduction and valence bands of an insulator,
a semiconductor, and a conductor.

where carriers are not permitted. However, as the atoms of a material are brought closer
together to form the crystal lattice structure, there is an interaction between atoms, which
will result in the electrons of a particular shell of an atom having slightly different energy
levels from electrons in the same orbit of an adjoining atom. The result is an expansion
of the fixed, discrete energy levels of the valence electrons of Fig. 1.6a to bands as shown
in Fig. 1.6b. In other words, the valence electrons in a silicon material can have varying
energy levels as long as they fall within the band of Fig. 1.6b. Figure 1.6b clearly reveals
that there is a minimum energy level associated with electrons in the conduction band and
a maximum energy level of electrons bound to the valence shell of the atom. Between the
two is an energy gap that the electron in the valence band must overcome to become a free
carrier. That energy gap is different for Ge, Si, and GaAs; Ge has the smallest gap and GaAs
the largest gap. In total, this simply means that:
An electron in the valence band of silicon must absorb more energy than one in the
valence band of germanium to become a free carrier. Similarly, an electron in the
valence band of gallium arsenide must gain more energy than one in silicon or
germanium to enter the conduction band.
This difference in energy gap requirements reveals the sensitivity of each type of
semiconductor to changes in temperature. For instance, as the temperature of a Ge sample
increases, the number of electrons that can pick up thermal energy and enter the conduction
band will increase quite rapidly because the energy gap is quite small. However, the number
of electrons entering the conduction band for Si or GaAs would be a great deal less. This
sensitivity to changes in energy level can have positive and negative effects. The design of
photodetectors sensitive to light and security systems sensitive to heat would appear to be
an excellent area of application for Ge devices. However, for transistor networks, where
stability is a high priority, this sensitivity to temperature or light can be a detrimental factor.
 The energy gap also reveals which elements are useful in the construction of light-emitting n-TYPE AND p-TYPE 7
devices such as light-emitting diodes (LEDs), which will be introduced shortly. The wider MATERIALS
the energy gap, the greater is the possibility of energy being released in the form of visible
or invisible (infrared) light waves. For conductors, the overlapping of valence and conduc-
tion bands essentially results in all the additional energy picked up by the electrons being
dissipated in the form of heat. Similarly, for Ge and Si, because the energy gap is so small,
most of the electrons that pick up sufficient energy to leave the valence band end up in the
conduction band, and the energy is dissipated in the form of heat. However, for GaAs the
gap is sufficiently large to result in significant light radiation. For LEDs (Section 1.9) the
level of doping and the materials chosen determine the resulting color.
Before we leave this subject, it is important to underscore the importance of understand-
ing the units used for a quantity. In Fig. 1.6 the units of measurement are electron volts (eV).
The unit of measure is appropriate because W (energy)= QV (as derived from the defining
equation for voltage: V = WIQ). Substituting the charge of one electron and a potential dif-
ference of 1 V results in an energy level referred to as one electron volt.
That is,
W= QV
= (1.6 X 10-19 C)(l V)
= 1.6 X 10-19 J
and

I 1 eV = 1.6 X 10-19 J (1.1)

1.5 n-TYPE AND p-TYPE MATERIALS

•
Because Si is the material used most frequently as the base (substrate) material in the con-
struction of solid-state electronic devices, the discussion to follow in this and the next few
sections deals solely with Si semiconductors. Because Ge, Si, and GaAs share a similar
covalent bonding, the discussion can easily be extended to include the use of the other
materials in the manufacturing process.
As indicated earlier, the characteristics of a semiconductor material can be altered sig-
nificantly by the addition of specific impurity atoms to the relatively pure semiconductor
material. These impurities, although only added at 1 part in 10 million, can alter the band
structure sufficiently to totally change the electrical properties of the material.
A semiconductor material that has been subjected to the doping process is called an
extrinsic material.
There are two extrinsic materials of immeasureable importance to semiconductor device
fabrication: n-type and p-type materials. Each is described in some detail in the following
subsections.

n-Type Material
Both n-type and p-type materials are formed by adding a predetermined number of impurity
atoms to a silicon base. An n-type material is created by introducing impurity elements that
have five valence electrons (pentavalent), such as antimony, arsenic, and phosphorus. Each is
a member of a subset group of elements in the Periodic Table of Elements referred to as Group
V because each has five valence electrons. The effect of such impurity elements is indicated in
Fig. 1.7 (using antimony as the impurity in a silicon base). Note that the four covalent bonds
are still present. There is, however, an additional fifth electron due to the impurity atom, which
is unassociated with any particular covalent bond. This remaining electron, loosely bound to
its parent (antimony) atom, is relatively free to move within the newly formed n-type material.
Since the inserted impurity atom has donated a relatively "free" electron to the structure:
Diffused impurities with five valence electrons are called donor atoms.
It is important to realize that even though a large number of free carriers have been estab-
lished in the n-type material, it is still electrically neutral since ideally the number of posi-
tively charged protons in the nuclei is still equal to the number of free and orbiting negatively
charged electrons in the structure.
8 SEMICONDUCTOR
DIODES

FIG. 1.7
Antimony impurity in n-type material.

The effect of this doping process on the relative conductivity can best be described
through the use of the energy-band diagram of Fig. 1.8. Note that a discrete energy level
(called the donor level) appears in the forbidden band with an Eg significantly less than that
of the intrinsic material. Those free electrons due to the added impurity sit at this energy
level and have less difficulty absorbing a sufficient measure of thermal energy to move into
the conduction band at room temperature. The result is that at room temperature, there are a
large number of carriers (electrons) in the conduction level, and the conductivity of the ma-
terial increases significantly. At room temperature in an intrinsic Si material there is about
one free electron for every 10 12 atoms. If the dosage level is 1 in 10 million (10 7), the ratio
10 12/107 = 105 indicates that the carrier concentration has increased by a ratio of 100,000: 1.

Energy

Conduction band t
E8 for intrinsic
materials
-r-r-1t Eg = considerably less than in Fig.
Donor energy level
l.6(b) for semiconductors

Valence band

FIG. 1.8
Effect of donor impurities on the energy band structure.

p-Type Material
The p-type material is formed by doping a pure germanium or silicon crystal with impurity
atoms having three valence electrons. The elements most frequently used for this purpose
are boron, gallium, and indium. Each is a member of a subset group of elements in the Peri-
odic Table of Elements referred to as Group III because each has three valence electrons.
The effect of one of these elements, boron, on a base of silicon is indicated in Fig. 1.9.
Note that there is now an insufficient number of electrons to complete the covalent bonds
of the newly formed lattice. The resulting vacancy is called a hole and is represented by a
small circle or a plus sign, indicating the absence of a negative charge. Since the resulting
vacancy will readily accept a free electron:
The diffused impurities with three valence electrons are called acceptor atoms.
The resulting p-type material is electrically neutral, for the same reasons described for
the n-type material.
 n-TYPE AND p-TYPE 9
MATERIALS

FIG. 1.9
Boron impurity in p-type material.

Electron versus Hole Flow

The effect of the hole on conduction is shown in Fig. 1.10. If a valence electron acquires
sufficient kinetic energy to break its covalent bond and fills the void created by a hole, then
a vacancy, or hole, will be created in the covalent bond that released the electron. There is,
therefore, a transfer of holes to the left and electrons to the right, as shown in Fig. 1. 10.
The direction to be used in this text is that of conventional flow, which is indicated by the
direction of hole flow.

(a) (c)
Hole flow

Electron flow
(b)

FIG. 1.10
Electron versus hole flow.

Majority and Minority Carriers

In the intrinsic state, the number of free electrons in Ge or Si is due only to those few elec-
trons in the valence band that have acquired sufficient energy from thermal or light sources
to break the covalent bond or to the few impurities that could not be removed. The vacan-
cies left behind in the covalent bonding structure represent our very limited supply of
holes. In an n-type material, the number of holes has not changed significantly from this
intrinsic level. The net result, therefore, is that the number of electrons far outweighs the
number of holes. For this reason:
In an n-type material (Fig. I.Ila) the electron is called the majority carrier and the
hole the minority carrier.
For the p-type material the number of holes far outweighs the number of electrons, as
shown in Fig. l.llb. Therefore:
In a p-type material the hole is the majority carrier and the electron is the minority carrier.
When the fifth electron of a donor atom leaves the parent atom, the atom remaining ac-
quires a net positive charge: hence the plus sign in the donor-ion representation. For similar
reasons, the minus sign appears in the acceptor ion.
10 SEMICONDUCTOR Donor ions Acceptor ions
DIODES
Majority
carriers
0

Minority Majority 0
carrier carriers

n-type p-type carrier

(a) (b)

FIG. 1.11
(a) n-type material; (b) p-type material.

The n- and p-type materials represent the basic building blocks of semiconductor devices.
We will find in the next section that the 'joining" of a single n-type material with a p-type ma-
terial will result in a semiconductor element of considerable importance in electronic systems.

1.6 SEMICONDUCTOR DIODE

•
Now that both n- and p-type materials are available, we can construct our first solid-state
electronic device: The semiconductor diode, with applications too numerous to mention, is
created by simply joining an n-type and a p-type material together, nothing more, just the
joining of one material with a majority carrier of electrons to one with a majority carrier of
holes. The basic simplicity of its construction simply reinforces the importance of the
development of this solid-state era.

No Applied Bias (V = 0 V)
At the instant the two materials are "joined" the electrons and the holes in the region of the
junction will combine, resulting in a lack of free carriers in the region near the junction, as
shown in Fig. 1.12a. Note in Fig. 1.12a that the only particles displayed in this region are
the positive and the negative ions remaining once the free carriers have been absorbed.
This region of uncovered positive and negative ions is called the depletion region due
to the "depletion" offree carriers in the region.
If leads are connected to the ends of each material, a two-terminal device results, as
shown in Figs. 1.12a and 1.12b. Three options then become available: no bias,forward
bias, and reverse bias. The term bias refers to the application of an external voltage across
the two terminals of the device to extract a response. The condition shown in Figs. 1.12a
and 1.12b is the no-bias situation because there is no external voltage applied. It is simply
a diode with two leads sitting isolated on a laboratory bench. In Fig. 1.12b the symbol for
a semiconductor diode is provided to show its correspondence with the p-n junction. In
each figure it is clear that the applied voltage is O V (no bias) and the resulting current is
0 A, much like an isolated resistor. The absence of a voltage across a resistor results in
zero current through it. Even at this early point in the discussion it is important to note the
polarity of the voltage across the diode in Fig. 1.12b and the direction given to the current.
Those polarities will be recognized as the defined polarities for the semiconductor diode.
If a voltage applied across the diode has the same polarity across the diode as in Fig. 1.12b,
it will be considered a positive voltage. If the reverse, it is a negative voltage. The same
standards can be applied to the defined direction of current in Fig. 1.12b.
Under no-bias conditions, any minority carriers (holes) in then-type material that find
themselves within the depletion region for any reason whatsoever will pass quickly into the
p-type material. The closer the minority carrier is to the junction, the greater is the attraction
for the layer of negative ions and the less is the opposition offered by the positive ions in
the depletion region of then-type material. We will conclude, therefore, for future discus-
sions, that any minority carriers of the n-type material that find themselves in the depletion
region will pass directly into the p-type material. This carrier flow is indicated at the top of
Fig. 1.12c for the minority carriers of each material.
 Depletion region SEMICONDUCTOR DIODE 11
,---A..---,

( 8 1.. , 8 _ (±) - -
- (±) (±)
0-0
8 G C, L
8 I.. 8 -
8 c v8 Metal contact

p n

l Iv =OmA
+ Vv=OV
lv:OmAJ
(no bias)

(a)

Minority carrier flow

+
~
Vv=OV
(no bias) I electron I hole

o ~
►I 0

Iv= OmA
/hole Ielectron

~ ~ Majority carrier flow

(b) (c)

FIG. 1.12
A p-n junction with no external bias: (a) an internal distribution of charge; (b) a diode symbol,
with the defined polarity and the current direction; (c) demonstration that the net carrier
flow is zero at the external terminal of the device when Vv = 0 V.

The majority carriers (electrons) of then-type material must overcome the attractive
forces of the layer of positive ions in the n-type material and the shield of negative ions in
the p-type material to migrate into the area beyond the depletion region of the p-type mate-
rial. However, the number of majority carriers is so large in the n-type material that there
will invariably be a small number of majority carriers with sufficient kinetic energy to pass
through the depletion region into the p-type material. Again, the same type of discussion
can be applied to the majority carriers (holes) of the p-type material. The resulting flow due
to the majority carriers is shown at the bottom of Fig. 1.12c.
A close examination of Fig. 1.12c will reveal that the relative magnitudes of the flow
vectors are such that the net flow in either direction is zero. This cancellation of vectors
for each type of carrier flow is indicated by the crossed lines. The length of the vector
representing hole flow is drawn longer than that of electron flow to demonstrate that the
two magnitudes need not be the same for cancellation and that the doping levels for each
material may result in an unequal carrier flow of holes and electrons. In summary, therefore:
In the absence of an applied bias across a semiconductor diode, the net flow of charge
in one direction is zero.
In other words, the current under no-bias conditions is zero, as shown in Figs. 1.12a
and 1.12b.

Reverse-Bias Condition (Vo < o V)

If an external potential of V volts is applied across the p-n junction such that the positive
terminal is connected to the n-type material and the negative terminal is connected to the
p-type material as shown in Fig. 1. 13, the number of uncovered positive ions in the deple-
tion region of the n-type material will increase due to the large number of free electrons
drawn to the positive potential of the applied voltage. For similar reasons, the number of
uncovered negative ions will increase in the p-type material. The net effect, therefore, is a
12 SEMICONDUCTOR ,..,__ I, Minority-carrier flow
DIODES /majority =" OA

- Vv +
+ o-----1►
.il1---o
-+--I,
p '-----y-----' n
Depletion region

L J - ~+
_11 + (Opposite)

Vv

(a) (b)

FIG. 1.13
Reverse-biased p-n junction: (a) internal distribution of charge under
reverse-bias conditions; (b) reverse-bias polarity and direction of reverse
saturation current.

widening of the depletion region. This widening of the depletion region will establish too
great a barrier for the majority carriers to overcome, effectively reducing the majority car-
rier flow to zero, as shown in Fig. 1.13a.
The number of minority carriers, however, entering the depletion region will not change,
resulting in minority-carrier flow vectors of the same magnitude indicated in Fig. 1.12c
with no applied voltage.
The current that exists under reverse-bias conditions is called the reverse saturation
current and is represented by ls.
The reverse saturation current is seldom more than a few microamperes and typically in
nA, except for high-power devices. The term saturation comes from the fact that it reaches its
maximum level quickly and does not change significantly with increases in the reverse-bias
potential, as shown on the diode characteristics of Fig. 1.15 for Vv < 0 V. The reverse-biased
conditions are depicted in Fig. 1.13b for the diode symbol and p-n junction. Note, in particu-
lar, that the direction of ls is against the arrow of the symbol. Note also that the uegative side of
the applied voltage is connected to the p-type material and the 12ositive side to then-type ma-
terial, the difference in underlined letters for each region revealing a reverse-bias condition.

Forward-Bias Condition (Va> o V)

A forward-bias or "on" condition is established by applying the positive potential to the
p-type material and the negative potential to the n-type material as shown in Fig. 1.14.
The application of a forward-bias potential Vv will "pressure" electrons in then-type mate-
rial and holes in the p-type material to recombine with the ions near the boundary and reduce
the width of the depletion region as shown in Fig. 1.14a. The resulting minority-carrier flow

~,.,
- - - - - - / maJorny
.
}
Iv= l m,jorlty - I,

e 8_8 8eg- (f) - <f> (±)

-~ - Cf) - -
+
e e_e :~ ~ Cf>~ +
0>-----1►
Vo
1-----0
p '--.,-J 11
Depletion region

~
~---
+ 11--_- - -~
~ ~
+~-
Vo (Similar)
(a) (b)

FIG. 1.14
Forward-biased p-n junction: (a) internal distribution of charge under forward-bias
conditions; (b) forward-bias polarity and direction of resulting current.
of electrons from the p-type material to then-type material (and of holes from then-type SEMICONDUCTOR DIODE 13
material to the p-type material) has not changed in magnitude (since the conduction level is
controlled primarily by the limited number of impurities in the material), but the reduction
in the width of the depletion region has resulted in a heavy majority flow across the junc-
tion. An electron of the n-type material now "sees" a reduced barrier at the junction due to
the reduced depletion region and a strong attraction for the positive potential applied to the
p-type material. As the applied bias increases in magnitude, the depletion region will con-
tinue to decrease in width until a flood of electrons can pass through the junction, resulting
in an exponential rise in current as shown in the forward-bias region of the characteristics
of Fig. 1.15. Note that the vertical scale of Fig. 1.15 is measured in milliamperes (although
some semiconductor diodes have a vertical scale measured in amperes), and the horizontal
scale in the forward-bias region has a maximum of 1 V. Typically, therefore, the voltage
across a forward-biased diode will be less than 1 V. Note also how quickly the current rises
beyond the knee of the curve.
It can be demonstrated through the use of solid-state physics that the general charac-
teristics of a semiconductor diode can be defined by the following equation, referred to as
Shockley's equation, for the forward- and reverse-bias regions:

I Iv = ls(eVv/nVr - 1) (A) (1.2)

where ls is the reverse saturation current

Vv is the applied forward-bias voltage across the diode
n is an ideality factor, which is a function of the operating conditions and physi-
cal construction; it has a range between 1 and 2 depending on a wide variety of
factors (n = I will be assumed throughout this text unless otherwise noted).
The voltage Vrin Eq. (1.1) is called the thennal voltage and is determined by

~
~ (V) (1.3)

where k is Boltzmann's constant = 1.38 X 10- 23 J/K

TK is the absolute temperature in kelvins = 273 + the temperature in °C
q is the magnitude of electronic charge = 1.6 X 10- 19 C

EXAMPLE 1.1 At a temperature of 27°C (common temperature for components in an

enclosed operating system), determine the thermal voltage Vr.
Solution: Substituting into Eq. (1.3), we obtain
T = 273 + °C = 273 + 27 = 300 K
kTK (1.38 X 10-23 J/K)(30 K)
V: - - - - - - - - - - - - - -
T - q - 1.6 X 10-19 C
= 25.875mV ~ 26mV
The thermal voltage will become an important parameter in the analysis to follow in this
chapter and a number of those to follow.

Initially, Eq. (1.2) with all its defined quantities may appear somewhat complex. How-
ever, it will not be used extensively in the analysis to follow. It is simply important at this
point to understand the source of the diode characteristics and which factors affect its shape.
A plot of Eq. (1.2) with ls = 10 pA is provided in Fig. 1.15 as the dashed line. If we
expand Eq. (1.2) into the following form, the contributing component for each region of
Fig. 1.15 can be described with increased clarity:
ID = I seVv/nVr - I s
For positive values of VD the first term of the above equation will grow very quickly and
totally overpower the effect of the second term. The result is the following equation, which
only has positive values and takes on the exponential format<? appearing in Fig. 1.16:
Iv ~ lseVv/nVr (Vv positive)
14 SEMICONDUCTOR ID (mA)
DIODES
I

20 .
I

I
19 I
I I I I I

I Actual commercially_ - -
18
I available unit
17 I

16 - - Eq. (l.11'- :
r--.
15 I
I
14 I
I
13
i
12
I
I
-
I Defined polarity and
11 -
direction for graph
10
,. I
-
9
I
I + VD
IJliil - -
I
8 ....-ID -
7
I
I k Forward-bias region -
~ I '---
6 (VD >0V, ID>0 mA)-
5
I ,I
4
3
J
2
I I
1
,• /
.,,. ,;,;
-40 -30 -20 -10 !J -----..._0.3 0.5 0.7 1 V
lOpA
I I
_ Reverse-bias region I I -----+ No-bias
.I
- 20pA
(VD<0V,JD=-ls) I I (VD=0V,ID=0mA)
30pA
40pA
I I I I
• 50pA
I I I I

FIG. 1.15
Silicon semiconductor diode characteristics.

The exponential curve of Fig. 1.16 increases very rapidly with increasing values of x.
Atx = 0, e0 = 1, whereas atx = 5, it jumps to greater than 148. Ifwe continued tox = 10,
the curve jumps to greater than 22,000. Clearly, therefore, as the value of x increases, the
curve becomes almost vertical, an important conclusion to keep in mind when we examine
the change in current with increasing values of applied voltage.

I I
.ex .ex I I I I
j

I - e5·5 =244. :=
l
200
I
5 I I
I

ii'
I

e1 = e =2.718
150
I
!'! 5 =148.4 - -
1 - 100
I
,,
0 ->- 1 ->->- 2 1,} I

- -
_ e0 = l
I •
'
50
e4 = 54.6
I I I

___,, -
>-
"'-
r,, V I ''k , ( e 3 =20.1
':9>-
I
--
I

1
I

2 -'/- 3 4
I
5
I
6 7 X ->-
1-. I J~ I I I I I
I -!- I I I I I I

FIG. 1.16
Plot of?.
 For negative values of Vv the exponential term drops very quickly below the level of/, SEMICONDUCTOR DIODE 15
and the resulting equation for / D is simply
(Vv negative)
Note in Fig. 1.15 that for negative values of Vv the current is essentially horizontal at
the level of - ls.
At V = 0 V, Eq. (1.2) becomes
Iv= I,<e 0 - 1) = ls(l - 1) = 0mA
as confirmed by Fig. 1.15.
The sharp change in direction of the curve at Vv = 0 Vis simply due to the change in
current scales from above the axis to below the axis. Note that above the axis the scale is in
milliamperes (mA), whereas below the axis it is in picoamperes (pA).
Theoretically, with all things perfect, the characteristics of a silicon diode should appear
as shown by the dashed line of Fig. 1.15. However, commercially available silicon diodes
deviate from the ideal for a variety of reasons including the internal "body" resistance and the
external "contact" resistance of a diode. Each contributes to an additional voltage at the same
current level, as determined by Ohm's law, causing the shift to the right witnessed in Fig. 1.15.
The change in current scales between the upper and lower regions of the graph was noted
earlier. For the voltage VD there is also a measurable change in scale between the right-hand
region of the graph and the left-hand region. For positive values of VD the scale is in tenths
of volts, and for the negative region it is in tens of volts.
It is important to note in Fig. 1.14b how:
The defined direction of conventional current for the positive voltage region matches
the arrowhead in the diode symbol.
This will always be the case for a forward-biased diode. It may also help to note that the
forward-bias condition is established when the bar representing the negative side of the
applied voltage matches the side of the symbol with the vertical bar.
Going back a step further by looking at Fig. 1.14b, we find a forward-bias condition is
established across a p-n junction when the positive side of the applied voltage is applied to
the p-type material (noting the correspondence in the letter p) and the negative side of the
applied voltage is applied to the n-type material (noting the same correspondence).
It is particularly interesting to note that the reverse saturation current of the commercial
unit is significantly larger than that of ls in Shockley's equation. In fact,
The actual reverse saturation current of a commercially available diode will normally
be measurably larger than that appearing as the reverse saturation current in
Shockley's equation.
This increase in level is due to a wide range of factors that include
- leakage currents
- generation of carriers in the depletion region
- higher doping levels that result in increased levels of reverse current
- sensitivity to the intrinsic level of carriers in the component materials by a squared
factor----double the intrinsic level, and the contribution to the reverse current could
increase by a factor of four.
- a direct relationship with the junction area----double the area of the junction, and
the contribution to the reverse current could double. High-power devices that have
larger junction areas typically have much higher levels of reverse current.
- temperature sensitivity-for every 5°C increase in current, the level of reverse sat-
uration current in Eq. 1.2 will double, whereas a 10°C increase in current will result
in doubling of the actual reverse current of a diode.
Note in the above the use of the terms reverse saturation current and reverse current. The
former is simply due to the physics of the situation, whereas the latter includes all the other
possible effects that can increase the level of current.
We will find in the discussions to follow that the ideal situation is for ls to be OA in the
reverse-bias region. The fact that it is typically in the range of 0.01 pA to 10 pA today as
compared to 0.1 µ,A to 1 µ,A a few decades ago is a credit to the manufacturing industry.
Comparing the common value of 1 nA to the I -µ,A level of years past shows an improve-
ment factor of 100,000.
16 SEMICONDUCTOR Breakdown Region
DIODES
Even though the scale of Fig. 1.15 is in tens of volts in the negative region, there is a point
where the application of too negative a voltage with the reverse polarity will result in a
sharp change in the characteristics, as shown in Fig. 1.17. The current increases at a very
rapid rate in a direction opposite to that of the positive voltage region. The reverse-bias
potential that results in this dramatic change in characteristics is called the breakdown
potential and is given the label VBv•

/
I \
I I
I I
I I
\ I
\ /-Zener
/ region

FIG. 1.17
Breakdown region.

As the voltage across the diode increases in the reverse-bias region, the velocity of the
minority carriers responsible for the reverse saturation current ls will also increase. Eventu-
ally, their velocity and associated kinetic energy (WK = ½mv2) will be sufficient to release
additional carriers through collisions with otherwise stable atomic structures. That is, an
ionization process will result whereby valence electrons absorb sufficient energy to leave the
parent atom. These additional carriers can then aid the ionization process to the point where
a high avalanche current is established and the avalanche breakdown region determined.
The avalanche region (VBv) can be brought closer to the vertical axis by increasing the
doping levels in the p- and n-type materials. However, as VBv decreases to very low levels,
such as -5 V, another mechanism, called Zener breakdown, will contribute to the sharp
change in the characteristic. It occurs because there is a strong electric field in the region
of the junction that can disrupt the bonding forces within the atom and "generate" carriers.
Although the Zener breakdown mechanism is a significant contributor only at lower levels
of VBv, this sharp change in the characteristic at any level is called the Zener region, and
diodes employing this unique portion of the characteristic of a Jrn junction are called Zener
diodes. They are described in detail in Section 1.15.
The breakdown region of the semiconductor diode described must be avoided if the
response of a system is not to be completely altered by the sharp change in characteristics
in this reverse-voltage region.
The maximum reverse-bias potential that can be applied before entering the break-
down region is called the peak inverse voltage (referred to simply as the PIV rating) or
the peak reverse voltage (denoted the PRV rating).
If an application requires a PIV rating greater than that of a single unit, a number of
diodes of the same characteristics can be connected in series. Diodes are also connected in
parallel to increase the current-carrying capacity.
In general, the breakdown voltage of GaAs diodes is about 10% higher those for silicon
diodes but after 200% higher than levels for Ge diodes.

Ge, Si, and GaAs

The discussion thus far has solely used Si as the base semiconductor material. It is now impor-
tant to compare it to the other two materials of importance: GaAs and Ge. A plot comparing
the characteristics of Si, GaAs, and Ge diodes is provided in Fig. 1.18. The curves are not
 SEMICONDUCTOR DIODE 17
:lv(mA!

30

25

20
Ge - Si GaAs
15

10

5
l
I I I

Vnv(GaAs) JO ov I

ls (GaAs) 0.3 ~o· .a l.Z V~(V)

Vic(Ge) '"' VK (Si) VK(GaAs)
5 pA
Vnv(Si)
lOpA
I, (Si)

Vnv(Ge)
1 µA
I, (Ge)

FIG. 1.18
Comparison of Ge, Si, and GaAs commercial diodes.

simply plots of Eq. 1.2 but the actual response of commercially available units. The total reverse TABLE 1.3
current is shown and not simply the reverse saturation current. It is immediately obvious that Knee Voltages VK
the point of vertical rise in the characteristics is different for each material, although the general
Semiconductor
shape of each characteristic is quite similar. Germanium is closest to the vertical axis and GaAs
is the most distant. As noted on the curves, the center of the knee (hence the K is the notation Ge 0.3
VK) of the curve is about 0.3 V for Ge, 0.7 V for Si, and 1.2 V for GaAs (see Table 1.3). Si 0.7
The shape of the curve in the reverse-bias region is also quite similar for each material, GaAs 1.2
but notice the measurable difference in the magnitudes of the typical reverse saturation
currents. For GaAs, the reverse saturation current is typically about 1 pA, compared to 10 pA
for Si and 1 µ,A for Ge, a significant difference in levels.
Also note the relative magnitudes of the reverse breakdown voltages for each material.
GaAs typically has maximum breakdown levels that exceed those of Si devices of the same
power level by about 10%, with both having breakdown voltages that typically extend be-
tween 50 V and 1 kV. There are Si power diodes with breakdown voltages as high as 20 kV.
Germanium typically has breakdown voltages ofless than 100 V, with maximums around
400 V. The curves of Fig. 1.18 are simply designed to reflect relative breakdown voltages
for the three materials. When one considers the levels of reverse saturation currents and
breakdown voltages, Ge certainly sticks out as having the least desirable characteristics.
A factor not appearing in Fig. 1.18 is the operating speed for each material-an impor- TABLE 1.4
tant factor in today's market. For each material, the electron mobility factor is provided Electron Mobility /1,n
in Table 1.4. It provides an indication of how fast the carriers can progress through the
material and therefore the operating speed of any device made using the materials. Quite Semiconductor
obviously, GaAs stands out, with a mobility factor more than five times that of silicon and Ge 3900
twice that of germanium. The result is that GaAs and Ge are often used in high-speed ap- Si 1500
plications. However, through proper design, careful control of doping levels, and so on, GaAs 8500
silicon is also found in systems operating in the gigahertz range. Research today is also
looking at compounds in groups 111-V that have even higher mobility factors to ensure that
industry can meet the demands of future high-speed requirements.
18 SEMICONDUCTOR
DIODES EXAMPLE 1.2 Using the curves of Fig 1.18:
a. Determine the voltage across each diode at a current of 1 mA.
b. Repeat for a current of 4 mA.
c. Repeat for a current of 30 mA.
d. Determine the average value of the diode voltage for the range of currents listed above.
e. How do the average values compare to the knee voltages listed in Table 1.3?

Solution:
a. Yv(Ge) = 0.2 V, Yv(Si) = 0.6 V, Vv (GaAs) = 1.1 V
b. Yv(Ge) = 0.3 V, Yv(Si) = 0.7 V, Vv (GaAs) = 1.2 V
c. Yv(Ge) = 0.42 V, Yv(Si) = 0.82 V, Vv (GaAs) = 1.33 V
d. Ge: Yav = (0.2 V + 0.3 V + 0.42 V)/3 = 0.307 V
Si: Yav = (0.6 V + 0.7 V + 0.82 V)/3 = 0.707 V
GaAs: Yav = (1.1 V + 1.2 V + 1.33 V)/3 = 1.21 V
e. Very close correspondence. Ge: 0.307 V vs. 0.3, V, Si: 0.707 V vs. 0.7 V, GaAs: 1.21 V
vs. 1.2 V.

Temperature Effects
Temperature can have a marked effect on the characteristics of a semiconductor diode, as
demonstrated by the characteristics of a silicon diode shown in Fig. 1.19:
In the forward-bias region the characteristics of a silicon diode shift to the left at a rate
of 2.5 m V per centigrade degree increase in temperature.

IIII
IIII
1-++++++-1-+-1-++++++++-1-+-1-++++++-l+-lf-H-I I I I l+++-1-+-1-++++++++-1-+-1-++++++-l-+-l-+-l
1-++++++-1-+-1-++++++++-1-+-1-++++++++-lf-+-1' [ D (mA) +++-1-+-1-++++++++-1-+-1-++++++-1-+-1-+-1
:=!=!=!!!=!=!=!=:=:=:=!=!!!=!=!=!=!~~!=!=!!!!=!=!=!~~:·shift to left= (100°C)(-2.5 mV/°C) = -0.35 V +++++-1--1
l-++-++-++-l-+-l-++++++-++-l-+-l-++-++-++-l+-11-+I-·+++++-11-¥.i'i_A =-i=E'-
,+- I+l ,,l-+++++-1-+-1-++++++++-1-+-H
1-++++++-1-+-1-++++++++-1-+-1-+++++++-+- 30 l++++-H+ : +-1-+-1-++++++++-11-1-1-++++-+
u! u ou;
::::::::::=:::::::::::::::::::::=~::::::::::=:::::::::::::::::::::::::~::::::::::=::::::::::::::::::::::::~:::::=::=::=:::::::::= tr) J ~ ~ V) ~
O
-·+--1+--1+,_+,_;_~l-+l-+l-++--++--++--l+--l+,_--:_;_~~l-+I-++--+-l+-l
1-++++++-1-+-l-l-+++++++-1-+-1-++++++++-11-1-1-+-++++ ~ , - ~ ~:~1-1-1-++++++-1-+-1-++++++++-1
1-++++++-1-+-l-l-+++++++-1-+-1-+++++++-+-25+++++-1-+-F-1-l-l-+-!i+-t+-l-+-1-++++++-1+-11-1-1-+++-1

1-++++++-1-+-l-l-+++++++-1-+-1-+++++++-+-20 +-++-1-+-l-l-r-+--ttt--1-<',+-1-+-1-++++++-1+-11-1-1-++++-+-1
1-++++++-1-+-1-1-+++++++-1-+-1-++++++-1+-11-1-1 Increasing : Decreasing +-++-11-1-1-++++++-1--1
+-+-+-+--+-+-+-+-+-+-+-+-+-+-+--+-+-+-+-+-+-+-+-+--+-+-+-+-+-+:1-1-:,tempe~atu~{ -- -~: temperature -+-+-+-+-+-+-+-+-+--+-+-+-+--1
+-+-+-+--+-+-+-+-+-+-+-+-+-+-+--+-+-+-+-+-+-+-+-+--+-+-+-+-+- 15 +-+--+-+-+-.......
1,+-+-++-+:• +-+--+-+-+-+-+-+-+-+-+--+-+-+-+-+-+-+-+-+-+-1
I I II
I I
I I •
+-+-+-+--+-+-+-+-+-+-+-+-+-+-+--+-+-+-+-+-+-+-+-+--+-+-+-+-+- l0 +-+--+-+-+-1-1-+-+1,+-+-~
: -+-+--1--1+-+-+-+-+-+--+-+-+-+-+-+-+-+-+-+-+-+-+
I I
I II ll
1 : R" Silicon diode at

+-+-+-+--+-+-+-+-+-t-+-+-+-+-+--+-+-+ls_,=-+-
0.1\,-'...
l +-
pA
+-+-+--++--++--+~~:_5
,_....,_-++--++--++--+-+_, , ;:::_H
#I-I"::;:::-:. : •--1-- room temperature -+-+-+-+-+-+--1--11-1-t--l

1-++++++-1-+-H 40 ,-++- 30 ,-++- 20 1- 101-+++-l-+-l.-',--l_

..,,_
,,,l-+~
"'•'+++++-1-+-1-++++++++-ll-l-l-++++-+-l
~1="!tiTl--t-rt"tt1-+tt+--ill1-1-~
Silicon diode at -......::::
J• '+:++:+:+:4~~1-ii~ ++ci.1
V +++++-1-+-·v~ cv )++++++++-+-+-+--1--1--1
1 +-1+-1--i
j,- +-1-++-++ l1-p1-
A+--++---t+--i~~-!"""...!"""-!""..:1-+1--+1--++--1.;+++++-1-+-1-++++++-1-+-H
room temperature~~t--~j",.!,.!~~;J.i'=1J='1ncreasing
l-++-+--+-+-+-+-+-+-+-+-+-++-+-++-1-+-H:!-•f-t,,emperature -t-1-++++++-1-+-1-++++++++-1-+-1-++++++-1+-11-1-H
1-++++++ Increasing ++..++++f-1-1-++++++-1-.Hl-l-1-++++++-1-+-1-++++++++-1-+-1-++++++-1-+-H
1-++++++ temperature +++++++-1-+-1-++++++++-1-+-1-++++++-1-+-1-1-+++++++-1-+-1-+-1
.
: 'r-'

--- -- -~ -
t-+-l-+-ll-+-l--1--1--1--1-+-+-+-+-Ht-++++,·~a:::~~~1~1. J¾Ac=t:::r:::r:::r:::rtttttttttttttl='l='l='l='~~~:::t:::t~
1-++++++-1-+-t-+,.l++++-ttt+-1-+-t-- -75°C +++++-l-l--il-l-l-++++++-l-+-l-++++++++-l-+-l-++++++-l-+-H
II

,I
1-+-+-+-+-+--+-+-+-+~ l-+-+-+-+-+0 ,
25°C
I I l-+-t-+-+-+-+-+--+-+-+-+-t--11-+-+-+-+-+--+-+-+-+-+-+-+-+-+-+-+--+-+-+-+-+-+-+-+-+--+-+-+-+-+-t--1

+-+-+-+-+-+--+-+-+-+-+-: f~5;S +-+--+-+-+-+-+-+-+-+-+--+-+--l--lt-+-l-+-+-+--+-+-+-+-+-t-+-+-+-+-+--+-+-+-+-+-+-+-+-+--+-+-+-+--l--lt-+-t-+-I

FIG. 1.19
Variation in Si diode characteristics with temperature change.
 An increase from room temperature (20°C) to 100°C (the boiling point of water) results SEMICONDUCTOR DIODE 19
in a drop of 80(2.5 mV) = 200 mV, or 0.2 V, which is significant on a graph scaled in
tenths of volts. A decrease in temperature has the reverse effect, as also shown in the figure:
In the reverse-bias region the reverse current of a silicon diode doubles for every I 0°C
rise in temperature.
For a change from 20°C to 100°C, the level of ls increases from 10 nA to a value of
2.56 µ,A, which is a significant, 256-fold increase. Continuing to 200°C would result in a
monstrous reverse saturation current of 2.62 mA. For high-temperature applications one
would therefore look for Si diodes with room-temperature ls closer to 10 pA, a level com-
monly available today, which would limit the current to 2.62 µA. It is indeed fortunate that
both Si and GaAs have relatively small reverse saturation currents at room temperature.
GaAs devices are available that work very well in the -200°C to +200°C temperature
range, with some having maximum temperatures approaching 400°C. Consider, for a mo-
ment, how huge the reverse saturation current would be if we started with a Ge diode with
a saturation current of 1 µ,A and applied the same doubling factor.
Finally, it is important to note from Fig. 1.19 that:
The reverse breakdown voltage of a semiconductor diode will increase or Russell Ohl (1898-1987)
decrease with temperature. American (Allentown, PA;
Holmdel, NJ; Vista, CA) Army
However, if the initial breakdown voltage is less than 5 V, the breakdown voltage may
Signal Corps, University of
actually decrease with temperature. The sensitivity of the breakdown potential to changes Colorado, Westinghouse, AT&T,
of temperature will be examined in more detail in Section 1.15. Bell Labs Fellow, Institute of
Radio Engineers-1955
(Courtesy of AT&T Archives
Summary History Center.)
A great deal has been introduced in the foregoing paragraphs about the construction of a Although vacuum tubes were
semiconductor diode and the materials employed. The characteristics have now been pre- used in all forms of communication
sented and the important differences between the response of the materials discussed. It is in the 1930s, Russell Ohl was deter-
now time to compare the p-n junction response to the desired response and reveal the pri- mined to demonstrate that the future
mary functions of a semiconductor diode. of the field was defined by semicon-
Table 1.5 provides a synopsis of material regarding the three most frequently used semi- ductor crystals. Germanium was not
conductor materials. Figure 1.20 includes a short biography of the first research scientist to immediately available for his
discover the p-n junction in a semiconductor material. research, so he turned to silicon, and
found a way to raise its level of
purity to 99.8%, for which he
received a patent. The actual discov-
TABLE 1.5 ery of the p-n junction, as often
The Current Commercial Use of Ge, Si, and GaAs happens in scientific research, was
the result of a set of circumstances
Ge: Germanium is in limited production due to its temperature sensitivity and high that were not planned. On February
reverse saturation current. It is still commercially available but is limited to 23, 1940, Ohl found that a silicon
some high-speed applications (due to a relatively high mobility factor) and crystal with a crack down the mid-
applications that use its sensitivity to light and heat such as photodetectors dle would produce a significant rise
and security systems. in current when placed near a source
Si: Without question the semiconductor used most frequently for the full range of of light. This discovery led to fur-
electronic devices. It has the advantage of being readily available at low cost ther research, which revealed that
and has relatively low reverse saturation currents, good temperature character- the purity levels on each side of the
istics, and excellent breakdown voltage levels. It also benefits from decades of crack were different and that a
enormous attention to the design of large-scale integrated circuits and process- barrier was formed at the junction
ing technology. that allowed the passage of current
GaAs: Since the early 1990s the interest in GaAs has grown in leaps and bounds, and it in only one direction-the first
will eventually take a good share of the development from silicon devices, solid-state diode had been identified
especially in very large scale integrated circuits. Its high-speed characteristics and explained. In addition, this sen-
are in more demand every day, with the added features of low reverse satura- sitivity to light was the beginning of
tion currents, excellent temperature sensitivities, and high breakdown voltages. the development of solar cells. The
More than 80% of its applications are in optoelectronics with the development results were quite instrumental in
of light-emitting diodes, solar cells, and other photodetector devices, but that the development of the transistor in
will probably change dramatically as its manufacturing costs drop and its use 1945 by three individuals also work-
in integrated circuit design continues to grow; perhaps the semiconductor ing at Bell Labs.
material of the future.
FlCi. 1.20
20 SEMICONDUCTOR
DIODES
1.7 IDEAL VERSUS PRACTICAL
•
In the previous section we found that a p-n junction will permit a generous flow of charge
when forward-biased and a very small level of current when reverse-biased. Both condi-
tions are reviewed in Fig. 1.21, with the heavy current vector in Fig. 1.21a matching the
direction of the arrow in the diode symbol and the significantly smaller vector in the oppo-
site direction in Fig. 1.21b representing the reverse saturation current.
An analogy often used to describe the behavior of a semiconductor diode is a mechanical
switch. In Fig. 1.21a the diode is acting like a closed switch permitting a generous flow of
charge in the direction indicated. In Fig. 1.21 b the level of current is so small in most cases
that it can be approximated as O A and represented by an open switch.

Vv Vv
+0 +0
IJiil 0 0
IJiil
-----+- ~
Iv ls

~
y 0---0

(a) (b)

FIG. 1.21
Ideal semiconductor diode: (a) forward-
biased; (b) reverse-biased.

In other words:
The semiconductor diode behaves in a manner similar to a mechanical switch in that it
can control whether current will flow between its two terminals.
However, it is important to also be aware that:
The semiconductor diode is different from a mechanical switch in the sense that when
the switch is closed it will only permit current to flow in one direction.
Ideally, if the semiconductor diode is to behave like a closed switch in the forward-bias
region, the resistance of the diode should be O !1. In the reverse-bias region its resistance
should be oofl to represent the open-circuit equivalent. Such levels of resistance in the forward-
and reverse-bias regions result in the characteristics of Fig. 1.22.

Ideal characteristics

0>---<1•-•~•----<0

0-----0 0------0
~
Actual characteristics
I,=" OmA

FIG. 1.22
Ideal versus actual semiconductor characteristics.
 The characteristics have been superimposed to compare the ideal Si diode to a real-world RESISTANCE LEVELS 21
Si diode. First impressions might suggest that the commercial unit is a poor impression of
the ideal switch. However, when one considers that the only major difference is that the
commercial diode rises at a level of 0.7 V rather than OV, there are a number of similarities
between the two plots.
When a switch is closed the resistance between the contacts is assumed to be O 0. At
the plot point chosen on the vertical axis the diode current is 5 mA and the voltage across
the diode is O V. Substituting into Ohm's law results in
Vv OV
Rp = - = -- = 0 0 (short-circuit equivalent)
Iv 5mA
In fact:
At any current level on the vertical line, the voltage across the ideal diode is O V and
the resistance is O 0.
For the horizontal section, if we again apply Ohm's law, we find
Vv 20V
RR= Iv = OmA ~ 00 0 (open-circuitequivalent)
Again:
Because the current is O mA anywhere on the horiwntal line, the resistance is
considered to be infinite ohms (an open-circuit) at any point on the axis.
Due to the shape and the location of the curve for the commercial unit in the forward-bias
region there will be a resistance associated with the diode that is greater than O0. However,
if that resistance is small enough compared to other resistors of the network in series with
the diode, it is often a good approximation to simply assume the resistance of the com-
mercial unit is O 0. In the reverse-bias region, if we assume the reverse saturation current
is so small it can be approximated as O mA, we have the same open-circuit equivalence
provided by the open switch.
The result, therefore, is that there are sufficient similarities between the ideal switch and
the semiconductor diode to make it an effective electronic device. In the next section the
various resistance levels of importance are determined for use in the next chapter, where
the response of diodes in an actual network is examined.

1.8 RESISTANCE LEVELS

•
As the operating point of a diode moves from one region to another the resistance of the
diode will also change due to the nonlinear shape of the characteristic curve. It will be dem-
onstrated in the next few paragraphs that the type of applied voltage or signal will define the
resistance level of interest. Three different levels will be introduced in this section, which
will appear again as we examine other devices. It is therefore paramount that their determi-
nation be clearly understood.

DC or Static Resistance
The application of a de voltage to a circuit containing a semiconductor diode will result in
an operating point on the characteristic curve that will not change with time. The resistance
of the diode at the operating point can be found simply by finding the corresponding levels
of Vv and Iv as shown in Fig. 1.23 and applying the following equation:

(1.4)

The de resistance levels at the knee and below will be greater than the resistance levels
obtained for the vertical rise section of the characteristics. The resistance levels in the
reverse-bias region will naturally be quite high. Since ohmmeters typically employ a rela-
tively constant-current source, the resistance determined will be at a preset current level
(typically, a few milliamperes).
22 SEMICONDUCTOR ! 0 (mA)
DIODES

----'"'1 0

FIG. 1.23
Determining the de resistance of a diode at a
particular operating point.

In general, therefore, the higher the current through a diode, the lower is the de resis-
tance level
Typically, the de resistance of a diode in the active (most utilized) will range from about
10 ll to 80 n.

EXAMPLE 1.3 Determine the de resistance levels for the diode of Fig. 1.24 at
a. Iv= 2 mA (low level)
b. Iv= 20 mA (high level)
c. Vv = -10 V (reverse-biased)

/0 (mA)

30

-silicon
20 ---------

10

-lOV 2
0 0.5 0.8 Vo (V)
- - - - - - ,~!µA

FIG. 1.24
Example 1.3.

Solution:
a. Atlv = 2 mA, Vv = 0.5 V (from the curve) and
Vv 0.5V
Rv = - = - - = 250 0
Iv 2mA
b. Atlv = 20 mA, Vv = 0.8 V (from the curve) and
Vv 0.8 V
Rv=-=--=40O
Iv 20mA
c. At Vv = -10 V, Iv= -Is= -1 µ,A (from the curve) and RESISTANCE LEVELS 23
Vv lOV
Rv =- = - - = 10 Mil
Iv 1 µ,A
clearly supporting some of the earlier comments regarding the de resistance levels of a
diode.

AC or Dynamic Resistance
Eq. (1.4) and Example 1.3 reveal that
the de resistance of a diode is independent of the shape of the characteristic in the
region surrounding the point of interest.
If a sinusoidal rather than a de input is applied, the situation will change completely. The
varying input will move the instantaneous operating point up and down a region of the char-
acteristics and thus defines a specific change in current and voltage as shown in Fig. 1.25.
With no applied varying signal, the point of operation would be the Q-point appearing on
Fig. 1.25, determined by the applied de levels. The designation Q-point is derived from the
word quiescent, which means "still or unvarying."

Diode characteristic ~

r------------ ~ Tangent line

l_----
Md (_)-point

~de operation)

FIG. 1.25
Defining the dynamic or ac resistance.

A straight line drawn tangent to the curve through the Q-point as shown in Fig. 1.26
will define a particular change in voltage and current that can be used to determine the ac
or dynamic resistance for this region of the diode characteristics. An effort should be made
to keep the change in voltage and current as small as possible and equidistant to either side
of the Q-point. In equation form,

(1.5)

where .:l signifies a finite change in the quantity.

The steeper the slope, the lower is the value of .:l Va for the same change in .:lia and the FIG. 1.26
lower is the resistance. The ac resistance in the vertical-rise region of the characteristic is Determining the ac resistance at a
therefore quite small, whereas the ac resistance is much higher at low current levels. Q-point.
In general, therefore, the lower the Q-point of operation (smaller current or lower
voltage), the higher is the ac resistance.
24 SEMICONDUCTOR
DIODES EXAMPLE 1.4 For the characteristics of Fig. 1.27:
a. Determine the ac resistance atlv = 2 mA.
b. Determine the ac resistance at Iv = 25 mA.
c. Compare the results of parts (a) and (b) to the de resistances at each current level.

lv(mA)

30 ------------------------

2 5 1 - - - - - - - - - - - - - - -, Aid
I
I
I
I
20 ------------------------ I

~ AVd

15

10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

'--y--1
AVd

FIG. 1.27
Example 1.4.

Solution:
a. For Iv = 2 mA, the tangent line at Iv = 2 mA was drawn as shown in Fig. 1.27 and a
swing of 2 mA above and below the specified diode current was chosen. At Iv = 4 mA,
Vv = 0.76 V, and at Iv= 0 mA, Vv = 0.65 V. The resulting changes in current and
voltage are, respectively,
Md = 4 mA - 0 mA = 4 mA
and Ll vd = 0.76 V - 0.65 V = 0.11 V
and the ac resistance is
LlVd 0.11 V
rd =- = - - = 27.5 0
Md 4mA
b. For Iv = 25 mA, the tangent line at Iv = 25 mA was drawn as shown in Fig. 1.27 and
a swing of 5 mA above and below the specified diode current was chosen. At Iv = 30 mA,
Vv = 0.8 V, and at Iv= 20 mA, Vv = 0.78 V. The resulting changes in current and
voltage are, respectively,
Md= 30mA - 20mA = lOmA
and LlVd = 0.8V - 0.78V = 0.02V
and the ac resistance is
_ Ll Vd _ 0.02 V _ O
rd - -- - --- - 2
Md lOmA
c. For Iv= 2 mA, Vv = 0.7 V and
Vv 0.7V
Rv = - = - - = 350 0
Iv 2mA
which far exceeds the rd of 27 .5 fl.
 For Iv= 25 mA, Vv = 0.79 V and RESISTANCE LEVELS 25
Vv 0.79V
Rv =- = - - = 31.62 fi
Iv 25mA
which far exceeds the rd of 2 n.

We have found the dynamic resistance graphically, but there is a basic definition in dif-
ferential calculus that states:
The derivative of a function at a point is equal to the slope of the tangent line drawn
at that point.
Equation (1.5), as defined by Fig. 1.26, is, therefore, essentially finding the derivative of
the function at the Q-point of operation. If we find the derivative of the general equation
(1.2) for the semiconductor diode with respect to the applied forward bias and then invert
the result, we will have an equation for the dynamic or ac resistance in that region. That is,
taking the derivative of Eq. (1.2) with respect to the applied bias will result in
d d
dVv (Iv) = dVv [Is(eVv/nVr - 1)]
and
dlv 1
-- = --;-;-(Iv + ls)
dVv nVr
after we apply differential calculus. In general, Iv >> ls in the vertical-slope section of
the characteristics and
dlv Iv
=
dVv nVr
Flipping the result to define a resistance ratio (R = Vil) gives
dVv nVr
- - = rd= -
dlv Iv
Substituting n = I and Yr == 26 mV from Example 1.1 results in

(1.6)

The significance of Eq. (1.6) must be clearly understood. It implies that

the dynamic resistance can be found simply by substituting the quiescent value of the
diode current into the equation.
There is no need to have the characteristics available or to worry about sketching tangent
lines as defined by Eq. (1.5). It is important to keep in mind, however, that Eq. (1.6) is
accurate only for values of Iv in the vertical-rise section of the curve. For lesser values of
Iv, n = 2 (silicon) and the value of rd obtained must be multiplied by a factor of 2. For
small values of Iv below the knee of the curve, Eq. (1.6) becomes inappropriate.
All the resistance levels determined thus far have been defined by the p-n junction and
do not include the resistance of the semiconductor material itself (called body resistance)
and the resistance introduced by the connection between the semiconductor material and the
external metallic conductor (called contact resistance). These additional resistance levels
can be included in Eq. (1.6) by adding a resistance denoted rB:

, 26mV
rd= - - - + rB ohms (1.7)
Iv

The resistance r' d, therefore, includes the dynamic resistance defined by Eq. (1.6) and
the resistance rB just introduced. The factor rB can range from typically 0.1 n for high-
power devices to 2 n for some low-power, general-purpose diodes. For Example 1.4 the ac
resistance at 25 mA was calculated to be 2 n. Using Eq. (1.6), we have
_ 26 mV _ 26 mV _ 0
rd-------104..u
Iv 25 mA •
26 SEMICONDUCTOR The difference of about 1 0 could be treated as the contribution of rs.
DIODES For Example 1.4 the ac resistance at 2 mA was calculated to be 27.5 0. Using Eq. (1.6)
but multiplying by a factor of 2 for this region (in the knee of the curve n = 2),

rd= 2( 2 \:v) = 2( ;=)

2 = 2(13 0) = 26 fl

The difference of 1.5 0 could be treated as the contribution due to rs.

In reality, determining rd to a high degree of accuracy from a characteristic curve using Eq.
( 1.5) is a difficult process at best and the results have to be treated with skepticism. At low lev-
els of diode current the factor rs is normally small enough compared to rd to permit ignoring
its impact on the ac diode resistance. At high levels of current the level of rs may approach that
of rd, but since there will frequently be other resistive elements of a much larger magnitude in
series with the diode, we will assume in this book that the ac resistance is determined solely
by rd, and the impact of rs will be ignored unless otherwise noted. Technological improve-
ments of recent years suggest that the level of rs will continue to decrease in magnitude and
eventually become a factor that can certainly be ignored in comparison to rd·
The discussion above centered solely on the forward-bias region. In the reverse-bias
region we will assume that the change in current along the ls line is nil from O V to the
Zener region and the resulting ac resistance using Eq. (1.5) is sufficiently high to permit
the open-circuit approximation.
Typically, the ac resistance of a diode in the active region will range from about 1 0 to 100 0.

Average AC Resistance
If the input signal is sufficiently large to produce a broad swing such as indicated in Fig.
1.28, the resistance associated with the device for this region is called the average ac resis-
tance. The average ac resistance is, by definition, the resistance determined by a straight
line drawn between the two intersections established by the maximum and minimum values
of input voltage. In equation form (note Fig. 1.28),

avdl
' a v = ¼ pt.topt.
(1.8)

For the situation indicated by Fig. 1.28,

Md= 17mA - 2mA = 15mA

lv(mA)

20

15

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

y
LlVd

FIG. 1.28
Determining the average ac resistance between indicated limits.
and Li Vd = 0.725 V - 0.65 V = 0.075 V DIODE EQUIVALENT 27
Li Vd 0.075 V CIRCUITS
with Tav = Lild = 15 mA = S fl
If the ac resistance (rd) were determined at/D = 2 mA, its value would be more than 5 0,
and if determined at 17 mA, it would be less. In between, the ac resistance would make the
transition from the high value at 2 mA to the lower value at 17 mA. Equation (1. 7) defines
a value that is considered the average of the ac values from 2 mA to 17 mA. The fact that
one resistance level can be used for such a wide range of the characteristics will prove quite
useful in the definition of equivalent circuits for a diode in a later section.
As with the de and ac resistance levels, the lower the level of currents used to determine
the average resistance, the higher is the resistance level.

Summary Table
Table 1.6 was developed to reinforce the important conclusions of the last few pages and
to emphasize the differences among the various resistance levels. As indicated earlier, the
content of this section is the foundation for a number of resistance calculations to be per-
formed in later sections and chapters.

TABLE 1.6
Resistance Levels

Special Graphical
Type Equation Characteristics Determination

DC or static Defined as a point on the

characteristics

ilVd 26mV
AC or dynamic rd=--=-- Defined by a tangent line
Md Iv at the Q-point

Average ac ilVdl Defined by a straight

rav = .:lid pt.to pt.
line between limits of
operation

1.9 DIODE EQUIVALENT CIRCUITS

•
An equivalent circuit is a combination of elements properly chosen to best represent the
actual terminal characteristics of a device or system in a particular operating region.
In other words, once the equivalent circuit is defined, the device symbol can be removed
from a schematic and the equivalent circuit inserted in its place without severely affecting
the actual behavior of the system. The result is often a network that can be solved using
traditional circuit analysis techniques.
28 SEMICONDUCTOR Piecewise-Linear Equivalent Circuit
DIODES
One technique for obtaining an equivalent circuit for a diode is to approximate the charac-
teristics of the device by straight-line segments, as shown in Fig. 1.29. The resulting equiv-
alent circuit is called a piecewise-linear equivalent circuit. It should be obvious from Fig.
1.29 that the straight-line segments do not result in an exact duplication of the actual char-
acteristics, especially in the knee region. However, the resulting segments are sufficiently
close to the actual curve to establish an equivalent circuit that will provide an excellent first
approximation to the actual behavior of the device. For the sloping section of the equiva-
lence the average ac resistance as introduced in Section 1.8 is the resistance level appearing
in the equivalent circuit of Fig. 1.28 next to the actual device. In essence, it defines the resis-
tance level of the device when it is in the "on" state. The ideal diode is included to establish
that there is only one direction of conduction through the device, and a reverse-bias condi-
tion will result in the open-circuit state for the device. Since a silicon semiconductor diode
does not reach the conduction state until VD reaches 0. 7 V with a forward bias (as shown in
Fig. 1.29), a battery VK opposing the conduction direction must appear in the equivalent
circuit as shown in Fig. 1.30. The battery simply specifies that the voltage across the device
must be greater than the threshold battery voltage before conduction through the device in
the direction dictated by the ideal diode can be established. When conduction is established
the resistance of the diode will be the specified value of rav·

I0 (mA)

10 --------------

rav

0 0.7 V 0.8 V V0 (V)

(VK)

FIG. 1.29
Defining the piecewise-linear equivalent
circuit using straight-line segments to approximate
the characteristic curve.

+
VK ,.,,--Ideal diode
Tav /

==> ~ 11 ~
___.. 0.7 V 10 Q
Iv

FIG. 1.30
Components of the piecewise-linear equivalent circuit.

Keep in mind, however, that VK in the equivalent circuit is not an independent voltage
source. If a voltmeter is placed across an isolated diode on the top of a laboratory bench, a
reading of 0. 7 V will not be obtained. The battery simply represents the horizontal offset of
the characteristics that must be exceeded to establish conduction.
The approximate level of rav can usually be determined from a specified operating
point on the specification sheet (to be discussed in Section 1.10). For instance, for a sili-
con semiconductor diode, if JF = 10 mA (a forward conduction current for the diode) at
Vv = 0.8 V, we know that for silicon a shift of 0.7 Vis required before the characteristics DIODE EQUIVALENT 29
rise, and we obtain CIRCUITS

dVdl 0.8V - 0.7V = 0.1 V = lOO

fav = dld pt. topt. lOmA - 0mA lOmA
as obtained for Fig. 1.29.
If the characteristics or specification sheet for a diode is not available the resistance r av
can be approximated by the ac resistance rd•

Simplified Equivalent Circ:uit

For most applications, the resistance rav is sufficiently small to be ignored in comparison to
the other elements of the network. Removing 'av from the equivalent circuit is the same as
implying that the characteristics of the diode appear as shown in Fig. 1.31. Indeed, this
approximation is frequently employed in semiconductor circuit analysis as demonstrated in
Chapter 2. The reduced equivalent circuit appears in the same figure. It states that a forward-
biased silicon diode in an electronic system under de conditions has a drop of 0. 7 V across
it in the conduction state at any level of diode current (within rated values, of course).

FIG. 1.31
Simplified equivalent circuit for the silicon semiconductor diode.

Ideal Equivalent Circ:uit

Now that rav has been removed from the equivalent circuit, let us take the analysis a step
further and establish that a 0.7-V level can often be ignored in comparison to the applied
voltage level. In this case the equivalent circuit will be reduced to that of an ideal diode as
shown in Fig. 1.32 with its characteristics. In Chapter 2 we will see that this approximation
is often made without a serious loss in accuracy.

+
0>-----~------0

~ ------------- Ideal diode

FIG. 1.32
Ideal diode and its characteristics.

In industry a popular substitution for the phrase "diode equivalent circuit" is diode model-
a model by definition being a representation of an existing device, object, system, and so on.
In fact, this substitute terminology will be used almost exclusively in the chapters to follow.

Summary Table
For clarity, the diode models employed for the range of circuit parameters and applications
are provided in Table 1.7 with their piecewise-linear characteristics. Each will be investi-
gated in greater detail in Chapter 2. There are always exceptions to the general rule, but it
30 SEMICONDUCTOR TABLE 1.7
DIODES Diode Equivalent Circuits (Models)

Type Conditions Model Characteristics

Piecewise-linear model <>+It!~

VK ' "'~ lde.11
oiode

Simplified model Rnetwork >> Tav ~ I j11--

_ ----l}llti-------o
o
VK ldc:il
diode

Ideal device Rnetwork

Enetwork
>>
>>
Tav
VK
0- ----..l}llt-- -o
Ideal
diode
L 0 v,,

is fairly safe to say that the simplified equivalent model will be employed most frequently
in the analysis of electronic systems, whereas the ideal diode is frequently applied in the
analysis of power supply systems where larger voltages are encountered.

1.10 TRANSITION AND DIFFUSION CAPACITANCE

It is important to realize that:
Every electronic or electrical device is frequency sensitive.
•
That is, the terminal characteristics of any device will change with frequency. Even the
resistance of a basic resistor, as of any construction, will be sensitive to the applied fre-
quency. At low to mid-frequencies most resistors can be considered fixed in value. How-
ever, as we approach high frequencies, stray capacitive and inductive effects start to play a
role and will affect the total impedance level of the element.
For the diode it is the stray capacitance levels that have the greatest effect. At low frequen-
cies and relatively small levels of capacitance the reactance of a capacitor, determined by
Xe = 1 /27TfC, is usually so high it can be considered infinite in magnitude, represented by
an open circuit, and ignored. At high frequencies, however, the level of Xe can drop to the
point where it will introduce a low-reactance "shorting" path. If this shorting path is across
the diode, it can essentially keep the diode from affecting the response of the network.
In the p-n semiconductor diode, there are two capacitive effects to be considered. Both
types of capacitance are present in the forward- and reverse-bias regions, but one so out-
weighs the other in each region that we consider the effects of only one in each region.
Recall that the basic equation for the capacitance of a parallel-plate capacitor is defined by
C = EA/ d, where Eis the permittivity of the dielectric (insulator) between the plates of areaA
separated by a distanced. In a diode the depletion region (free of carriers) behaves essentially
like an insulator between the layers of opposite charge. Since the depletion width (d) will in-
crease with increased reverse-bias potential, the resulting transition capacitance will decrease,
as shown in Fig. 1.33. The fact that the capacitance is dependent on the applied reverse-bias
potential has application in a number of electronic systems. In fact, in Chapter 16 the varactor
diode will be introduced whose operation is wholly dependent on this phenomenon.
This capacitance, called the transition ( Cr), barriers, or depletion region capacitance, is
determined by

C(O)
Cr = _(l_+_!V_R_/V_K_!)-n (1.9)
 C(pF) REVERSE RECOVERY :n
15 TIME
:
t

,, ,
I

10
J
Cr I
I
,,........... ~er+ Cn = Cv
5 ./
....
~

-25 -20 -15 -10 -5 0 +0.25 +0.5 (V)

FIG. 1.33
Transition and diffusion capacitance versus applied bias for a silicon diode.

where C(O) is the capacitance under no-bias conditions and VR is the applied reverse bias
potential. The power n is ½ or ½ depending on the manufacturing process for the diode.
Although the effect described above will also be present in the forward-bias region, it
is overshadowed by a capacitance effect directly dependent on the rate at which charge is
injected into the regions just outside the depletion region. The result is that increased levels
of current will result in increased levels of diffusion capacitance ( Cv) as demonstrated by
the following equation:

(1.10)

where TT is the minority carrier lifetime-the time is world take for a minority carrier such
as a hole to recombine with an electron in the n-type material. However, increased levels
of current result in a reduced level of associated resistance (to be demonstrated shortly),
and the resulting time constant (T = RC), which is very important in high-speed applica-
tions, does not become excessive.
In general, therefore,
the transition capacitance is the predominant capacitive effect in the reverse-bias
region whereas the diffusion capacitance is the predominant capacitive effect in the
forward-bias region.
FIG. 1.34
The capacitive effects described above are represented by capacitors in parallel with the Including the effect of the transition
ideal diode, as shown in Fig. 1.34. For low- or mid-frequency applications (except in the or diffusion capacitance on the
power area), however, the capacitor is normally not included in the diode symbol. semiconductor diode.

1. 11 REVERSE RECOVERY TIME

There are certain pieces of data that are normally provided on diode specification sheets
provided by manufacturers. One such quantity that has not been considered yet is the
reverse recovery time, denoted by trr- In the forward-bias state it was shown earlier that
•
there are a large number of electrons from the n-type material progressing through the
p-type material and a large number of holes in the n-type material-a requirement for con-
duction. The electrons in the p-type material and holes progressing through the n-type
material establish a large number of minority carriers in each material. If the applied volt-
age should be reversed to establish a reverse-bias situation, we would ideally like to see the
diode change instantaneously from the conduction state to the nonconduction state. How-
ever, because of the large number of minority carriers in each material, the diode current
will simply reverse as shown in Fig. 1.35 and stay at this measurable level for the period of
time ts (storage time) required for the minority carriers to return to their majority-carrier
state in the opposite material. In essence, the diode will remain in the short-circuit state
with a current /reverse determined by the network parameters. Eventually, when this storage
phase has passed, the current will be reduced in level to that associated with the nonconduc-
tion state. This second period of time is denoted by tt (transition interval). The reverse recov-
ery time is the sum of these two intervals: trr = ts + tt. This is an important consideration in
32 SEMICONDUCTOR
DIODES
Change of state (on - oft)
/forward applied at t = t1

/ Desired response

t,

'--t±-~I ts

t,,
t,

FIG. 1.35
Defining the reverse recovery time.

high-speed switching applications. Most commercially available switching diodes have a

trr in the range of a few nanoseconds to 1 µ,s. Units are available, however, with a trr of
only a few hundred picoseconds (10- 12 s).

1.12 DIODE SPECIFICATION SHEETS

Data on specific semiconductor devices are normally provided by the manufacturer in one
of two forms. Most frequently, they give a very brief description limited to perhaps one
•
page. At other times, they give a thorough examination of the characteristics using graphs,
artwork, tables, and so on. In either case, there are specific pieces of data that must be
included for proper use of the device. They include:
1. The forward voltage VF (at a specified current and temperature)
2. The maximum forward current IF (at a specified temperature)
3. The reverse saturation current IR (at a specified voltage and temperature)
4. The reverse-voltage rating [PIV or PRV or V(BR), where BR comes from the term
"breakdown" (at a specified temperature)]
5. The maximum power dissipation level at a particular temperature
6. Capacitance levels
7. Reverse recovery time trr
8. Operating temperature range
Depending on the type of diode being considered, additional data may also be provided,
such as frequency range, noise level, switching time, thermal resistance levels, and peak
repetitive values. For the application in mind, the significance of the data will usually be
self-apparent. If the maximum power or dissipation rating is also provided, it is understood
to be equal to the following product:

I Pvmax = Vvlv (1.11)

where Iv and Vv are the diode current and voltage, respectively, at a particular point of
operation.
If we apply the simplified model for a particular application (a common occurrence), we
can substitute Vv = VT= 0.7 V for a silicon diode in Eq.(1.11) and determine the resulting
power dissipation for comparison against the maximum power rating. That is,

I Pctissipated ~ (0.7 V)lv I (1.12)

The data provided for a high-voltage/low-leakage diode appear in Figs. 1.36 and 1.37. This
example would represent the expanded list of data and characteristics. The term rectifier is
applied to a diode when it is frequently used in a rectification process, described in Chapter 2.
Specific areas of the specification sheet are highlighted in blue, with letters correspond-
ing to the following description:
A The data sheet highlights the fact that the silicon high-voltage diode has a minimum
reverse-bias voltage of 125 Vat a specified reverse-bias current.
B Note the wide range of temperature operation. Always be aware that data sheets typi- DIODE SPECIFICATION ]3
cally use the centigrade scale, with 200°C = 392°F and -65°C = -85°F. SHEETS
C The maximum power dissipation level is given by Pv = Vvlv = 500 mW = 0.5 W.
The effect of the linear derating factor of 3.33 mW!°C is demonstrated in Fig. 1.37a.
Once the temperature exceeds 25°C the maximum power rating will drop by 3.33 mW
for each 1°C increase in temperature. At a temperature of 100°C, which is the boiling
point of water, the maximum power rating has dropped to one half of its original value.
An initial temperature of 25°C is typical inside a cabinet containing operating elec-
tronic equipment in a low-power situation.
D The maximum sustainable current is 500 mA. The plot of Fig. 1.37b reveals that the
forward current at 0.5 Vis about 0.01 mA, but jumps to 1 mA (100 times greater) at
about 0.65 V. At 0.8 V the current is more than 10 mA, and just above 0.9 V it is close

DIFFUSED SILICON PLANAR

A - + - - - • BV ... 125 V (MIN) @ 100 pA (BA Y73) DO-35 OUTLINE

ABSOLUTE MAXIMUM RATINGS (Note I)

Temperatures 1
I7 MIN

-·~i
,:25.40)
Storage Temperature Range -65°C to +200°C
B ___,,___ _ _ _ _ Maximum Junction Operating Temperature +175°C
Lead Temperature +260°C
Power Dissipation (Note 2)
C - - - - - - < t - - - - - - - Maximum Total Power Dissipation at 25°C Ambient 500mW
~ j40()56) 0.180(4.57)

Linear Power Derating Factor (from 25°C) 3.33mW/°C

Io
D----1t-------IF
Maximum Voltage and Currents
WlV Working Inverse Voltage

Average Rectified Current

Continuous Forward Current
BAY73 IOOV

200mA
500mA
0.021 (0.5)))
---DIA
0.019 (0.483)
_[l 0.075 (1.91)
--DIA
0.060(152)

ir Peak Repetitive Forward Current 600mA NOTES:

Copper clad steel leads. tin plated
ir (surge) Peak Forward Surge Current Gold plated leads available
Pulse Width= I s l.0A Hennetically sealed glass package
Pulse Width= I µs 4.0A Package weight is 0.14 gram

ELECTRICAL CHARACTERISTICS (25°C Ambient Temperature unless otherwise noted)

BAY73
SYMBOL CHARACTERISTIC UNITS TEST CONDITIONS
MIN MAX
E ------<f-------t--V F - ~ Forward Voltage-------< 0.85 1.00 V IF=200mA
0.81 0.94 V IF= IOOmA
0.78 0.88 V IF= 50mA
0.69 0.80 V IF= !OmA
0.67 0.75 V IF= 5.0 mA
0.60 0.68 V IF=l.0mA
F----,1-----+--IR-f-- Reverse Current-----1------1 500 nA VR =20 V, TA= 125°C
1.0 µA VR =100 V, TA= 125°C
0.2 nA VR =20V, TA =25°C
O.S nA VR =100 V, TA= 25°C

BV Breakdown Voltage 125 V IR= IOOµA

G ---11-----+--C-e--Capacitance------+----t 5.0 pF VR =0, f= 1.0 MHz
H I r r - ~ Reverse Recovery Time---+------, 3.0 µs IF = 10 mA, VR = 35 V
RL =1.0 to 100 kQ
CL = IO pF, JAN 256

NOTES
I These ratings are limiting values above which the serviceability of the diode may be impaired.
2 These are steady state limits. The factory should be consuhed on applications involving pulses or low duty-cycle operation.

FIG. 1.36
Electrical characteristics of a high-voltage, low-leakage diode.
 FORWARD VOLTAGE VERSUS REVERSE VOLT AGE VERSUS
POWER DERATING CURVE FORWARD CURRENT REVERSE CURRENT
500 1000 1.0

~
1 V,
C:
mA o.s [L:r c..J::;J=:.i2a:=j:~
.s 1:
1 300 ~
:::,
u
<I.)
0.2

0.1
'i3 i'.:
~
<I.)
200 <I.)

2: ~ 0.05
0:::
C:
I
Cl
100 ~ 0.02
......
~
0.01 - - -··- - - - - - -
25 50 75 100125150175 200 0.2 0.4 0.6 0.8 1.0 1.2 25 50 75 100 125
TA -Ambient temperature - °C VF - Forward voltage - volts V R - Reverse voltage - volts
(a) (b) (c)

REVERSE CURRENT VERSUS CAPACITANCE VERSUS DYNAMIC IMPEDANCE VERSUS

TEMPERATURE REVERSE VOLTAGE FORWARD CURRENT
SK
I <!'.
100 · · ···;,,znLW
lK It t--+--+---+··· + - 5.0 4-, E I kHz..)
~
I 10
-t---t---
c T,;_:
'!
,. ; 4.0
I /:
u 1:
~
+······il ·········t···· · T T 13.0
·I
~
:::,

]
u
1.0 son
,7 u
20 I ~
-.,. 0
u. 0.1
u I
1.0
i ~
I
0 0.01 ,._..___._........_-'---''-'---'--'-'--'---'-'-'"---
25 50 75 100 125 150 -16 -12 -8.0 -4.0 0 0 1.0 10 100 lK !OK
TA - Ambient temperature - °C VR - Reverse voltage - volts Rv - Dynamic impedance - n
(d) (e) (t)

FIG. 1.37
Terminal characteristics of a high-voltage diode.

to 100 mA. The curve of Fig. 1.37b certainly looks nothing like the characteristic
curves appearing in the last few sections. This is a result of using a log scale for the
current and a linear scale for the voltage.
Log scales are often used to provide a broader range of values for a variable in a
limited amount of space.
If a linear scale was used for the current, it would be impossible to show a range
of values from 0.01 mA to 1000 mA. If the vertical divisions were in 0.01-mA incre-
ments, it would take 100,000 equal intervals on the vertical axis to reach 1000 mA. For
the moment recognize that the voltage level at given levels of current can be found by
using the intersection with the curve. For vertical values above a level such as 1.0 mA,
the next level is 2 mA, followed by 3 mA, 4 mA, and 5 mA. The levels of 6 mA to 10 mA
can be determined by simply dividing the distance into equal intervals (not the true
distribution, but close enough for the provided graphs). For the next level it would be
10 mA, 20 mA, 30 mA, and so on. The graph of Fig. 1.37b is called a semi-log plot to
reflect the fact that only one axis uses a log scale. A great deal more will be said about
log scales in Chapter 9.
E The data provide a range of Vp (forward-bias voltages) for each current level. The
higher the forward current, the higher is the applied forward bias. At 1 mA we find Vp
can range from 0.6 V to 0.68 V, but at 200 mA it can be as high as 0.85 V to 1.00 V.
For the full range of current levels with 0.6 V at 1 mA and 0.85 V at 200 mA it is cer-
tainly a reasonable approximation to use 0.7 V as the average value.
F The data provided clearly reveal how the reverse saturation current increases with
applied reverse bias at a fixed temperature. At 25°C the maximum reverse-bias cur-
rent increases from 0.2 nA to 0.5 nA due to an increase in reverse-bias voltage by the
same factor of 5. At 125 °C it jumps by a factor of 2 to the high level of 1 µ,A. Note the
34
 extreme change in reverse saturation current with temperature as the maximum cur- SEMICONDUCTOR DIODE
rent rating jumps from 0.2 nA at 25°C to 500 nA at 125°C (at a fixed reverse-bias NOTATION
voltage of 20 V). A similar increase occurs at a reverse-bias potential of 100 V. The
semi-log plots of Figs. 1.37c and 1.37d provide an indication of how the reverse satu-
ration current changes with changes in reverse voltage and temperature. At first
glance Fig. 1.37c might suggest that the reverse saturation current is fairly steady for
changes in reverse voltage. However, this can sometimes be the effect of using a log
scale for the vertical axis. The current has actually changed from a level of 0.2 nA to
a level of 0.7 nA for the range of voltages representing a change of almost 6 to 1. The
dramatic effect of temperature on the reverse saturation current is clearly displayed in
Fig. 1.37d. At a reverse-bias voltage of 125 V the reverse-bias current increases from
a level of about 1 nA at 25°C to about 1 µ,A at 150°C, an increase of a factor of 1000
over the initial value.
Temperature and applied reverse bias are very important factors in designs sensitive
to the reverse saturation current.
G As shown in the data listing and on Fig. l.37e, the transition capacitance at a reverse-
bias voltage of 0 V is 5 pF at a test frequency of 1 MHz. Note the severe change in
capacitance level as the reverse-bias voltage is increased. As mentioned earlier, this
sensitive region can be put to good use in the design of a device (V aractor; Chapter 16)
whose terminal capacitance is sensitive to the applied voltage.
H The reverse recovery time is 3 µ,s for the test conditions shown. This is not a fast time
for some of the current high-performance systems in use today. However, for a variety
of low- and mid-frequency applications it is acceptable.
The curves of Fig. l.37f provide an indication of the magnitude of the ac resistance of the
diode versus forward current. Section 1.8 clearly demonstrated that the dynamic resistance
of a diode decreases with increase in current. As we go up the current axis of Fig. l.37f it
is clear that if we follow the curve, the dynamic resistance will decrease. At 0.1 mA it is
close to 1 kll; at 10 mA, 10 ll; and at 100 mA, only 1 ll; this clearly supports the earlier
discussion. Unless one has had experience reading log scales, the curve is challenging to
read for levels between those indicated because it is a log-log plot. Both the vertical axis
and the horizontal axis employ a log scale.
The more one is exposed to specification sheets, the "friendlier" they will become, es-
pecially when the impact of each parameter is clearly understood for the application under
investigation.

1.1:S SEMICONDUCTOR DIODE NOTATION

•
The notation most frequently used for semiconductor diodes is provided in Fig. 1.38. For
most diodes any marking such as a dot or band, as shown in Fig. 1.38, appears at the cath-
ode end. The terminology anode and cathode is a carryover from vacuum-tube notation.
The anode refers to the higher or positive potential, and the cathode refers to the lower or
negative terminal. This combination of bias levels will result in a forward-bias or "on"
condition for the diode. A number of commercially available semiconductor diodes appear
in Fig. 1.39.

_,..,or •, K, etc.

FIG. 1.38
Semiconductor diode notation.
 hl
General purpose diode Surface mount high-power PIN diode Power (stud) diode Power (planar) diode

Beam lead pin diode Flat chip surface mount diode Power diode Power (disc, puck) diode

FIG. 1.39
Various types ofjunction diodes.

1. 14 DIODE TESTING
The condition of a semiconductor diode can be determined quickly using (1) a digital dis-
play meter (DDM) with a diode checking function, (2) the ohmmeter section of a multime-
ter, or (3) a curve tracer.
•
8
- -
.,... ~
BK PRECISION" - -
Diode Checking Fundion
TooL tUT
A digital display meter with a diode checking capability appears in Fig. 1.40. Note the
small diode symbol at the top right of the rotating dial. When set in this position and
hooked up as shown in Fig. 1.41a, the diode should be in the "on" state and the display will
provide an indication of the forward-bias voltage such as 0.67 V (for Si). The meter has an
internal constant-current source (about 2 mA) that will define the voltage level as indicated
in Fig. 1.41b. An OL indication with the hookup of Fig. 1.41a reveals an open (defective)
diode. If the leads are reversed, an OL indication should result due to the expected open-
circuit equivalence for the diode. In general, therefore, an OL indication in both directions
is an indication of an open or defective diode.

FIG. 1.40 Red lead! Black lead

Digital display meter. (Courtesy of (VQ) ! (COM)
B&K Precision Corporation.) 2 t - - - ----M
IIJ,,I _.)
0 0.67 V

(a) (b)

FIG. 1.41
Checking a diode in the forward-bias state.

Ohmmeter Testing
In Section 1.8 we found that the forward-bias resistance of a semiconductor diode is quite
low compared to the reverse-bias level. Therefore, if we measure the resistance of a diode
36
using the connections indicated in Fig. 1.42, we can expect a relatively low level. The result- DIODE TESTING J7
ing ohmmeter indication will be a function of the current established through the diode by the
internal battery (often 1.5 V) of the ohmmeter circuit. The higher the current, the lower is the
resistance level. For the reverse-bias situation the reading should be quite high, requiring a (Ohmmeter)
high resistance scale on the meter, as indicated in Fig. 1.42b. A high resistance reading in
both directions indicates an open (defective-device) condition, whereas a very low resis-
Red lead
(VD)
t t
Relatively low R
Black lead
(COM)
tance reading in both directions will probably indicate a shorted device.
+ ►I -
(a)
Curve Tracer
Relatively high R
The curve tracer of Fig. 1.43 can display the characteristics of a host of devices, including Black lead I I Red lead
the semiconductor diode. By properly connecting the diode to the test panel at the bottom (COM) + + (VO)
center of the unit and adjusting the controls, one can obtain the display of Fig. 1.44. Note
that the vertical scaling is 1 mA/div, resulting in the levels indicated. For the horizontal axis ►I +
the scaling is 100 mV/div, resulting in the voltage levels indicated. For a 2-mA level as (b)

defined for a DDM, the resulting voltage would be about 625 mV = 0.625 V. Although the FICi. 1.42
instrument initially appears quite complex, the instruction manual and a few moments of Checking a diode with an
exposure will reveal that the desired results can usually be obtained without an excessive ohmmeter.
amount of effort and time. The display of the instrument will appear on more than one occa-
sion in the chapters to follow as we investigate the characteristics of the variety of devices.

FICi. 1.43
Curve tracer. (© Agilent Technologies, Inc. Reproduced with
Permission, Courtesy ofAgilent Technologies, Inc.)

Vertical
!UmA per div.
I
9mA mA
8mA
Horizontal
7mA per div.
JOO

II
6mA mV
,-
SmA

4mA M<ep
I
3mA

2mA
I
I B or 8m
lmA per div.

OmA
V
OV O.IV 0.2V 0.3V 0.4V 0.5V 0.6V 0.7V 0.8V 0.9V 1.0V - - - -

FICi. 1.44
Curve tracer response to IN4007 silicon diode.
38 SEMICONDUCTOR
DIODES
1.15 ZENER DIODES
•
The Zener region of Fig. 1.45 was discussed in some detail in Section 1.6. The characteristic
drops in an almost vertical manner at a reverse-bias potential denoted Vz. The fact that
the curve drops down and away from the horizontal axis rather than up and away for the
positive-VD region reveals that the current in the Zener region has a direction opposite to
that of a forward-biased diode. The slight slope to the curve in the Zener region reveals that
there is a level of resistance to be associated with the Zener diode in the conduction mode.
This region of unique characteristics is employed in the design of Zener diodes, which
have the graphic symbol appearing in Fig. 1.46a. The semiconductor diode and the Zener
diode are presented side by side in Fig. 1.46 to ensure that the direction of conduction of
each is clearly understood together with the required polarity of the applied voltage. For
the semiconductor diode the "on" state will support a current in the direction of the arrow
in the symbol. For the Zener diode the direction of conduction is opposite to that of the
arrow in the symbol, as pointed out in the introduction to this section. Note also that the
polarity of Vv and Vz are the same as would be obtained if each were a resistive element
as shown in Fig. 1.46c.

Vz
+ + +
0

(a) (b) (c)

FIC. 1.45 FIC. 1.46

Reviewing the Zener region. Conduction direction: (a) Zener diode;
(b) semiconductor diode;
(c) resistive element.

The location of the Zener region can be controlled by varying the doping levels. An in-
crease in doping that produces an increase in the number of added impurities, will decrease
the Zener potential. Zener diodes are available having Zener potentials of 1.8 V to 200 V
with power ratings from ¼ W to 50 W. Because of its excellent temperature and current
capabilities, silicon is the preferred material in the manufacture of Zener diodes.
It would be nice to assume the Zener diode is ideal with a straight vertical line at the
Zener potential. However, there is a slight slope to the characteristics requiring the piece-
wise equivalent model appearing in Fig. 1.47 for that region. For most of the applications
appearing in this text the series resistive element can be ignored and the reduced equivalent
model of just a de battery of Vz volts employed. Since some applications of Zener diodes
swing between the Zener region and the forward-bias region, it is important to understand
the operation of the Zener diode in all regions. As shown in Fig. 1.47, the equivalent model
for a Zener diode in the reverse-bias region below Vz is a very large resistor (as for the
standard diode). For most applications this resistance is so large it can be ignored and the
open-circuit equivalent employed. For the forward-bias region the piecewise equivalent is
the same as described in earlier sections.
The specification sheet for a 10-V, 500-mW, 20% Zener diode is provided as Table 1.8,
and a plot of the important parameters is given in Fig. 1.48. The term nominal used in the
specification of the Zener voltage simply indicates that it is a typical average value. Since this
is a 20% diode, the Zener potential of the unit one picks out of a lot (a term used to describe a
package of diodes) can be expected to vary as 10 V + 20%, or from 8 V to 12 V. Both 10%
and 50% diodes are also readily available. The test current lzT is the current defined by the
 lz

I
I ,J _12v_i
+l_

rz
-
+l
~0.7V

-r
_,,,,.,-0.7V

____________________ --r25 10 µA= IR

mA = ,,,
Vz

- ,,= S.5ll =Z,, ------------{u-12.5 mA

rz = ii Vz
- - - - - !!._I'!,_ - - - - - - - - - - - - -1I I7M = 32 mA

FIG. 1.47
Zener diode characteristics with the equivalent model for each region.

TABLE 1.8
Electrical Characteristics (25°C Ambient Temperature)

Zener Maximum Maximum Maximum Maximum

Voltage Test Dynamic Knee Reverse Test Regulator Typical
Nominal Current Impedance Impedance Current Voltage Current Temperature
Vz lzT ZzTatlzT Zzxatlzx IR at VR VR lzM Coefficient
(V) (mA) (0) (0) (mA) (µA) (V) (mA) (%/OC)

10 12.5 8.5 700 0.25 10 7.2 32 +0.072

Temperature coefficient (Tc) Dynamic impedance (rz)

versus Zener current versus Zener current
+0.12 1 kQ
II , ....
9
~ +0.08 ,_1J~
I II
a
I
500

200
-. - ......
~ lOV ',,..
I

.)
s
8u
+0.04

0
~
"
u
§
"O
Q.)

.§
100
50

20
' ... , ...
' -'
, !'I
..... ' ...
I'll'\.
- ......
N.i.6V-
r-,. ,

" -~
u
~ -0.04 ___ .,.. -~ 10
~ ~ 3.6V 5 r-..... ,_"'r-,.. lOV
11
~
J
C:

-0.08
I Ill
6
II 2
~
-0.12 II 1
24v' .....
1

0.01 0.05 0.1 0.5 1 5 10 50100 0.1 0.2 0.5 1 2 5 10 20 50 100

Zener current / 2 - (mA) Zener current / 2 - (mA)

(a) (b)

FIG. 1.48
Electrical characteristics for a 10-V, 500-m W Zener diode.
]9
40 SEMICONDUCTOR ¼-power level. It is the current that will define the dynamic resistance ZzT and appears in
DIODES the general equation for the power rating of the device. That is,

(1.13)

Substituting lzT into the equation with the nominal Zener voltage results in
P2= = 4/ZTVz = 4(12.5 mA)(lO V) = 500 mW
which matches the 500-mW label appearing above. For this device the dynamic resistance
is 8.5 0, which is usually small enough to be ignored in most applications. The maximum
knee impedance is defined at the center of the knee at a current of lzK = 0.25 mA. Note
that in all the above the letter Tis used in subscripts to indicate test values and the letter K
to indicate knee values. For any level of current below 0.25 mA the resistance will only get
larger in the reverse-bias region. The knee value therefore reveals when the diode will start
to show very high series resistance elements that one may not be able to ignore in an appli-
cation. Certainly 500 0 = 0.5 kO may be a level that can come into play. At a reverse-bias
voltage the application of a test voltage of 7 .2 V results in a reverse saturation current of
10 µ,A, a level that could be of some concern in some applications. The maximum regulator
current is the maximum continuous current one would want to support in the use of the
Zener diode in a regulator configuration. Finally, we have the temperature coefficient
(Tc) in percent per degree centigrade.
The Zener potential of a Zener diode is very sensitive to the temperature of operation.
The temperature coefficient can be used to find the change in Zener potential due to a
change in temperature using the following equation:

=
LlV /V
z z
Tc X 100%/°C (%/°C) (1.14)
T1 - To

where T1 is the new temperature level

T0 is room temperature in an enclosed cabinet (25°C)
Tc is the temperature coefficient
and Vz is the nominal Zener potential at 25°C.
To demonstrate the effect of the temperature coefficient on the Zener potential, consider
the following example.

EXAMPLE 1.5 Analyze the 10-V Zener diode described by Table 1. 7 if the temperature is
increased to 100°C (the boiling point of water).
Solution: Substituting into Eq. (1.14), we obtain
TcVz
Ll Vz = 100% (T1 - To)

= (0.072%/°C)(lO V) (lO0oC - 25oC)

100%
and Ll Vz = 0.54 V
The resulting Zener potential is now
Vz' = Vz + 0.54 V = 10.54 V
which is not an insignificant change.

It is important to realize that in this case the temperature coefficient was positive. For Zener
diodes with Zener potentials less than 5 V it is very common to see negative temperature
coefficients, where the Zener voltage drops with an increase in temperature. Figure 1.48a
provides a plot of T versus Zener current for three different levels of diodes. Note that the
3.6-V diode has a negative temperature coefficient, whereas the others have positive values.
The change in dynamic resistance with current for the Zener diode in its avalanche re-
gion is provided in Fig. 1.48b. Again, we have a log-log plot, which has to be carefully read.
Initially it would appear that there is an inverse linear relationship between the dynamic LIGHT-EMITTING DIODES 41
resistance because of the straight line. That would imply that if one doubles the current, one
cuts the resistance in half. However, it is only the log-log plot that gives this impression,
because if we plot the dynamic resistance for the 24-V Zener diode versus current using
linear scales we obtain the plot of Fig. 1.49, which is almost exponential in appearance.
Note on both plots that the dynamic resistance at very low currents that enter the knee of
the curve is fairly high at about 200 0. However, at higher Zener currents, away from the
knee, at, say 10 mA, the dynamic resistance drops to about 5 0.

1 1-Anode-1
i 8
L i-Cathode-l
FIG. 1.49
Zener terminal identification and symbols.

The terminal identification and the casing for a variety of Zener diodes appear in Fig.
1.49. Their appearance is similar in many ways to that of the standard diode. Some areas of
application for the Zener diode will be examined in Chapter 2.

1.16 LIGHT-EMITTING DIODES

•
The increasing use of digital displays in calculators, watches, and all forms of instrumenta-
tion has contributed to an extensive interest in structures that emit light when properly
biased. The two types in common use to perform this function are the light-emitting diode
(LED) and the liquid-crystal display (LCD). Since the LED falls within the family of p-n
junction devices and will appear in some of the networks of the next few chapters, it will
be introduced in this chapter. The LCD display is described in Chapter 16.
As the name implies, the light-emitting diode is a diode that gives off visible or invis-
ible (infrared) light when energized. In any forward-biased p-n junction there is, within the
structure and primarily close to the junction, a recombination of holes and electrons. This
recombination requires that the energy possessed by the unbound free electrons be trans-
ferred to another state. In all semiconductor p-n junctions some of this energy is given off
in the form of heat and some in the form of photons.
In Si and Ge diodes the greater percentage of the energy converted during recombina-
tion at the junction is dissipated in the form of heat within the structure, and the emitted
light is insignificant.
For this reason, silicon and germanium are not used in the construction of LED devices.
On the other hand:
Diodes constructed of GaAs emit light in the infrared (invisible) wne during the
recombination process at the p-njunction.
Even though the light is not visible, infrared LEDs have numerous applications where
visible light is not a desirable effect. These include security systems, industrial processing,
optical coupling, safety controls such as on garage door openers, and in home entertainment
centers, where the infrared light of the remote control is the controlling element.
Through other combinations of elements a coherent visible light can be generated. Table 1.9
provides a list of common compound semiconductors and the light they generate. In addi-
tion, the typical range of forward bias potentials for each is listed.
The basic construction of an LED appears in Fig. 1.50 with the standard symbol used
for the device. The external metallic conducting surface connected to the p-type material is
smaller to permit the emergence of the maximum number of photons of light energy when
the device is forward-biased. Note in the figure that the recombination of the injected carri-
ers due to the forward-biased junction results in emitted light at the site of the recombination.
42 SEMICONDUCTOR TABLE 1.9
DIODES Light-Emitting Diodes

Typical Forward
Color Construction Voltage (V)

Amber AllnGaP 2.1

Blue GaN 5.0
Green GaP 2.2
Orange GaAsP 2.0
Red GaAsP 1.8
White GaN 4.1
Yellow AllnGaP 2.1

There will, of course, be some absorption of the packages of photon energy in the structure
itself, but a very large percentage can leave, as shown in the figure.

c-/f
+0
~ -----□
llJJJI
(-) ID Vo

(b)

"'Metal
contact

(a)

FIG. 1.50
(a) Process of electroluminescence in the LED; (b) graphic symbol.

Just as different sounds have different frequency spectra (high-pitched sounds generally
have high-frequency components, and low sounds have a variety oflow-frequency compo-
nents), the same is true for different light emissions.
The frequency spectrum for infrared light extends from about 100 THz (T = tera =
1012) to 400 THz, with the visible light spectrum extending from about 400 to 750 THz.
It is interesting to note that invisible light has a lower frequency spectrum than visible
light.
In general, when one talks about the response of electroluminescent devices, one refer-
ences their wavelength rather than their frequency.
The two quantities are related by the following equation:

IA / I (ml
(1.15)

where c = 3 X 108 mis (the speed oflight in a vacuum)

f = frequency in Hertz
A = wavelength in meters.
 LIGHT-EMITTING DIODES 43
EXAMPLE 1.6 Using Eq. (1.15), find the range of wavelength for the frequency range of
visible light (400 THz-750 THz).
Solution:

c = 3 X 108~[ lO:m] = 3 X 1017 nm/s

c 3 X 1017 nm/s 3 X 10 17 nm/s
A = - = ----- - - - - - - = 750 nm
f 400 THz 400 X 1012 Hz
, __C -_ 3 X 1017 nm/s 3 X 10 17 nm/s
/l - - - - - - - = 400 nm
f 750THz 750 X 1012 Hz
400 nm to 750 nm

Note in the above example the resulting inversion from higher frequency to smaller wave-
length. That is, the higher frequency results in the smaller wavelength. Also, most charts
use either nanometers (nm) or angstrom (A) units. One angstrom unit is equal to 10- 10 m.
The response of the average human eye as provided in Fig. 1.51 extends from about
350 nm to 800 nm with a peak near 550 nm.
It is interesting to note that the peak response of the eye is to the color green, with red and
blue at the lower ends of the bell curve. The curve reveals that a red or a blue LED must
have a much stronger efficiency than a green one to be visible at the same intensity. In other
words, the eye is more sensitive to the color green than to other colors. Keep in mind that
the wavelengths shown are for the peak response of each color. All the colors indicated on
the plot will have a bell-shaped curve response, so green, for example, is still visible at 600
nm, but at a lower intensity level.

Luminosity (Lm/w)

700

600

500

400
~ ULTRAVIOLET INFRARED ....-

300 Yellow
Amber

200

100

0
0 100 400 500 600 700 800 900 A (nm)

FIG. 1.51
Standard response curve of the human eye, showing the eye's response to light energy
peaks at green and falls offfor blue and red.

In Section 1.4 it was mentioned briefly that GaAs with its higher energy gap of 1.43 eV
made it suitable for electromagnetic radiation of visible light, whereas Si at 1.1 eV resulted pri-
marily in heat dissipation on recombination. The effect of this difference in energy gaps can be
44 SEMICONDUCTOR explained to some degree by realizing that to move an electron from one discrete energy level
DIODES to another requires a specific amount of energy. The amount of energy involved is given by

~ (1.16)
~
with X 10- 19 J]
Eg = joules (J) [1 eV = 1.6
h = Planck's constant= 6.626 X 10- 34 J • s.
c = 3 X 108 mis
A = wavelength in meters
If we substitute the energy gap level of 1.43 e V for GaAs into the equation, we obtain the
following wavelength:

1.43 eV[
1.6 X 10-19
l eV
J] = 2.288 X 10- 19 J

he (6.626 X 10-34 J • s)(3 X 108 mis)

and A - - - -------------
- Eg - 2.288 X 10-19 J
= 869nm
For silicon, with Eg = l. 1 eV
A= 1130nm
which is well beyond the visible range of Fig. 1.51.
The wavelength of 869 nm places GaAs in the wavelength zone typically used in infrared
devices. For a compound material such as GaAsP with a band gap of 1.9 e V the resulting
wavelength is 654 nm, which is in the center of the red zone, making it an excellent com-
pound semiconductor for LED production. In general, therefore:
The wavelength and frequency of light of a specific color are directly related to the
energy band gap of the material.
A first step, therefore, in the production of a compound semiconductor that can be used
to generate light is to come up with a combination of elements that will generate the desired
energy band gap.
The appearance and characteristics of a subminiature high-efficiency red LED manufac-
tured by Hewlett-Packard are given in Fig. 1.52. Note in Fig. 1.52b that the peak forward
current is 60 mA, with 20 mA the typical average forward current. The text conditions
listed in Fig. 1.52c, however, are for a forward current of 10 mA. The level of Vv under
forward-bias conditions is listed as VF and extends from 2.2 V to 3 V. In other words, one
can expect a typical operating current of about 10 mA at 2.3 V for good light emission, as
shown in Fig. l .52e. In particular, note the typical diode characteristics for an LED, permit-
ting similar analysis techniques to be described in the next chapter.
Two quantities yet undefined appear under the heading Electrical/Optical Characteristics
at TA= 25°C. They are the axial luminous intensity (Iv) and the luminous efficacy (11v)- Light
intensity is measured in candelas. One candela (cd) corresponds to a light flux of 41r lumens
(lm) and is equivalent to an illumination of 1 footcandle on a l-ft2 area 1 ft from the light
source. Even if this description may not provide a clear understanding of the candela as a unit of
measure, it should be enough to allow its level to be compared between similar devices. Figure
l .52f is a normalized plot of the relative luminous intensity versus forward current. The term
normalized is used frequently on graphs to give comparisons of response to a particular level.
A normalized plot is one where the variable of interest is plotted with a specific level
defined as the reference value with a magnitude of one.
In Fig. l.52f the normalized level is taken at IF = 10 mA. Note that the relative lumi-
nous intensity is 1 at IF = 10 mA. The graph quickly reveals that the intensity of the light
is almost doubled at a current of 15 mA and is almost three times as much at a current of
20 mA. It is important to therefore note that:
The light intensity of an LED will increase with forward current until a point of
saturation arrives where any further increase in current will not effectively increase
the level of illumination.
 Absolute Maximum Ratings at TA = 25°C
High-Efficiency Red
Parameter 4160 Units

Power dissipation 120 mW

Average forward current 20[!] mA
Peak forward current 60 mA
Operating and storage temperature range -55°C to 100°c
Lead soldering temperature 230°C for 3 s
[1.6 mm (0.063 in.) from body]
NOTE: I. Derate from 50°C at 0.2 mV/°C.

(b)

(a)

ElectricaVOptical Characteristics at TA = 25°C

High-Efficiency Red
4160
Symbol Description Min. Typ. Max. Units Test Conditions
IF= 10 mA
Iv Axial luminous
intensity 1.0 3.0 med
20112 Included angle
between half
luminous intensity
points 80 degree Note 1
A.peak Peak wavelength 635 nm Measurement
at peak
Ad Dominant wavelength 628 nm Note 2
Ts Speed of response 90 ns
C Capacitance 11 pF VF=0;f=IMhz
0Jc Thermal resistance 120 oc/W Junction to
cathode lead at
0.79 mm (0.031
in.) from body
VF Forward voltage 2.2 3.0 V Ip= 10 mA
BVR Reverse breakdown
voltage 5.0 V IR= 100 µ,A
'T/v Luminous efficacy 147 lm/W Note 3
NOTES:
I. 0 112 is the off-axis angle at which the luminous intensity is half the axial luminous intensity.
2. The dominant wavelength, Ad, is derived from the CIE chromaticity diagram and represents the single
wavelength that defines the color of the device.
3. Radiant intensity, I,, in watts/steradian, may be found from the equation I, = IJ'T/v, where Iv is the
luminous intensity in candelas and 'T/v is the luminous efficacy in lumens/watt.
(c)

FIG. 1.52
Hewlett-Packard subminiature high-efficiency red solid-state lamp: (a) appearance; (b) absolute maximum ratings; (c) electrical/optical
characteristics; (d) relative intensity versus wavelength; (e) forward current versus forward voltage; (f) relative luminous intensity
versus forward current; (g) relative efficiency versus peak current; (h) relative luminous intensity versus angular displacement.

For instance, note in Fig. 1.52g that the increase in relative efficiency starts to level off
as the current exceeds 50 mA.
The term efficacy is, by definition, a measure of the ability of a device to produce the
desired effect. For the LED this is the ratio of the number of lumens generated per applied
watt of electrical power.
The plot of Fig. 1.52d supports the information appearing on the eye-response curve of
Fig. 1.51. As indicated above, note the bell-shaped curve for the range of wavelengths that
will result in each color. The peak value of this device is near 630 nm, very close to the
peak value of the GaAsP red LED. The curves of green and yellow are only provided for
reference purposes.
45
 20
,,,GaAsPRed I J
<t: TA= 25 C
High efficiency s
I 15
Red 1:Q.)
t::
::,
u
10
1::
"'
~ j
0 1 - - ~~ ~'.'.'.'.:__l__ _ _ ~ ~ - ~ :___---=~=""""'..;;:::~ ~==-- - -_J "" I
5
.....,c..
500 550 600 650 700 750
Wavelength-nm
,)
0
0 0.5 1.0 1.5 2.0 2.5 3.0
(d)
VF - Forward voltage - V

(e)

. - --
1.6
3.0 1.5
0 ,,
c~
.§
"O
;>-..,,::
1.4 ,,
~s 1.3
~ ~·
-~ ~ 2.0 t----+--------,t------+-.,~ ----, :gs 1.2
"'
::,
0
-
~

"'
::::~
"'"'
"' "O
>
1.1
I
/
"§] ·µ -~
Q.)
1.0
,.,:g~ J
..= ~ 1.0 t - - - - + - -____,.,C...----+------l
c,:"' a
0.9
.... I
-E"' Eo 5
0 0.8

~
"' =
-- 0.7
0·6 o 10 20 30 40 50 60
80°

9o't---+-+--t------f'':='
5 10 15 20 20° 40° 60° 80° 100°
IF - Forward current - mA /peak- Peak current - mA
(f) (g) (h)

FIG. 1.52
Continued.

Figure 1.52h is a graph of light intensity versus angle measured from 0° (head on)
to 90° (side view). Note that at 40° the intensity has already dropped to 50% of the
head-on intensity.
One of the major concerns when using an LED is the reverse-bias breakdown voltage,
which is typically between 3 V and 5 V (an occasional device has a 10-V level).
This range of values is significantly less than that of a standard commercial diode,
where it can extend to thousands of volts. As a result one has to be acutely aware of this
severe limitation in the design process. In the next chapter one protective approach will be
introduced.
In the analysis and design of networks with LEDs it is helpful to have some idea of the
voltage and current levels to be expected.
For many years the only colors available were green, yellow, orange, and red, permitting
the use of the average values of VF= 2 V and IF= 20 mAfor obtaining an approximate
operating level.
However, with the introduction of blue in the early 1990s and white in the late 1990s the
magnitude of these two parameters has changed. For blue the average forward bias voltage
can be as high as 5 V, and for white about 4.1 V, although both have a typical operating
current of 20 mA or more. In general, therefore:
Assume an average forward-bias voltage of 5 V for blue and 4 V for white LEDs at
currents of 20 mA to initiate an analysis of networks with these types of LEDs.
Every once in a while a device is introduced that seems to open the door to a slue of
possibilities. Such is the case with the introduction of white LEDs. The slow start for white
LEDs is primarily due to the fact that it is not a primary color like green, blue, and red.
Every other color that one requires, such as on a TV screen, can be generated from these
three colors (as in virtually all monitors available today). Yes, the right combination of
these three colors can give white-hard to believe, but it works. The best evidence is the
46
human eye, which only has cones sensitive to red, green, and blue. The brain is responsible LIGHT-EMITTING DIODES 47
for processing the input and perceiving the "white" light and color we see in our everyday
lives. The same reasoning was used to generate some of the first white LEDs, by combining
the right proportions of a red, a green, and a blue LED in a single package. Today, however,
most white LEDs are constructed of a blue gallium nitride LED below a film of yttrium-
aluminum garnet (YAG) phosphor. When the blue light hits the phosphor, a yellow light is
generated. The mix of this yellow emission with that of the central blue LED forms a white
light-incredible, but true.
Since most of the lighting for homes and offices is white light, we now have another
option to consider versus incandescent and fluorescent lighting. The rugged characteristics
of LED white light along with lifetimes that exceed 25,000 hours, clearly suggest that
this will be a true competitor in the near future. Various companies are now providing
replacement LED bulbs for almost every possible application. Some have efficacy ratings
as high as 135.7 lumens per watt, far exceeding the 25 lumens per watt of a few years
ago. It is forecast that 7 W of power will soon be able to generate 1,000 Im of light, which
exceeds the illumination of a 60 W bulb and can run off four D cell batteries. Imagine
the same lighting with less than 1/8 the power requirement. At the present time entire of-
fices, malls, street lighting, sporting facilities, and so on are being designed using solely
LED lighting. Recently, LEDs are the common choice for flashlights and many high-end
automobiles due to the sharp intensity at lower de power requirements. The tube light of
Fig. 1.53a replaces the standard fluorescent bulb typically found in the ceiling fixtures of
both the home and industry. Not only do they draw 20% less energy while providing 25%
additional light but they also last twice as long as a standard fluorescent bulb. The flood
light of Fig. 1.53b draws 1. 7 watts for each 140 lumens of light resulting in an enormous
90% savings in energy compared to the incandescent variety. The chandelier bulbs of Fig.
1.53c have a lifetime of 50,000 hours and only draw 3 watts of power while generating
200 lumens of light.

(a) (b) (c)

FIG. 1.53
LED residential and commercial lighting.

Before leaving the subject, let us look at a seven-segment digital display housed in a
typical dual in-line integrated circuit package as shown in Fig. 1.54. By energizing the
proper pins with a typical 5-V de level, a number of the LEDs can be energized and the
desired numeral displayed. In Fig. 1.54a the pins are defined by looking at the face of
the display and counting counterclockwise from the top left pin. Most seven-segment
displays are either common-anode or common-cathode displays, with the term anode
referring to the defined positive side of each diode and the cathode referring to the nega-
tive side. For the common-cathode option the pins have the functions listed in Fig. 1.54b
and appear as in Fig. 1.54c. In the common-cathode configuration all the cathodes are
connected together to form a common point for the negative side of each LED. Any LED
with a positive 5 V applied to the anode or numerically numbered pin side will tum on
and produce light for that segment. In Fig. 1.54c, 5 V has been applied to the terminals
that generate the numeral 5. For this particular unit the average forward tum-on voltage
is 2.1 Vat a current of 10 mA.
Various LED configurations are examined in the next chapter.
48 SEMICONDUCTOR COMMON CATHODE

il
DIODES PIN# FUNCTION
1. Anode f
2. ANODEg

. ~--
• 1

. L'.
• f
14•
b•
Fl
3.
4.
5.
6.
NOPIN
COMMON CATHODE
NOPIN
ANODEe

•el g
0.630"
fl c •
• ,,,_.,_~ _. _J_
1.0875"
I
7.
8.
9.
ANODEd
ANODEc
ANODEd
10. NOPIN
•7 d 8• ___l 11. NOPIN
12. COMMON CATHODE

Lo.803 .. J 13.
14.
ANODEb
ANODEa

(a) (b)

Computer control
sv svsv sv sv
s-=- I I s-=-s

21~13 14

~-.. 4~,113

5 10
12

6 ...-ii 9 - - ~
~--◄ 7 8

(c)

FIG. 1.54
Seven-segment display: (a)face with pin idenfication; (b) pin function; (c) displaying the numeral 5.

1. 17 SUMMARY
Important Conclusions and Conc:epts •
1. The characteristics of an ideal diode are a close match with those of a simple switch
except for the important fact that an ideal diode can conduct in only one direction.
2. The ideal diode is a short in the region of conduction and an open circuit in the
region of nonconduction.
3. A semiconductor is a material that has a conductivity level somewhere between that
of a good conductor and that of an insulator.
4. A bonding of atoms, strengthened by the sharing of electrons between neighboring
atoms, is called covalent bonding.
5. Increasing temperatures can cause a significant increase in the number of free elec-
trons in a semiconductor material.
6. Most semiconductor materials used in the electronics industry have negative tem-
perature coefficients; that is, the resistance drops with an increase in temperature.
7. Intrinsic materials are those semiconductors that have a very low level of impurities,
whereas extrinsic materials are semiconductors that have been exposed to a doping
process.
8. An n-type material is formed by adding donor atoms that have five valence electrons
to establish a high level of relatively free electrons. In an n-type material, the electron
is the majority carrier and the hole is the minority carrier.
9. Ap-type material is formed by adding acceptor atoms with three valence electrons to
establish a high level of holes in the material. In a p-type material, the hole is the
majority carrier and the electron is the minority carrier.
10. The region near the junction of a diode that has very few carriers is called the deple-
tion region.
11. In the absence of any externally applied bias, the diode current is zero.
12. In the forward-bias region the diode current increases exponentially with increase in
voltage across the diode.
13. In the reverse-bias region the diode current is the very small reverse saturation cur- COMPUTER ANALYSIS 49
rent until Zener breakdown is reached and current will flow in the opposite direction
through the diode.
14. The reverse saturation currentls will just about double in magnitude for every 10-fold
increase in temperature.
15. The de resistance of a diode is determined by the ratio of the diode voltage and cur-
rent at the point of interest and is not sensitive to the shape of the curve. The de resis-
tance decreases with increase in diode current or voltage.
16. The ac resistance of a diode is sensitive to the shape of the curve in the region of inter-
est and decreases for higher levels of diode current or voltage.
17. The threshold voltage is about 0.7 V for silicon diodes and 0.3 V for germanium diodes.
18. The maximum power dissipation level of a diode is equal to the product of the diode
voltage and current.
19. The capacitance of a diode increases exponentially with increase in the forward-bias
voltage. Its lowest levels are in the reverse-bias region.
20. The direction of conduction for a Zener diode is opposite to that of the arrow in the
symbol, and the Zener voltage has a polarity opposite to that of a forward-biased diode.
21. Light emitting diodes (LEDs) emit light under forward-bias conditions but require 2
V to 4 V for good emission.

Equations
kT
Iv = ls(eVv/nVr - 1) Vr=- k = 1.38 X 10-23 J/K
q
VK ~ 0.7 V (Si)
VK ~ 1.2 V (GaAs)
VK ~ 0.3 V (Ge)
Vv
Rv=-
lv
LlVd 26mV
rd=--= - - -
Ll/d Iv

Tav =
avdl
Llld pt.to pt.

Pvm,,. = Vvlv

1. 18 COMPUTER ANALYSIS
•
Two software packages designed to analyze electronic circuits will be introduced and applied
throughout the text. They include Cadence OrCAD, version 16.3 (Fig. 1.55), and Multi-
sim, version 11.0.1 (Fig. 1.56). The content was written with sufficient detail to ensure that
the reader will not need to reference any other computer literature to apply both programs.

FIG. 1.55 FIG. 1.56

Cadence OrCAD Design package version 16.3. Multisim 11.0.1.
(Photo by Dan Trudden/Pearson.) (Photo by Dan Trudden/Pearson.)
50 SEMICONDUCTOR Those of you who have used either program in the past will find that the changes are minor
DIODES and appear primarily in the front end and in the generation of specific data and plots.
The reason for including two programs stems from the fact that both are used throughout
the educational community. You will find that the OrCAD software has a broader area of
investigation but the Multisim software generates displays that are a better match to the
actual laboratory experience.
The demo version of OrCAD is free from Cadence Design Systems, Inc., and can be
downloaded directly from the EMA Design Automation, Inc., web site, info@emaeda.com.
Multisim must be purchased from the National Instruments Corporation using their web
site, ni.com/multisim.
In previous editions, the OrCAD package was referred to as a PSpice program primarily
because it is a subset of a more sophisticated version used extensively in industry called
SPICE. The result is the use of the term PSpice in the descriptions to follow when initiating
an analysis using the OrCAD software.
The downloading process for each software package will now be introduced along with
the general appearance of the resulting screen.

OrCAD
Installation:
Insert the OrCAD Release 16.3 DVD into the disk drive to open the Cadence OrCAD
16.3 software screen.
Select Demo Installation and the Preparing Setup dialog box will open, followed by
the message Welcome to the Installation Wizard for OrCAD 16.3 Demo. Select
Next, and the License Agreement dialog box opens. Choose I accept and select
Next, and the Choose Destination dialog box will open with Install OrCAD 16.3
Demo Accept C:\OrCAD\OrCAD_16.3 Demo.
Select Next, and the Start Copying Files dialog box opens. Choose Select again, and
the Ready to Install Program dialog box opens. Click Install, and the Installing
Crystal Report Xii box will appear. The Setup dialog box opens with the prompt:
Setup status installs program. The Install Wizard is now installing the OrCAD
16.3 Demo.
At completion, a message will appear: Searching for and adding programs to the
Windows firewall exception list. Generating indexes for Cadence Help. This
may take some time.
When the process has completed, select Finish and the Cadence OrCAD 16.3 screen
will appear. The software has been installed.

Screen Icon: The screen icon can be established (if it does not appear automatically) by
applying the following sequence. START-All Programs-Cadence-OrCAD 16.3 Demo-
OrCAD Capture CIS Demo, followed by a right-click of the mouse to obtain a listing
where Send to is chosen, followed by Desktop (create shortcut). The OrCAD icon will
then appear on the screen and can be moved to the appropriate location.

Folder Creation: Starting with the OrCAD opening screen, right-click on the Start
option at the bottom left of the screen. Then choose Explore followed by Hard Drive
(C:). Then place the mouse on the folder listing, and a right-click will result in a listing in
which New is an option. Choose New followed by Folder, and then type in OrCAD 11.3
in the provided area of the screen, followed by a right-click of the mouse. A location for all
the files generated using OrCAD has now been established.

Multisim
Installation:
Insert the Multisim disk into the DVD disk drive to obtain the Autoplay dialog box.
Then select Always do this for software and games, followed by the selection of
Auto-run to open the NI Circuit Design Suite 11.0 dialog box.
 Enter the full name to be used and provide the serial number. (The serial number
appears in the Certificate of Ownership document that came with the NI Circuit
Design Suite packet.)
Selecting Next will result in the Destination Directory dialog box from which one will
Accept the following: C:\Program Files(X86) National Instruments\. Select Next
to open the Features dialog box and then select NI Circuit Design Suite 11.0.1
Education.
Selecting Next will result in the Product Notification dialog box with a succeeding
Next resulting in the License Agreement dialog box. A left-click of the mouse on I
accept can then be followed by choosing Next to obtain the Start Installation dialog
box. Another left-click and the installation process begins, with the progress being
displayed. The process takes between 15 and 20 minutes.
At the conclusion of the installation, you will be asked to install the NI Elvismx driver
DVD. This time Cancel will be selected, and the NI Circuit Design Suite 11.0.1
dialog box will appear with the following message: NI Circuit Design Suite 11.0.1
has been installed. Click Finish, and the response will be to restart the computer to
complete the operation. Select Restart, and the computer will shut down and start up
again, followed by the appearance of the Multisim Screen dialog box.
Select Activate and then Activate through secure Internet connection, and the Acti-
vation Wizard dialog box will open. Enter the serial number followed by Next to
enter all the information into the NI Activation Wizard dialog box. Selecting Next
will result in the option of Send me an email confirmation of this activation. Select
this option and the message Product successfully activated will appear. Selecting
Finish will complete the process.

Screen Icon: The process described for the OrCAD program will produce the same
results for Multisim.

Folder Creation: Following the procedure introduced above for the OrCAD program, a
folder labeled OrCAD 16.3 was established for the Multisim files.
The computer section of the next chapter will cover the details of opening both the
OrCAD and Multisim analysis packages, setting up a specific circuit, and generating a
variety of results.

PROBLEMS
*Note: Asterisks indicate more difficult problems.
1.3 Covalent Bonding and Intrinsic Materials
•
1. Sketch the atomic structure of copper and discuss why it is a good conductor and how its struc-
ture is different from that of germanium, silicon, and gallium arsenide.
2. In your own words, define an intrinsic material, a negative temperature coefficient, and cova-
lent bonding.
3. Consult your reference library and list three materials that have a negative temperature coeffi-
cient and three that have a positive temperature coefficient.

1.4 Energy Levels

4. a. How much energy in joules is required to move a charge of 12 µ,C through a difference in
potential of 6 V?
b. For part (a), find the energy in electron-volts.
5. If 48 eV of energy is required to move a charge through a potential difference of 3.2 V, deter-
mine the charge involved.
6. Consult your reference library and determine the level of Eg for GaP, ZnS, and GaAsP, three semi-
conductor materials of practical value. In addition, determine the written name for each material.

1.5 a-Type and p-Type Materials

7. Describe the difference between n-type and p-type semiconductor materials.
8. Describe the difference between donor and acceptor impurities.
9. Describe the difference between majority and minority carriers.
SEMICONDUCTOR 10. Sketch the atomic structure of silicon and insert an impurity of arsenic as demonstrated for
DIODES silicon in Fig. 1.7.
11. Repeat Problem 10, but insert an impurity of indium.
12. Consult your reference library and find another explanation of hole versus electron flow. Using
both descriptions, describe in your own words the process of hole conduction.

1.6 Semicondudor Diode

13. Describe in your own words the conditions established by forward- and reverse-bias conditions
on a p-n junction diode and how the resulting current is affected.
14. Describe how you will remember the forward- and reverse-bias states of the p-n junction
diode. That is, how will you remember which potential (positive or negative) is applied to
which terminal?
15. a. Determine the thermal voltage for a diode at a temperature of 20°C.
b. For the sarne diode of part (a), find the diode current using Eq. 1.2 if ls = 40 nA, n = 2 (low
value of Vv), and the applied bias voltage is 0.5 V.
16. Repeat Problem 15 for T = 100°C (boiling point of water). Assume that ls has increased to 5.0 f-LA.
17. a. Using Eq. (1.2), determine the diode current at 20°C for a silicon diode with n = 2, ls =
0.1 f-LA at a reverse-bias potential of-10 V.
b. Is the result expected? Why?
18. Given a diode current of 8 rnA and n = l, find ls if the applied voltage is 0.5 V and the tem-
perature is room temperature (25°C).
*19. Given a diode current of 6 rnA, Vr = 26 mV, n = l, and ls= 1 nA, find the applied voltage Vv.
20. a. Plot the function y = ex for x from 0 to 10. Why is it difficult to plot?
b. What is the value of y = ~ at x = 0?
c. Based on the results of part (b), why is the factor -1 important in Eq. (1.2)?
21. In the reverse-bias region the saturation current of a silicon diode is about 0.1 f-LA (T = 20°C).
Determine its approximate value if the temperature is increased 40°C.
22. Compare the characteristics of a silicon and a germanium diode and determine which you would
prefer to use for most practical applications. Give some details. Refer to a manufacturer's listing
and compare the characteristics of a germanium and a silicon diode of similar maximum ratings.
23. Determine the forward voltage drop across the diode whose characteristics appear in Fig. 1.19 at
temperatures of -75°C, 25°C, 125°C and a current of 10 rnA. For each temperature, determine the
level of saturation current. Compare the extremes of each and comment on the ratio of the two.

1.7 Ideal versus Practical

24. Describe in your own words the meaning of the word ideal as applied to a device or a system.
25. Describe in your own words the characteristics of the ideal diode and how they determine the
on and off states of the device. That is, describe why the short-circuit and open-circuit equiva-
lents are appropriate.
26. What is the one important difference between the characteristics of a simple switch and those
of an ideal diode?

1.8 Resistance Levels

27. Determine the static or de resistance of the commercially available diode of Fig. 1.15 at a for-
ward current of 4 rnA.
28. Repeat Problem 27 at a forward current of 15 rnA and compare results.
29. Determine the static or de resistance of the commercially available diode of Fig. 1.15 at a reverse
voltage of - 10 V. How does it compare to the value determined at a reverse voltage of - 30 V?
30. Calculate the de and ac resistances for the diode of Fig. 1.15 at a forward current of 10 rnA and
compare their magnitudes.
31. a. Determine the dynamic (ac) resistance of the commercially available diode of Fig. 1.15 at a
forward current of 10 rnA using Eq. (1.5).
b. Determine the dynamic (ac) resistance of the diode of Fig. 1.15 at a forward current of 10 rnA
using Eq. (1.6).
c. Compare solutions of parts (a) and (b).
32. Using Eq. (1.5), determine the ac resistance at a current of 1 rnA and 15 rnA for the diode of
Fig. 1.15. Compare the solutions and develop a general conclusion regarding the ac resistance
and increasing levels of diode current.
 33. Using Eq. (1.6), determine the ac resistance at a current of 1 mA and 15 mA for the diode of
Fig. 1.15. Modify the equation as necessary for low levels of diode current. Compare to the
solutions obtained in Problem 32.
34. Determine the average ac resistance for the diode of Fig. 1.15 for the region between 0.6 V
and 0.9 V.
35. Determine the ac resistance for the diode of Fig. 1.15 at 0.75 V and compare it to the average
ac resistance obtained in Problem 34.

1.9 Diode Equivalent Circuits

36. Find the piecewise-linear equivalent circuit for the diode of Fig. 1. 15. Use a straight-line seg-
ment that intersects the horizontal axis at 0.7 V and best approximates the curve for the region
greater than 0.7 V.
37. Repeat Problem 36 for the diode of Fig. 1.27.
38. Find the piecewise-linear equivalent circuit for the germanium and gallium arsenide diodes of
Fig. 1.18.

1.1 o Transition and Diffusion Capacitance

*39. a. Referring to Fig. 1.33, determine the transition capacitance at reverse-bias potentials of
- 25 V and -10 V. What is the ratio of the change in capacitance to the change in voltage?
b. Repeat part (a) for reverse-bias potentials of -10 V and -1 V. Determine the ratio of the
change in capacitance to the change in voltage.
c. How do the ratios determined in parts (a) and (b) compare? What does this tell you about
which range may have more areas of practical application?
40. Referring to Fig. 1.33, determine the diffusion capacitance at 0 V and 0.25 V.
41. Describe in your own words how diffusion and transition capacitances differ.
42. Determine the reactance offered by a diode described by the characteristics of Fig. 1.33 at a
forward potential of 0.2 V and a reverse potential of -20 V if the applied frequency is 6 MHz.
43. The no-bias transition capacitance of a silicon diode is 8 pF with VK = 0.7 V and n = 1/2.
What is the transition capacitance if the applied reverse bias potential is 5 V?
44. Find the applied reverse bias potential if the transition capacitance of a silicon diode is 4 pF but
the no-bias level is 10 pF with n = 1/3 and VK = 0.7 V.

1.11 Reverse Recovery Time

45. Sketch the waveform for i of the network of Fig. 1.57 if t1 = 2ts and the total reverse recovery
time is 9 ns.

v,

10
+

IOkQ

(}
I

FIG. 1.57
Problem 45.

1.12 Diode Specification Sheets

*46. Plot Ip versus Vp using linear scales for the diode of Fig. 1.37. Note that the provided graph
employs a log scale for the vertical axis (log scales are covered in Sections 9.2 and 9.3).
47. a. Comment on the change in capacitance level with increase in reverse-bias potential for the
diode of Fig. 1.37.
b. What is the level of C(0)?
c. Using VK = 0.7 V, find the level ofn in Eq. 1.9.
48. Does the reverse saturation current of the diode of Fig. 1.37 change significantly in magnitude
for reverse-bias potentials in the range -25 V to -100 V?
SEMICONDUCTOR *49. For the diode of Fig. 1.37 determine the level of IR at room temperature (25°C) and the boiling
DIODES point of water (100°C). Is the change significant? Does the level just about double for every
10°C increase in temperature?
50. For the diode of Fig. 1.37, determine the maximum ac (dynamic) resistance at a forward cur-
rent of 0.1, 1.5, and 20 mA. Compare levels and comment on whether the results support con-
clusions derived in earlier sections of this chapter.
51. Using the characteristics of Fig. 1.37, determine the maximum power dissipation levels for the
diode at room temperature (25°C) and 100°C. Assuming that VF remains fixed at 0.7 V, how
has the maximum level of IF changed between the two temperature levels?
52. Using the characteristics of Fig. 1.37, determine the temperature at which the diode current
will be 50% of its value at room temperature (25°C).

1.15 Zener Diodes

53. The following characteristics are specified for a particular Zener diode: Vz = 29 V, VR = 16.8 V,
lzr = 10 mA, IR = 20 µ,A, and IZM = 40 mA. Sketch the characteristic curve in the manner
displayed in Fig. 1.47.
*54. At what temperature will the 10-V Zener diode of Fig. 1.47 have a nominal voltage of 10. 75 V?
(Hint: Note the data in Table 1.7.)
55. Determine the temperature coefficient of a 5-V Zener diode (rated 25°C value) if the nominal
voltage drops to 4.8 V at a temperature of 100°C.
56. Using the curves of Fig. 1.48a, what level of temperature coefficient would you expect for a
20-V diode? Repeat for a 5-V diode. Assume a linear scale between nominal voltage levels and
a current level of 0.1 mA.
57. Determine the dynamic impedance for the 24-V diode at Iz = 10 mA for Fig. 1.48b. Note that
it is a log scale.
*58. Compare the levels of dynamic impedance for the 24-V diode of Fig. 1.48b at current levels of
0.2, 1, and 10 mA. How do the results relate to the shape of the characteristics in this region?

1.16 Light-Emitting Diodes

59. Referring to Fig. 1.52e, what would appear to be an appropriate value of VK for this device?
How does it compare to the value of VK for silicon and germanium?
60. Given that Eg = 0.67 eV for germanium, find the wavelength of peak solar response for the
material. Do the photons at this wavelength have a lower or higher energy level?
61. Using the information provided in Fig. 1.52, determine the forward voltage across the diode if
the relative luminous intensity is 1.5.
*62. a. What is the percentage increase in relative efficiency of the device of Fig. 1.52 if the peak
current is increased from 5 mA to 10 mA?
b. Repeat part (a) for 30 mA to 35 mA (the same increase in current).
c. Compare the percentage increase from parts (a) and (b). At what point on the curve would
you say there is little to be gained by further increasing the peak current?
63. a. If the luminous intensity at 0° angular displacement is 3.0 med for the device of Fig. 1.52,
at what angle will it be 0.75 med?
b. At what angle does the loss of luminous intensity drop below the 50% level?
*64. Sketch the current derating curve for the average forward current of the high-efficiency red
LED of Fig. 1.52 as determined by temperature. (Note the absolute maximum ratings.)
CHAPTER OBJECTIVES
•
• Understand the concept of load-line analysis and how it is applied to diode networks.
• Become familiar with the use of equivalent circuits to analyze series, parallel, and
series-parallel diode networks.
• Understand the process ofrectification to establish a de level from a sinusoidal ac
input.
• Be able to predict the output response of a clipper and clamper diode configuration.
• Become familiar with the analysis of and the range of applications for Zener diodes.

2.1 INTRODUCTION
•
The construction, characteristics, and models of semiconductor diodes were introduced in
Chapter 1. This chapter will develop a working knowledge of the diode in a variety of
configurations using models appropriate for the area of application. By chapter's end, the
fundamental behavior pattern of diodes in de and ac networks should be clearly under-
stood. The concepts learned in this chapter will have significant carryover in the chapters
to follow. For instance, diodes are frequently employed in the description of the basic con-
struction of transistors and in the analysis of transistor networks in the de and ac domains.
This chapter demonstrates an interesting and very useful aspect of the study of a field
such as electronic devices and systems:
Once the basic behavior of a device is understood, its function and response in an
infinite variety of configurations can be examined.
In other words, now that we have a basic knowledge of the characteristics of a diode
along with its response to applied voltages and currents, we can use this knowledge to ex-
amine a wide variety of networks. There is no need to reexamine the response of the device
for each application.
In general:
The analysis of electronic circuits can follow one of two paths: using the actual
characteristics or applying an approximate model for the device.
For the diode the initial discussion will include the actual characteristics to clearly dem-
onstrate how the characteristics of a device and the network parameters interact. Once there
is confidence in the results obtained, the approximate piecewise model will be employed to
verify the results found using the complete characteristics. It is important that the role and
the response of various elements of an electronic system be understood without continually
56 DIODE APPLICATIONS having to resort to lengthy mathematical procedures. This is usually accomplished through
the approximation process, which can develop into an art itself. Although the results ob-
tained using the actual characteristics may be slightly different from those obtained using a
series of approximations, keep in mind that the characteristics obtained from a specification
sheet may be slightly different from those of the device in actual use. In other words, for
example, the characteristics of a 1N4001 semiconductor diode may vary from one element
to the next in the same lot. The variation may be slight, but it will often be sufficient to
justify the approximations employed in the analysis. Also consider the other elements of the
network: Is the resistor labeled 100 n exactly 100 il? Is the applied voltage exactly 10 V or
perhaps 10.08 V? All these tolerances contribute to the general belief that a response deter-
mined through an appropriate set of approximations can often be "as accurate" as one that
employs the full characteristics. In this book the emphasis is toward developing a working
knowledge of a device through the use of appropriate approximations, thereby avoiding an
unnecessary level of mathematical complexity. Sufficient detail will normally be provided,
however, to permit a detailed mathematical analysis if desired.

2.2 LOAD-LINE ANALYSIS

•
The circuit of Fig. 2.1 is the simplest of diode configurations. It will be used to describe the
analysis of a diode circuit using its actual characteristics. In the next section we will replace
the characteristics by an approximate model for the diode and compare solutions. Solving
the circuit of Fig. 2.1 is all about finding the current and voltage levels that will satisfy
both the characteristics of the diode and the chosen network parameters at the same time.

In (mA)

In
---+-

r
+ VD
+ +
E-=- R VR

0 Vn (V)
"II" (a) (b)

FIG. 2.1
Series diode configuration: (a) circuit; (b) characteristics.

In Fig. 2.2 the diode characteristics are placed on the same set of axes as a straight line
defined by the parameters of the network. The straight line is called a load line because the
intersection on the vertical axis is defined by the applied load R. The analysis to follow is
therefore called load-line analysis. The intersection of the two curves will define the solu-
tion for the network and define the current and voltage levels for the network.
Before reviewing the details of drawing the load line on the characteristics, we need to
determine the expected response of the simple circuit of Fig. 2.1. Note in Fig. 2.1 that the
effect of the "pressure" established by the de supply is to establish a conventional current
in the direction indicated by the clockwise arrow. The fact that the direction of this current
has the same direction as the arrow in the diode symbol reveals that the diode is in the
"on" state and will conduct a high level of current. The polarity of the applied voltage has
resulted in a forward-bias situation. With the current direction established, the polarities
for the voltage across the diode and resistor can be superimposed. The polarity of VD and
the direction of Iv clearly reveal that the diode is indeed in the forward-bias state, result-
ing in a voltage across the diode in the neighborhood of 0. 7 V and a current on the order
of 10 mA or more.
 LOAD-LINE ANALYSIS 57

~ Characteristics (device)
E
R

-------- Load line (network)

0 E

FIG. 2.2
Drawing the load line and finding the point of operation.

The intersections of the load line on the characteristics of Fig. 2.2 can be determined by
first applying Kirchhoff s voltage law in the clockwise direction, which results in
+E - Vv - VR = 0

or IE= Vv + IvR I (2.1)

The two variables of Eq. (2.1), Vv and Iv, are the same as the diode axis variables of
Fig. 2.2. This similarity permits plotting Eq. (2.1) on the same characteristics of Fig. 2.2.
The intersections of the load line on the characteristics can easily be determined if one
simply employs the fact that anywhere on the horizontal axis Iv = 0 A and anywhere on
the vertical axis Vv = 0 V.
If we set Vv = 0 Vin Eq. (2.1) and solve for Iv, we have the magnitude of Iv on the
vertical axis. Therefore, with Vv = 0 V, Eq. (2.1) becomes
E = Vv + IvR
= OV + IvR

and Iv= -
R
El Vv=OV
(2.2)

as shown in Fig. 2.2. If we set Iv= 0 A in Eq. (2.1) and solve for Vv, we have the magni-
tude of Vv on the horizontal axis. Therefore, with Iv = 0 A, Eq. (2.1) becomes
E = Vv + IvR
= Vv + (OA)R

and (2.3)

as shown in Fig. 2.2. A straight line drawn between the two points will define the load line
as depicted in Fig. 2.2. Change the level of R (the load) and the intersection on the vertical
axis will change. The result will be a change in the slope of the load line and a different
point of intersection between the load line and the device characteristics.
We now have a load line defined by the network and a characteristic curve defined by the
device. The point of intersection between the two is the point of operation for this circuit.
By simply drawing a line down to the horizontal axis, we can determine the diode voltage
VvQ, whereas a horizontal line from the point of intersection to the vertical axis will provide
the level of IvQ• The currentlv is actually the current through the entire series configuration
of Fig. 2.la. The point of operation is usually called the quiescent point (abbreviated "Q-
point") to reflect its "still, unmoving" qualities as defined by a de network.
58 DIODE APPLICATIONS The solution obtained at the intersection of the two curves is the same as would be ob-
tained by a simultaneous mathematical solution of
E Vv
Iv= R -R [derivedfromEq. (2.1)]

and
Since the curve for a diode has nonlinear characteristics, the mathematics involved would
require the use of nonlinear techniques that are beyond the needs and scope of this book.
The load-line analysis described above provides a solution with a minimum of effort and a
"pictorial" description of why the levels of solution for VvQ and IvQ were obtained. The
next example demonstrates the techniques introduced above and reveals the relative ease
with which the load line can be drawn using Eqs. (2.2) and (2.3).

EXAMPLE 2.1 For the series diode configuration of Fig. 2.3a, employing the diode char-
acteristics of Fig. 2.3b, determine:
a. VvQ and IvQ·
b. VR.

In (mA)

20
18
+ VD
16
_.._ Si
14
12
ID 10
+ + 8
E-=-lOV R 0.5 k!l VR 6
4
2

0 0.5 0.8 Vn (V)

-=- (a) (b)

FIG. 2.3
(a) Circuit; (b) characteristics.

Solution:
a. Eq. (2.2): Iv=- El lOV
=--=20mA
R Vv=OV 0.5 kil
Eq. (2.3): Vv = El1v=OA = 10 V
The resulting load line appears in Fig. 2.4. The intersection between the load line and
the characteristic curve defines the Q-point as
VvQ ~ 0.78V
IvQ ~ 18.SmA
The level of Vv is certainly an estimate, and the accuracy of Iv is limited by the chosen
scale. A higher degree of accuracy would require a plot that would be much larger and
perhaps unwieldy.
b. VR = E - Vv = lOV - 0.78V = 9.22V

As noted in the example above,

the load line is determined solely by the applied network, whereas the characteristics
are defined by the chosen device.
Changing the model we use for the diode will not disturb the network so the load line to
be drawn will be exactly the same as appearing in the example above.
Since the network of Example 2.1 is a de network the Q-point of Fig.2.4 will remain
fixed with VvQ = 0.78 V and IvQ = 18.5 mA. In Chapter 1 a de resistance was defined at
any point on the characteristics by Rvc = Vv/Iv.
 Iv (mA) LOAD-LINE ANALYSIS 59
E
R"--
20
lvQ = 18.5 mA 18 -
16
14
12
10
8
6
4
2

0 0.5\1 2 3 4 5 6 7 8 9 10 V0 (V)
(E)
VvQ= 0.78 V

FIG. 2.4
Solution to Example 2.1.

Using the Q-point values, the de resistance for Example 2.1 is

Rv = VvQ = 0.78 V = 42.16 0
18.5 mAlvQ
An equivalent network (for these operating conditions only) can then be drawn as shown
in Fig. 2.5.

42.16!1

+
E-=- lOV

FIG. 2.5
Network quivalent to Fig. 2.4.

The current
E lOV lOV
I---- - - - - 18.SmA
D - Rv +R 42.16 n + 500 n 542.16 n

V ----
RE (500 O)(lOV)
and
R - Rv +R 42.16 n + 500 n = 9•22 v

matching the results of Example 2.1.

In essence, therefore, once a de Q-point has been determined the diode can be replaced
by its de resistance equivalent. This concept of replacing a characteristic by an equivalent
model is an important one and will be used when we consider ac inputs and equivalent models
for transistors in the chapters to follow. Let us now see what effect different equivalent
models for the diode will have on the response in Example 2.1

EXAMPLE 2.2 Repeat Example 2.1 using the approximate equivalent model for the sili-
con semiconductor diode.
Solution: The load line is redrawn as shown in Fig. 2.6 with the same intersections as
defined in Example 2.1. The characteristics of the approximate equivalent circuit for the
diode have also been sketched on the same graph. The resulting Q-point is
VvQ =
0.7V
lvQ = 18.SmA
60 DIODE APPLICATIONS Iv (mA)

20
Iv/= 18.5 mA
16
14
12
10
8
6
4
2

0 0.5\1 2 3 4 5 6 7 8 9 10 Yv (V)
VvQ=0.7V

FIG. 2.6
Solution to Example 2.1 using the diode approximate model.

The results obtained in Example 2.2 are quite interesting. The level of lvQ is exactly the
same as obtained in Example 2.1 using a characteristic curve that is a great deal easier to
draw than that appearing in Fig. 2.4. The Vv = 0. 7 V here and the 0. 78 V from Example
2.1 are of a different magnitude to the hundredths place, but they are certainly in the same
neighborhood if we compare their magnitudes to the magnitudes of the other voltages of
the network.
For this situation the de resistance of the Q-point is
VvQ 0.7V
Rv = - = - - - = 37.840
lvQ 18.5 mA
which is still relatively close to that obtained for the full characteristics.
In the next example we go a step further and substitute the ideal model. The results will
reveal the conditions that must be satisfied to apply the ideal equivalent properly.

EXAMPLE 2.3 Repeat Example 2.1 using the ideal diode model.
Solution: As shown in Fig. 2.7, the load line is the same, but the ideal characteristics
now intersect the load line on the vertical axis. The Q-point is therefore defined by
VvQ = OV
lvQ = 20mA

Iv (mA)

Q-point
lvQ=20mA 20
18
16
14
12
10
\
8
6
4
2
IJll ⇒
-
Iv

0 ~1 2 3 4 5 6 7 8 9 10 Yv (V)

FIG. 2.7
Solution to Example 2.1 using the ideal diode model.
The results are sufficiently different from the solutions of Example 2.1 to cause some con- SERIES DIODE 61
cern about their accuracy. Certainly, they do provide some indication of the level of voltage CONFIGURATIONS
and current to be expected relative to the other voltage levels of the network, but the addi-
tional effort of simply including the 0. 7-V offset suggests that the approach of Example 2.2
is more appropriate.
Use of the ideal diode model therefore should be reserved for those occasions when
the role of a diode is more important than voltage levels that differ by tenths of a volt and
in those situations where the applied voltages are considerably larger than the threshold
voltage VK· In the next few sections the approximate model will be employed exclusively
since the voltage levels obtained will be sensitive to variations that approach VK· In later
sections the ideal model will be employed more frequently since the applied voltages will
frequently be quite a bit larger than VK and the authors want to ensure that the role of the
diode is correctly and clearly understood.
In this case,
VvQ OV
Rv = - - = - - - = 0 0 (or a short-circuit equivalent)
lvQ 20mA

2.] SERIES DIODE CONFIGURATIONS

•
In the last section we found that the results obtained using the approximate piecewise-linear
equivalent model were quite close, if not equal, to the response obtained using the full
characteristics. In fact, if one considers all the variations possible due to tolerances, tem-
perature, and so on, one could certainly consider one solution to be "as accurate" as the
other. Since the use of the approximate model normally results in a reduced expenditure of
time and effort to obtain the desired results, it is the approach that will be employed in this
book unless otherwise specified. Recall the following:
The primary purpose of this text is to develop a general knowledge of the behavior,
capabilities, and possible areas of application of a device in a manner that will
minimize the need for extensive mathematical developments.
For all the analysis to follow in this chapter it is assumed that
The forward resistance of the diode is usually so small compared to the other series
elements of the network that it can be ignored.
This is a valid approximation for the vast majority of applications that employ diodes.
Using this fact will result in the approximate equivalents for a silicon diode and an ideal
diode that appear in Table 2.1. For the conduction region the only difference between
the silicon diode and the ideal diode is the vertical shift in the characteristics, which is
accounted for in the equivalent model by a de supply of 0.7 V opposing the direction of
forward current through the device. For voltages less than 0.7 V for a silicon diode and OV
for the ideal diode the resistance is so high compared to other elements of the network that
its equivalent is the open circuit.
For a Ge diode the offset voltage is 0.3 V and for a GaAs diode it is 1.2 V. Otherwise
the equivalent networks are the same. For each diode the label Si, Ge, or GaAs will appear
along with the diode symbol. For networks with ideal diodes the diode symbol will appear
as shown in Table 2.1 without any labels.
The approximate models will now be used to investigate a number of series diode con-
figurations with de inputs. This will establish a foundation in diode analysis that will carry
over into the sections and chapters to follow. The procedure described can, in fact, be ap-
plied to networks with any number of diodes in a variety of configurations.
For each configuration the state of each diode must first be determined. Which diodes are
"on" and which are "off'? Once determined, the appropriate equivalent can be substituted
and the remaining parameters of the network determined.
In general, a diode is in the "on" state if the cu"ent established by the applied sources
is such that its direction matches that of the arrow in the diode symbol, and VD =::: 0. 7 V
for silicon, Vv ==:: 0.3 V for germanium, and Vv ==:: 1.2 V for gallium arsenide.
For each configuration, mentally replace the diodes with resistive elements and note the
resulting current direction as established by the applied voltages ("pressure"). If the resulting
62 DIODE APPLICATIONS TABLE 2.1
Approximate and Ideal Semiconductor Diode Models.

Silicon:
+ 0.7V-
►I
~~Si

~ ___._►I
0
===;> ---0 0---
Iv=OA
Iv=OA
Si

Ideal:
Iv +ov- + Vv=OV-

~--►I
===;> ~
~
Iv Iv

~
Vv

___._►I
Iv=OA
===> ---0
Iv=OA
0---

direction is a "match" with the arrow in the diode symbol, conduction through the diode will
occur and the device is in the "on" state. The description above is, of course, contingent on
Si
the supply having a voltage greater than the "turn-on" voltage (VK) of each diode.
If a diode is in the "on" state, one can either place a 0.7-V drop across the element or
+
E redraw the network with the VK equivalent circuit as defined in Table 2.1. In time the prefer-
R
ence will probably simply be to include the 0. 7-V drop across each "on" diode and to draw a
diagonal line through each diode in the "off' or open state. Initially, however, the substitu-
tion method will be used to ensure that the proper voltage and current levels are determined.
The series circuit of Fig. 2.8 described in some detail in Section 2.2 will be used to
FICi. 2.8 demonstrate the approach described in the above paragraphs. The state of the diode is first
Series diode configuration. determined by mentally replacing the diode with a resistive element as shown in Fig. 2.9a.
The resulting direction of/ is a match with the arrow in the diode symbol, and since E > VK,
the diode is in the "on" state. The network is then redrawn as shown in Fig. 2.9b with the
appropriate equivalent model for the forward-biased silicon diode. Note for future refer-
ence that the polarity of VD is the same as would result if in fact the diode were a resistive
element. The resulting voltage and current levels are the following:

(2.4)

+
I ?''
E-=-
- I--

R
+
VR
+r1
E-=-
+ Vv-

0.7V

R
7t IR

+
VR

(a) -=- (b)

FICi. 2.9
(a) Determining the state of the diode of Fig. 2.8; (b) substituting the
equivalent model for the "on" diode of Fig. 2.9a.
 SERIES DIODE 63
(2.5) CONFIGURATIONS

(2.6)

In Fig. 2.10 the diode of Fig. 2.7 has been reversed. Mentally replacing the diode with
a resistive element as shown in Fig. 2.11 will reveal that the resulting current direction
does not match the arrow in the diode symbol. The diode is in the "off' state, resulting in
the equivalent circuit of Fig. 2.12. Due to the open circuit, the diode current is OA and the
voltage across the resistor R is the following:
VR =IRR= IvR = (OA)R = OV

if',,V, --I -
+ Vn=E

lt
+
E
Si

R
+
VR
:~r R
+
VR
+lF
E~ R
IR

+
VR

T "='

FIC. 2.10 FIC. 2.11 FIC. 2.12

Reversing the diode of Fig. 2.8. Determining the state of the diode Substituting the equivalent nwdel
of Fig. 2.10. for the "off' diode of Fig. 2.10.

The fact that VR = 0 V will establish E volts across the open circuit as defined by Kirchhoff s
voltage law. Always keep in mind that under any circumstances-de, ac instantaneous
values, pulses, and so on-Kirchhoffs voltage law must be satisfied!

EXAMPLE 2.4 For the series diode configuration of Fig. 2.13, determine Vv, VR, and Iv.

+ Vn

lt
+
E~8V
r Si

R
IR

2.2 kQ VR
+

FIC. 2.13
Circuit for Example 2.4.

Solution: Since the applied voltage establishes a current in the clockwise direction to
match the arrow of the symbol and the diode is in the "on" state,
Vv = 0.7V
VR = E - Vv = 8 V - 0.7 V = 7.3 V
VR 7.3 V
Iv= IR= R= 2 _2 kD ~ 3.32mA
64 DIODE APPLICATIONS
EXAMPLE 2.5 Repeat Example 2.4 with the diode reversed.
Solution: Removing the diode, we find that the direction of I is opposite to the arrow in
the diode symbol and the diode equivalent is the open circuit no matter which model is
employed. The result is the network of Fig. 2.14, where ID = 0 A due to the open circuit.
Since VR = IRR, we have VR = (O)R = 0 V. Applying Kirchhoff's voltage law around
the closed loop yields
E - Vv - VR = 0
and Vv = E - VR = E - 0 = E = 8V

lv=OA

+
l vv-
E-=-sv R
+
2.2 kQ VR

FIC. 2.14
Determining the unknown quantities for
Example 2.5.

In particular, note in Example 2.5 the high voltage across the diode even though it is an
"off' state. The current is zero, but the voltage is significant. For review purposes, keep the
following in mind for the analysis to follow:
An open circuit can have any voltage across its terminals, but the current is always OA.
A short circuit has a 0-V drop across its terminals, but the current is limited only by the
surrounding network.
In the next example the notation of Fig. 2.15 will be employed for the applied voltage. It
is a common industry notation and one with which the reader should become very familiar.
Such notation and other defined voltage levels are treated further in Chapter 4.

E=+lOVO E=-5 VO

FIC. 2.15
Source notation.

+O.SV

lvf
+ EXAMPLE 2.6 For the series diode configuration of Fig. 2.16, determine Vv, VR, and ID·
Si Vv
Solution: Although the "pressure" establishes a current with the same direction as the
+ arrow symbol, the level of applied voltage is insufficient to tum the silicon diode "on."
R l.2k.Q VR The point of operation on the characteristics is shown in Fig. 2.17, establishing the open-
circuit equivalent as the appropriate approximation, as shown in Fig. 2.18. The resulting
voltage and current levels are therefore the following:
-=-
Iv= OA
FIC. 2.16
Series diode circuit for VR =IRR= IvR = (OA)l.2kf! = OV
Example 2.6. and Vv = E = O.SV
 SERIES DIODE 65
CONFIGURATIONS
+O.SV

I{~=OmA
Vv =O.SV

0 / 0.7V
Vv =0.5 V

FICi. 2.17 FICi. 2.18

Operating point with E = 0.5 V. Determining In, VR, and Vnfor
the circuit of Fig. 2. I 6.

EXAMPLE 2.7 Determine V0 and Iv for the series circuit of Fig. 2.19.
Solution: An attack similar to that applied in Example 2.4 will reveal that the resulting
current has the same direction as the arrowheads of the symbols of both diodes, and the
network of Fig. 2.20 results because E = 12 V > (0.7 V + 1.8 V [Table 1.8]) = 2.5 V.
Note the redrawn supply of 12 V and the polarity of V0 across the 680-D resistor. The
resulting voltage is
V0 = E - VK1 - VK2 = 12 V - 2.5 V = 9.5 V
VR Vo 9.5V
and Iv = IR =- = - = - - = 13.97 mA
R R 680 D

"II"

FICi. 2.19 FICi. 2.20

Circuit for Example 2.7. Determining the unknown quantities for
Example 2. 7.

EXAMPLE 2.8 Determine ID, Vv 2, and V0 for the circuit of Fig. 2.21.
Solution: Removing the diodes and determining the direction of the resulting current /
result in the circuit of Fig. 2.22. There is a match in current direction for one silicon diode
but not for the other silicon diode. The combination of a short circuit in series with an open
circuit always results in an open circuit and Iv= 0 A, as shown in Fig. 2.23.

E-=-( '
- I-
+L "v'';',~"v'\,'\1
-I -

R 5.6kQ
+
V
-
I= 0

-r
0

"II"
-=-

FICi. 2.21 FICi. 2.22 FICi. 2.23

Circuit for Example 2.8. Determining the state of the diodes Substituting the equivalent state for
of Fig. 2.21. the open diode.
66 DIODE APPLICATIONS

FIG. 2.24
Determining the unknown quantities for the
circuit of Example 2.8.

The question remains as to what to substitute for the silicon diode. For the analysis to
follow in this and succeeding chapters, simply recall for the actual practical diode that when
Iv = 0 A, Vv = 0 V (and vice versa), as described for the no-bias situation in Chapter 1.
The conditions described by Iv= 0 A and Vv, = 0 V are indicated in Fig. 2.24. We have
V0 =IRR= IvR = (OA)R = OV
and Vv2 = Vopencircuit = E = 20 V
Applying Kirchhoffs voltage law in a clockwise direction gives
E - Vv, - Vv 2 - V0 = 0
and Vv 2 = E - Vv, - V0 = 20 V - 0 - 0
= 20V
with

EXAMPLE 2.9 Determine I, V1, V2 , and V0 for the series de configuration of Fig. 2.25.

+ v, -
R,
E 1 = lOV Vo
~
4.7kQ Si
+
R2 2.2kQ V2

E 2 =-5V

FIG. 2.25
Circuit for Example 2.9.

Solution: The sources are drawn and the current direction indicated as shown in Fig. 2.26.
The diode is in the "on" state and the notation appearing in Fig. 2.27 is included to indicate
this state. Note that the "on" state is noted simply by the additional Vv = 0.7 Von the
figure. This eliminates the need to redraw the network and avoids any confusion that may

R, + 0.7 V -

( ' 4.7 kQ +
+ v1 - +
+
E1-=.. IO V
2.2 kQ R2 V2
A v,,

-1 5Y-=-
I
":'
E2
+
J.
FIG. 2.26 FIG. 2.27
Determining the state of the diode for the Determining the unknown quantities for the network
network of Fig. 2.25. of Fig. 2.25. KVL, Kirchhoff voltage loop.
result from the appearance of another source. As indicated in the introduction to this sec- PARALLEL AND 67
tion, this is probably the path and notation that one will take when a level of confidence SERIES-PARALLEL
CONFIGURATIONS
has been established in the analysis of diode configurations. In time the entire analysis will
be performed simply by referring to the original network. Recall that a reverse-biased
diode can simply be indicated by a line through the device.
The resulting current through the circuit is
E1+E2-Vv 10V+5V-0.7V 14.3 V
I=------
R1 + R2 4.7 kfl + 2.2 kfl 6.9kfl
- 2.07mA
and the voltages are
= IR1 = (2.07 mA)(4.7 kfl) =
V1 9.73 V
V2 = IR2 = (2.07 mA)(2.2 kfl) = 4.55 V
Applying Kirchhoff's voltage law to the output section in the clockwise direction results in
+ V2 - V0 = 0
-E2
and V0 = V2 - E2 = 4.55 V - 5 V = - 0.45 V
The minus sign indicates that V0 has a polarity opposite to that appearing in Fig. 2.25.

2.4 PARALLEL AND SERIES-PARALLEL

CONFIGURATIONS
The methods applied in Section 2.3 can be extended to the analysis of parallel and series-
parallel configurations. For each area of application, simply match the sequential series of
steps applied to series diode configurations.
•
EXAMPLE 2.10 Determine V0 , Ii, Iv,, and IVz for the parallel diode configuration of Fig. 2.28.

+ VR -
11
.!J.._ 0.33k.Q ~ 0.33 kQ
+ r--4V~r-----<i>---~....---o +
+
R {In, {[Dz
+
R + i/D1

-1
E-=- lOV D, Si Vo E-=-IOY o.7v-=-

-r
Dz Si

FICi. 2.28 FICi. 2.29

Networkfor Example 2.10. Determining the unknown quantities for
the network of Example 2.10.

Solution: For the applied voltage the "pressure" of the source acts to establish a current
through each diode in the same direction as shown in Fig. 2.29. Since the resulting current
direction matches that of the arrow in each diode symbol and the applied voltage is greater
than 0.7 V, both diodes are in the "on" state. The voltage across parallel elements is always
the same and
V0 = 0.7V
The current is
VR E - Vv lOV - 0.7V
Ii = - = --- = ----- = 28.18 mA
R R 0.33kfl
Assuming diodes of similar characteristics, we have
Ii 28.18mA
Iv I = Iv z = -2 = - -
2 - = 14•09 mA
68 DIODE APPLICATIONS This example demonstrates one reason for placing diodes in parallel. If the current rat-
ing of the diodes of Fig. 2.28 is only 20 mA, a current of 28.18 mA would damage the
device if it appeared alone in Fig. 2.28. By placing two in parallel, we limit the current to
a safe value of 14.09 mA with the same terminal voltage.
+8V

R EXAMPLE 2.11 In this example there are two LEDs that can be used as a polarity detec-
tor. Apply a positive source voltage and a green light results. Negative supplies result in a
red light. Packages of such combinations are commercially available.
Find the resistor R to ensure a current of 20 mA through the "on" diode for the configu-
Red>'= Green ration of Fig. 2.30. Both diodes have a reverse breakdown voltage of 3 V and an average
>'" ?' tum-on voltage of 2 V.
Solution: The application of a positive supply voltage results in a conventional current
that matches the arrow of the green diode and turns it on.
The polarity of the voltage across the green diode is such that it reverse biases the red
FICi. 2.30 diode by the same amount. The result is the equivalent network of Fig. 2.31.
Network for Example 2.11. Applying Ohm's law, we obtain
I = 20 mA = E - VLED 8V - 2V
+8V
R R
{20mA 6V
and R = - - = 300 0
20mA
R
Note that the reverse breakdown voltage across the red diode is 2 V, which is fine for an
LED with a reverse breakdown voltage of 3 V.
However, if the green diode were to be replaced by a blue diode, problems would
develop, as shown in Fig. 2.32. Recall that the forward bias required to tum on a blue diode
+

y -=- 2V is about 5 V. The result would appear to require a smaller resistor R to establish the current
of 20 mA. However, note that the reverse bias voltage of the red LED is 5 V, but the
reverse breakdown voltage of the diode is only 3 V. The result is the voltage across the red
LED would lock in at 3 Vas shown in Fig. 2.33. The voltage across R would be 5 V and
the current limited to 20 mA with a 250 n resistor but neither LED would be on.

FICi. 2.31
Operating conditions for the
network of Fig. 2.30. +8V +8V

R -3V R

+ +

FICi. 2.32 FICi. 2.33

Network of Fig. 2.31 Demonstrating damage to the red LED if the
with a blue diode. reverse breakdown voltage is exceeded.

A simple solution to the above is to add the appropriate resistance level in series with
each diode to establish the desired 20 mA and to include another diode to add to the
reverse-bias total reverse breakdown voltage rating, as shown in Fig. 2.34. When the blue
LED is on, the diode in series with the blue LED will also be on, causing a total voltage
drop of 5.7 V across the two series diodes and a voltage of 2.3 V across the resistor R 1,
establishing a high emission current of 19 .17 mA. At the same time the red LED diode and
 8V PARALLEL AND 69
SERIES-PARALLEL
CONFIGURATIONS
rI IR,-- 8 V120n
- 5.7 V -
-
19 17 mA
·

+ +
Si 0.7V
+ 5.7V
Red~ 5V

"II" "II"

FIG. 2.34
Protective measure for the red LED of Fig. 2.33.

its series diode will also be reverse biased, but now the standard diode with a reverse
breakdown voltage of 20 V will prevent the full reverse-bias voltage of 8 V from appear-
ing across the red LED. When forward biased, the resistor R2 will establish a current of
19.63 mA to ensure a high level of intensity for the red LED.

12V
EXAMPLE 2.12 Determine the voltage V0 for the network of Fig. 2.35.
Solution: Initially, it might appear that the applied voltage will tum both diodes "on"
because the applied voltage ("pressure") is trying to establish a conventional current
through each diode that would suggest the "on" state. However, if both were on, there Si ~green
would be more than one voltage across the parallel diodes, violating one of the basic rules
of network analysis: The voltage must be the same across parallel elements.
The resulting action can best be explained by remembering that there is a period of
build-up of the supply voltage from 0 V to 12 V even though it may take milliseconds or
2.2 kQ
microseconds. At the instant the increasing supply voltage reaches 0.7 V the silicon diode will
tum "on" and maintain the level of 0.7 V since the characteristic is vertical at this voltage-the
current of the silicon diode will simply rise to the defined level. The result is that the volt-
age across the green LED will never rise above 0. 7 V and will remain in the equivalent "II"

open-circuit state as shown in Fig. 2.36.

FIG. 2.35
The result is Networkfor Example 2.12.
V0 = 12V - 0.7V = 11.3V

green LED

2.2 V

2.2 kQ

"II"

FIG. 2.36
Determining V0 for the network of
Fig. 2.35.
70 DIODE APPLICATIONS
EXAMPLE 2.13 Determine the currents Ji, [z, and lv2 for the network of Fig. 2.37.

~I
+ 0.7 v-

D1
+
E-=- 20V Si Dz
lv2
r______________ Ja
R2
5.6k0
"II" - V2 +

FIG. 2.37 FIG. 2.38

Network for Example 2.13. Determining the unknown quantities for
Example 2.13.

Solution: The applied voltage (pressure) is such as to tum both diodes on, as indicated
by the resulting current directions in the network of Fig. 2.38. Note the use of the abbrevi-
ated notation for "on" diodes and that the solution is obtained through an application of
techniques applied to de series-parallel networks. We have
VK2 0.7V
Ji=-=--=0.212mA
R1 3.3k0
Applying Kirchhoff's voltage law around the indicated loop in the clockwise direction
yields
-V2 +E - VK1 - VK2 = 0
and V2 = E - VK1 - VK2 = 20V - 0.7V - 0.7V = 18.6V
Y2 18.6V
with [z = - = 5 6 "' = 3.32 mA
R2 . ku
At the bottom node a,
lv2 +Ii= 12
and lv 2 = 12 - /1 = 3.32 mA - 0.212 mA ~ 3.11 mA

(1) E= lOV
1
Si

D1
2.5 AND/OR GATES
The tools of analysis are now at our disposal, and the opportunity to investigate a computer
configuration is one that will demonstrate the range of applications of this relatively sim-
ple device. Our analysis will be limited to determining the voltage levels and will not
•
Si include a detailed discussion of Boolean algebra or positive and negative logic.
(0) ov Vo
The network to be analyzed in Example 2.14 is an OR gate for positive logic. That is, the
2 10-V level of Fig. 2.39 is assigned a "1" for Boolean algebra and the 0-V input is assigned
D2
a "O." An OR gate is such that the output voltage level will be a 1 if either or both inputs is
R 1 kQ a 1. The output is a Oif both inputs are at the O level.
The analysis of AND/OR gates is made easier by using the approximate equivalent for
a diode rather than the ideal because we can stipulate that the voltage across the diode must
"II"
be 0.7 V positive for the silicon diode to switch to the "on" state.
FIG. 2.39 In general, the best approach is simply to establish a "gut" feeling for the state of the
Positive logic OR gate. diodes by noting the direction and the "pressure" established by the applied potentials. The
analysis will then verify or negate your initial assumptions.

EXAMPLE 2.14 Determine V0 for the network of Fig. 2.39.

Solution: First note that there is only one applied potential; 10 V at terminal 1. Terminal 2
with a 0-V input is essentially at ground potential, as shown in the redrawn network of
Fig. 2.40. Figure 2.40 "suggests" that D 1 is probably in the "on" state due to the applied 10 V, AND/OR GATES 71
whereas D 2 with its "positive" side at O V is probably "off." Assuming these states will
result in the configuration of Fig. 2.41.

+
D1
+ll:11 VK

.7V
V0 = E- VK= 9.3 V (a I level)
o---+--+----o
+ Vo E-=-lOV

11
E-=-lOV
Dz

] "II"
ov
"II"
R

"II"
lkQ

"II"
I
R lkQ

FIG. 2.40 FIG. 2.41

Redrawn network of Fig. 2.39. Assumed diode states for Fig. 2.40.

The next step is simply to check that there is no contradiction in our assumptions. That is,
note that the polarity across D 1 is such as to turn it on and the polarity across D 2 is such as to
turn it off. For D 1 the "on" state establishes V0 at V0 = E - VD = 10 V - 0. 7 V = 9.3 V.
With 9.3 Vat the cathode(-) side of D 2 and OVat the anode (+)side, D 2 is definitely in the
"off' state. The current direction and the resulting continuous path for conduction further
confirm our assumption that D 1 is conducting. Our assumptions seem confirmed by the
resulting voltages and current, and our initial analysis can be assumed to be correct. The out-
put voltage level is not 10 V as defined for an input of 1, but the 9.3 Vis sufficiently large to
be considered a I level. The output is therefore at a I level with only one input, which suggests
that the gate is an OR gate. An analysis of the same network with two 10-V inputs will result
in both diodes being in the "on" state and an output of 9.3 V. A 0-V input at both inputs will
not provide the 0. 7 V required to turn the diodes on, and the output will be a Odue to the 0-V
output level. For the network of Fig. 2.41 the current level is determined by
E - Vv lOV - 0.7V
I= - - - ----- = 9.3 mA
R lkll

(1) Si
EXAMPLE 2.15 Determine the output level for the positive logic AND gate of Fig. 2.42. E 1 = lOV
An AND gate is one where a 1 output is only obtained when a 1 input appears at each and 1
D1
every input.
(0) Si
Solution: Note in this case that an independent source appears in the grounded leg of the Ez=OV o---tlll----t---0 Vo
network. For reasons soon to become obvious, it is chosen at the same level as the input 2
Dz
logic level. The network is redrawn in Fig. 2.43 with our initial assumptions regarding the R lkQ
state of the diodes. With 10 V at the cathode side of D 1 it is assumed that D 1 is in the "off'
state even though there is a 10-V source connected to the anode of D 1 through the resistor. +
E -=-10v
-i
FIG. 2.42
Positive logic AND gate.

FIG. 2.43
Substituting the assumed states for the diodes of Fig. 2.42.
72 DIODE APPLICATIONS However, recall that we mentioned in the introduction to this section that the use of the
approximate model will be an aid to the analysis. For D 1, where will the 0.7 V come from
if the input and source voltages are at the same level and creating opposing "pressures"?
D 2 is assumed to be in the "on" state due to the low voltage at the cathode side and the
availability of the 10-V source through the 1-kll resistor.
For the network of Fig. 2.43 the voltage at V0 is 0.7 V due to the forward-biased diode
D 2. With 0.7 Vat the anode of D 1 and 10 Vat the cathode, D 1 is definitely in the "off'
state. The current/ will have the direction indicated in Fig. 2.43 and a magnitude equal to
E - VK lOV - 0.7V
/=-- ----=93mA
R I kll •

The state of the diodes is therefore confirmed and our earlier analysis was correct. Al-
though not O V as earlier defined for the O level, the output voltage is sufficiently small to
be considered a Olevel. For the AND gate, therefore, a single input will result in a 0-level
output. The remaining states of the diodes for the possibilities of two inputs and no inputs
will be examined in the problems at the end of the chapter.

2.6 SINUSOIDAL INPUTS; HALF-WAVE

RECTIFICATION
•
The diode analysis will now be expanded to include time-varying functions such as the
sinusoidal waveform and the square wave. There is no question that the degree of diffi-
culty will increase, but once a few fundamental maneuvers are understood, the analysis
will be fairly direct and follow a common thread.
The simplest of networks to examine with a time-varying signal appears in Fig. 2.44. For
the moment we will use the ideal model (note the absence of the Si, Ge, or GaAs label) to
ensure that the approach is not clouded by additional mathematical complexity.

V;
+
+ +
V; R Vo
0

1 cycle
"II"
V; = Vrn sin (J]f
FIG. 2.44
Half-wave rectifier.

Over one full cycle, defined by the period T of Fig. 2.44, the average value (the algebraic
sum of the areas above and below the axis) is zero. The circuit of Fig. 2.44, called a half-wave
rectifier, will generate a waveform vO that will have an average value of particular use in the
ac-to-dc conversion process. When employed in the rectification process, a diode is typically
referred to as a rectifier. Its power and current ratings are typically much higher than those
of diodes employed in other applications, such as computers and communication systems.
During the interval t = 0 - T /2 in Fig. 2.44 the polarity of the applied voltage vi is such
as to establish "pressure" in the direction indicated and turn on the diode with the polarity
appearing above the diode. Substituting the short-circuit equivalence for the ideal diode will
result in the equivalent circuit of Fig. 2.45, where it is fairly obvious that the output signal
is an exact replica of the applied signal. The two terminals defining the output voltage are
connected directly to the applied signal via the short-circuit equivalence of the diode.
For the period T /2 - T, the polarity of the input vi is as shown in Fig. 2.46, and the
resulting polarity across the ideal diode produces an "off' state with an open-circuit equiva-
lent. The result is the absence of a path for charge to flow, and v0 = iR = (O)R = 0 V for
the period T /2 - T. The input vi and the output v0 are sketched together in Fig. 2.47 for
comparison purposes. The output signal v0 now has a net positive area above the axis over
 + SINUSOIDAL INPUTS; 73
+ + + + Vo
HALF-WAVE

~
RECTIFICATION
V; R Vo ---+- V; R vo=vi

0 T
"II" "II"
2

FIG. 2.45
Conduction region (O- T /2).

+ 0------0
+ + Vo

V;

+
) R Vo ---+-

+
V; R v 0 =0V

0 T
2
V0

I
=0V

T
"II" "II"

FIG. 2.46
Nonconduction region (T / 2 - T).

V;

t Vdc=OV
0

0
T---,
FIG. 2.47
Half-wave rectified signal.

a full period and an average value determined by

I Vctc = 0.318 Vm I half-wave (2.7)

The process of removing one-half the input signal to establish a de level is called half-
wave rectification.
The effect of using a silicon diode with VK = 0.7 Vis demonstrated in Fig. 2.48 for the
forward-bias region. The applied signal must now be at least 0.7 V before the diode can turn
"on." For levels ofv; less than 0.7 V, the diode is still in an open-circuit state and Va= 0 V,
as shown in the same figure. When conducting, the difference between Va and v; is a fixed

V; R
0 I 11 T
I II 2
~
Offset due to VK

FIG. 2.48
Effect of VK on half-wave rectified signal.
74 DIODE APPLICATIONS level of VK = 0.1 V and v0 = vi - VK, as shown in the figure. The net effect is a reduction
in area above the axis, which reduces the resulting de voltage level. For situations where
Vm >> VK, the following equation can be applied to determine the average value with a
relatively high level of accuracy.

(2.8)
In fact, if Vm is sufficiently greater than VK, Eq. (2.7) is often applied as a first approxi-
mation for Vac-

EXAMPLE 2.16
a. Sketch the output vO and determine the de level of the output for the network of Fig. 2.49.
b. Repeat part (a) if the ideal diode is replaced by a silicon diode.
c. Repeat parts (a) and (b) if Vm is increased to 200 V, and compare solutions using Eqs.
(2.7) and (2.8).

V; + +
V; R 2 kn Vo

0 T t

FIC. 2.49
Network for Example 2.16.

Solution:
a. In this situation the diode will conduct during the negative part of the input as shown in
Fig. 2.50, and vO will appear as shown in the same figure. For the full period, the de level is
Vac = -0.318Vm = -0.318(20V) = -6.36V
The negative sign indicates that the polarity of the output is opposite to the defined
polarity of Fig. 2.49.

V; - ... + Vo

+
V; o,kQ +
Vo
0 T
2
T

FIC. 2.50
Resulting v0for the circuit of Example 2.16.

b. For a silicon diode, the output has the appearance of Fig. 2.51, and
Vac ~ -0.318(Vm - 0.7 V) = -0.318(19.3 V) ~ -6.14 V
The resulting drop in de level is 0.22 V, or about 3.5%.
c. Eq. (2.7): Vac = -0.318 Vm = -0.318(200V) = -63.6V
0 T
2 Eq. (2.8): Vac = -0.318(Vm - VK) = -0.318(200 V - 0.7 V)
\ = -(0.318)(199.3 V) = - 63.38 V
20 V - 0.7 V = 19.3 V
which is a difference that can certainly be ignored for most applications. For part (c) the
FIC. 2.51 offset and drop in amplitude due to VK would not be discernible on a typical oscillo-
Effect of VK on output of scope if the full pattern is displayed.
Fig. 2.50.
PIV (PRY) FULL-WAVE 75
RECTIFICATION
The peak inverse voltage (PIV) [or PRV (peak reverse voltage)] rating of the diode is of
primary importance in the design of rectification systems. Recall that it is the voltage rat-
ing that must not be exceeded in the reverse-bias region or the diode will enter the Zener
avalanche region. The required PIV rating for the half-wave rectifier can be determined
from Fig. 2.52, which displays the reverse-biased diode of Fig. 2.44 with maximum applied
voltage. Applying Kirchhoffs voltage law, it is fairly obvious that the PIV rating of the
diode must equal or exceed the peak value of the applied voltage. Therefore,

I PIV rating ~ Vm I half-wave rectifier (2.9)

_ V(PIV) +
~~ I = ~ 0----o t
Vm J R V0 =IR=(0)R=0V

+ +
FIG. 2.52
Determining the required PIV rating for the
half-wave rectifier.

2.7 FULL-WAVE RECTIFICATION

Bridge Network •
The de level obtained from a sinusoidal input can be improved 100% using a process
called full-wave rectification. The most familiar network for performing such a function
appears in Fig. 2.53 with its four diodes in a bridge configuration. During the period t = 0
to T/2 the polarity of the input is as shown in Fig. 2.54. The resulting polarities across the
ideal diodes are also shown in Fig. 2.54 to reveal that D2 and D 3 are conducting, whereas
D 1 and D 4 are in the "off' state. The net result is the configuration of Fig. 2.55, with its
indicated current and polarity across R. Since the diodes are ideal, the load voltage is
v0 = vi, as shown in the same figure.

0
V;

T
+

V;
r
V;

FIG. 2.53 FIG. 2.54

Full-wave bridge rectifier. Network of Fig. 2.53 for the period
0-,. T / 2 of the input voltage v;.

V;

FIG. 2.55
Conduction path for the positive region of v;.
76 DIODE APPLICATIONS For the negative region of the input the conducting diodes are D 1 and D4 , resulting in the
configuration of Fig. 2.56. The important result is that the polarity across the load resistor R
is the same as in Fig. 2.54, establishing a second positive pulse, as shown in Fig. 2.56. Over
one full cycle the input and output voltages will appear as shown in Fig. 2.57.

V;

0 0 T T
2

FIG. 2.56
Conduction path for the negative region of v;.

V;

0 T 0 T Tt
2

FIG. 2.57
Input and output waveforms for a full-wave rectifier.

Since the area above the axis for one full cycle is now twice that obtained for a half-wave
system, the de level has also been doubled and
Yac = 2[Eq. (2.7)] = 2(0.318Ym)

or I Yac = 0.636 Vm I full-wave (2.10)

If silicon rather than ideal diodes are employed as shown in Fig. 2.58, the application of
Kirchhoff's voltage law around the conduction path results in
V; - VK - V0 - VK = 0
and

+ + VK=0.7V
~-
V;
0 T T t
2

FIG. 2.58
Determining V0 m.Jor silicon diodes in the bridge configuration.

The peak value of the output voltage v0 is therefore

V0 max = Vm - 2VK
For situations where Vm >> 2VK, the following equation can be applied for the average
value with a relatively high level of accuracy:

(2.11)

Then again, if Vm is sufficiently greater than 2VK, then Eq. (2.10) is often applied as a first
approximation for Yac•
PIV The required PIV of each diode (ideal) can be determined from Fig. 2.59 obtained at FULL-WAVE 77
the peak of the positive region of the input signal. For the indicated loop the maximum RECTIFICATION
voltage across R is Vm and the PIV rating is defined by

I PIV ~ Vm I full-wave bridge rectifier (2.12)

PIV
+

Center-Tapped Transformer
A second popular full-wave rectifier appears in Fig. 2.60 with only two diodes but requir-
ing a center-tapped (CT) transformer to establish the input signal across each section of the
secondary of the transformer. During the positive portion of vi applied to the primary of the
transformer, the network will appear as shown in Fig. 2.61 with a positive pulse across FICi. 2.59
each section of the secondary coil. D 1 assumes the short-circuit equivalent and D 2 the Determining the required PN for
open-circuit equivalent, as determined by the secondary voltages and the resulting current the bridge configuration.
directions. The output voltage appears as shown in Fig. 2.61.

V;

0 +

FICi. 2.60
Center-tapped transformer full-wave rectifier.

1:2

0 T R 0 T
2 2

FICi. 2.61
Network conditions for the positive region of v;.

During the negative portion of the input the network appears as shown in Fig. 2.62, revers-
ing the roles of the diodes but maintaining the same polarity for the voltage across the load re-
sistor R. The net effect is the same output as that appearing in Fig. 2.57 with the same de levels.

V; Vo

,~, ,~, Vm

0
I
I \

T
2
\

T t ;=1
+ + 0
I
I \
\

T
2
T t

1)
FICi. 2.62
Network conditions for the negative region of v;.
78 DIODE APPLICATIONS PIV The network of Fig. 2.63 will help us determine the net PIV for each diode for this
full-wave rectifier. Inserting the maximum voltage for the secondary voltage and Vm as
established by the adjoining loop results in
PIY +
PIV = Vsecondary + VR
= Vm + Vm
and PIV ~ 2 Vm CT transformer, full-wave rectifier (2.13)

EXAMPLE 2.17 Determine the output waveform for the network of Fig. 2.64 and calcu-
FIC. 2.63 late the output de level and the required PIV of each diode.
Determining the PIV level for
the diodes of the CT transformer
full-wave rectifier. V;
+

2kQ
V;
0 T t

2 kQ 2kQ

FIC. 2.64
Bridge network for Example 2.17.

V;
+ + Vo
+ + 2 kQ Vo
lOV
V; 2 kQ
V;
0 T 0 T
2 kQ 2
2

FIC. 2.65 FIC. 2.66

Network of Fig. 2.64 for the positive region of v;. Redrawn network of Fig. 2.65.

Solution: The network appears as shown in Fig. 2.65 for the positive region of the input
voltage. Redrawing the network results in the configuration of Fig. 2.66, where Va = ½v; or
Va max = -21 V;max = -21 (10V) = 5V,asshowninFig.2.66.Forthenegativepartoftheinput,
the roles of the diodes are interchanged and Va appears as shown in Fig. 2.67.
The effect of removing two diodes from the bridge configuration is therefore to reduce
the available de level to the following:
0 T T
Vctc = 0.636(5 V) = 3.18 V
2 or that available from a half-wave rectifier with the same input. However, the PIV as deter-
FIC. 2.67 mined from Fig. 2.59 is equal to the maximum voltage across R, which is 5 V, or half of
Resulting output for Example 2.17. that required for a half-wave rectifier with the same input.

2.8 CLIPPERS
•
The previous section on rectification gives clear evidence that diodes can be used to change
the appearance of an applied waveform. This section on clippers and the next on dampers
will expand on the wave-shaping abilities of diodes.
Clippers are networks that employ diodes to "clip" away a portion of an input signal
without distorting the remaining part of the applied waveform.
 The half-wave rectifier of Section 2.6 is an example of the simplest form of diode clipper- CLIPPERS 79
one resistor and a diode. Depending on the orientation of the diode, the positive or negative
region of the applied signal is "clipped" off.
There are two general categories of clippers: series and parallel. The series configura-
tion is defined as one where the diode is in series with the load, whereas the parallel variety
has the diode in a branch parallel to the load.

Series
The response of the series configuration of Fig. 2.68a to a variety of alternating waveforms
is provided in Fig. 2.68b. Although first introduced as a half-wave rectifier (for sinusoidal
waveforms), there are no boundaries on the type of signals that can be applied to a clipper.

V; Vo V; Vo

V
...._ ...._
+ + 0
V; R Vo

-V
-=-
(a) (b)

FIG. 2.68
Series clipper.

V;

rl 1 V
I_ +

T V; R Vo

FIG. 2.69
Series clipper with a de supply.

The addition of a de supply to the network as shown in Fig. 2.69 can have a pronounced
effect on the analysis of the series clipper configuration. The response is not as obvious
because the de supply can aid or work against the source voltage, and the de supply can be
in the leg between the supply and output or in the branch parallel to the output.
There is no general procedure for analyzing networks such as the type in Fig. 2.69, but
there are some things one can do to give the analysis some direction.
First and most important:
1. Take careful note of where the output voltage is defined.
In Fig. 2.69 it is directly across the resistor R. In some cases it may be across a combi-
nation of series elements.
Next:
2. Try to develop an overall sense of the response by simply noting the "pressure"
established by each supply and the effect it will have on the conventional current
direction through the diode.
In Fig. 2.69, for instance, any positive voltage of the supply will try to turn the diode on
by establishing a conventional current through the diode that matches the arrow in the
diode symbol. However, the added de supply V will oppose that applied voltage and try to
keep the diode in the "off' state. The result is that any supply voltage greater than V volts
will turn the diode on and conduction can be established through the load resistor. Keep in
mind that we are dealing with an ideal diode for the moment, so the turn-on voltage is
simply O V. In general, therefore, for the network of Fig. 2.69 we can conclude that the
80 DIODE APPLICATIONS diode will be on for any voltage vi that is greater than V volts and off for any lesser voltage.
For the "off' condition, the output would be OV due to the lack of current, and for the "on"
condition it would simply be v0 = vi - Vas determined by Kirchhoff's voltage law.
3. Determine the applied voltage (transition voltage) that will result in a change of
state for the diode from the "ofr' to the "on" state.
This step will help to define a region of the applied voltage when the diode is on and
when it is off. On the characteristics of an ideal diode this will occur when Vv = 0 V and
Iv= 0 mA. For the approximate equivalent this is determined by finding the applied volt-
age when the diode has a drop of 0.7 V across it (for silicon) and Iv = 0 mA.
This exercise was applied to the network of Fig. 2.69 as shown in Fig. 2.70. Note the
substitution of the short-circuit equivalent for the diode and the fact that the voltage across
the resistor is O V because the diode current is OmA. The result is vi - V = 0, and so

(2.14)

is the transition voltage.

V vd=OV
~ II1---0-+--o----_ _,-+----o+

V;

FIC. 2.70
Determining the transition level for the circuit of Fig. 2.69.

This permits drawing a line on the sinusoidal supply voltage as shown in Fig. 2.71 to
FIC. 2.71 define the regions where the diode is on and off.
Using the transition voltage to For the "on" region, as shown in Fig. 2.72, the diode is replaced by a short-circuit
define the "on" and "off" regions. equivalent, and the output voltage is defined by

(2.15)

For the "off' region, the diode is an open circuit, Iv = 0 mA, and the output voltage is

4. It is often helpful to draw the output waveform directly below the applied voltage
using the same scales for the horizontal axis and the vertical axis.
Using this last piece of information, we can establish the 0-V level on the plot of Fig. 2.73
FIC. 2.72
for the region indicated. For the "on" condition, Eq. (2.15) can be used to find the output
Determining v0 for the diode
in the "on" state. voltage when the applied voltage has its peak value:
V 0 peak = Vm - V
and this can be added to the plot of Fig. 2.73. It is then simple to fill in the missing section
of the output curve.

T t EXAMPLE 2.18 Determine the output waveform for the sinusoidal input of Fig. 2.74.

v; = V (diodes change state) Solution:

Step 1: The output is again directly across the resistor R.
FIC. 2.73 Step 2: The positive region of vi and the de supply are both applying "pressure" to tum the
Sketching the waveform of v0 using diode on. The result is that we can safely assume the diode is in the "on" state for the entire
the results obtained for vO above range of positive voltages for vi. Once the supply goes negative, it would have to exceed
and below the transition level. the de supply voltage of 5 V before it could tum the diode off.
 V=5 V CLIPPERS 81
~ II1-+----t---......---o
+ +

FIG. 2.74
Series clipper for Example 2.18.

Step 3: The transition model is substituted in Fig. 2.75, and we find that the transition
from one state to the other will occur when
vi+ 5V = 0V
or

-=-
FIG. 2.75
Determining the transition level for the clipper of Fig. 2.74.

Step 4: In Fig. 2.76 a horizontal line is drawn through the applied voltage at the transition
level. For voltages less than -5 V the diode is in the open-circuit state and the output is 0
V, as shown in the sketch of Va. Using Fig. 2. 76, we find that for conditions when the diode
is on and the diode current is established the output voltage will be the following, as deter-
mined using Kirchhoff s voltage law:

20 V; + 5 V = 20 V + 5 V = 25 V

5V v0 = 0 V + 5 V = 5 V

T- t 0
\ 2T \ T
Transition v0 =-5V+5V=0V
voltage

FIG. 2.76
Sketching v 0 for Example 2.18.

The analysis of clipper networks with square-wave inputs is actually easier than with si-
nusoidal inputs because only two levels have to be considered. In other words, the network
can be analyzed as if it had two de level inputs with the resulting Va plotted in the proper
time frame. The next example demonstrates the procedure. 20

EXAMPLE 2.19 Find the output voltage for the network examined in Example 2.18 if the
0 T IT
2 - --1-0 ___.
applied signal is the square wave of Fig. 2.77.
Solution: For vi = 20 V (0 - T /2) the network of Fig. 2.78 results. The diode is in the FIG. 2.77
short-circuit state, and Va = 20 V + 5 V = 25 V. For vi = -10 V the network of Fig. 2.79 Applied signal for Example 2.19.
82 DIODE APPLICATIONS results, placing the diode in the "off' state, and v0 = iRR = (O)R = 0 V. The resulting
output voltage appears in Fig. 2.80.

+
lOV-=-
- f ll+ sv
+
25V

+ ov
0 T T
-=- 2

FIG. 2.78 FIG. 2.79 FIG. 2.80

v0 atv; = +20V. v0 atv;=-JOV. Sketching v0 for Example 2.19.

Note in Example 2.19 that the clipper not only clipped off 5 V from the total swing, but
also raised the de level of the signal by 5 V.

Parallel
The network of Fig. 2.81 is the simplest of parallel diode configurations with the output for
the same inputs of Fig. 2.68. The analysis of parallel configurations is very similar to that
applied to series configurations, as demonstrated in the next example.

+ R +
V;

---+-
0 0 0

-V -V -V ------

FIG. 2.81
Response to a parallel clipper.

EXAMPLE 2.20 Determine v0 for the network of Fig. 2.82.

Solution:
Step 1: In this example the output is defined across the series combination of the 4-V sup-
ply and the diode, not across the resistor R.

vi

+ R +

V; Vo
0 +
4V
-16 ------
0 -r
v-=-
0

FIG. 2.82
Example 2.20.
Step 2: The polarity of the de supply and the direction of the diode strongly suggest that CLIPPERS 83
the diode will be in the "on" state for a good portion of the negative region of the input
signal. In fact, it is interesting to note that since the output is directly across the series com-
bination, when the diode is in its short-circuit state the output voltage will be directly
across the 4-V de supply, requiring that the output be fixed at 4 V. In other words, when VR=OV
the diode is on the output will be 4 V. Other than that, when the diode is an open circuit,
the current through the series network will be OmA and the voltage drop across the resistor
will be OV. That will result in v0 = vi whenever the diode is off.
+

V;
id=4 VJ=OV
+

Vo
Step 3: The transition level of the input voltage can be found from Fig. 2.83 by substitut- +
ing the short-circuit equivalent and remembering the diode current is OmA at the instant of V - 4V
transition. The result is a change in state when 0
-I 0

vi= 4V
FIG. 2.83
Step 4: In Fig. 2.84 the transition level is drawn along with v0 = 4 V when the diode is Determining the transition level
on. For vi ~ 4 V, v0 = 4 V, and the waveform is simply repeated on the output plot. for Example 2.20.

16 -
_ _ _ _ 4 V transition level
0 T

vR = iRR = i~ = (0) R = 0 V
R

+ +
0 T T
2
VKit°"7~ Vo

FIG. 2.84
v-=-4v
Sketching v0 for Example 2.20. O>--------I--------<O
FIG. 2.85
Determining the transition level for
To examine the effects of the knee voltage VKofa silicon diode on the output response, the network of Fig. 2.82.
the next example will specify a silicon diode rather than the ideal diode equivalent.

EXAMPLE 2.21 Repeat Example 2.20 using a silicon diode with VK = 0.7 V. + +
-;;;;-Q.7V
Solution: The transition voltage can first be determined by applying the condition ia = 0 A
at va = Vv = 0.7 V and obtaining the network of Fig. 2.85. Applying Kirchhoffs voltage
V;
~ It
law around the output loop in the clockwise direction, we find that -=-4v
Vi+ VK - V = 0 0 I-
---+------<a

and vi= V - VK = 4 V - 0.7 V = 3.3 V FIG. 2.86

Determining v0 for the diode of
For input voltages greater than 3.3 V, the diode will be an open circuit and v0 = vi. For
Fig. 2.82 in the "on" state.
input voltages less than 3.3 V, the diode will be in the "on" state and the network of Fig. 2.86
results, where
v0 = 4V- 0.7V = 3.3V
The resulting output waveform appears in Fig. 2.87. Note that the only effect of VK was to
16 V -
drop the transition level to 3.3 from 4 V.
3.3 V
0 T T
2
There is no question that including the effects of VK will complicate the analysis some-
what, but once the analysis is understood with the ideal diode, the procedure, including the FIG. 2.87
effects of VK, will not be that difficult. Sketching v0 for Example 2.21.
 Simple Series Clippers (Ideal Diodes)
POSITIVE NEGATIVE

+
+ ~ o
R Vo Q V; R +
Vo 0
~
t
0-0-------<J -~ D-<1--------<J

Biased Series Clippers (ideal Diodes)

rl II....._---------<>+
V
Vo

rillV - + (Vm - V)
R Vo 0 V; R VO Q t-<----'-- - -
-V _______ _ _/
-V

-(¼i+ V) -=-
~ I I l-+---ia------<1>----0
Vo

V
+ ~II+
V
+
V; R VO V; R
--- --- ---
V
0
- (¼, - V)
-=-
V
0 1---------
Simple Parallel Clippers (Ideal Diodes)

+ R
V; 0~
+
Vo

0-0------+---<J

Biased Parallel Clippers (Ideal Diodes)

~v-=-
0--4-..-_------I+----<D

~
V;-I Vo
0
~
v--:;;;-- -V v-=-
-;;.1. +T :;- o--.;------I--o
-=-
0

FIC. 2.88
Clipping circuits.

84
Summary CLAMPERS 85

A variety of series and parallel clippers with the resulting output for the sinusoidal input
are provided in Fig. 2.88. In particular, note the response of the last configuration, with its
ability to clip off a positive and a negative section as determined by the magnitude of the
de supplies.

2.9 CLAMPERS
•
The previous section investigated a number of diode configurations that clipped off a por-
tion of the applied signal without changing the remaining part of the waveform. This sec-
tion will examine a variety of diode configurations that shift the applied signal to a
different level.
A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a
waveform to a different de level without changing the appearance of the applied signal.
Additional shifts can also be obtained by introducing a de supply to the basic structure.
The chosen resistor and capacitor of the network must be chosen such that the time constant
determined by T = RC is sufficiently large to ensure that the voltage across the capacitor
does not discharge significantly during the interval the diode is nonconducting. Through-
out the analysis we assume that for all practical purposes the capacitor fully charges or
discharges in five time constants.
The simplest of clamper networks is provided in Fig. 2.89. It is important to note that
the capacitor is connected directly between input and output signals and the resistor and the
diode are connected in parallel with the output signal.
Clamping networks have a capacitor connected directly from input to output with a
resistive element in parallel with the output signal. The diode is also in parallel with the
output signal but may or may not have a series de supply as an added element.

V;
C
V
r +
V; R Vo
0 T T
2
-V
':'

FIG. 2.89
Clamper.

There is a sequence of steps that can be applied to help make the analysis straightfor-
ward. It is not the only approach to examining dampers, but it does offer an option if dif-
ficulties surface.
Step 1: Start the analysis by examining the response of the portion of the input signal
that will forward bias the diode.
Step 2: During the period that the diode is in the "on" state, assume that the capac-
itor will charge up instantaneously to a voltage level determined by the surrounding
network. C
For the network of Fig. 2.89 the diode will be forward biased for the positive portion of
the applied signal. For the interval Oto T/2 the network will appear as shown in Fig. 2.90. +
The short-circuit equivalent for the diode will result in Va = 0 V for this time interval, as
shown in the sketch of Va in Fig. 2.92. During this same interval of time, the time constant
determined by T = RC is very small because the resistor R has been effectively "shorted
out" by the conducting diode and the only resistance present is the inherent (contact, wire)
resistance of the network. The result is that the capacitor will quickly charge to the peak
value of V volts as shown in Fig. 2.90 with the polarity indicated. FIG. 2.90
Step 3: Assume that during the period when the diode is in the "off" state the capac- Diode "on" and the capacitor
itor holds on to its established voltage level. charging to V volts.
86 DIODE APPLICATIONS Step 4: Throughout the analysis, maintain a continual awareness of the location and
defined polarity for v0 to ensure that the proper levels are obtained.
C When the input switches to the -V state, the network will appear as shown in Fig. 2.91,
~ - +
with the open-circuit equivalent for the diode determined by the applied signal and stored

v1-:J
~v :
- 0
voltage across the capacitor-both "pressuring" current through the diode from cathode to
anode. Now that R is back in the network the time constant determined by the RC product
is sufficiently large to establish a discharge period Sr, much greater than the period
+ -
T /2 - T, and it can be assumed on an approximate basis that the capacitor holds onto all
its charge and, therefore, voltage (since V = Q/C) during this period.
Since v0 is in parallel with the diode and resistor, it can also be drawn in the alternative
FICi. 2.91 position shown in Fig. 2.91. Applying Kirchhoffs voltage law around the input loop results in
Determining v0 with the diode "off."
-V- V- V 0 = 0
and V0 = -2V
v ---- The negative sign results from the fact that the polarity of 2V is opposite to the polarity
defined for v0 • The resulting output waveform appears in Fig. 2.92 with the input signal.
The output signal is clamped to 0 V for the interval 0 to T/2 but maintains the same total
0 T T t swing (2V) as the input.
2 Step 5: Check that the total swing of the output matches that of the input.
This is a property that applies for all clamping networks, giving an excellent check on
the results obtained.

EXAMPLE 2.22 Determine vO for the network of Fig. 2.93 for the input indicated.
0 T T t
2
f = I 000 Hz
C = lµF

-2V
O>-------<l•------------0
+ +
0
FICi. 2.92
R lOOkQ V0
Sketching v 0 for the network of
Fig. 2.91.

-20 -
T--1
C FICi. 2.93
~--+,------------0+ Applied signal and network for Example 2.22.
Ve

20V R 100k!1 V0
+ Solution: Note that the frequency is 1000 Hz, resulting in a period of 1 ms and an inter-
v-=-sv val of 0.5 ms between levels. The analysis will begin with the period t 1 - t2 of the input
+ signal since the diode is in its short-circuit state. For this interval the network will appear
FICi. 2.94
as shown in Fig. 2.94. The output is across R, but it is also directly across the 5-V battery
Determining vO and Vc with the if one follows the direct connection between the defined terminals for vO and the battery
diode in the "on" state. terminals. The result is vO = 5 V for this interval. Applying Kirchhoff's voltage law around
the input loop results in
-20V + Ve - 5V = 0
and Ve= 25 V
+ The capacitor will therefore charge up to 25 V. In this case the resistor R is not shorted
R V0 out by the diode, but a Thevenin equivalent circuit of that portion of the network that
includes the battery and the resistor will result in RTh = 0 n with £Th = V = 5 V. For
the period t 2 - t 3 the network will appear as shown in Fig. 2.95.
The open-circuit equivalent for the diode removes the 5-V battery from having any
KVL effect on v0, and applying Kirchhoff's voltage law around the outside loop of the network
results in
FICi. 2.95
Determining v0 with the diode + 10 V + 25 V - V = 0 0

in the "off'' state. and V 0 = 35 V

 The time constant of the discharging network of Fig. 2.95 is determined by the product CLAMPERS 87
RC and has the magnitude
T = = (100 kf!)(0.1 µ,F) = 0.01 s = 10 ms
RC
The total discharge time is therefore 5T = 5(10 ms) = 50 ms.
Since the interval t2 - t3 will only last for 0.5 ms, it is certainly a good approximation
that the capacitor will hold its voltage during the discharge period between pulses of the
input signal. The resulting output appears in Fig. 2.96 with the input signal. Note that the
output swing of 30 V matches the input swing as noted in step 5.

Vo

35

T
10

0 t3 t4 ~ 30V
30V

-20
_l 5

0 t)
'2 t3 t4
1 ~ t--+--+--+------<1----0+

FIG. 2.96 R
V; and v0 for the clamper of Fig. 2.93.

EXAMPLE 2.23 Repeat Example 2.22 using a silicon diode with VK = 0.7 V.
Solution: For the short-circuit state the network now takes on the appearance of Fig. FIG. 2.97
2.97, and v0 can be determined by Kirchhoff's voltage law in the output section: Determining vO and Vc with the
diode in the "on" state.
+5 V - 0.7 V - Va =0
and Va= 5V - 0.7V = 4.3V
For the input section Kirchhoff's voltage law results in ~24.3 V
+ +
-20 V + Ye+ 0.7 V - 5 V = 0
and Ve= 25 V - 0.7 V = 24.3 V IOV
For the period t2 - t3 the network will now appear as in Fig. 2.98, with the only change
being the voltage across the capacitor. Applying Kirchhoff's voltage law yields
+lOV + 24.3V - Va= 0
and Va= 34.3 V
FIG. 2.98
The resulting output appears in Fig. 2.99, verifying the statement that the input and output Determining v0 with the diode
swings are the same. in the open state.

34.3 V

30V

FIG. 2.99
Sketching v0 for the clamper of Fig. 2.93
with a silicon diode.
 Clamping Networks

------0+
VI

V + ~C
T
0 t---+--,-- R 0 1-...-----.-- ,r -. - R
t 2V
-V 2V
0 1-----' -__,_----l.--J~
-=- -2V 1

Vo \,~)

~C + ~C +
t2V
v,
V; R V
0 0 V; R Vo

2V 1
Vi 1
-=- 1 -=-
0

Vo V0

~C + rl C
+
V; R Vo 0
-V, --rt V;
v,-=- R Vo t 2V
2V + 0
1 -=- -V,

FICi. 2.100
Clamping circuits with ideal diodes (5r = 5RC >> T / 2 ).

A number of clamping circuits and their effect on the input signal are shown in
Fig. 2.100. Although all the waveforms appearing in Fig. 2.100 are square waves, clamp-
ing networks work equally well for sinusoidal signals. In fact, one approach to the analysis
of clamping networks with sinusoidal inputs is to replace the sinusoidal signal by a square
wave of the same peak values. The resulting output will then form an envelope for the
sinusoidal response as shown in Fig. 2.101 for a network appearing in the bottom right of
Fig. 2.100.

+
V; R
0 0
-;;;;-lOV -lOV
-20V +

FICi. 2.101
Clamping network with a sinusoidal input.

2.10 NETWORKS WITH A DC AND AC SOURCE

•
The analysis thus far has been limited to circuits with a single de, ac, or square wave input.
This section will expand that analysis to include both an ac and a de source in the same
configuration. In Fig. 2.102 the simplest of two-source networks has been constructed.
88
For such a system it is especially important that the Superposition Theorem can be applied. NETWORKS WITH A DC 89
That is, AND AC SOURCE

The response of any network with both an ac and a de source can be found by finding
the response to each source independently and then combining the results.
+ i v ~ Si
v, '\, 2VP---P
DC Sourc:e
The network is redrawn as shown in Fig. 2.103 for the de source. Note that the ac source was
removed by simply replacing it with a short-circuit equivalent to the condition Vs= 0 V.
~1
E -=-10v
Using the approximate equivalent circuit for the diode, the output voltage is
VR = E - Vv = 10 V - 0.7 V = 9.3 V
9.3V FIG. 2.102
and the currents are Iv= IR= - - = 4.65mA Network with a de and ac supply.
2kil

+ 0.7V -
AC Sourc:e
The de source is also replaced by a short-circuit equivalent, as shown in Fig. 2.104. The
diode will be replaced by the ac resistance, as determined by Eq. 1.5 in Chapter 1-the
current in the equation being the quiescent or de value. For this case, R 2k0,
+
_ 26 m V _ 26 m V _ 5 5 n E -=-10v
rd - - --- - . 9u
Iv 4.65mA

FIG. 2.103
Applying superposition to determine
+ ~ Si effects of the de source.
2V '\, D

ov
-2V

FIG. 2.104
Determing the response of vR to the applied ac source.

Replacing the diode by this resistance will result in the circuit of Fig. 2.105. For the peak
value of the applied voltage, the peak values of vR and vv will be
2 kil (2 V) == 1. 99 V
VRpeak = 2 kil + 5.59 il
and VDpeak = Vspeak - VRpeak = 2 V - 1.99 V = 0.01 V = 10 mV

5.59!1

FIG. 2.105
Replacing the diode of Fig. 2.104 by its
equivalent ac resistance.

Combining the results of the de and ac analysis will result in the waveforms of Fig. 2.106
for VR and VD-
 VR

12
11
10
VRQ=9.3V
9
8
7
6
5 de shift
4
3
2
1

0 2 3 4 5 6 t (ms) 2 3 4 5 6 t (ms)
(a) (b)

FICi. 2.106
(a) VR and (b) vvfor the network of Fig. 2.102.

Note that the diode has an important impact on the resulting output voltage vR but very
little impact on the ac swing.
For comparison purposes the same system will now be analyzed using the actual charac-
teristics and a load-line analysis. In Fig. 2.107 the de load line has been drawn as described
in Section 2.2. The resulting de current is now slightly less due to a voltage drop across the
diode that is slightly more than the approximate value of 0.7 V. For the peak value of the
inputvoltagetheloadlinewillhaveintersectionsofE= 12Vand/ = = fi
= 6mA.For 1!X
the negative peak the intersections are at 8 V and 4 mA. Take particular note of the region
of the diode characteristics traversed by the ac swing. It defines the region for which the
diode resistance was determined in the analysis above. In this case, however, the quiescent
value of de current is ~4.6 mA so the new ac resistance is
26mV
ra = -- = 5.65D
4.6mA
which is very close to the above value.

Iv (mA)

si..c---1+ -,;,,,i.r-+---+---11---+---+--+---+---+--+---+---1
lvQ = 4.6 rnA
4 ~ -11--11-- ---""~ --+--F,-~lo.c-- - + - - - - l l - - - + - - - l - - + - - l - - + - - - + - - - - l

0 1 2 3 4 5 6 7 8 9 11 12
--11-- change in v D

FICi. 2.107
Shifting load line due to Vs source.
90
 In any event, it is now clear that the change in diode voltage for this region is very small, ZENER DIODES 91
resulting in minimum impact on the output voltage. In general, the diode had a strong im-
pact on the de level of the output voltage but very little impact on the ac swing of the output.
The diode was clearly close to ideal for the ac voltage and 0. 7 V off for the de level. This is
all due primarily to the almost vertical rise of the diode once conduction is fully established
through the diode. In most cases, diodes in the "on" state that are in series with loads will
have some effect on the de level but very little effect on the ac swing if the diode is fully
conducting for the full cycle.
For the future, when dealing with diodes and an ac signal the de level through the diode
is first determined and the ac resistance level determined by Eq. 1.3. This ac resistance can
then be substituted in place of the diode for the required analysis.

2.11 ZENER DIODES

•
The analysis of networks employing Zener diodes is quite similar to the analysis of semi-
conductor diodes in previous sections. First the state of the diode must be determined,
followed by a substitution of the appropriate model and a determination of the other
unknown quantities of the network. Figure 2.108 reviews the approximate equivalent cir-
cuits for each region of a Zener diode assuming the straight-line approximations at each
break point. Note that the forward-bias region is included because occasionally an applica-
tion will skip into this region also.

- :i_ + ~
0.7V
0 ..._ 0 0----0 0----0

o
+► ¼-
_.,.,,.,,,,- a
,___±_J1-
~ ~ I 1----o

Vz
ov

FIC. 2.108
Approximate equivalent circuits for the Zener diode in the three possible 40V
regions of application.

The first two examples will demonstrate how a Zener diode can be used to establish
reference voltage levels and act as a protection device. The use of a Zener diode as a regu-
lator will then be described in detail because it is one of its major areas of application. A
regulator is a combination of elements designed to ensure that the output voltage of a supply
remains fairly constant.
vo2
Vz, 6V

EXAMPLE 2.24 Determine the reference voltages provided by the network of Fig. 2.109, vo,
which uses a white LED to indicate that the power is on. What is the level of current Si

through the LED and the power delivered by the supply? How does the power absorbed by
the LED compare to that of the 6-V Zener diode? V2z 3.3 V

Solution: First we have to check that there is sufficient applied voltage to turn on all the
"II"
series diode elements. The white LED will have a drop of about 4 V across it, the 6-V and
3.3-V Zener diodes have a total of 9.3 V, and the forward-biased silicon diode has 0.7 V, FIC. 2.109
for a total of 14 V. The applied 40 V is then sufficient to turn on all the elements and, one Reference setting circuit for
hopes, establish a proper operating current. Example 2.24.
92 DIODE APPLICATIONS Note that the silicon diode was used to create a reference voltage of 4 V because
V01 = V2i + VK = 3.3 V + 0.7V = 4.0V
Combining the voltage of the 6-V Zener diode with the 4 V results in
V0 z = V01 + Vz, = 4 V + 6 V = 10 V
Finally, the 4 V across the white LED will leave a voltage of 40 V - 14 V = 26 V across
the resistor, and
VR 40V - Vaz - VLEo 40V - lOV - 4 V 26V
I = I =- = - - - - - - = - - - - - - - = - - = 20 mA
R LED R 1.3 kll 1.3 kll 1.3 kll
which should establish the proper brightness for the LED.
The power delivered by the supply is simply the product of the supply voltage and cur-
rent drain as follows:
Ps = Els = EIR = (40 V)(20 mA) = 800 mW
The power absorbed by the LED is
PLEo = VLEoILEo = (4 V)(20 mA) = 80 mW
and the power absorbed by the 6-V Zener diode is
Pz = Vzlz = (6 V)(20 mA) = 120 mW
The power absorbed by the Zener diode exceeds that of the LED by 40 mW.

EXAMPLE 2.25 The network of Fig. 2.110 is designed to limit the voltage to 20 V during
the positive portion of the applied voltage and to O V for a negative excursion of the
applied voltage. Check its operation and plot the waveform of the voltage across the sys-
tem for the applied signal. Assume the system has a very high input resistance so it will not
affect the behavior of the network.

V;
R

+
V; 20V

Si

FIG. 2.110
Controlling network for Example 2.25.

Solution: For positive applied voltages less than the Zener potential of 20 V the Zener
diode will be in its approximate open-circuit state, and the input signal will simply distrib-
ute itself across the elements, with the majority going to the system because it has such a
high resistance level.
Once the voltage across the Zener diode reaches 20 V the Zener diode will tum on as
shown in Fig. 2.11 la and the voltage across the system will lock in at 20 V. Further
increases in the applied voltage will simply appear across the series resistor with the volt-
age across the system and the forward-biased diode remaining fixed at 20 V and 0.7 V,
respectively. The voltage across the system is fixed at 20 V, as shown in Fig. 2.llla,
because the 0. 7 V of the diode is not between the defined output terminals. The system is
therefore safe from any further increases in applied voltage.
For the negative region of the applied signal the silicon diode is reverse biased and
presents an open circuit to the series combination of elements. The result is that the full
negatively applied signal will appear across the open-circuited diode and the negative volt-
age across the system locked in at OV, as shown in Fig. 2.111 b.
The voltage across the system will therefore appear as shown in Fig. 2.111 c.
 R R ZENER DIODES 93

,,>W.7V
~
v,_T WV ,,aWV V;<20.7V
+ +

~ III + O

0.7V

(a) (b)

V·
60V ;'

I
,, X
I
;,,....,,

\
V

\
0

20V

0 \ I
\ I
\ '
.
' ', _..., /

(c)

FIC. 2.111
Response of the network of Fig. 2.110 to the application of a 60-V sinusoidal signal.

The use of the Zener diode as a regulator is so common that three conditions surrounding
the analysis of the basic Zener regulator are considered. The analysis provides an excellent
opportunity to become better acquainted with the response of the Zener diode to different R
operating conditions. The basic configuration appears in Fig. 2.112. The analysis is first
for fixed quantities, followed by a fixed supply voltage and a variable load, and finally a vz
fixed load and a variable supply. + +
V;-=- Vz RL
PZM
V1 and R Fixed
The simplest of Zener diode regulator networks appears in Fig. 2.112. The applied de volt- "II"

age is fixed, as is the load resistor. The analysis can fundamentally be broken down into
FIC. 2.112
two steps. Basic Zener regulator.
I. Determine the state of the Zener diode by removing it from the network and
calculating the voltage across the resulting open circuit.
R
Applying step 1 to the network of Fig. 2.112 results in the network of Fig. 2.113, where
an application of the voltage divider rule results in
+ +
v;-=- V
(2.16)

If V 2='. Vz, the Zener diode is on, and the appropriate equivalent model can be substituted. "II"
If V < Vz, the diode is off, and the open-circuit equivalence is substituted.
FIC. 2.113
2. Substitute the appropriate equivalent circuit and solve for the desired unknowns. Determining the state of the
For the network of Fig. 2.112, the "on" state will result in the equivalent network of Zener diode.
Fig. 2.114. Since voltages across parallel elements must be the same, we find that

(2.17)
94 DIODE APPLICATIONS IR R
-~----~--~-----=1IL
Vz It
+ + +
Vi-=- --=-vz
PZM

V;

FIG. 2.114
Substituting the Zener equivalent for the
"on" situation.

The Zener diode current must be determined by an application of Kirchhoff's current law.
That is,
IR= Iz +h
and Iz = IR - h I (2.18)

where

and

The power dissipated by the Zener diode is determined by

I Pz = Vzlz I (2.19)

that must be less than the PzM specified for the device.
Before continuing, it is particularly important to realize that the first step was employed
only to determine the state of the Zener diode. If the Zener diode is in the "on" state, the
voltage across the diode is not V volts. When the system is turned on, the Zener diode will
tum on as soon as the voltage across the Zener diode is Vz volts. It will then "lock in" at
this level and never reach the higher level of V volts.

EXAMPLE 2.26
a. For the Zener diode network of Fig. 2.115, determine VL, VR, Iz, and Pz.
b. Repeat part (a) with RL = 3 k!1.

+ Vii
R

I kQ
+'z +
+
V,--
- 16V Vz = 10 V RL 1.2 kQ VL

PzM=30mW

FIG. 2.115
Zener diode regulator for Example 2.26.

Solution:
a. Following the suggested procedure, we redraw the network as shown in Fig. 2.116.
Applying Eq. (2.16) gives
RLV- 1.2 k!1(16 V)
V= ' -----= 873V
R + RL 1 k!1 + l.2k!1 .
 +
R

I k11
- !'z 71h
JR

+
+
ZENER DIODES 95

Rl 1.2kn Vl
V;-=- 16 V V

FICi. 2.116
Determining V for the regulator of Fig. 2.115.

Since V = 8.73 Vis less than Vz = 10 V, the diode is in the "off" state, as shown on
the characteristics of Fig. 2.117. Substituting the open-circuit equivalent results in the iz (mA)
+ Vz -
same network as in Fig. 2.116, where we find that
VL = V = 8.73V
M
Vz=lOV
VR = Vi - VL = 16V - 8.73V = 7.27V
\ 0 Vz
Iz = OA
8.73 V
and Pz = Vzlz = Yz(OA) = OW
b. Applying Eq. (2.16) results in
FICi. 2.117
V =RLV; 3 kfl(l6 V) = 12 V Resulting operating point for the
R + RL 1 kfl + 3 kfl network of Fig. 2.115.
Since V = 12 Vis greater than Vz= 10 V, the diode is in the "on" state and the net-
work of Fig. 2.118 results. Applying Eq. (2.17) yields
VL = Vz = lOV
and VR = V; - VL = 16 V - 10 V = 6 V
VL lOV
with h = RL = 3kfl = 3.33mA
VR 6V
and IR = R= I kfl = 6 mA
so that Iz = IR - h [Eq. (2.18)]
= 6mA - 3.33mA
= 2.67mA

I ill
+

FICi. 2.118
Network of Fig. 2.115 in the "on" state.

The power dissipated is

Pz = Vzlz = (10 V)(2.67 mA) = 26.7 mW
which is less than the specified PZM = 30 mW.

Fixed V;, Variable R1

Due to the offset voltage Vz, there is a specific range of resistor values (and therefore load cur-
rent) that will ensure that the Zener is in the "on" state. Too small a load resistance RL will result
ina voltage VL across the load resistor less than Vz, and the Zener device will be in the "off'' state.
96 DIODE APPLICATIONS To determine the minimum load resistance of Fig. 2.112 that will tum the Zener diode
on, simply calculate the value of RL that will result in a load voltage VL = Vz. That is,
RLV;
VL= V z = - - -
RL + R
Solving for RL, we have

(2.20)

Any load resistance value greater than the RL obtained from Eq. (2.20) will ensure that the
Zener diode is in the "on" state and the diode can be replaced by its Vz source equivalent.
The condition defined by Eq. (2.20) establishes the minimum RL, but in tum specifies
the maximum h as

(2.21)

Once the diode is in the "on" state, the voltage across R remains fixed at

(2.22)

and / R remains fixed at

(2.23)

The Zener current

(2.24)

resulting in a minimum lz when his a maximum and a maximum/z when his a minimum
value, since / R is constant.
Since [z is limited to /ZM as provided on the data sheet, it does affect the range of RL and
therefore h- Substituting IZM for lz establishes the minimum h as

(2.25)

and the maximum load resistance as

(2.26)

EXAMPLE 2.27
a. For the network of Fig. 2.119, determine the range of RL and h that will result in VRL
being maintained at 10 V.
b. Determine the maximum wattage rating of the diode.

lkQ

+ R

V, = SOY Vz=IOY
IZM= 32 mA

FICi. 2.119
Voltage regulator for Example 2.27.
Solution: ZENER DIODES 97
a. To determine the value of RL that will tum the Zener diode on, apply Eq. (2.20):
R . = RVz = (1 kll)(lO V) = 10 kll = 250 O
4mn V; - Vz 50V - lOV 40
The voltage across the resistor R is then determined by Eq. (2.22):
VR = V; - Vz = 50 V - 10 V = 40 V
and Eq. (2.23) provides the magnitude of IR:
VR 40V
IR = R= I kll = 40 mA

The minimum level of his then determined by Eq. (2.25):

I4mn = IR - IZM = 40mA - 32mA = 8mA
with Eq. (2.26) determining the maximum value of RL:
Yz lOV
RL =- = - - = 1.25 kO
max I4mn 8 mA
A plot of VL versus RL appears in Fig. 2.120a and for VL versus h in Fig. 2.120b.

IOV IOV

o 2so n 1.25 kQ 0 8mA 40mA

(a) (b)

FIC. 2.120
VL versus RL and hfor the regulator of Fig. 2.119.

b. Pmax = VzIZM
= (10 V)(32 mA) = 320 mW

Fixed R1, Variable V;

For fixed values of RL in Fig. 2.112, the voltage V; must be sufficiently large to tum the
Zener diode on. The minimum turn-on voltage V; = V;min is determined by
RLV;
VL = Vz = - - -
RL + R

(RL + R)Vz
and V;min = __ R_L__ (2.27)

The maximum value of V; is limited by the maximum Zener current IzM• Since IzM =
IR-h,

(2.28)

Since his fixed at VzfRL and IZM is the maximum value of Iz, the maximum V; is
defined by

(2.29)
98 DIODE APPLICATIONS
EXAMPLE 2.28 Determine the range of values of Vi that will maintain the Zener diode of
Fig. 2.121 in the "on" state.

+ 220n
R
-- ilz
JR

71/L

+
vi Vz = 20 V 1.2 kQ VL
RL
lzM =60mA

FIG. 2.121
Regulator for Example 2.28.

Solution:
Eq. (2.27):
= (RL + R)Vz = (1200 !1 + 220 !1)(20 V) = 23 _67 V
vimin RL 1200 n
VL Vz 20V
IL = - = - = - - = 16.67 mA
RL RL l.2kil
Eq. (2.28): IRmax = lzM + h = 60 mA + 16.67 mA
= 76.67mA
Eq. (2.29): Vimax = 1RmaxR + Vz
= (76.67 mA)(0.22 kil) + 20 V
= 16.87V + 20V
= 36.87V
A plot of VL versus Vi is provided in Fig. 2.122.

20V

0 10 I 40 V;
23.67 V 36.87 V

FIG. 2.122
VL versus V;for the regulator of Fig. 2.121.

The results of Example 2.28 reveal that for the network of Fig. 2.121 with a fixed RL,
the output voltage will remain fixed at 20 V for a range of input voltage that extends from
23.67 V to 36.87 V.

2.12 VOLTAGE-MULTIPLIER CIRCUITS

•
Voltage-multiplier circuits are employed to maintain a relatively low transformer peak
voltage while stepping up the peak output voltage to two, three, four, or more times the
peak rectified voltage.
Voltage Doubler VOLTAGE-MULTIPLIER 99
CIRCUITS
The network of Fig. 2.123 is a half-wave voltage doubler. During the positive voltage half-
cycle across the transformer, secondary diode D 1 conducts (and diode D2 is cut off), charg-
ing capacitor C1 up to the peak rectified voltage (Vm)- Diode D 1 is ideally a short during
this half-cycle, and the input voltage charges capacitor C1 to Vm with the polarity shown in
Fig. 2.124a. During the negative half-cycle of the secondary voltage, diode D 1 is cut off
and diode D 2 conducts charging capacitor C2 . Since diode D 2 acts as a short during the
negative half-cycle (and diode D 1 is open), we can sum the voltages around the outside
loop (see Fig. 2.124b):
-Vm - VC1 + VC2 = 0
-Vm - Vm + Vc2 = 0
from which we obtain

FIC. 2.123
Half-wave voltage doubler.

Diode D 2 Diode D 2
/ nonconducting / conducting

Diode D 1 Diode D 1
conducting nonconducting

(a) (b)

FIC. 2.124
Double operation, showing each half-cycle of operation: (a) positive half-cycle;
(b) negative half-cycle.

On the next positive half-cycle, diode D 2 is nonconducting and capacitor C2 will discharge
through the load. If no load is connected across capacitor C2, both capacitors stay
charged-C1 to Vm and C2 to 2 Vm· If, as would be expected, there is a load connected to
the output of the voltage doubler, the voltage across capacitor C2 drops during the positive
half-cycle (at the input) and the capacitor is recharged up to 2Vm during the negative half-
cycle. The output waveform across capacitor C2 is that of a half-wave signal filtered by a
capacitor filter. The peak inverse voltage across each diode is 2 Vm·
Another doubler circuit is the full-wave doubler of Fig. 2.125. During the positive
half-cycle of transformer secondary voltage (see Fig. 2.126a) diode D 1 conducts, charging
capacitor C1 to a peak voltage Vm· Diode D 2 is nonconducting at this time.
During the negative half-cycle (see Fig. 2.126b) diode D2 conducts, charging capacitor
Cz, while diode D 1 is nonconducting. If no load current is drawn from the circuit, the volt-
age across capacitors C1 and C2 is 2Vm· If load current is drawn from the circuit, the voltage
across capacitors C1 and C2 is the same as that across a capacitor fed by a full-wave rectifier
circuit. One difference is that the effective capacitance is that of C1 and C2 in series, which
is less than the capacitance of either C1 or C2 alone. The lower capacitor value will provide
poorer filtering action than the single-capacitor filter circuit.
100 DIODE APPLICATIONS
D1 +

~II \-:'.11 vm C1

2Vm

+
vm C2

FIG. 2.125
Full-wave voltage doubler.

Di/ Conducting / Nonconducting

,------,>----~---7
+ I
I

J
C1 1+

~II +
=i=1- v.,
1
I

:+
C2==F Vm
1-
1
I
----~---_J
Dz ~Nonconducting D z ~ Conducting

(a) (b)

FIG. 2.126
Alternate half-cycles of operation for full-wave voltage doubler.

The peak inverse voltage across each diode is 2Vm, as it is for the filter capacitor circuit.
In summary, the half-wave or full-wave voltage-doubler circuits provide twice the peak
voltage of the transformer secondary while requiring no center-tapped transformer and only
2Vm PIV rating for the diodes.

Voltage Tripler and Quadrupler

Figure 2.127 shows an extension of the half-wave voltage doubler, which develops three
and four times the peak input voltage. It should be obvious from the pattern of the circuit

~-------Tripler (3Vm)--------

C3

1~ - - - - Doubler(2Vm) - - - ~

,~ - - - - - - - - - - Quadrupler(4Vm) - - - - - - - - - - ~1

FIG. 2.127
Voltage tripler and quadrupler.
connection how additional diodes and capacitors may be connected so that the output volt- PRACTICAL 101
age may also be five, six, seven, and so on, times the basic peak voltage CVm)- APPLICATIONS
In operation, capacitor C1 charges through diode D 1 to a peak voltage Vm during the posi-
tive half-cycle of the transformer secondary voltage. Capacitor C2 charges to twice the peak
voltage, 2Vm, developed by the sum of the voltages across capacitor C1 and the transformer
during the negative half-cycle of the transformer secondary voltage.
During the positive half-cycle, diode D 3 conducts and the voltage across capacitor C2
charges capacitor C3 to the same 2Vm peak voltage. On the negative half-cycle, diodes D2
and D4 conduct with capacitor C3, charging C4 to 2Vm·
The voltage across capacitor C2is 2Vm, across C1 and C3 itis 3Vm, and across C2 and C4it
is 4 Vm· If additional sections of diode and capacitor are used, each capacitor will be charged
to 2Vm. Measuring from the top of the transformer winding (Fig. 2.127) will provide odd
multiples of Vm at the output, whereas measuring the output voltage from the bottom of the
transformer will provide even multiples of the peak voltage Vm·
The transformer rating is only Vm, maximum, and each diode in the circuit must be rated
at 2Vm PIV. If the load is small and the capacitors have little leakage, extremely high de
voltages may be developed by this type of circuit, using many sections to step up the de
voltage.

2.1 J PRAOICAL APPLICATIONS

•
The range of practical applications for diodes is so broad that it would be virtually impos-
sible to consider all the options in one section. However, to develop some sense for the use
of the device in everyday networks, a number of common areas of application are intro-
duced below. In particular, note that the use of diodes extends well beyond the important
switching characteristic that was introduced earlier in this chapter.

Rectification
Battery chargers are a common household piece of equipment used to charge everything
from small flashlight batteries to heavy-duty, marine, lead-acid batteries. Since all are
plugged into a 120-V ac outlet such as found in the home, the basic construction of each is
quite similar. In every charging system a transformer must be included to cut the ac volt-
age to a level appropriate for the de level to be established. A diode (also called rectifier)
arrangement must be included to convert the ac voltage, which varies with time, to a fixed
de level such as described in this chapter. Some de chargers also include a regulator to
provide an improved de level (one that varies less with time or load). Since the car battery
charger is one of the most common, it will be described in the next few paragraphs.
The outside appearance and the internal construction of a Sears 6/2 AMP Manual Bat-
tery Charger are provided in Fig. 2.128. Note in Fig. 2.128b that the transformer (as in most
chargers) takes up most of the internal space. The additional air space and the holes in the
casing are there to ensure an outlet for the heat that develops due to the resulting current
levels.
The schematic of Fig. 2.129 includes all the basic components of the charger. Note first
that the 120 V from the outlet are applied directly across the primary of the transformer.
The charging rate of 6 A or 2 A is determined by the switch, which simply controls how
many windings of the primary will be in the circuit for the chosen charging rate. If the
battery is charging at the 2-A level, the full primary will be in the circuit, and the ratio of
the turns in the primary to the turns in the secondary will be a maximum. If it is charging
at the 6-A level, fewer turns of the primary are in the circuit, and the ratio drops. When
you study transformers, you will find that the voltage at the primary and secondary is
directly related to the turns ratio. If the ratio from primary to secondary drops, then the
voltage drops also. The reverse effect occurs if the turns on the secondary exceed those
on the primary.
The general appearance of the waveforms appears in Fig. 2.129 for the 6-A charging
level. Note that so far, the ac voltage has the same wave shape across the primary and the
secondary. The only difference is in the peak value of the waveforms. Now the diodes take
 (b)

FICi. 2.128
Battery charger: (a) external appearance; (b) internal construction.

/Peak= 18 V
f V V V \ -12V

120Vac
/j +
Positive clamp
of charger

Transformer
(step-down)
~0-----------,

Circuit Current
breaker meter Negative clamp
of charger

FICi. 2.129
Electrical schematic for the battery charger of Fig. 2.128.

over and convert the ac waveform, which has zero average value (the waveform above
equals the waveform below), to one that has an average value (all above the axis) as shown
in the same figure. For the moment simply recognize that diodes are semiconductor elec-
tronic devices that permit only conventional current to flow through them in the direction
indicated by the arrow in the symbol. Even though the waveform resulting from the diode
action has a pulsing appearance with a peak value of about 18 V, it will charge the 12-V
battery whenever its voltage is greater than that of the battery, as shown by the shaded area.
102
Below the 12-V level the battery cannot discharge back into the charging network because PRACTICAL 103
the diodes permit current flow in only one direction. APPLICATIONS
In particular, note in Fig. 2.128b the large plate that carries the current from the rectifier
(diode) configuration to the positive terminal of the battery. Its primary purpose is to pro-
vide a heat sink (a place for the heat to be distributed to the surrounding air) for the diode
configuration. Otherwise the diodes would eventually melt down and self-destruct due to
the resulting current levels. Each component of Fig. 2.129 has been carefully labeled in
Fig. 2.128b for reference.
When current is first applied to a battery at the 6-A charge rate, the current demand, as
indicated by the meter on the face of the instrument, may rise to 7 A or almost 8 A. However,
the level of current will decrease as the battery charges until it drops to a level of 2 A or 3 A.
For units such as this that do not have an automatic shutoff, it is important to disconnect
the charger when the current drops to the fully charged level; otherwise, the battery will
become overcharged and may be damaged. A battery that is at its 50% level can take as long
as 10 hours to charge, so one should not expect it to be a IO-minute operation. In addition, if
a battery is in very bad shape, with a lower than normal voltage, the initial charging current
may be too high for the design. To protect against such situations, the circuit breaker will
open and stop the charging process. Because of the high current levels, it is important that
the directions provided with the charger be carefully read and applied.
In an effort to compare the theoretical world with the real world, a load (in the form of
a headlight) was applied to the charger to permit a viewing of the actual output waveform.
It is important to note and remember that a diode with zero current through it will not
display its rectifying capabilities. In other words, the output from the charger of Fig. 2.129
will not be a rectified signal unless a load is applied to the system to draw current through
the diode. Recall from the diode characteristics that when / D = 0 A, VD = 0 V.
By applying the headlamp as a load, however, sufficient current is drawn through the
diode for it to behave like a switch and convert the ac waveform to a pulsating one as
shown in Fig. 2.130 for the 6-A setting. First note that the waveform is slightly distorted
by the nonlinear characteristics of the transformer and the nonlinear characteristics of the
diode at low currents. The waveform, however, is certainly close to what is expected when
we compare it to the theoretical patterns of Fig. 2.129. The peak value is determined from
the vertical sensitivity as
Ypeak = (3.3 divisions)(5 V/division) = 16.5 V vs. the 18 V of Fig. 1.129

-__,,--.,
2 ms/div

FICi. 2.130
Pulsating response of the charger of Fig. 2.129
to the application of a headlamp as a load.

with a de level of
Yctc = 0.636Ypeak = 0.636(16.5 V) = 10.49 V
A de meter connected across the load registered 10.41 V, which is very close to the theo-
retical average (de) level of 10.49 V.
One may wonder how a charger having a de level of 10.49 V can charge a 12-V battery
to a typical level of 14 V. It is simply a matter of realizing that (as shown in Fig. 2.130) for
a good deal of each pulse, the voltage across the battery will be greater than 12 V and the
battery will be charging-a process referred to as trickle charging. In other words, charg-
ing does not occur during the entire cycle, but only when the charging voltage is more than
the voltage of the battery.
104 DIODE APPLICATIONS Protective Configurations
Diodes are used in a variety of ways to protect elements and systems from excessive volt-
ages or currents, polarity reversals, arcing, and shorting, to name a few. In Fig. 2.131a, the
switch on a simple RL circuit has been closed, and the current will rise to a level deter-
mined by the applied voltage and series resistor R as shown on the plot. Problems arise
when the switch is quickly opened as in Fig. 2.131 b to essentially tell the circuit that the
current must drop to zero almost instantaneously. You will remember from your basic
circuits courses, however, that the inductor will not permit an instantaneous change in cur-
rent through the coil. A conflict results, which will establish arcing across the contacts of
the switch as the coil tries to find a path for discharge. Recall also that the voltage across
an inductor is directly related to the rate of change in current through the coil (vL = L diLfdt).
When the switch is opened, it is trying to dictate that the current change almost instanta-
neously, causing a very high voltage to develop across the coil that will then appear across
the contacts to establish this arcing current. Levels in the thousands of volts will develop
across the contacts, which will soon, if not immediately, damage the contacts and thereby
the switch. The effect is referred to as an "inductive kick." Note also that the polarity of the
voltage across the coil during the "build-up" phase is opposite to that during the "release"
phase. This is due to the fact that the current must maintain the same direction before and
after the switch is opened. During the "build-up" phase, the coil appears as a load, whereas
during the release phase, it has the characteristics of a source. In general, therefore, always
keep in mind that
Trying to change the current through an inductive element too quickly may result in an
inductive kick that could damage surrounding elements or the system itself.

Vcontact

---------~-~-----
R
R

5-r = s(j)
(a) (b)

FICi. 2.131
(a) Transient phase of a simple RL circuit; (b) arcing that results across a switch when opened in series with an RL circuit.

In Fig. 2.132a the simple network above may be controlling the action of a relay.
When the switch is closed, the coil will be energized, and steady-state current levels will
be established. However, when the switch is opened to deenergize the network, we have
the problem introduced above because the electromagnet controlling the relay action will
appear as a coil to the energizing network. One of the cheapest but most effective ways to
protect the switching system is to place a capacitor (called a "snubber") across the terminals
of the coil as shown in Fig. 2.132b. When the switch is opened, the capacitor will initially
appear as a short to the coil and will provide a current path that will bypass the de supply
and switch. The capacitor has the characteristics of a short (very low resistance) because of
the high-frequency characteristics of the surge voltage, as shown in Fig. 2.131b. Recall that
the reactance of a capacitor is determined by Xe = 1/2nfC, so the higher the frequency, the
less is the resistance. Normally, because of the high surge voltages and relatively low cost, ce-
ramic capacitors of about 0.01 µ,Fare used. You don't want to use large capacitors because
the voltage across the capacitor will build up too slowly and will essentially slow down the
 R

11==Relay

C=0.01 µF
(a) (b) (c)

FICi. 2.132
(a) Inductive characteristics of a relay; (b) snubber protection for the configuration ofpart (a);
(c) capacitive protection for a switch.

performance of the system. The resistor of 100 n in series with the capacitor is introduced
solely to limit the surge current that will result when a change in state is called for. Often,
the resistor does not appear because of the internal resistance of the coil as established by
many turns of fine wire. On occasion, you may find the capacitor across the switch as shown
in Fig. 2.132c. In this case, the shorting characteristics of the capacitor at high frequencies
will bypass the contacts with the switch and extend its life. Recall that the voltage across a
capacitor cannot change instantaneously. In general, therefore,
Capacitors in parallel with inductive elements or across switches are often there to act
as protective elements, not as typical network capacitive elements.
Finally, the diode is often used as a protective device for situations such as above. In
Fig. 2.133, a diode has been placed in parallel with the inductive element of the relay con-
figuration. When the switch is opened or the voltage source quickly disengaged, the polarity
of the voltage across the coil is such as to turn the diode on and conduct in the direction
indicated. The inductor now has a conduction path through the diode rather than through
the supply and switch, thereby saving both. Since the current established through the coil
must now switch directly to the diode, the diode must be able to carry the same level of
.----
Protective
diode
current that was passing through the coil before the switch was opened. The rate at which
the current collapses will be controlled by the resistance of the coil and the diode. It can
be reduced by placing an additional resistor in series with the diode. The advantage of the
diode configuration over that of the snubber is that the diode reaction and behavior are not FICi. 2.133
frequency dependent. However, the protection offered by the diode will not work if the ap- Diode protection for an RL circuit.
plied voltage is an alternating one such as ac or a square wave since the diode will conduct
for one of the applied polarities. For such alternating systems, the "snubber" arrangement
would be the best option.
In the next chapter we will find that the base-to-emitter junction of a transistor is
forward-biased. That is, the voltage VsE of Fig. 2.134a will be about 0.7 V positive. To
prevent a situation where the emitter terminal would be made more positive than the base
terminal by a voltage that could damage the transistor, the diode shown in Fig. 2.134a
is added. The diode will prevent the reverse-bias voltage VEB from exceeding 0.7 V. On

B npn
+ transistor
VEB VBE
Limit._/' +
to0.7V ~-----oE

(a) (b)

FICi. 2.134
(a) Diode protection to limit the emitter-to-base voltage of a
transistor; (b) diode protection to prevent a reversal in
collector current.
106 DIODE APPLICATIONS occasion, you may also find a diode in series with the collector terminal of a transistor as
shown in Fig. 2.134b. Normal transistor action requires that the collector be more positive
than the base or emitter terminal to establish a collector current in the direction shown.
However, if a situation arises where the emitter or base terminal is at a higher potential
than the collector terminal, the diode will prevent conduction in the opposite direction. In
general, therefore,
Diodes are often used to prevent the voltage between two points from exceeding 0. 7 V
or to prevent conduction in a particu'lar direction.
As shown in Fig. 2.135, diodes are often used at the input terminals of systems such
as op-amps to limit the swing of the applied voltage. For the 400-m V level the signal
will pass undisturbed to the input terminals of the op-amp. However, if the voltage
jumps to a level of 1 V, the top and bottom peaks will be clipped off before appearing at
the input terminals of the op-amp. Any clipped-off voltage will appear across the series
resistor R 1.

V;

400mV
V;

0 + D1
0.7V 0.7V
+

c:::v-
-400mV ------
V

V;

lV
lV
700mV

0
0

-700mV

-900mV

FICi. 2.135
Diode control of the input swing to an op-amp or a high-input-impedance network.

The controlling diodes of Fig. 2.135 may also be drawn as shown in Fig. 2.136 to control
the signal appearing at the input terminals of the op-amp. In this example, the diodes are act-
ing more like shaping elements than as limiters as in Fig. 2.135. However, the point is that
The p'lacement of elements may change, but their function may still be the same. Do
not expect every network to appear exactly as you studied it for the first time.
In general, therefore, don't always assume that diodes are used simply as switches. There
is a wide variety of uses for diodes as protective and limiting devices.
 .:. PRACTICAL 107
APPLICATIONS
D2

V;
R,
D,

"II"

V;
R,
D, D2

"II" "II"

(a)

lOV
6.7V

.:.

V; + /_0.7V
Dz
0.7V

R1
0 D,
;,v} 6.7V
V

+6V

(b)

FICi. 2.136
(a) Alternate appearances for the network of Fig. 2.135; (b) establishing random levels of control
with separate de supplies.

Polarity Insurance
There are numerous systems that are very sensitive to the polarity of the applied voltage.
For instance, in Fig. 2.137a, assume for the moment that there is a very expensive piece of
equipment that would be damaged by an incorrectly applied bias. In Fig. 2.137b the correct
applied bias is shown on the left. As a result, the diode is reverse-biased, but the system
works just fine-the diode has no effect. However, if the wrong polarity is applied as

Required
+ 4V - 15.3 V +
+ R + 12 V 12 V
16V 12 V $ 16V $
system system
+ +
Diode polarity protection Diode open Diode conducting
(a) (b) (c)

FICi. 2.137
(a) Polarity protection for an expensive, sensitive piece of equipment; (b) correctly applied polarity;
(c) application of the wrong polarity.
 108 DIODE APPLICATIONS shown in Fig. 2.137c, the diode will conduct and ensure that no more than 0.7 V will
appear across the terminals of the system, protecting it from excessive voltages of the
wrong polarity. For either polarity, the difference between the applied voltage and the load

r;; Defined pol~ty

for sens1tlve
movement
or diode voltage will appear across the series source or network resistance.
In Fig. 2.138 a sensitive measuring movement cannot withstand voltages greater than
- 1 V of the wrong polarity. With this simple design the sensitive movement is protected from
----0----t / l-------0--- voltages of the wrong polarity of more than 0.7 V.

Controlled Battery-Powered Backup

L Protective diode
In numerous situations a system should have a backup power source to ensure that the
FICi. 2.138
system will still be operational in case of a loss of power. This is especially true of security
Protection for a sensitive meter
systems and lighting systems that must turn on during a power failure. It is also important
movement.
when a system such as a computer or a radio is disconnected from its ac-to-dc power con-
version source to a portable mode for traveling. In Fig. 2.139 the 12-V car radio operating
off the 12-V de power source has a 9-V battery backup system in a small compartment in
the back of the radio ready to take over the role of saving the clock mode and the channels
stored in memory when the radio is removed from the car. With the full 12 V available
from the car, D 1 is conducting, and the voltage at the radio is about 11.3 V. D 2 is reverse-
biased (an open circuit), and the reserve 9-V battery inside the radio is disengaged.
However, when the radio is removed from the car, D 1 will no longer be conducting because
the 12-V source is no longer available to forward-bias the diode. However, D 2 will be
forward-biased by the 9-V battery, and the radio will continue to receive about 8.3 V to
maintain the memory that has been set for components such as the clock and the channel
selections.

1--,--➔~>---+-+----+-D--'1'--.---~ Internal
Automobile + electronics
electrical 12V
system
12V

-=- Car radio

FICi. 2.139
Backup system designed to prevent the loss of memory in a
car radio when the radio is removed from the car.

Polarity Detector
Through the use of LEDs of different colors, the simple network of Fig. 2.140 can be used
to check the polarity at any point in a de network. When the polarity is as indicated for the
applied 6 V, the top terminal is positive, D 1 will conduct along with LEDl, and a green
light will result. Both D 2 and LED2 will be back-biased for the above polarity. However,
if the polarity at the input is reversed, D2 and LED2 will conduct, and a red light will
appear, defining the top lead as the lead at the negative potential. It would appear that the

+ ---Al~-..
6V

FICi. 2.140
Polarity detector using diodes and LEDs.
network would work without diodes D 1 and D 2. However, in general, LEDs do not like to PRACTICAL 109
be reverse-biased because of sensitivity built in during the doping process. Diodes D 1 and APPLICATIONS
D 2 offer a series open-circuit condition that provides some protection to the LEDs. In the
forward-bias state, the additional diodes D 1 and D2 reduce the voltage across the LEDs to
more common operating levels.

Displays
Some of the primary concerns of using electric light bulbs in exit signs are their limited
lifetime (requiring frequent replacement); their sensitivity to heat, fire, and so on; their
durability factor when catastrophic accidents occur; and their high voltage and power
requirements. For this reason LEDs are often used to provide the longer life span, higher
durability levels, and lower demand voltage and power levels (especially when the reserve
de battery system has to take over).
In Fig. 2.141 a control network determines when the EXIT light should be on. When it
is on, all the LEDs in series will be on, and the EXIT sign will be fully lit. Obviously, if
one of the LEDs should bum out and open up, the entire section will tum off. However,
this situation can be improved by simply placing parallel LEDs between every two points.
Lose one, and you will still have the other parallel path. Parallel diodes will, of course, reduce
the current through each LED, but two at a lower level of current can have a luminescence
similar to one at twice the current. Even though the applied voltage is ac, which means that
the diodes will tum on and off as the 60-Hz voltage swings positive and negative, the per-
sistence of the LEDs will provide a steady light for the sign.

EXIT
FICi. 2.141
EXIT sign using LEDs.

Setting Voltage Reference Levels R

7.4 V
Diodes and Zeners can be used to set reference levels as shown in Fig. 2.142. The net- + + 4.6V- +
work, through the use of two diodes and one Zener diode, is providing three different 0.7V
voltage levels. 6.7V
+
12V 0.7V
Establishing a Voltage Level Insensitive to the Load Current 6V
As an example that clearly demonstrates the difference between a resistor and a diode in a +
voltage-divider network, consider the situation of Fig. 2.143a, where a load requires about 6V
6 V to operate properly but a 9-V battery is all that is available. For the moment let us
assume that operating conditions are such that the load has an internal resistance of 1 kll.
Using the voltage-divider rule, we can easily determine that the series resistor should be FICi. 2.142
470 0 (commercially available value) as shown in Fig. 2.143b. The result is a voltage Providing different reference levels
across the load of 6.1 V, an acceptable situation for most 6-V loads. However, if the operat- using diodes.
ing conditions of the load change and the load now has an internal resistance of only 600 0,
the load voltage will drop to about 4.9 V, and the system will not operate correctly. This
sensitivity to the load resistance can be eliminated by connecting four diodes in series with
the load as shown in Fig. 2.143c. When all four diodes conduct, the load voltage will be
110 DIODE APPLICATIONS ? R

J+ +
Variable +
4700 +

r- r-
-=-9v 6V load _ lkO(9V) "'
-=-9v 1 kfl VRL- 1 kO+470O- 6.1 V

-=-
(a) (b)

+0.7 V- +0.7 V- +0.7 V-+0.7 V-

(c)

FIC. 2.143
(a) How to drive a 6-V load with a 9-V supply (b) using afzxed resistor value.
(c) Using a series combination of diodes.

about 6.2 V, irrespective of the load impedance (within device limits, of course)-the sen-
sitivity to the changing load characteristics has been removed.

AC Regulator and Square-Wave Generator

Two back-to-back Zeners can also be used as an ac regulator as shown in Fig. 2.144a. For
the sinusoidal signal v; the circuit will appear as shown in Fig. 2.144b at the instant
v; = 10 V. The region of operation for each diode is indicated in the adjoining figure.
Note that Z 1 is in a low-impedance region, whereas the impedance of Z2 is quite large, cor-
responding to the open-circuit representation. The result is that v0 = v; when v; = 10 V.
The input and the output will continue to duplicate each other until v; reaches 20 V. Then
Zi will "tum on" (as a Zener diode), whereas Z1 will be in a region of conduction with a
resistance level sufficiently small compared to the series 5-kll resistor to be considered a

V; Vo
+ 5kQ +
Z1
V; 20-V< Vo
rot Zeners rot
Z2

(a)

I
5kO +
Z1
+ 20V
V; = lOV-=-
+ 0 V
Z2

(b)

FIC. 2.144
Sinusoidal ac regulation: (a) 40-V peak-to-peak sinusoidal ac regulator;
(b) circuit operation at v; = 10 V.
short circuit. The resulting output for the full range of v; is provided in Fig. 2.144a. Note SUMMARY 111
that the waveform is not purely sinusoidal, but its root mean square (rms) value is lower
than that associated with a full 22-V peak signal. The network is effectively limiting the
rms value of the available voltage. The network of Fig. 2.144b can be extended to that of a
simple square-wave generator (due to the clipping action) if the signal v; is increased to
perhaps a 50-V peak with 10-V Zeners as shown in Fig. 2.145 with the resulting output
waveform.

V;

Vo
+ 5k0 + +
Z1
V;
lOV
Vo
10-V\
0 21t rot Zeners + -lOV
Z2

FICi. 2.145
Simple square-wave generator.

2.14 SUMMARY
Important Conclusions and Concepts •
1. The characteristics of a diode are unaltered by the network in which it is employed.
The network simply determines the point of operation of the device.
2. The operating point of a network is determined by the intersection of the network
equation and an equation defining the characteristics of the device.
3. For most applications, the characteristics of a diode can be defined simply by the
threshold voltage in the forward-bias region and an open circuit for applied volt-
ages less than the threshold value.
4. To determine the state of a diode, simply think of it initially as a resistor, and find
the polarity of the voltage across it and the direction of conventional current through
it. If the voltage across it has a forward-bias polarity and the current has a direction
that matches the arrow in the symbol, the diode is conducting.
5. To determine the state of diodes used in a logic gate, first make an educated guess
about the state of the diodes, and then test your assumptions. If your estimate is
incorrect, refine your guess and try again until the analysis verifies the conclusions.
6. Rectification is a process whereby an applied waveform of zero average value is
changed to one that has a de level. For applied signals of more than a few volts, the
ideal diode approximations can normally be applied.
7. It is very important that the PIV rating of a diode be checked when choosing a diode
for a particular application. Simply determine the maximum voltage across the diode
under reverse-bias conditions, and compare it to the nameplate rating. For the typical
half-wave and full-wave bridge rectifiers, it is the peak value of the applied signal. For
the CT transformer full-wave rectifier, it is twice the peak value (which can get quite
high).
8. Clippers are networks that "clip" away part of the applied signal either to create a
specific type of signal or to limit the voltage that can be applied to a network.
9. Clampers are networks that "clamp" the input signal to a different de level. In any
event, the peak-to-peak swing of the applied signal will remain the same.
10. Zener diodes are diodes that make effective use of the Zener breakdown potential of
an ordinary p-n junction characteristic to provide a device of wide importance and
application. For Zener conduction, the direction of conventional flow is opposite to
the arrow in the symbol. The polarity under conduction is also opposite to that of
the conventional diode.
112 DIODE APPLICATIONS 11. To determine the state of a Zener diode in a de network, simply remove the Zener
from the network, and determine the open-circuit voltage between the two points
where the Zener diode was originally connected. If it is more than the Zener poten-
tial and has the correct polarity, the Zener diode is in the "on" state.
12. A half-wave or full-wave voltage doubler employs two capacitors; a tripler, three
capacitors; and a quadrupler, four capacitors. In fact, for each, the number of diodes
equals the number of capacitors.

Equations
Approximate:
Silicon: VK = 0.7V; Iv is determined by network.
Germanium: VK = 0.3V; Iv is determined by network.
Gallium arsenide: VK = 1.2V; Iv is determined by network.
Ideal:
Iv is determined by network.
For conduction:

Half-wave rectifier:
Yctc = 0.318Ym
Full-wave rectifier:
Yctc = 0.636Vm

2.15 COMPUTER ANALYSIS

Cadenc:e OrCAD •
Series Diode Configuration In the previous chapter the OrCAD 16.3 folder was estab-
lished as the location for our projects. This section will define the name of our project, set
up the software for the analysis to be performed, describe how to build a simple circuit,
and, finally, perform the analysis. The coverage will be quite extensive since this will be
the first true exposure to the mechanics associated with using the software package. In the
chapters to follow you will find the analysis can be performed quite rapidly to obtain
results that confirm the long-hand solutions.
Our first project can now be initiated by double-clicking on the OrCAD Capture CIS
Demo icon on the screen, or you can use the sequence Start-AU Programs-Cadence-
OrCAD 16.3 Demo. The resulting screen has only a few active keys on the top toolbar.
The first at the top left is the Create document key (or you can use the sequence File-New-
Project). Selecting the key will result in a New Project dialog box, in which the Name of
the project must be entered. For our purposes we will choose OrCAD 2-1 as shown in the
heading of Fig. 2.146, and select Analog or Mixed AID (to be used for all the analyses of
this text). Note at the bottom of the dialog box that the Location appears as C:\OrCAD
16.3 as set earlier. Click OK, and another dialog box will appear titled Create PSpice
Project. Select Create a blank project (again, for all the analyses to be performed in this
text). Click OK, and additional keys will be turned on along with additional toolbars. A
Project Manager Window will appear with OrCAD 2-1 as its heading. The new project
listing will appear with an icon and an associated + sign in a small square. Clicking on
the + sign will take the listing a step further to SCHEMA TICl. Click + again (to the left
of SCHEMATIC!), and PAGEl will appear; clicking on a - sign will reverse the pro-
cess. Double-clicking on PAGEl will create a working window titled SCHEMATIC!:
PAGEl, revealing that a project can have more than one schematic file and more than one
associated page. The width and the height of the window can be adjusted by grabbing an
edge to obtain a double-headed arrow and dragging the border to the desired location. Either
window on the screen can be moved by clicking on the top heading to make it dark blue
and then dragging it to any location.
 COMPUTER ANALYSIS 113
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i SCHEMAT IC1 OrCAD •

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Ill

FICi. 2.146
Cadence OrCAD analysis of a series diode configuration.

Now we are ready to build the simple circuit of Fig. 2.146. Select the Place part key (the
top key on the far right vertical toolbar that looks like an integrated circuit with a positive
sign in the bottom right corner) to obtain the Place Part dialog box. Since this is the first
circuit to be constructed, we must ensure that the parts appear in the list of active libraries.
Go to Libraries and select the Add Library key (looks like a dashed rectangular box with
a yellow star in the top left corner). The result is a Browse File in which analog.olb can
be selected, followed by Open to place it in the active list of Libraries. Repeat the process
to add the eval.olb and source.olb libraries. All three libraries will be required to build the
networks appearing in this text. However, it is important to realize that:
Once the library files have been selected, they will appear in the active listing for each
new project without having to add them each time-a step, such as the Folder step
above, that does not have to be repeated with each similar project.
Click the small x in the top right corner of the dialog box to remove the Place Part dialog
box. We can now place components on the screen. For the de voltage source, first select the
Place Part key and then select SOURCE in the library listing. Under Part List, a list of
available sources will appear; select VDC for this project. Once VDC has been selected, its
symbol, label, and value will appear on the picture window at the bottom left of the dialog
box. Click the Place Part key on the top of the dialog box, and the VDC source will follow
the cursor across the screen. Move it to a convenient location, left-click the mouse, and it
will be set in place as shown in Fig. 2.146.
Since a second source is present in Fig. 2.146, move the cursor to the general area of the
second source and click it in place. Since this is the last source to appear in the network,
execute a right click of the mouse and select End Mode. Choosing this option will end the
procedure, leaving the last source in a red dashed box. The fact that it is red indicates that
it is still in the active mode and can be operated on. One more click of the mouse, and the
second source will be in place and the red active status removed. The second source can
be rotated 180° to match Fig. 2.146 by first clicking the source to make it red (active) to
obtain a long list of options and select Rotate. Since each rotation only turns it 90° coun-
terclockwise, two rotations will be required. The rotations can also be accomplished using
the sequence Ctrl-R.
One of the most important steps in the procedure is to ensure that a 0-V ground poten-
tial is defined for the network so that voltages at any point in the network have a reference
point. The result is a requirement that every network must have a ground defined. For our
purposes, the 0/SOURCE option will be our choice when the GND key is selected. It is
obtained by selecting the ground symbol in the middle of the far right toolbar to obtain the
Place Ground dialog box. Scroll down until 0/SOURCE is selected and click OK. The
result is a ground that can be placed anywhere on the screen. As with the voltage source,
114 DIODE APPLICATIONS multiple grounds can be added by simply going from one point to another. The process is
ended with a right click and the End Mode option.
The next step will be to place the resistors of the network of Fig. 2.146. This is accom-
plished by selecting the Place Part key again and then selecting the ANALOG library.
Scrolling the options, note that R will appear and should be selected. Click the Place Part
key, and the resistor will appear next to the cursor on the screen. Move it to the desired
location and click it in place. The second resistor can be placed by simply moving to the
general area of its location in Fig. 2.146 and clicking it in place. Since there are only two
resistors, the process can be ended by making a right click of the mouse and selecting End
Mode. The second resistor will have to be rotated to the vertical position using the same
procedure described for the second voltage source.
The last element to be placed is the diode. Selecting the Place Part keypad will again
result in the Place Part dialog box, in which the EVAL library is chosen from the Libraries
listing. Then type D under Part heading and select D14148 under Part List followed by
the Place Part command to place on the screen in the same manner described for the source
and resistors.
Now that all the components are on the screen you may want to move them to positions
corresponding directly with Fig. 2.146. This is accomplished by simply clicking on the
element and holding the left-click down as you move the element.
All the required elements are on the screen, but they need to be connected. This is ac-
complished by selecting the Place wire key, which looks like a step, near the top of the
toolbar to the left of the toolbar with the Place Part key. The result is a crosshair with a
center that should be placed at the point to be connected. Place the crosshair at the top of the
voltage source, and left-click it once to connect it to that point. Then draw a line to the end
of the next element, and click the mouse again when the crosshair is at the correct point. A
red line will result with a square at each end to confirm that the connection has been made.
Then move the crosshair to the other elements, and build the circuit. Once everything is
connected, a right click will provide the End Mode option. Don't forget to connect the
source to ground as shown in Fig. 2.146.
Now we have all the elements in place, but their labels and values are wrong. To change
any parameter, simply double-click on the parameter (the label or the value) to obtain the
Display Properties dialog box. Type in the correct label or value, click OK, and the quan-
tity is changed on the screen. The labels and values can be moved by simply clicking on
the center of the parameter until it is closely surrounded by the four small squares and then
dragging it to the new location. Another left click, and it is deposited in its new location.
Finally, we can initiate the analysis process, called Simulation, by selecting the New
Simulation Profile key near the top left of the display-it resembles a data page with a
star in the top right comer. A New Simulation dialog box will result that first asks for
the Name of the simulation. OrCAD 2-1 is entered, and none is left in the Inherit From
request. Then select Create, and a Simulation Setting dialog box will appear in which
Analysis-Analysis Type-Bias Point is sequentially selected. Click OK, and select the Run
key (which looks like an isolated arrowhead in a green background) or choose PSpice-Run
from the menu bar. An Output Window will result that appears to be somewhat inactive.
It will not be used in the current analysis, so close (X) the window, and the circuit of Fig.
2.146 will appear with the voltage and current levels of the network. The voltage, current,
or power levels can be removed (or replaced) from the display by simply selecting the
V, I, or W in the third toolbar from the top. Individual values can be removed by simply
selecting the value and pressing the Delete key. Resulting values can be moved by simply
left-clicking the value and dragging it to the desired location.
The results of Fig. 2.146 show that the current through the series configuration is
2.081 mA through each element, compared to the 2.072 mA of Example 2.9. The voltage
across the diode is 218.8 mV - (-421.6 mV) ~ 0.64 V, compared to the 0.7 V applied
in the long-hand solution of Example 2.9. The voltage across R 1 is 10 V - 218.8 mV ~
9.78 V, compared to 9.74 Vin the long-hand solution. The voltage across the resistor R2
is 5 V - 421.6 mV ~ 4.58 V, compared to 4.56 Vin Example 2.9.
To understand the differences between the two solutions, one must be aware that the diode
has internal characteristics that affect its behavior such as the reverse saturation current and
its resistance levels at different current levels. Those characteristics can be viewed through
the sequence Edit-PSpice Model resulting in the PSpice Model Editor Demo dialog box.
You will find that the default value of the reverse saturation current is 2.682 nA-a quantity COMPUTER ANALYSIS 115
that can have an important effect on the characteristics of the device. If we choose ls =
3.5E-15A (a value determined by trial and error) and delete the other parameters for the
device, a new simulation of the network will result in the response of Fig. 2.147. Now the
current through the circuit is 2.072 mA, which is an exact match with the result of Example
2.9. The voltage across the diode is 260.2 mV + 440.9 mV ~ 0.701 V, or essentially
0.7 V, and the voltage across each resistor is exactly as obtained in the long-hand solution.
In other words, by choosing this value of reverse saturation current, we created a diode with
characteristics that permitted the approximation that Vv = 0.7 V when in the "on" state.

OrCAD Caplurt' OS· Oetw EdillUl1 • I/ • (SCHE ...

!;di! '[Jew Iools _e1ace Macro ~pice /1,ccesso ries
Window !::!c!p cadence ~

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Ill
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FICi. 2.147
The circuit of Fig. 2.146 reexamined with ls set at 3.5E-15A.

The results can also be viewed in tabulated form by selecting PSpice at the head of the
screen followed by View Output File. The result is the listing of Fig. 2.148 (modified to
conserve space), which includes the CIRCUIT DESCRIPTION with all the components
of the network, the Diode MODEL PARAMETERS with the chosen Is value, and the
INITIAL TRANSIENT SOLUTION with the de voltage levels, current levels, and total
power dissipation.
The analysis is now complete for the diode circuit of interest. Granted, there was a wealth
of information provided to establish and investigate this rather simple network. However,
the vast majority of this material will not be repeated in the PSpice examples to follow,
which will have a dramatic effect on the length of the descriptions. For practice purposes,
it is suggested that other examples in this chapter be checked using PSpice and that the
exercises at the end of the chapter be investigated to develop confidence in applying the
software package.

Diode Charaderistlcs The characteristics of the D 1N4148 diode used in the above analysis
will now be obtained using a few maneuvers somewhat more sophisticated than those
employed in the first example. The process begins by first building the network of Fig.
2.149 using the procedures just described. Note in particular that the source is labeled E and
set at 0V (its initial value). Next the New Simulation Profile icon is selected from the tool-
bar to obtain the New Simulation dialog box. For the Name, Fig. 2-150 is entered since it
is the location of the graph to be obtained. Create is then selected and the Simulation Set-
tings dialog box will appear. Under Analysis Type, DC Sweep is chosen because we want
to sweep through a range of values for the source voltage. When DC Sweep is selected a list
of options will simultaneously appear in the right-hand region of the dialog box, requiring
that some choices be made. Since we plan to sweep through a range of voltages, the Sweep
variable is a Voltage source. Its name must be entered as E as appearing in Fig. 2.149. The
sweep will be Linear (equal space between data points) with a Start value of 0 V, End
Value of 10 V, and an Increment of0.01 V. After making all the entries, click OK and the
116 DIODE APPLICATIONS **** CIRCUIT DESCRIPf!ON

*Analysis directives:
.TRAN 0 IOOOns 0
.PROBE V(alias(*)) !(alias(*))
W(alias(*)) D(alias(*)) NOISE(alias(*))
.INC "..ISCHEMAT!Cl.net"

**** INCLUDING SCHEMAT!Cl.net ****

• source ORCAD2-2
V_El N00103 0 IOVdc
V_E2 0 N00099 5Vde
R_Rl N00103 N00204 4.7k TC=O,0
R_R2 N00099 NOOl 85 2.2k TC=O,0
D_Dl N00204 N00185 DIN4148

**** Diode MODEL PARAMETERS

DIN4148
IS 2.000000E-15

**** INITIAL TRANSIENT SOLUTION TEMPERATURE= 27.000 DEG C

NODE VOLTAGE
(N00099) -5.0000
(N00103) 10.0000
(N00185) -.4455
(N00204) .2700

VOLTAGE SOURCE CURRENTS

NAME CURRENT
V_El -2.070E-03
V_E2 -2.070E-03

TOTAL POWER DISSIPATION 3.1 IE--02 WATTS

FIG. 2.148
Output file for PSpice Windows analysis of the circuit of Fig. 2.147.

RUN PSpice option can be selected. The analysis will be performed with the source voltage
changing from 0 V to 10 Vin 1000 steps (as resulting from the division of 10 V/0.01 V).
The result, however, is simply a graph with a horizontal scale from 0 V to 10 V.
Since the plot we want is of Iv versus Vv, we must change the horizontal (x-axis) to Vv.
This is accomplished by selecting Plot and then Axis Settings. An Axis Settings dialog
box will appear, in which choices have to be made. If Axis Variables is selected, an X-Axis

OrCAD Capture OS - Demo Edmon - V- (SCI-L

i File Edit View Tools Place

Opoon~ Wmduw Ht:lp

lk

E D1
OVdc U1N4148

FIG. 2.149
Network for obtaining the characteristics of the DJN4148 diode.
Variable dialog box will appear with a list of variables that can be chosen for the x-axis. COMPUTER ANALYSIS 117
Vl(Dl) will be selected since it represents the voltage across the diode. If we then select
OK, the Axis Settings dialog box will return, where User Defined is selected under the
Data Range heading. User Defined is chosen because it will allow us to limit the graph
to a range of OV to 1 V since the "on" voltage of the diode should be around 0. 7 V. After
entering the 0---1 V range, selecting OK will result in a graph with Vl(Dl) as the x variable
with a range of OV to 1 V. The horizontal axis now seems to be set for the desired plot.
We must now tum our attention to the vertical axis, which should be the diode current.
Choosing Trace followed by Add Trace will result in an Add Trace dialog box in which
I(Dl) will appear as one of the possibilities. Selecting I(Dl) will also cause it to appear as
the Trace Expression at the bottom of the dialog box. Selecting OK will then result in the
diode characteristics of Fig. 2.150, clearly showing a steep rise around 0.7 V.

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FIC. 2.150
Characteristics of the DJN4148 diode.

If we tum back to the PSpice Model Editor for the diode and change ls to 3.5E-15A as
in the previous example, the curve will shift to the right. Similar procedures will be used to
obtain the characteristic curves for a variety of elements to be introduced in later chapters.

Multisim
Fortunately, there are a number of similarities between Cadence OrCAD and Multisim.
Then again, there are a number of differences also, but the saving point is that once you
become proficient in the use of one software package, the other will be much easier to learn.
For those users familiar with the earlier versions of Multisim, you will find that the new
version has a minimum of changes, permitting an easy transition to the new procedures.
Once the Multisim icon is chosen, a screen will appear with a vast array of toolbars. The
content of each and the name of each can be found through the sequence View-toolbars.
The result is a long vertical list of available toolbars. The content and location of each can
be found by simply selecting or deleting a toolbar and noting the effect on the full screen.
For our purposes the Standard, View, Main, Components, Simulation Switch, Simula-
tion and Instruments will be used.
When using Multisim you have a choice between using "virtual" or "real" components.
Virtual components are those that can be given any value when you build the network. The
term real comes from the fact that the resulting list is a list of standard component values
that can be purchased from a supplier. Finding a component is initiated by first selecting
the second keypad (from the left) on the component toolbar that looks like a resistor. As
you approach the key, the label Place Basic will appear. Once it is chosen, the Select a
118 DIODE APPLICATIONS Component dialog box will appear that contains a subset titled Family. Third down on
that list is a RATED_VIRTUAL option with a resistor symbol. When this is selected a list
of components including RESISTOR_RA TED, CAPACITOR_RATED, INDUCTOR_
RA TED, and a variety of others will appear. If RESISTOR-RA TED is selected, a resistor
symbol will appear under the Symbol heading. Note that the resistor docs not have a specific
value. If we now select OK and place it on the screen in much the same way we did for
the OrCAD introduction, you will find that the value was automatically labeled Rl with
a value of 1 kll. In order to place another resistor the same sequence must be followed,
but this time the resistor will automatically be called R2 but with the same value of 1 kll.
This labeling process will continue in the same manner with the same 1-kll value for as
many resistors as you place. As was done with OrCAD, the resistor labels and values can
be changed quite easily. Of course, if the chosen resistor is a standard value then it can be
found directly under the RESISTOR listing of "real" components.
We are now ready to build the diode network of Example 2.13 so we can compare
results. The diodes chosen will be commercially available under the "real" listing. In this
case two 1N4009 diodes were found by first selecting the keypad Place Diode to the
right of the Place Basic keypad to obtain the Select a Component dialog box. Then the
sequence Family-DIODE-1N4009-OK will result in a diode on the screen labeled Dl
with 1N4009 below the symbol, as shown in Fig. 2.151. Next we can place the resistors
on the screen by going to the RESISTOR option and typing in the value of one of the
resistors, in this case, the 3.3-kll resistor in the area provided at the top of the resistor
listing. This certainly removes the need to scroll through the list looking for a particular
resistor. Once found and placed, it will appear as Rl with a value of 3.3 kll. The same
procedure will result in a second resistor called R2 with a value of 5.6 kll. In each case the
elements are initially placed closest to where they will end up. The de voltage source is
found by going to the Place Source keypad, which is the first keypad in the Component
toolbar. Under Family, POWER SOURCES is selected, followed by DC_POWER.
Click OK and a voltage source will appear on the screen with the label Vl at a level of
12 V. The last circuit element to be set on the screen is the ground, which is accomplished
by going back to the Place Source option and, after selecting POWER SOURCES,
choosing "ground" under the Component listing. Click OK and the ground can be placed
anywhere on the screen.

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FIG. 2.151
Verifying the results of Example 2.13 using Multisim.

Now that all the components are on the screen, they must be placed and labeled properly.
For each component, simply selecting the device will create a blue dashed box around it to
indicate it is in the active mode. When clicked to establish this condition, it can be moved
to any location on the screen. To rotate an element, establish the active mode and apply
Crtl-R to rotate it 90 degrees. Each application of this process will rotate it an additional
90 degrees. Changing a label simply requires double-clicking the label of interest to create
a small blue box around it and produce a dialog box for the change. For the source, a dia-
log box labeled DC_POWER will result, in which the heading Label is selected and the
reIDEs retyped as E. Click OK and the label E will appear. The same procedure can change
the value to 20 V, although in this case the Value heading is chosen and the units are chosen
using the scroll at the right of the entered value.
The next step is to determine what quantities are to be measured and how to measure
them. For this network a multimeter will be used to measure the current through the resistor
Rl. The multimeter is found at the top of the Instrument toolbar. After selection it can be
placed on the screen in the same manner as the other elements. Double-clicking the meter
will then result in the Multimeter-XXMl dialog box, in which A is selected to set the mul-
timeter as an ammeter. In addition, the DC box (a straight line) must be selected because
we are dealing with de voltages. The current through the diode D1 and the voltage across
the resistor R2 will be found using Indicators, which are found as the tenth option to the
right on the Component toolbar. The software symbol looks like an LED with a red dashed
figure eight inside. Click on this option and a Select a Component dialog box will appear.
Under Family, select AMMETER and then take note of the Component listing and the
four options for the orientation of the indicator. For our analysis the AMMETER_H will
be chosen since the plus sign or entering point for the current is on the left for the diode
Dl. Click OK and the indicator can be placed to the left of the diode Dl. For the voltage
across the resistor R2, the option VOLTMETER_HR is chosen so the polarity matches
that across the resistor.
Finally, all the components and meters must be connected. This is accomplished by
simply placing the cursor at the end of an element until a small circle and a set of crosshairs
appear to designate the starting point. Once these are in place, click the location and an x
will appear at the terminal. Then move to the end of the other element and left-click the
mouse again-a red connecting wire will automatically appear with the most direct route
between the two elements. The process is called Automatic Wiring.
Now that all the components are in place it is time to initiate the analysis of the circuit,
an operation that can be performed in one of three ways. One option is to select Simulate
at the head of the screen followed by Run. The next is the green arrow in the Simulation
toolbar. The last is to simply toggle the switch at the head of the screen to the 1 position. In
each case a solution appears in the indicators after a few seconds that seems to flicker over
time. This flickering simply indicates the software package is repeating the analysis over
time. To accept the solution and stop the continuing simulation, either toggle the switch to
the Oposition or select the lightning bolt keypad again.
The current through the diode is 3.349 mA, which compares well with the 3.32 mA in
Example 2.13. The voltage across the resistor R2 is 18.722 V, which is close to the 18.6 V
of the same example. After the simulation, the multimeter can be displayed as shown in
Fig. 2.151 by double-clicking on the meter symbol. By clicking anywhere on the meter, the
top portion is dark blue, and the meter can be moved to any location by simply clicking on
the blue region and dragging it to the desired location. The current of 193.285 µ,A is very
close to the 212 µ,A of Example 2.13. The differences are primarily due to the fact that each
diode voltage is assumed to be 0. 7 V, whereas in fact it is different for each diode of Fig.
2.151 since the current through each is different. In all, however, the Multisim solution is
a very close match with the approximate solution of Example 2.13.

PROBLEMS
*Note: Asterisks indicate more difficult problems.
2.2 Load-Line Analysis
•
1. a. Using the characteristics of Fig. 2.152b, determine Iv, Vv, and VR for the circuit of Fig. 2.152a.
b. Repeat part (a) using the approximate model for the diode, and compare results.
c. Repeat part (a) using the ideal model for the diode, and compare results.
2. a. Using the characteristics of Fig. 2.152b, determine Iv and Vv for the circuit of Fig. 2.153.
b. Repeat part (a) with R = 0.47 k!l.
c. Repeat part (a) with R = 0.68 k!l.
d. Is the level of Vv relatively close to 0.7 Vin each case?
How do the resulting levels of Iv compare? Comment accordingly.
DIODE APPLICATIONS + Vv
Si

+
+
E-=-12V R 0.75 kfl VR

(a)

Iv (mA)

- 3

>- 25

>- 2

>- 15

- 1

- 5

- 0 -
0.7V
1
1
-- 2 -
I
3 -- 4 -
I
5 -- 6 - I
7 -- 8 -
I
- 9 -- 10 -
I
~ 11 - -- 12 V (V
I I I- --1 I I I f I I

(b)

FIC. 2.152
Problems 1 and 2.

3. Determine the value of R for the circuit of Fig. 2.153 that will result in a diode current of
10 mA if E = 1 V. Use the characteristics of Fig. 2.152b for the diode.
4. a. Using the approximate characteristics for the Si diode, determine Vn, In, and VR for the
circuit of Fig. 2.154.
b. Perform the same analysis as part (a) using the ideal model for the diode.
c. Do the results obtained in parts (a) and (b) suggest that the ideal model can provide a good
approximation for the actual response under some conditions?

Si
~ Si +
+ + E-=- 30V R 1.5 kfl VR
R 0.2kfl VR

FIC. 2.153 FIC. 2.154

Problems 2 and 3. Problem 4.
2.3 Series Diode Configurations PROBLEMS
5. Determine the current I for each of the configurations of Fig. 2.155 using the approximate
equivalent model for the diode.

20V

--
10n
11 +
Si I Si
Si
Si i,
+
10 n A -=- 10 V 10n

20Q Si

(a)

(b) (c)

FIG. 2.155
Problem 5.

6. Determine V0 and Iv for the networks of Fig. 2.156.

- 5V +8V
o--.rv"'.,.-- ----o Vo
Si 1.2 ki1

2.2 kQ 4 .7 kQ

Si

-6V
(a) (b)

FIG. 2.156
Problems 6 and 49.

*7. Determine the level of V0 for each network of Fig. 2.157.

12 y Si Ge 2 kQ +10 V 1.2 kQ Si
o>--------,~-----~t-----'VV\,-----<I.,...____..o V 0

4.7kQ
"=" lOk!l

lOV

(a) (b)

FIG. 2.157
Problem 7.

*8. Determine V0 and Iv for the networks of Fig. 2.158.

-- Iv

Si
Vo

I
t lOmA 2.2 kQ 2.2 k!l
+20V

i -=- -=-
(a) (b)

FIG. 2.158
Problem 8.
DIODE APPLICATIONS *9. Determine V01 and V0 z for the networks of Fig. 2.159.

GaAs 3.3 kQ

T T
(a) (b)

FIG. l.159
Problem 9.

2.4 Parallel and Series-Parallel Configurations

10. Determine V0 and In for the networks of Fig. 2.160.

20V

12V Si tlo
-~-----o vo Ge Si

GaAs Vo
4.7kQ

2.2 kQ

T 4V

(a) (b)

FIG. l.160
Problems IO and 50.

*11. Determine V0 and /for the networks of Fig. 2.161.

lV +16 V

Si
t'
Si GaAs Si Si

---------o Vo -------o vo
lkQ 4.7 kO

--4V

(a) (b)

FIG. l.161
Problem II.

12. Determine V0 l' V0 z, and /for the network of Fig. 2.162.

*13. Determine V0 and In for the network of Fig. 2.163.
 V,,l

Ikn 0.47 kn
v,,2 ID/ 2kn

+ t' Si
20V Si Si +IOV -----ovo

2 kQ

"II" "II"

FIG. 2.162 FIG. 2.163

Problem 12. Problems 13 and 51.

2.5 AND/OR Gates

14. Determine V0 for the network of Fig. 2.39 with OVon both inputs.
15. Determine V0 for the network of Fig. 2.39 with 10 V on both inputs.
16. Determine V0 for the network of Fig. 2.42 with OV on both inputs.
17. Determine V0 for the network of Fig. 2.42 with 10 V on both inputs.
18. Determine V0 for the negative logic OR gate of Fig. 2.164.
19. Determine V0 for the negative logic AND gate of Fig. 2.165.

-SV -SV

Si Si

ov ov
Vo Vo
Si Si

1 kn 2.2 kn

"II" -SV

FIG. 2.164 FIG. 2.165

Problem 18. Problem 19.

20. Determine the level of V0 for the gate of Fig. 2.166.

21. Determine V0 for the configuration of Fig. 2.167.

!OV sv
Si Si

10 V sv
o-----1-------<I-----O VO 0-----1---1----oVo
Si Ge

I kn 2.2 kn

!OV "II"

FIG. 2.166 FIG. 2.167

Problem 20. Problem 21.

2.6 Sinusoidal Inputs; Half-Wave Rectification

22. Assuming an ideal diode, sketch v;, vd, and id for the half-wave rectifier of Fig. 2.168. The
input is a sinusoidal waveform with a frequency of 60 Hz. Determine the profit value of v; from
the given de level.
23. Repeat Problem 22 with a silicon diode (VK = 0.7 V).
24. Repeat Problem 22 with a 10 kfl load applied as shown in Fig. 2.169. Sketch vL and iL.
DIODE APPLICATIONS + vd
Ideal Yctc=2 V
IJliil

I •low =irL
ViO

id +
~ 0
V; 2kQ VL
,w

FICi. 2.168 FICi. 2.169

Problems 22 through 24. Problem 24.

25. For the network of Fig. 2.170, sketch v O and determine Yctc·
*26. For the network of Fig. 2.171, sketch v 0 and iR.

2k0

+ +
V; = 120 V (rms) Ideal

FICi. 2.170 FICi. 2.171

Problem 25. Problem 26.

*27. a. Given Pmax = 14 mW for each diode at Fig. 2.172, determine the maximum current rating
of each diode (using the approximate equivalent model).
b. Determine Imax for the parallel diodes.
c. Determine the current through each diode at V;max using the results of part (b ).
d. If only one diode were present, which would be the expected result?

Si

+
V; Si 4.7ill 68kQ

FICi. 2.172
Problem 27.

2.7 Full-Wave Rectification

28. A full-wave bridge rectifier with a 120-V rms sinusoidal input has a load resistor of 1 kil.
a. If silicon diodes are employed, what is the de voltage available at the load?
b. Determine the required PIV rating of each diode.
c. Find the maximum current through each diode during conduction.
d. What is the required power rating of each diode?
29. Determine v 0 and the required PIV rating of each diode for the configuration of Fig. 2.173. In
addition, determine the maximum current through each diode.

V; - - - - - - o Vo
+
2.2ill

-=-
FICi. 2.173
Problem 29.
*30. Sketch v 0 for the network of Fig. 2.174 and determine the de voltage available. PROBLEMS

V;

-=-
FICi. 2.174
Problem 30.

*31. Sketch v 0 for the network of Fig. 2.175 and determine the de voltage available.

V;

FICi. 2.175
Problem 31.

2.8 Clippers
32. Determine vO for each network of Fig. 2.176 for the input shown.

v, Si 8V Ideal

+ +
~II 1--_-------0+
100 kQ v, 2kQ

-20V

FICi. 2.176
Problem 32.

33. Determine vO for each network of Fig. 2.177 for the input shown.

V;
Si Si 4V
12 V
V; c,---IIJ,l--1_111~ Vo

f°'°
l.8kQ

-12V -=-
(a) (b)

FICi. 2.177
Problem 33.
DIODE APPLICATIONS *34. Determine vO for each network of Fig. 2.178 for the input shown.

v,
20V
4 V + Ideal Ideal
o--=1 IIt---...-----.----<J
+ +
V; 1 kQ 2.2 kQ

-5 V +5 V

(a) (b)

FIG. 2.178
Problem 34.

*35. Determine vO for each network of Fig. 2.179 for the input shown.

~
ill
Si
+ V· ~AA__Jllrv
,~2~~~
3V

- 0

V; + V0 Si

-r
- 4V

0 0

(a) (b)

FIG. 2.179
Problem 35.

36. Sketch iR and v O for the network of Fig. 2.180 for the input shown.

VI lOkQ

+ _. +
iR
Si Si
V;
+ Vo

0
-r
5.3 V-=- 7.3V-;;;;-

+I 0

FIG. 2.180
Problem 36.

2.9 Clampers
37. Sketch v 0 for each network of Fig. 2.181 for the input shown.

C C

~
20V Vi o---lt----1>-----....--0 Vo
+
Ideal
Ideal R V
0 R
0

-20 V
(a) (b)

FIG. 2.181
Problem 37.
 38. Sketch v 0 for each network of Fig. 2.182 for the input shown.

V; C C
o---f o---f +
+ + +
Ideal
V; Ideal R Vo V; + R Vo

£..=_20V

(a) (b)

FIG. 2.182
Problem 38.

*39. For the network of Fig. 2.183:

a. Calculate 5-r.
b. Compare 5-r to half the period of the applied signal.
c. Sketch v0 •

V; C
12V o------1->---------U
+ 0.1 µF +
Si

~2V
-12 V +
f= 1 kHz

FIG. 2.183
Problem 39.

*40. Design a clamper to perform the function indicated in Fig. 2.184.

V;

+30V
Ideal diodes
20V

+ +
Design
0

-lOV

-20V

FIG. 2.184
Problem 40.

*41. Design a clamper to perform the function indicated in Fig. 2.185.

V; Silicon diodes Vo

lOV
2.7V
+ +
0
V; Design Vo
0

-IOV
-17.3 V

FIG. 2.185
Problem 41.
DIODE APPLICATIONS 2.1 o Zener Diodes
*42. a. Determine VL> h, [z, and IR for the network of Fig. 2.186 if RL = 180 n.
b. Repeat part (a) if RL = 470 n.
c. Determine the value of RL that will establish maximum power conditions for the Zener diode.
d. Determine the minimum value of RL to ensure that the Zener diode is in the "on" state.

Rs

+ ~
IR
220Q
fz 7lh+
20V
Vz = lOV
RL VL
Pzmax =400mW

FIG. 2.186
Problem 42.

*43. a. Design the network of Fig. 2.187 to maintain VL at 12 V for a load variation (h) from 0 mA
to 200 mA. That is, determine Rs and Vz.
b. Determine Pz max for the Zener diode of part (a).
*44. For the network of Fig. 2.188, determine the range of V; that will maintain VL at 8 V and not
exceed the maximum power rating of the Zener diode.

16 vo--"'"""...---.--7+ h V; o---<,V,""',-----.-----,
91 Q
Vz V2 =8 V
RL 0.22 kQ
Pzmax = 400 mW

l "II"

FIG. 2.187 FIG. 2.188

Problem 43. Problems 44 and 52.

45. Design a voltage regulator that will maintain an output voltage of 20 V across a 1-kil load with
an input that will vary between 30 V and 50 V. That is, determine the proper value of Rs and
the maximum current IZM.
46. Sketch the output of the network of Fig. 2.145 if the input is a 50-V square wave. Repeat for a
5-V square wave.

2.11 Voltage-Multiplier Circuits

47. Determine the voltage available from the voltage doubler of Fig. 2.123 if the secondary voltage
of the transformer is 120 V (rms).
48. Determine the required PIV ratings of the diodes of Fig. 2.123 in terms of the peak secondary
voltage Vm.

2.14 Computer Analysis

49. Perform an analysis of the network of Fig. 2.156b using PSpice Windows.
50. Perform an analysis of the network of Fig. 2.161 b using PSpice Windows.
51. Perform an analysis of the network of Fig. 2.162 using PSpice Windows.
52. Perform a general analysis of the Zener network of Fig. 2.188 using PSpice Windows.
53. Repeat Problem 49 using Multisim.
54. Repeat Problem 50 using Multisim.
55. Repeat Problem 51 using Multisim.
56. Repeat Problem 52 using Multisim.
CHAPTER OBJECTIVES
• Become familiar with the basic construction and operation of the Bipolar
Junction Transistor.
•
• Be able to apply the proper biasing to insure operation in the active region.
• Recognize and be able to explain the characteristics of an npn or pnp transistor.
• Become familiar with the important parameters that define the
response of a transistor.
• Be able to test a transistor and identify the three terminals.

3.1 INTRODUCTION
•
During the period 1904 to 1947, the vacuum tube was the electronic device of interest and
development. In 1904, the vacuum-tube diode was introduced by J. A. Fleming. Shortly
thereafter, in 1906, Lee De Forest added a third element, called the control grid, to the
Dr. William Shockley (seated);
Dr. John Bardeen (left); Dr. Walter
vacuum diode, resulting in the first amplifier, the triode. In the following years, radio H. Brattain. (Courtesy of AT&T
Archives and History Center.)
and television provided great stimulation to the tube industry. Production rose from
Dr. Shockley Born: London,
about 1 million tubes in 1922 to about 100 million in 1937. In the early 1930s the four-
England, 1910
element tetrode and the five-element pentode gained prominence in the electron-tube PhD Harvard,
industry. In the years to follow, the industry became one of primary importance, and rapid 1936
advances were made in design, manufacturing techniques, high-power and high-frequency Dr. Bardeen Born: Madison,
applications, and miniaturization. Wisconsin, 1908
On December 23, 1947, however, the electronics industry was to experience the advent PhD Princeton,
of a completely new direction of interest and development. It was on the afternoon of this 1936
day that Dr. S. William Shockley, Walter H. Brattain, and John Bardeen demonstrated the Dr. Brattain Born: Arnoy,
amplifying action of the first transistor at the Bell Telephone Laboratories as shown in China, 1902
Fig. 3.1. The original transistor (a point-contact transistor) is shown in Fig. 3.2. The ad- PhD University
of Minnesota,
vantages of this three-terminal solid-state device over the tube were immediately obvious:
1928
It was smaller and lightweight; it had no heater requirement or heater loss; it had a rugged
construction; it was more efficient since less power was absorbed by the device itself; it All shared the Nobel Prize in 1956
was instantly available for use, requiring no warm-up period; and lower operating voltages for this contribution.
were possible. Note that this chapter is our first discussion of devices with three or more
terminals. You will find that all amplifiers (devices that increase the voltage, current, or FIG. 3.1
Coinventors of the first transistor
power level) have at least three terminals, with one controlling the flow or potential between
at Bell Laboratories.
the other two.
no BIPOLAR JUNCTION
TRANSISTORS
3.2 TRANSISTOR CONSTRUCTION
•
The transistor is a three-layer semiconductor device consisting of either two n- and one
p-type layers of material or two p- and one n-type layers of material. The former is called
an npn transistor, and the latter is called a pnp transistor. Both are shown in Fig. 3.3 with
the proper de biasing. We will find in Chapter 4 that the de biasing is necessary to establish
the proper region of operation for ac amplification. The emitter layer is heavily doped,
with the base and collector only lightly doped. The outer layers have widths much greater
than the sandwiched p- or n-type material. For the transistors shown in Fig. 3.2 the ratio of
the total width to that of the center layer is 0.150/0.001 = 150: 1. The doping of the sand-
wiched layer is also considerably less than that of the outer layers (typically, 1:10 or less).
FICi. 3.2 This lower doping level decreases the conductivity (increases the resistance) of this mate-
The first transistor. (Courtesy of rial by limiting the number of "free" carriers.
AT&T Archives and History Center.) For the biasing shown in Fig. 3.3 the terminals have been indicated by the capital letters
E for emitter, C for collector, and B for base. An appreciation for this choice of notation will
0.150 in. develop when we discuss the basic operation of the transistor. The abbreviation BJT, from
bipolar junction transistor, is often applied to this three-terminal device. The term bipolar
0.001 in.
reflects the fact that holes and electrons participate in the injection process into the oppo-
i i sitely polarized material. If only one carrier is employed (electron or hole), it is considered
E n C a unipolar device. The Schottky diode of Chapter 16 is such a device.
p p

+ - + -
I I,_______._____. I I
VEE Vee
3.3 TRANSISTOR OPERATION
•
The basic operation of the transistor will now be described using the pnp transistor of Fig. 3 .3a.
The operation of the npn transistor is exactly the same if the roles played by the electron and
hole are interchanged. In Fig. 3.4a the pnp transistor has been redrawn without the base-to-
(a)
collector bias. Note the similarities between this situation and that of the forward-biased diode
in Chapter 1. The depletion region has been reduced in width due to the applied bias, resulting
in a heavy flow of majority carriers from the p- to then-type material.
0.150 in.
Let us now remove the base-to-emitter bias of the pnp transistor of Fig. 3.3a as shown
in Fig. 3.4b. Consider the similarities between this situation and that of the reverse-biased
diode of Section 1.6. Recall that the flow of majority carriers is zero, resulting in only a
E C minority-carrier flow, as indicated in Fig. 3.4b. In summary, therefore:
One p-n junction of a transistor is reverse-biased, whereas the other is forward-biased.
B

- + - + + Majority carriers + Minority carriers

II~ - II
VEE Vee
E
(b)

FICi. 3.3
Types of transistors: (a) pnp; Depletion region
1- B
B+
Depletion region
(b) npn.

~ - + 11--------- _ _ _+__,II---~
VEE Vee
(a) (b)

FICi. 3.4
Biasing a transistor: (a)forward-bias; (b) reverse-bias.

In Fig. 3.5 both biasing potentials have been applied to a pnp transistor, with the resulting
majority- and minority-carrier flows indicated. Note in Fig. 3.5 the widths of the depletion
regions, indicating clearly which junction is forward-biased and which is reverse-biased.
As indicated in Fig. 3.5, a large number of majority carriers will diffuse across the forward-
biased p-n junction into the n-type material. The question then is whether these carriers will
contribute directly to the base current In or pass directly into the p-type material. Since the
sandwiched n-type material is very thin and has a low conductivity, a very small number of
 + Majority carriers + Minority carriers COMMON-BASE UI
p p CONFIGURATION

Depletion regions
t/B
~___,+II,__-_ _ ___,+ I I---~
VEE Vee

FIG. 3.5
Majority and minority carrier flow of a pnp
transistor.

these carriers will take this path of high resistance to the base terminal. The magnitude of the
base current is typically on the order of microamperes, as compared to milliamperes for the
emitter and collector currents. The larger number of these majority carriers will diffuse across
the reverse-biased junction into the p-type material connected to the collector terminal as indi-
cated in Fig. 3.5. The reason for the relative ease with which the majority carriers can cross the
reverse-biased junction is easily understood if we consider that for the reverse-biased diode
the injected majority carriers will appear as minority carriers in the n-type material. In other
words, there has been an injection of minority carriers into the n-type base region material.
Combining this with the fact that all the minority carriers in the depletion region will cross the
reverse-biased junction of a diode accounts for the flow indicated in Fig. 3.5.
Applying Kirchhoff's current law to the transistor of Fig. 3.5 as if it were a single node,
we obtain

I IE= le+ IB I (3.1)

and find that the emitter current is the sum of the collector and base currents. The collector
current, however, comprises two components-the majority and the minority carriers as
B
indicated in Fig. 3.5. The minority-current component is called the leakage current and is
given the symbol Ico Uc current with emitter terminal Open). The collector current, there- (a)
fore, is determined in total by

le = 1cmajority + Icommority (3.2)

For general-purpose transistors, le is measured in milliamperes and Ico is measured in
microamperes or nanoamperes. Ico, like ls for a reverse-biased diode, is temperature sen-
sitive and must be examined carefully when applications of wide temperature ranges are
considered. It can severely affect the stability of a system at high temperature if not con-
sidered properly. Improvements in construction techniques have resulted in significantly
lower levels of Ico, to the point where its effect can often be ignored. 11
VEE "II'

3.4 COMMON-BASE CONFIGURATION

•
The notation and symbols used in conjunction with the transistor in the majority of texts
and manuals published today are indicated in Fig. 3.6 for the common-base configuration
with pnp and npn transistors. The common-base terminology is derived from the fact that
E
~
IE
~
le
C

the base is common to both the input and output sides of the configuration. In addition, the
base is usually the terminal closest to, or at, ground potential. Throughout this text all cur-
B
rent directions will refer to conventional (hole) flow rather than electron flow. The result is
that the arrows in all electronic symbols have a direction defined by this convention. Recall (b)
that the arrow in the diode symbol defined the direction of conduction for conventional
FIG. 3.6
current. For the transistor:
Notation and symbols used with the
The arrow in the graphic symbol defines the direction of emitter current (conventional common-base configuration: (a) pnp
flow) through the device. transistor; (b) npn transistor.
U2 BIPOLAR JUNCTION All the current directions appearing in Fig. 3.6 are the actual directions as defined by the
TRANSISTORS choice of conventional flow. Note in each case that/E = le + ls. Note also that the applied
biasing (voltage sources) are such as to establish current in the direction indicated for each
branch. That is, compare the direction of IE to the polarity of VEE for each configuration
and the direction of le to the polarity of Vcc-
To fully describe the behavior of a three-terminal device such as the common-base am-
plifiers of Fig. 3.6 requires two sets of characteristics---0ne for the driving point or input
parameters and the other for the output side. The input set for the common-base amplifier
as shown in Fig. 3.7 relates an input current (le) to an input voltage (VsE) for various levels
of output voltage (Vcs)-

IE (mA)

Vc8 = 20 V

8
Vc 8 = 10 V
7
6 Vc8 = 1 V

5
4

3
2

0 0.2 0.4 0.6 0.8 1.0 VBE (V)

FIG. 3.7
Input or driving point characteristics for a
common-base silicon transistor amplifier.

The output set relates an output current (/c) to an output voltage (Vcs) for various levels
of input current (IE) as shown in Fig. 3.8. The output or collector set of characteristics has
three basic regions of interest, as indicated in Fig. 3.8: the active, cutoff, and saturation

le (mA)

---------Active region (unshaded area)------#-,

7 mA
7

6 mA
6

5 mA
5

4 mA
4

3 mA
3

2mA
2

Ico=ICBo
0
t
-1 0 t 10 20 30 /40 Yes (V)
Cutoff region BVCBo

FIG. 3.8
Output or collector characteristics for a common-base transistor amplifier.
regions. The active region is the region normally employed for linear (undistorted) ampli- COMMON-BASE U3
fiers. In particular: CONFIGURATION

In the active region the base-emitter junction is forward-biased, whereas the collector-
base junction is reverse-biased.
The active region is defined by the biasing arrangements of Fig. 3.6. At the lower end of
the active region the emitter current (IE) is zero, and the collector current is simply that due to
J6
E o------eIE =0

01°\lco
C

the reverse saturation current/co, as indicated in Fig. 3.9. The current/co is so small (micro-
amperes) in magnitude compared to the vertical scale of le (milliamperes) that it appears on
B f Emitter
open
Collector to base
virtually the same horizontal line as le = 0. The circuit conditions that exist when IE = 0 for
the common-base configuration are shown in Fig. 3.9. The notation most frequently used FIC. 3.9
for Ico on data and specification sheets is, as indicated in Fig. 3.9, lcso (the collector-to- Reverse saturation current.
base current with the emitter leg open). Because of improved construction techniques, the
level of lcso for general-purpose transistors in the low- and mid-power ranges is usually
so low that its effect can be ignored. However, for higher power units I CBO will still appear
in the microampere range. In addition, keep in mind that lcso, like ls, for the diode (both
reverse leakage currents) is temperature sensitive. At higher temperatures the effect of lcso
may become an important factor since it increases so rapidly with temperature.
Note in Fig. 3.8 that as the emitter current increases above zero, the collector current
increases to a magnitude essentially equal to that of the emitter current as determined by
the basic transistor-current relations. Note also the almost negligible effect of VCB on the
collector current for the active region. The curves clearly indicate that a first approximation
to the relationship between IE and le in the active region is given by

(3.3)
As inferred by its name, the cutoff region is defined as that region where the collector
current is 0 A, as revealed on Fig. 3.8. In addition:
In the cutoff region the base-emitter and collector-base junctions of a transistor are
both reverse-biased.
The saturation region is defined as that region of the characteristics to the left of
Vcs = 0 V. The horizontal scale in this region was expanded to clearly show the dramatic
change in characteristics in this region. Note the exponential increase in collector current
as the voltage VCB increases toward 0 V.
In the saturation region the base-emitter and collector-base junctions are forward-biased.
The input characteristics of Fig. 3.7 reveal that for fixed values of collector voltage (Vcs),
as the base-to-emitter voltage increases, the emitter current increases in a manner that closely
resembles the diode characteristics. In fact, increasing levels of VCB have such a small effect
on the characteristics that as a first approximation the change due to changes in VCB can be
ignored and the characteristics drawn as shown in Fig. 3 .1 0a. If we then apply the piecewise-
linear approach, the characteristics of Fig. 3.10b result. Taking it a step further and ignoring
the slope of the curve and therefore the resistance associated with the forward-biased junction
results in the characteristics of Fig. 3.10c. For the analysis to follow in this book the equivalent
model of Fig. 3.10c will be employed for all de analysis of transistor networks. That is, once a
transistor is in the "on" state, the base-to-emitter voltage will be assumed to be the following:

I VsE == 0.7V I (3.4)

In other words, the effect of variations due to VCB and the slope of the input characteristics
will be ignored as we strive to analyze transistor networks in a manner that will provide a
good approximation to the actual response without getting too involved with parameter
variations of less importance.
It is important to fully appreciate the statement made by the characteristics of Fig. 3.10c.
They specify that with the transistor in the "on" or active state the voltage from base to
emitter will be 0.7 Vat any level of emitter current as controlled by the external network.
In fact, at the first encounter of any transistor configuration in the de mode, one can now
immediately specify that the voltage from base to emitter is 0. 7 V if the device is in the
active region-a very important conclusion for the de analysis to follow.
 IE (mA) IE (mA) IE(mA)

8 8 8

7 7 7
-------AnyVcs
6 6 6

5 5 5
~ ~
4 4 4

3 3 3

2 2 2

/ 2 0.7 V /i 0.7 V

0 0.2 0.4 0.6 0.8 ½iE (V) 0 0.2 0.4 0.6 0.8 ½iE (V) 0 0.2 0.4 0.6 0.8 VBE (V)

(a) (b) (c)

FIG. 3.10
Developing the equivalent model to be employed for the base-to-emitter region of an amplifier in the de mode.

EXAMPLE 3.1
a. Using the characteristics of Fig. 3.8, determine the resulting collector current if IE= 3 mA
and VcB = 10 V.
b. Using the characteristics of Fig. 3.8, determine the resulting collector current if IE
remains at 3 mA but VCB is reduced to 2 V.
c. Using the characteristics of Figs. 3.7 and 3.8, determine VBEif le= 4 mA and VCB = 20 V.
d. Repeat part (c) using the characteristics of Figs. 3.8 and 3.10c.

Solution:
a. The characteristics clearly indicate that le ~ IE = 3 mA.
b. The effect of changing VCB is negligible and I c continues to be 3 mA.
c. From Fig. 3.8, le ~ le = 4 mA. On Fig. 3.7 the resulting level of VBE is about 0.74 V.
d. Again from Fig. 3.8, le ~ le= 4 mA. However, on Fig. 3.10c, VBE is 0.7 V for any
level of emitter current.

Alpha (a)
DC Mode In the de mode the levels of le and IE due to the majority carriers are related by
a quantity called alpha and defined by the following equation:

I a,, ~~ I (3.5)

where le and le are the levels of current at the point of operation. Even though the charac-
teristics of Fig. 3.8 would suggest that a = 1, for practical devices alpha typically extends
from 0.90 to 0.998, with most values approaching the high end of the range. Since alpha is
defined solely for the majority carriers, Eq. (3.2) becomes

I le = aIE + ICBo I (3.6)

For the characteristics of Fig. 3.8 when le = 0 mA, le is therefore equal to ICBo, but as
mentioned earlier, the level of ICBo is usually so small that it is virtually undetectable on
the graph of Fig. 3.8. In other words, when/E = 0mAonFig. 3.8, le also appears to be 0mA
for the range of VCB values.
1]4
AC Mode For ac situations where the point of operation moves on the characteristic COMMON-BASE ns
curve, an ac alpha is defined by CONFIGURATION

O'.ac = Mel
~ (3.7)
JE VCB~constant

The ac alpha is formally called the common-base, short-circuit, amplification factor, for
reasons that will be more obvious when we examine transistor equivalent circuits in
Chapter 5. For the moment, recognize that Eq. (3.7) specifies that a relatively small change
in collector current is divided by the corresponding change in le with the collector-to-base
voltage held constant. For most situations the magnitudes of aac and adc are quite close,
permitting the use of the magnitude of one for the other. The use of an equation such as
(3.7) will be demonstrated in Section 3.6.

Biasing
The proper biasing of the common-base configuration in the active region can be deter-
mined quickly using the approximation le ~ le and assuming for the moment that
18 ~ 0 µ,A. The result is the configuration of Fig. 3.11 for the pnp transistor. The arrow of
the symbol defines the direction of conventional flow for IE ~ le. The de supplies are
then inserted with a polarity that will support the resulting current direction. For the npn
transistor the polarities will be reversed.

E C

18 = 0 µA

~ --+--1 I I1-_- ----'--- +--i I I....__ _,

VEE Vee

FIG. 3.11
Establishing the proper biasing
management for a common-base pnp
transistor in the active region.

Some students feel that they can remember whether the arrow of the device symbol is
pointing in or out by matching the letters of the transistor type with the appropriate letters
of the phrases "pointing in" or "not pointing in." For instance, there is a match between
the letters npn and the italic letters of not pointing in and the letters pnp with pointing in.

Breakdown Region
As the applied voltage VCB increases there is a point where the curves take a dramatic
upswing in Fig. 3.8. This is due primarily to an avalanche effect similar to that described
for the diode in Chapter 1 when the reverse-bias voltage reached the breakdown region.
As stated earlier the base-to-collector junction is reversed biased in the active region, but
there is a point where too large a reverse-bias voltage will lead to the avalanche effect.
The result is a large increase in current for small increases in the base-to-collector
voltage. The largest permissible base-to-collector voltage is labeled BVcso as shown
in Fig. 3.8. It is also referred to as V(BR)CBO as shown on the characteristics of Fig. 3.23 to
be discussed later. Note in each of the above notations the use of the uppercase letter O to
represent that the emitter leg is in the open state (not connected). It is important to remem-
ber when taking note of this data point that this limitation is only for the common-base
configuration. You will find in the common-emitter configuration that this limiting volt-
age is quite a bit less.
n& BIPOLAR JUNCTION
TRANSISTORS
3.5 COMMON-EMITTER CONFIGURATION
The most frequently encountered transistor configuration appears in Fig. 3.12 for the pnp
and npn transistors. It is called the common-emitter configuration because the emitter is
common to both the input and output terminals (in this case common to both the base and
•
collector terminals). Two sets of characteristics are again necessary to describe fully the
behavior of the common-emitter configuration: one for the input or base-emitter circuit
and one for the output or collector-emitter circuit. Both are shown in Fig. 3.13.

__..
le

1
p
+ .._Ia ":"Vee
-==■ Vee B +
p

"II'

__..
le
C C

Ia
Bo-_..
_ _...,, .._
Ia
B

/Et
E E

(a) (b)

FIG. 3.12
Notation and symbols used with the common-emitter configuration: (a) npn transistor;
(b) pnp transistor.

le (mA)
8 h - ,...,..-,""TTTTTT"",..-,-,...,..-,""TTTTTT""r-r...,..-,""TTTTTT",..-,-,rr-,-r-r,

la (µA)
VCE=lV
VCE= lOV
100
(Saturation region) 90 VCE = 20 V
80
70
60
50
40
30
20
10
la =0µA
0 5 15 20 VCE (V) 0 0.2 0.4 0.6 0.8 1.0 VaE (V)
(Cutoff region)
IcEO= f3Icao

(a) (b)

FIG. 3.13
Characteristics of a silicon transistor in the common-emitter configuration: (a) collector characteristics; (b) base characteristics.
 The emitter, collector, and base currents are shown in their actual conventional current COMMON-EMITTER U7
direction. Even though the transistor configuration has changed, the current relations devel- CONFIGURATION
oped earlier for the common-base configuration are still applicable. That is, le = le + IB
and/c = a/E.
For the common-emitter configuration the output characteristics are a plot of the output
current (le) versus output voltage (VcE) for a range of values of input current (/B)- The input
characteristics are a plot of the input current (/B) versus the input voltage (VBE) for a range
of values of output voltage (VcE)-
Note that on the characteristics of Fig. 3.14 the magnitude of IB is in microamperes,
compared to milliamperes of/C· Consider also that the curves of/B are not as horizontal as
those obtained for IE in the common-base configuration, indicating that the collector-to-
emitter voltage will influence the magnitude of the collector current.
The active region for the common-emitter configuration is that portion of the upper-right
quadrant that has the greatest linearity, that is, that region in which the curves for IB are
nearly straight and equally spaced. In Fig. 3.14a this region exists to the right of the verti-
cal dashed line at VcEsat and above the curve for IB equal to zero. The region to the left of
VCEsat is called the saturation region.
In the active region of a common-emitter amplifier, the base-emitter junction is
forward-biased, whereas the collector-base junction is reverse-biased.
You will recall that these were the same conditions that existed in the active region of
the common-base configuration. The active region of the common-emitter configuration
can be employed for voltage, current, or power amplification.
The cutoff region for the common-emitter configuration is not as well defined as for the
common-base configuration. Note on the collector characteristics of Fig. 3.14 that le is not
equal to zero when IB is zero. For the common-base configuration, when the input current
IE was equal to zero, the collector current was equal only to the reverse saturation current
lea, so that the curve le = 0 and the voltage axis were, for all practical purposes, one.
The reason for this difference in collector characteristics can be derived through the
proper manipulation of Eqs. (3.3) and (3.6). That is,
Eq. (3.6): le = ale + lcBo
Substitution gives Eq. (3.3): le = a(Ic + IB) + ICBo
a/B ICBo
Rearranging yields Ic=--+-- (3.8)
1-a 1-a
If we consider the case discussed above, where IB = 0 A, and substitute a typical value
of a such as 0.996, the resulting collector current is the following:
a(0 A) lcBo
le = 1- a + 1 - 0.996
lcBO
= 0.004 = 250/cBO

If IcBo were 1 µ,A, the resulting collector current with IB = 0 A would be 250(1 µ,A) =
0.25 mA, as reflected in the characteristics of Fig. 3.14.
For future reference, the collector current defined by the condition IB = 0 µ,A will be
assigned the notation indicated by the following equation:

IcEo =- -I
ICBo (3.9)
1- a ls=O µ,A

In Fig. 3.13 the conditions surrounding this newly defined current are demonstrated with
its assigned reference direction.
For linear (least distortion) amplification purposes, cutofffor the common-emitter
configuration will be defined by le = /CEo-
In other words, the region below IB = 0 µ,A is to be avoided if an undistorted output
signal is required.
When employed as a switch in the logic circuitry of a computer, a transistor will have
two points of operation of interest: one in the cutoff and one in the saturation region. The
na BIPOLAR JUNCTION ! 8 (µA)
TRANSISTORS

100
90
80
70
60
50
40
C 30
20
Bo---- 10
Ia =0 CEO

0 0.2 0.4
tt~~:::0~%"""''"'
0.610.8 VaECV)
F 0.7V

FICi. 3.14 FICi. 3.15

Circuit conditions related to IcEO· Piecewise-linear equivalent for the
diode characteristics of Fig. 3.13b.

cutoff condition should ideally be l c = 0 mA for the chosen VCE voltage. Since l CEO is typi-
cally low in magnitude for silicon materials, cutoff will exist for switching purposes when
ls= 0 µA or le = lCEofor silicon transistors only. For germanium transistors, however,
cutofffor switching purposes will be defined as those conditions that exist when le = lcso-
This condition can normally be obtained for germanium transistors by reverse-biasing the
base-to-emitter junction a few tenths of a volt.
Recall for the common-base configuration that the input set of characteristics was ap-
proximated by a straight-line equivalent that resulted in VsE = 0.7 V for any level of le
greater than 0 mA. For the common-emitter configuration the same approach can be taken,
resulting in the approximate equivalent of Fig. 3.15. The result supports our earlier conclu-
sion that for a transistor in the "on" or active region the base-to-emitter voltage is 0.7 V. In
this case the voltage is fixed for any level of base current.

EXAMPLE3.2
a. Using the characteristics of Fig. 3.13, determine le at ls= 30 µ,A and VCE = 10 V.
b. Using the characteristics of Fig. 3.13, determine le at VsE = 0.7 V and VCE = 15 V.

Solution:
a. At the intersection of ls= 30 µ,A and VCE = 10 V, le= 3.4 mA.
b. Using Fig. 3.13b, we obtain ls= 20 µ,A at the intersection of VsE = 0.7 V and VCE =
15 V (between VCE= 10 V and 20 V). From Fig. 3.13a we find that le= 2.5 mA at the
intersection of ls = 20 µ,A and VCE= 15 V.

Beta (/J)
DC Mode In the de mode the levels of le and ls are related by a quantity called beta and
defined by the following equation:

~ (3.10)
~
where le and ls are determined at a particular operating point on the characteristics. For
practical devices the level of f3 typically ranges from about 50 to over 400, with most in the
midrange. As for a, the parameter f3 reveals the relative magnitude of one current with
respect to the other. For a device with a /3 of 200, the collector current is 200 times the
magnitude of the base current.
 On specification sheets f3ctc is usually included as hFE with the italic letter h derived from COMMON-EMITTER U9
an ac hybrid equivalent circuit to be introduced in Chapter 5. The subscript FE is derived CONFIGURATION
fromforward-current amplification and common-emitter configuration, respectively.

AC Mode For ac situations an ac beta is defined as follows:

f3ac = /1/ Mel (3.11)

B VCE=constant

The formal name for f3ac is common-emitter,forward-current, amplification factor. Since the
collector current is usually the output current for a common-emitter configuration and the base
current is the input current, the term amplification is included in the nomenclature above.
Equation (3.11) is similar in format to the equation for aac in Section 3.4. The procedure
for obtaining aac from the characteristic curves was not described because of the difficulty
of actually measuring changes of/candle on the characteristics. Equation (3.11 ), however,
can be described with some clarity, and, in fact, the result can be used to find aac using an
equation to be derived shortly.
On specification sheets f3ac is normally referred to as hte· Note that the only difference
between the notation used for the de beta, specifically, f3ctc = hFE, is the type of lettering
for each subscript quantity.
The use of Eq. (3.11) is best described by a numerical example using an actual set of
characteristics such as appearing in Fig. 3.13a and repeated in Fig. 3.17. Let us determine
f3ac for a region of the characteristics defined by an operating point of IB = 25 µ,A and VCE
= 7.5 Vas indicated on Fig. 3.16. The restriction of VCE = constant requires that a vertical
line be drawn through the operating point at VCE = 7.5 V. At any location on this vertical
line the voltage VcE is 7.5 V, a constant. The change in /B(/1/B) as appearing in Eq. (3.11)
is then defined by choosing two points on either side of the Q-point along the vertical axis
of about equal distances to either side of the Q-point. For this situation the IB = 20 µ,A and
30 µ,A curves meet the requirement without extending too far from the Q-point. They also

le (mA)
9

8 ~ 90µA
I
,y ~
7

6
l~
J,, -
?1fl
~

~ ~
i-

--- -
_a- ..!:' 80µA
+-

70µA

60µA
;r

~ ?-
LI
_!:l;;t. - ;_i

- -+ - 50µA
I
5

~
.-:-:- H
n - 40µA
......
4 +- .LL!

r'A' IB2-,- 30µA

f
_ _ L. ~

A .----- - :P. -t- -- ~ 25 ~ ,.-l

Q-pt. 20µA
~ j...

2 /Bl
I
I"" /....! I
lOµA
1
I I I
,~ .... w

,. ~ + □ IB=OµA
I
0 5 / 10 15 20 25 VeE (V)
VCE =7.5V

FIG. 3.16
Determining f3ac and f3ctcfrom the collector characteristics.
140 BIPOLAR JUNCTION define levels of In that are easily defined rather than require interpolation of the level of In
TRANSISTORS between the curves. It should be mentioned that the best determination is usually made by
keeping the chosen /:J,./n as small as possible. At the two intersections of/n and the vertical axis,
the two levels of/e can be determined by drawing a horizontal line over to the vertical axis and
reading the resulting values of le. The resulting f3ac for the region can then be determined by

/3ac =
Mel
/:J,./
1e2 - 1e1
/ - /
n VCE~constant n2 n1
3.2mA - 2.2mA 1 mA
30 µ.,A - 20 µ.,A 10 µ.,A
= 100
The solution above reveals that for an ac input at the base, the collector current will be
about 100 times the magnitude of the base current.
If we determine the de beta at the Q-point, we obtain
le 2.7mA
/3ac = - = - - = 108
In 25 µ.,A
Although not exactly equal, the levels of f3ac and /3ac are usually reasonably close and
are often used interchangeably. That is, if f3ac is known, it is assumed to be about the same
magnitude as /3ac, and vice versa. Keep in mind that in the same lot (large number of transis-
tors manufactured at the same time), the value of f3ac will vary somewhat from one transistor
to the next even though each transistor has the same number code. The variation may not
be significant, but for the majority of applications, it is certainly sufficient to validate the
approximate approach above. Generally, the smaller the level of ICEo, the closer are the
magnitudes of the two betas. Since the trend is toward lower and lower levels of ICEo,
the validity of the foregoing approximation is further substantiated.
If the characteristics of a transistor are approximated by those appearing in Fig. 3.17,
the level of f3ac would be the same in every region of the characteristics. Note that the step
in In is fixed at 10 µ.,A and the vertical spacing between curves is the same at every point in
the characteristics-namely, 2 mA. Calculating the f3ac at the Q-point indicated results in
Mel 9mA - 7mA 2mA
f3ac = din VCE~constant = 45 µ.,A - 35 µ.,A = 10 µ.,A = 200

Is= 60 µA

11121-------------------
- 18 = 50 µA
101--------------------
81-----------.... ;. . _________
9 - - - - - - - - - -- -- - -lQ-poinr
ls=40µA

7 - - - - - - - - - - - - - - - ~ I8 = 30 µA
6 i-----------... 1 _ _ _ _ _ _ _ _ _ _ __

4s-1-_________. ._____________
3-
Is= 10 µA
21----------..---------------
1-
I
I
fs=0µAUcw=0µA)
I I I I /
() 5 15

FIG. 3.17
Characteristics in which f3ac is the same everywhere and f3ac = f3ctc·
Determining the de beta at the same Q-point results in
le 8mA
/3a = - = - - = 200
c In 40 µ.,A
revealing that if the characteristics have the appearance of Fig. 3 .17, the magnitudes of f3ac and
f3ac will be the same at every point on the characteristics. In particular, note that ICEo = 0 µ.,A.
 Although a true set of transistor characteristics will never have the exact appearance of COMMON-EMITTER 141
Fig. 3.17, it does provide a set of characteristics for comparison with those obtained from CONFIGURATION
a curve tracer (to be described shortly).
For the analysis to follow, the subscript de or ac will not be included with f3 to avoid
cluttering the expressions with unnecessary labels. For de situations it will simply be rec-
ognized as /3ac and for any ac analysis as f3ac· If a value of /3 is specified for a particular
transistor configuration, it will normally be used for both the de and ac calculations.
A relationship can be developed between f3 and a using the basic relationships in-
troduced thus far. Using /3 = Ie/IB, we have IB = Ie//3, and from a= le/le we have
IE= le/a. Substituting into
le= le+ IB
le le
we have -=[ +-
a e f3
and dividing both sides of the equation by I e results in
1 1
- = 1+-
a /3
or f3 = a/3 + a = (/3 + 1)a

so that
IF p! I I
(3.12)

or (3.13)

In addition, recall that

IeBo
IeEo = -1- -
- a
but using an equivalence of
1
--={3+1
1- a
derived from the above, we find that
IeEO = (/3 + 1)/CBo

or I IeEo ~ f3IeBo I (3.14)

as indicated on Fig. 3.13a. Beta is a particularly important parameter because it provides a

direct link between current levels of the input and output circuits for a common-emitter
configuration. That is,

(3.15)

and since IE= le+ IB

= f3/B + /B
we have I IE = (/3 + 1)/B I (3.16)

Both of the equations above play a major role in the analysis in Chapter 4.

Biasing
The proper biasing of a common-emitter amplifier can be determined in a manner similar
to that introduced for the common-base configuration. Let us assume that we are presented
with an npn transistor such as shown in Fig. 3.18a and asked to apply the proper biasing to
place the device in the active region.
 ,,. fc7
.,.
==>
-
IB
~

~ IE
le
==>

~IE
+
-=-Vee

(a) (b) (c)

FIG. 3.18
Determining the proper biasing arrangement for a common-emitter npn transistor configuration.

The first step is to indicate the direction of IE as established by the arrow in the tran-
sistor symbol as shown in Fig. 3.18b. Next, the other currents are introduced as shown,
keeping in mind Kirchhoff's current law relationship: le + IB = IE. That is, IE is the sum
of le and IB and both le and IB must enter the transistor structure. Finally, the supplies are
introduced with polarities that will support the resulting directions of IB and le as shown
in Fig. 3.18c to complete the picture. The same approach can be applied to pnp transistors.
If the transistor of Fig. 3.18 was a pnp transistor, all the currents and polarities of Fig.
3.18c would be reversed.

Breakdown Region
As with the common-base configuration, there is a maximum collector-emitter voltage that
can be applied and still remain in the active stable region of operation. In Fig. 3.19 the
characteristics of Fig. 3.8 have been extended to demonstrate the impact on the character-
istics at high levels of VCE· At high levels of base current the currents almost climb verti-
cally, whereas at lower levels a region develops that seems to back up on itself. This region
is particularly noteworthy because an increase in current is resulting in a drop in voltage-
totally different from that of any resistive element where an increase in current results in
an increase in potential drop across the resistor. Regions of this nature are said to have a

le (mA)

0 5 10 15 20
BVCEo

FIG. 3.19
Examining the breakdown region of a transistor in the common-emitter
configuration.
142
negative-resistance characteristic. Although the concept of a negative resistance may COMMON-COLLECTOR 143
seem strange at this point, this text will introduce devices and systems that rely on this type CONFIGURATION
of characteristic to perform their desired task.
The recommended maximum value for a transistor under normal operating conditions is
labeledBVcEo as shown in Fig. 3.19 or V(BR)CEO as shown in Fig. 3.23. ltis less thanBVcso
and in fact, is often half the value of BVCBO· For this breakdown region there are two reasons
for the dramatic change in the curves. One is the avalanche breakdown mentioned for the
common-base configuration, whereas the other, called punch-through, is due to the Early
Effect, to be introduced in Chapter 5. In total the avalanche effect is dominant because any
increase in base current due to the breakdown phenomena will be increase the resulting
collector current by a factor beta. This increase in collector current will then contribute to
the ionization (generation of free carriers) process during breakdown, which will cause a
further increase in base current and even higher levels of collector current.

3.6 COMMON-COLLECTOR CONFIGURATION

The third and final transistor configuration is the common-collector configuration, shown
in Fig. 3.20 with the proper current directions and voltage notation. The common-collector
configuration is used primarily for impedance-matching purposes since it has a high input
•
impedance and low output impedance, opposite to that of the common-base and common-
emitter configurations.

/Ei E /Et E

1
p n
,.,_

1
IB + IB

B
-=- VEE ~

B
-;;;;-VEE
p + n +
VBB-;;;;- VBB-=-
+ C fc C tic

-=- -=-
,.,_
IE IE
~
E E

JB JB
~ ~
B B

lei
let

C C

(a) (b)

FIG. 3.20
Notation and symbols used with the common-collector configuration: (a) pnp transistor;
(b) npn transistor.

A common-collector circuit configuration is provided in Fig. 3.21 with the load resistor
connected from emitter to ground. Note that the collector is tied to ground even though the
transistor is connected in a manner similar to the common-emitter configuration. From a
design viewpoint, there is no need for a set of common-collector characteristics to choose
the parameters of the circuit of Fig. 3.21. It can be designed using the common-emitter
characteristics of Section 3.5. For all practical purposes, the output characteristics of the
common-collector configuration are the same as for the common-emitter configuration. For
the common-collector configuration the output characteristics are a plot of IE versus V CE -=-
for a range of values of 18 . The input current, therefore, is the same for both the common- FIG. 3.21
emitter and common-collector characteristics. The horizontal voltage axis for the common- Common-collector configuration
collector configuration is obtained by simply changing the sign of the collector-to-emitter used for impedance-matching
voltage of the common-emitter characteristics. Finally, there is an almost unnoticeable purposes.
144 BIPOLAR JUNCTION change in the vertical scale of le of the common-emitter characteristics if le is replaced
TRANSISTORS by IE for the common-collector characteristics (since a == 1). For the input circuit of the
common-collector configuration the common-emitter base characteristics are sufficient for
obtaining the required information.

3.7 LIMITS OF OPERATION

•
For each transistor there is a region of operation on the characteristics that will ensure that
the maximum ratings are not being exceeded and the output signal exhibits minimum dis-
tortion. Such a region has been defined for the transistor characteristics of Fig. 3.22. All of
the limits of operation are defined on a typical transistor specification sheet described in
Section 3.8.
Some of the limits of operation are self-explanatory, such as maximum collector
current (normally referred to on the specification sheet as continuous collector current)
and maximum collector-to-emitter voltage (often abbreviated as BVCEO or V(BR)eEo on
the specification sheet). For the transistor of Fig. 3.22, /emax was specified as 50 mA and
BVCEo as 20 V. The vertical line on the characteristics defined as VCE,a, specifies the
minimum VCE that can be applied without falling into the nonlinear region labeled the
saturation region. The level of VCE,., is typically in the neighborhood of the 0.3 V speci-
fied for this transistor.
The maximum dissipation level is defined by the following equation:

(3.17)

Saturation
region

18 = 0 µA

5 10 IS 20 Va; (V)
Cutoff BVrno
region

FIG. 3.22
Defining the linear (undistorted) region of operation for a transistor.

For the device of Fig. 3.22, the collector power dissipation was specified as 300 mW.
The question then arises of how to plot the collector power dissipation curve specified by
the fact that
Pema:x = VCEie = 300mW
or VCEie = 300mW
At lc,,.ax At any point on the characteristics the product of VCE and / c must be equal to TRANSISTOR 145
300 mW. If we choose / c to be the maximum value of 50 mA and substitute into the rela- SPECIFICATION SHEET
tionship above, we obtain
VCEic = 300mW
VcE(50mA) = 300mW
300mW = 6V
VCE = 50mA

At Vamax As a result we find that if le = 50 mA, then VCE = 6 Von the power dissipa-
tion curve as indicated in Fig. 3.22. If we now choose VCE to be its maximum value of
20 V, the level of le is the following:
(20 V)/c = 300 mW
300mW
le= 20V = lSmA
defining a second point on the power curve.

At le = }lc,,,ax If we now choose a level of le in the midrange such as 25 mA and solve

for the resulting level of VCE, we obtain
VcE(25 mA) = 300 mW
300mW
and VCE = - - - = 12V
25mA
as also indicated in Fig. 3.22.
A rough estimate of the actual curve can usually be drawn using the three points defined
above. Of course, the more points one has, the more accurate is the curve, but a rough es-
timate is normally all that is required.
The cutoff region is defined as that region below le = ICEo• This region must also be
avoided if the output signal is to have minimum distortion. On some specification sheets
only IcBo is provided. One must then use the equation IcEo = f3lcBo to establish some
idea of the cutoff level if the characteristic curves are unavailable. Operation in the result-
ing region of Fig. 3.22 will ensure minimum distortion of the output signal and current and
voltage levels that will not damage the device.
If the characteristic curves are unavailable or do not appear on the specification sheet
(as is often the case), one must simply be sure that le, VCE, and their product VCEic fall
into the following range:

ICEo ~ le ~ fem,,.
VCEsat ~ VCE ~ VCEmax (3.18)
VCEic ~ Pc=

For the common-base characteristics the maximum power curve is defined by the following
product of output quantities:

I Pc= = VCBic I (3.19)

3.8 TRANSISTOR SPECIFICATION SHEET

•
Since the specification sheet is the communication link between the manufacturer and
user, it is particularly important that the information provided be recognized and correctly
understood. Although all the parameters have not been introduced, a broad number will
now be familiar. The remaining parameters will be introduced in the chapters that follow.
Reference will then be made to this specification sheet to review the manner in which the
parameter is presented.
The information provided as Fig. 3.23 is provided by the Fairchild Semiconductor
Corporation. The 2N4123 is a general-purpose npn transistor with the casing and terminal
146 BIPOLAR JUNCTION identification appearing in the top-right comer of Fig. 3.23a. Most specification sheets are
TRANSISTORS broken down into maximum ratings, thermal characteristics, and electrical characteristics.
The electrical characteristics are further broken down into "on," "off," and small-signal
characteristics. The "on" and "off" characteristics refer to de limits, whereas the small-
signal characteristics include the parameters of importance to ac operation.
Note in the maximum rating list that VcEmax = VcEo = 30 V with /cmax = 200 mA.
The maximum collector dissipation Pcmax = Pv = 625 mW. The derating factor under
the maximum rating specifies that the maximum rating must be decreased 5 mW for every
1° rise in temperature above 25°C. In the "off" characteristics IcBo is specified as 50 nA

MAXIMUM RATINGS
Rating Symbol 2N4123 Unit
Collector-Emitter Voltage VCEo 30 Vdc FAIRCHILD
Collector-Base Voltage Vrno 40 Vdc SEMICONDUCTOR TM

Emitter-Base Voltage VEBo 5.0 Vdc

Collector Current - Continuous le 200 mAdc
Total Device Dissipation @ TA = 25"C Pn 625 mW
Derate above 25 ·c 5.0 mw·c
Operating and Storage Junction Tj,Ts1g -55 to +150 ·c
Temperature Range

THERMAL CHARACTERISTICS
Characteristic Symbol Max Unit
Thermal Resistance, Junction to Case ·cw
General Purpose
Re,e 83.3
Thermal Resistance, Junction to Ambient 200 'CW
Transistor
Re,A
NPN Silicon
ELECTRICAL CHARACTERISTICS (TA= 25°C unless otherwise noted)
Characteristic Symbol Min Max Unit
OFF CHARACTERISTICS
Collector-Emitter Breakdown Voltage (1) V(BR)CEO 30 Vdc
(le = 1.0 mAdc, IF 0)
Collector-Base Breakdown Voltage V(BR)eBO 40 Vdc
(le = 10 µAde, IE = 0)
Emitter-Base Breakdown Voltage V(BR)EBO 5.0 - Vdc
(IE= 10 µAde, le = 0)
Collector Cutoff Current Imo - 50 nAdc
(Vrn = 20Vdc, IE= 0)
Emitter Cutoff Current IEBO - 50 nAdc
(VBE = 3.0 Vdc, le= 0)
ON CHARACTERISTICS
DC Current Gain(!)
(le= 2.0 mAdc, VCE = l.0Vdc) hFE 50 150 -
(le= 50 mAdc, VCE = l.0Vdc) 25 -
Collector-Emitter Saturation Voltage(!) VCE(sat) - 0.3 Vdc
(le= 50 mAdc, lB = 5.0 mAdc)
Base-Emitter Saturation Voltage(!) VBE(sat) - 0.95 Vdc
(le= 50 mAdc, IB = 5.0 mAdc)
SMALL-SIGNAL CHARACTERISTICS
Current-Gain - Bandwidth Product fT 250 MHz
(le= 10 mAdc, Vrn = 20 Vdc, f = 100 MHz)
Output Capacitance Cobo - 4.0 pF
(Vrn = 5.0Vdc, IE= 0, f = 100MHz)
Input Capacitance Cibo - 8.0 pF
(VBE = 0.5Vdc, le= 0, f= lO0kHz)
Collector-Base Capacitance C,b - 4.0 pF
(IE= 0, Vrn = 5.0 V, f = 100 kHz)
Small-Signal Current Gain hte 50 200 -
(le= 2.0 mAdc, VCE= lOVdc, f = 1.0 kHz)
Current Gain - High Frequency
(le= 10 mAdc, Vrn = 20 Vdc, f = 100 MHz) hte 2.5 - -
(le = 2.0 mAdc, V CE= 10 V, f = 1.0 kHz) 50 200
Noise Figure NF - 6.0 dB
(le= 100 µAde, VCE = 5.0Vdc, Rs= l.0kohm, f = 1.0 kHz)
(1) Pulse Test: Pulse Width= 300 µs. Duty Cycle= 2.0%

(a)

FIG. 3.23
Transistor specification sheet.
 h PARAMETERS
VCE = 10 V,f = 1 kHz, TA = 25°C
Figure 1 - Current Gain Figure 3 - Capacitance
10
300
7.0
200

-
__ . -- --- .... - - G:'
8
0)
C)
5.0
- ...
--
~
~~

--
_,Ciba

....... -
3.0
·I
~ Cobo
/ i--,.... r---. ...
,_ """'-- ...
u 2.0
~""~
50

30
-- - -
1.0
0.1 0.2 0.5 1.0 2.0 5.0 0.1 0.2 0.3 0.5 0.71.0 2.0 3.0 5.0 7.0 10 20 3040
le, Collector current (mA) Reverse bias voltage (V)
(b) (d)

STATIC CHARACTERISTICS
Figure 2 - DC Current Gain
2.0
T1 = +125° C VCE= l V
~
0)
-~ _.,. i--
........,- ...........
.......
1.0 +25°C ........
'"1$a - --
-
-- -
l' -
5 0.7
.
--- -...
-- --
·ca 0.5 -55° C 1""1111
Of)

=0.3
0)
I::
i---
..-- r--.. ..
""i...
'Ill

,.
8 ~
u
Cl 0.2
~
' ,. "\ ~ ~
~
'
,-t::
0.1
0.1 0.2 0.3 0.5 0.7 1.0 2.0 3.0 5.0 7.0 10 20 30 50 70 100 200
le, Collector current (mA)
(c)

AUDIO SMALL SIGNAL CHARACTERISTICS

NOISE FIGURE
(VeE = 5 Vdc, TA = 25°C)
Bandwidth= 1.0 Hz

Figure 5 - Frequency Variations

12
Figure 4 - Switching Times \
I I I I I I
200
'-. 10 \ Source resistance = 200 Q
' / le=lmA

-- " ' -
ts
I I I . I
100 ... \.
~
Source resistance = 200 Q
8
70
'' ' ..... 0) \.. '\ / le=0.5mA

''-' ... "- !3Of) I I I II

50 '\. / '\.
....,
~
/ Source resistance = 1 k Q
! ..... ..::: 6
r-,, I' .,,v

'-
/ td 0) le =50µA
8 30
.... ~ r.... ~I"' !/ / t r ,~ V j ....
~ .... ~
R
. ,- . .
v
I' ""
----
20 4 ,_ ~
1, :-- V ""io...
...
Vee=3V .... ~ /
tf
....
.......
.,,,.,,, ~
~

\
""r-- ~

10.0 lellB=lO
"""" "'"' ~
2
_ Source resistance = 500 Q
7.0 VEB (off) = 0.5 V
I
le= 100 µAj
5.0 0
1.0 2.0 3.0 5.0 20 30 50 100 200 0.1 0.2 0.4 2 4 10 20 40 100
le, Collector current (mA) f, Frequency (kHz)

(e) (t)

FIG. 3.23
Continued.
 Figure 6 - Source Resistance
14
I
,
- J= 1 kHz ' '
/ / / /
J
Figure 7 - Input Impedance
12
I/ I / ,/ 20 .........
le= 1 mA / / .........
~10 '
~ le =0.5 mA 1,,
~/
/
V ,
J
/
/
~
10
--
"!3 8
/ 1,r"-----
5.0
" / / ..........
JJ
bl)
..:::
._,,, V "-Y
....'
/ / le=50µA
j" 6 I",..
.,. . v / "-- /
' - -........ _,__
2.0 '-
--.. ""~
~ 4 .... / i.,-
/ " lOOµA
[ 1.0
'
le ..s
2 .l o.5

0 0.2
0.1 0.2 0.4 1.0 2.0 4.0 10 20 40 100 0.1 0.2 0.5 1.0 2.0 5.0 10
Rs , Source resistance (k Q) le, Collector current (mA)

(g) (h)

Figure 8 - Voltage Feedback Ratio Figure 9 - Output Admittance

10 100

{- 7.0
''
~
,J

.g 5.0 '-
'- L/ " '
e /
_.,. 3.0
,
g
] 2.0 ' I' ""i-,.. )I
~

~
~ ~.,
j ....... i..,.,
~~

0 1.0
>
,..,,~ 0.7
0.5 1.0
0.1 0.2 0.5 1.0 2.0 5.0 10 0.1 0.2 0.5 1.0 2.0 5.0 10
le, Collector current (mA) le, Collector current (mA)

(i) (i)

FIG. 3.23
Continued.

and in the "on" characteristics VCE,., = 0.3 V. The level of hFE has a range of 50 to 150 at
le = 2 mA and VCE = 1 V and a minimum value of 25 at a higher current of 50 mA at the
same voltage.
The limits of operation have now been defined for the device and are repeated below
in the format of Eq. (3.18) using hFE = 150 (the upper limit) and lcEO == f3lcBo = (150)
(50 nA) = 7.5 µ,A. Certainly, for many applications the 7.5 µ,A= 0.0075 mA can be con-
sidered to be 0 mA on an approximate basis.

Limits of Operation
7.5 µ,A~ le~ 200mA
0.3V ~ VCE ~ 30V
VCElc ~ 650mW

/J Variation
In the small-signal characteristics the level of hfe (/3ac) is provided along with a plot of how
it varies with collector current in Fig. 3.23b. In Fig. 3.23c the effect of temperature and
collector current on the level of hFE (f3dc) is demonstrated. At room temperature (25°C),
note that hFE (f3dc) is a maximum value of 1 in the neighborhood of about 8 mA. As le
increases beyond this level, hFE drops off to one-half the value with le equal to 50 mA. It
also drops to this level if le decreases to the low level of0.15 mA. Since this is a normalized
148
curve, if we have a transistor with f3ctc = hFE = 120 at room temperature (25°C), the TRANSISTOR TESTING 149
maximum value at 8 mA is 120. At le= 50 mA it has dropped to about 0.52 and hfe =
(0.52)120 = 62.4. In other words, normalizing reveals that the actual level of hFE at any
level of le has been divided by the maximum value of hFE at that temperature and
le= 8 mA. Note also that the horizontal scale of Fig. 3.23(c) is a log scale. Log scales are
examined in depth in Chapter 9. You may want to look back at the plots of this section
when you find time to review the first few sections of Chapter 9.

Capacitance Variation The capacitance Ciba and Caba of Fig. 3.23(d) are the input and
output capacitance levels, respectively, for the transistor in the common-base configura-
tion. Their level is such that their impact can be ignored except for relatively high frequen-
cies. Otherwise, they can be approximated by open circuits in any de or ac analysis.

Switching Times Figure 3.23(e) includes the important parameters that define the
response of a transistor to an input that switches from the "off" to "on" state or vice versa.
Each parameter will be discussed in detail in Section 4.15.

Noise Figures Versus Frequency and Source Resistance The noise figure is a measure of
the additional disturbance that is added to the desired signal response of an amplifier. In
Fig. 3.23(f) the dB level of the noise figure is displayed for a wide frequency response at
particular levels of source resistance. The lowest levels occur at the highest frequencies for
the variety of collector currents and source resistance. As the frequency drops the noise
figure increases with a strong sensitivity to the collector current.
In Fig. 3.23(g) the noise figure is plotted for various levels of source resistance and
collector current. For each current level the higher the source resistance, the higher the
noise figure.

Hybrid Parameters Figures 3.23(b), (h), (i), and (j) provide the components of a hybrid
equivalent model for the transistor that will be discussed in detail in Chapter 5. In each case,
note that the variation is plotted against the collector current-a defining level for the equiv-
alent network. For most applications the most important parameters are hfe and hie· The
higher the collector current, the higher the magnitude of hfe and the lower the level of hie· As
indicated above, all the parameters will be discussed in detail in Sections 5.19-5.21.
Before leaving this description of the characteristics, note that the actual collector char-
acteristics are not provided. In fact, most specification sheets provided by manufacturers
fail to provide the full characteristics. It is expected that the data provided are sufficient to
use the device effectively in the design process.

3.9 TRANSISTOR TESTING

•
As with diodes, there are three routes one can take to check a transistor: use of a curve
tracer, a digital meter, and an ohmmeter.

Curve Tracer
The curve tracer of Fig. 1.43 will provide the display of Fig. 3.24 once all the controls have
been properly set. The smaller displays to the right reveal the scaling to be applied to the
characteristics. The vertical sensitivity is 2 mA/div, resulting in the scale shown to the left
of the monitor's display. The horizontal sensitivity is 1 V/div, resulting in the scale shown
below the characteristics. The step function reveals that the curves are separated by a dif-
ference of 10 µ,A, starting at O µ,A for the bottom curve. The last scale factor provided can
be used to quickly determine the f3ac for any region of the characteristics. Simply multiply
the displayed factor by the number of di visions between / B curves in the region of interest.
For instance, let us determine f3ac at a Q-point of le= 7 mA and VCE = 5 V. In this region
of the display, the distance between In curves is frj of a division, as indicated on Fig. 3.25.
Using the factor specified, we find that

/3 ac =_2__wv(
10
200
div )=1so
150 BIPOLAR JUNCTION 20mA ·'
TRANSISTORS
I8mA Vertical
I
80 µA per div
2mA
I6mA
.... 70µA
I4mA I .... Horizontal

I2mA , / 60 uA

50 µA
per div
IV

lOmA ,__,
401µA
8mA Per Step
r I
30 µA 10 µA
6mA
20 µA
4mA /3 or gm
I
10 µA per div
2mA 200
I
0µA
0mA j

0V IV 2V 3V 4V SV 6V 7V 8V 9V lOV

FIG. 3.24
Curve tracer response to 2N3904 npn transistor.

HOU>

.3 E, E, - /
le= 8mA
.L
I82 =40µA

<
=mdiv / Q-point
Uc=1 mA, VcE=5V)
[BJ= 30 µA

FIG. 3.25
Determining f3acfor the transistor characteristics of Fig. 3.24 at le = 7 mA
andVcE= 5V.

(a)
Using Eq. (3.11) gives

f3ac = Mel
/j.J
8.2 mA - 6.4 mA
40 JLA - 30 JLA
B VCE=constant

Transistor = l.SmA = 180

JFET lOJLA
SCR verifying the determination above.

s,oA TRAHSISTOA TISTIII

Transistor Testers
There is a variety of transistor testers available. Some are simply part of a digital meter as
shown in Fig. 3.26a that can measure a variety of levels in a network. Others, such as that
in Fig. 3.26, are dedicated to testing a limited number of elements. The meter of Fig. 3.26b
can be used to test transistors, JFETs (Chapter 6), and SCRs (Chapter 17) in and out of the
•••
El( PRECISION ~- • .~ circuit. In all cases the power must first be turned off to the circuit in which the element
appears to ensure that the internal battery of the tester is not damaged and to provide a cor-
(b)
rect reading. Once a transistor is connected, the switch can be moved through all the pos-
FIG. 3.26 sible combinations until the test light comes on and identifies the terminals of the transistor.
Transistor testers: (a) digital meter; The tester will also indicate an OK if the npn or pnp transistor is operating properly.
(b) dedicated tester. (Courtesy of Any meter with a diode-checking capability can also be used to check the status of a
B+K Precision Corporation.) transistor. With the collector open the base-to-emitter junction should result in a low voltage
of about 0.7 V with the red (positive) lead connected to the base and the black (negative) TRANSISTOR CASING 151
lead connected to the emitter. A reversal of the leads should result in an OL indication to AND TERMINAL
IDENTIFICATION
represent the reverse-biased junction. Similarly, with the emitter open, the forward- and
reverse-bias states of the base-to-collector junction can be checked.
LowR

Ohmmeter
An ohmmeter or the resistance scales of a digital multimeter (DMM) can be used to check
the state of a transistor. Recall that for a transistor in the active region the base-to-emitter
junction is forward-biased and the base-to-collector junction is reverse-biased. Essentially,
therefore, the forward-biased junction should register a relatively low resistance, whereas
the reverse-biased junction shows a much higher resistance. For an npn transistor, the FICi. 3.27
Checking the forward-biased base-to-
forward-biased junction (biased by the internal supply in the resistance mode) from base to
emitter junction of an npn transistor.
emitter should be checked as shown in Fig. 3.27 and result in a reading that will typically
fall in the range of 100 fl to a few kilohms. The reverse-biased base-to-collector junction
HighR
(again reverse-biased by the internal supply) should be checked as shown in Fig. 3.28 with
a reading typically exceeding 100 kfl. For a pnp transistor the leads are reversed for each D
junction. Obviously, a large or small resistance in both directions (reversing the leads) for C
either junction of an npn or pnp transistor indicates a faulty device.
If both junctions of a transistor result in the expected readings, the type of transistor can
also be determined by simply noting the polarity of the leads as applied to the base-emitter
junction. If the positive (+) lead is connected to the base and the negative lead (-) to the
emitter, a low resistance reading would indicate an npn transistor. A high resistance reading E
would indicate a pnp transistor. Although an ohmmeter can also be used to determine the FICi. 3.28
leads (base, collector, and emitter) of a transistor, it is assumed that this determination can Checking the reverse-biased
be made by simply looking at the orientation of the leads on the casing. base-to-collector junction of an npn
transistor.

3.10 TRANSISTOR CASING AND

TERMINAL IDENTIFICATION
•
After the transistor has been manufactured using one of the techniques described in Appen-
dix A, leads of, typically, gold, aluminum, or nickel are then attached and the entire struc-
ture is encapsulated in a container such as that shown in Fig. 3.29. Those with the
heavy-duty construction are high-power devices, whereas those with the small can (top
hat) or plastic body are low- to medium-power devices.

(a) (b) (c)

FICi. 3.29
Various types of general-purpose or switching transistors: (a) low power; (b) medium power;
(c) medium to high power.

Whenever possible, the transistor casing will have some marking to indicate which leads
are connected to the emitter, collector, or base of a transistor. A few of the methods com-
monly used are indicated in Fig. 3.30.
The internal construction of a TO-92 package in the Fairchild line appears in Fig. 3.31.
Note the very small size of the actual semiconductor device. There are gold bond wires, a
copper frame, and an epoxy encapsulation.
152 BIPOLAR JUNCTION EBC
White
TRANSISTORS dot

~
~
C(case)

B
E
C

C
I EB C
C

FIG. 3.30
Transistor terminal identification.

Axial molding
compound injection

Epoxy package
Copper frame
Locking tabs

(a) (b) (c)

FIG. 3.31
Internal construction of a Fairchild transistor in a TO-92 package.

Four (quad) individual pnp silicon transistors can be housed in the 14-pin plastic dual-in-
line package appearing in Fig. 3.32a. The internal pin connections appear in Fig. 3.32b. As
with the diode IC package, the indentation in the top surface reveals the number 1 and 14 pins.

(Top View)
C B C

C B E NC
NC - No internal connection
(a) (b)
FIG. 3.32
Type Q2T2905 Texas Instruments quad pnp silicon transistor:
(a) appearance; (b) pin connections.

3.11 TRANSISTOR DEVELOPMENT

•
As mentioned in Section 1.1, Moore's law predicts that the transistor count of an inte-
grated circuit will double every 2 years. First presented in a paper by Gordon E. Moore in
1965, the prediction has had an amazing accuracy level. A plot of the transistor count ver-
sus years appearing in Fig. 3.33 is almost linear through the years. The amazing number of
two billion transistors in a single integrated circuit using 45 nm lines is really beyond com-
prehension. A 1 in. line contains more than 564,000 of the 45 nm lines of construction used
in ICs today. Try to draw 100 lines in a 1 in. width using a pencil-almost impossible. The
relative dimensions of drawing 45 nm lines in a 1 in. width would be like drawing a line
 Transistor count TRANSISTOR 153
DEVELOPMENT
10,000,000,000
2 billion level
1,000,000,000

100 million level

100,000,000

10,000,000

1 million level
1,000,000

l
Log scale
100,000

10,000
10,000 level

1000

I
I
I
10 I
1Moore's paper presented
I

1 '----'------'------'-----'------'------'-----
1960 1965 1970 1980 1990 2000 2010 Year
Linear scale _ _ _.,..

FIG. 3.33
Transistor IC count versus time for the period I 960 to the present.

with a width of 1 in. across a highway that is almost 9 miles long.* Although there is con-
tinuing talk that Moore's law will eventually suffer from density, performance, reliability,
and budget comers, the general consensus of the industrial community is that Moore's law
will continue to be applicable for the next decade or two. Although silicon continues to be
the leading fabrication material, there is a family of semiconductors referred to as III V
compound semiconductors (the three and five referring to the number of valence elec-
trons in each element) that are making important inroads into future development. One in
particular is indium gallium arsenide, or InGaAs, which has improved transport character-
istics. Others include GaAIAs, AIGaN, and AllnN, which are all being developed for
increased speed, reliability, stability, reduced size, and improved fabrication techniques.
Currently the Intel® Core™ i7 Quad Core processor has over 730 million transistors
with a clock speed of 3.33 GHz in a package slightly larger than a 1.6" square. Recent
developments by Intel include their Tukwila processor that will house over two billion
transistors. Interestingly enough, Intel continues to employ silicon in its research develop-
ment of transistors that will be 30% smaller and 25% faster than today's fastest transistors
using 20 nm technology. IBM, in concert with the Georgia Institute of Technology, has
developed a silicon-germanium transistor that can operate at frequencies exceeding 500
GHz-an enormous increase over current standards.
Innovation continues to be the backbone of this ever-developing field, with one Swedish
team introducing ajunctionless transistor primarily to simplify the manufacturing process.
Another has introduced carbon nanotubes (a carbon molecule in the form of a hollow
cylinder that has a diameter about 1/50,000 the width of a human hair) as a path toward
faster, smaller, and cheaper transistors. Hewlett Packard is developing a Crossbar Latch
transistor that employs a grid of parallel conducting and signal wires to create junctions
that act as switches.
The question was often asked many years ago: Where can the field go from here? Obvi-
ously, based on what we see today, there seems to be no limit to the innovative spirit of
individuals in the field as they search for new directions of investigation.

*In metric units, it would be like drawing more than 220,000 lines in a I-cm length or a I-cm width line across
a highway over 2.2 km long.
154 BIPOLAR JUNCTION
TRANSISTORS
3.12 SUMMARY
Important Conclusions and Concepts •
1. Semiconductor devices have the following advantages over vacuum tubes: They are
(1) of smaller size, (2) more lightweight, (3) more rugged, and (4) more efficient. In
addition, they have (1) no warm-up period, (2) no heater requirement, and (3) lower
operating voltages.
2. Transistors are three-terminal devices of three semiconductor layers having a base or
center layer a great deal thinner than the other two layers. The outer two layers are
both of either n- or p-type materials, with the sandwiched layer the opposite type.
3. One p-n junction of a transistor is forward-biased, whereas the other is reverse-
biased.
4. The de emitter current is always the largest current of a transistor, whereas the base
current is always the smallest. The emitter current is always the sum of the other two.
5. The collector current is made up of two components: the majority component and
the minority current (also called the leakage current).
6. The arrow in the transistor symbol defines the direction of conventional current flow
for the emitter current and thereby defines the direction for the other currents of the
device.
7. A three-terminal device needs two sets of characteristics to completely define its
characteristics.
8. In the active region of a transistor, the base-emitter junction is forward-biased,
whereas the collector-base junction is reverse-biased.
9. In the cutoff region the base-emitter and collector-base junctions of a transistor are
both reverse-biased.
10. In the saturation region the base-emitter and collector-base junctions are forward-
biased.
11. On an average basis, as a first approximation, the base-to-emitter voltage of an operat-
ing transistor can be assumed to be 0.7 V.
12. The quantity alpha (a) relates the collector and emitter currents and is always close to
one.
13. The impedance between terminals of a forward-biased junction is always relatively
small, whereas the impedance between terminals of a reverse-biased junction is usu-
ally quite large.
14. The arrow in the symbol of an npn transistor points out of the device (not pointing
in), whereas the arrow points in to the center of the symbol for a pnp transistor
(pointing in).
15. For linear amplification purposes, cutoff for the common-emitter configuration will
be defined by le = leEO•
16. The quantity beta (/3) provides an important relationship between the base and collec-
tor currents, and is usually between 50 and 400.
17. The de beta is defined by a simple ratio of de currents at an operating point,
whereas the ac beta is sensitive to the characteristics in the region of interest. For
most applications, however, the two are considered equivalent as a first approximation.
18. To ensure that a transistor is operating within its maximum power level rating, simply
find the product of the collector-to-emitter voltage and the collector current, and
compare it to the rated value.

Equations
IE= le+ In, le = 1 emajority + 1eommonty' VnE ~ 0.7V

O'.ctc = -,
le
le
O'.ac
Mel
= /:,./ , IeEo -I
leno
=-
1 - a ls~o,,A
E VCB=constant

= Mel
le
a=---
/3
f3ctc = In' f3ac !:,./ ,
/3 + 1
B VcE=constant

le= f3ln, le= (/3 + 1)/n, Pemax = VeEie

3.13 COMPUTER ANALYSIS
Cadenc:e OrCAD •
Since the transistor characteristics were introduced in this chapter, it seems appropriate
COMPUTER ANALYSIS 155

that a procedure for obtaining those characteristics using PSpice Windows should be exam-
ined. The transistors are listed in the EVAL library and start with the letter Q. The library
includes two npn transistors, two pnp transistors, and two Darlington configurations. The
fact that there is a series of curves defined by the levels of/n will require that a sweep of/n
values (a nested sweep) occur within a sweep of collector-to-emitter voltages. This is
unnecessary for the diode, however, since only one curve would result.
First, the network in Fig. 3.34 is established using the same procedure as defined in
Chapter 2. The voltage Vee will establish our main sweep, whereas the voltage Vnn will
determine the nested sweep. For future reference, note the panel at the top right of the menu
bar with the scroll control when building networks. This option allows you to retrieve ele-
ments that have been used in the past. For instance, if you placed a resistor a few elements
ago, simply return to the scroll bar and scroll until the resistor R appears. Click the location
once, and the resistor will appear on the screen.

file _!;dit Y; ew Iools !'.lace _Macro P,Spice

l\ccessories Qptions Window !:ielp
cadence
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FIG. 3.34
Network employed to obtain the collector
characteristics of the Q2N2222 transistor.

Once the network is established as appearing in Fig. 3.34, select the New Simulation
Profile key and insert OrCAD 3-1 as the Name. Then select Create to obtain the Simula-
tion Settings dialog box. The Analysis type will be DC Sweep, with the Sweep variable
being a Voltage Source. Insert VCC as the name for the swept voltage source and select
Linear for the sweep. The Start value is OV, the End value 10 V, and the Increment 0.01 V.
It is important not to select x in the top right corner of the box to leave the settings
control. We must first enter the nested sweep variable by selecting Secondary Sweep and
inserting VBB as the voltage source to be swept. Again, it will be a Linear sweep, but now
the starting value will be 2. 7 V to correspond with an initial current of 20 µ,A as determined by
Vnn - VnE 2.7V - 0.7V
In= Rn lOOkO - 20µ,A
The End value is 10.7 V to correspond with a current of 100 µ,A. The Increment is set
at 2 V, corresponding to a change in base current of 20 µ,A. Both sweeps are now set, but
before leaving the dialog box be sure both sweeps are enabled by a check in the box
next to each sweep. Often after entering the second sweep, the user fails to establish the
second sweep before leaving the dialog box. Once both are selected, leave the dialog box
and select Run PSpice. The result will be a graph with a voltage VCC varying from O V
156 BIPOLAR JUNCTION fl SCHEMATIC1-OrCAD3-1 plot 2 - PSp1ce ND Demo - [OrCAD3-1 plot 2 (active)]
TRANSISTORS
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FIG. 3.35
Collector characteristics for the transistor of Fig. 3.34.

to 10 V. To establish the various l curves, apply the sequence Trace-Add Trace to obtain
the Add Trace dialog box. Select IC(Ql), the collector current of the transistor for the
vertical axis. An OK, and the characteristics will appear. Unfortunately, however, they
extend from -10 mA to +20 mA on the vertical axis. This can be corrected by the sequence
Plot-Axis Settings, which again will result in the Axis Settings dialog box. Select Y-Axis
and under Data Range choose User Defined and set the range as 0-20 mA. An OK, and
the plot of Fig. 3.35 will appear. Labels on the plot can be added using the production ver-
sion of OrCAD.
The first curve at the bottom of Fig. 3.35 represents lB = 20 µ,A. The curve above is lB =
40 µ,A, the next 60 µ,A, and so on. If we choose a point in the middle of the characteristics
defined by VeE = 4 V and lB = 60 µ,A as shown in Fig. 3.35 f3 can be determined from
le 11 mA
f3 = - = -- = 183.3
lB 60 µ,A
Like the diode, the other parameters of the device will have a noticeable effect on the oper-
ating conditions. If we return to the transistor specifications using Edit-PSpice Model to
obtain the PSpice Model Editor Demo dialog box, we can delete all the parameters except
the Bf value. Be sure to leave the parentheses surrounding the value of Bf during the dele-
tion process. When you exit the box the Model Editor/16.3 dialog box will appear asking
you to save changes. It was saved as OrCAD 3-1 and the circuit was simulated again to
obtain the characteristics of Fig. 3.36 following another adjustment of the range of the
vertical axis.
Note first that the curves are all horizontal, meaning the element is void of any resistive
characteristics. In addition, the equal spacing of the curves throughout reveals that beta is the
same everywhere. At the intersection of VCE = 4 V and lB = 60 µ,A, the new value of f3 is
le 14.6mA
/3 = - = - - - = 243.3
lB 60 µ,A
The real value of the above analysis is to recognize that even though beta may be provided,
the actual performance of the device will be very dependent on its other parameters.
Assume an ideal device is always a good starting point, but an actual network provides a
different set ofresults.
~ SCI-IEMATICl-OrC/\D 3-1 - PSp,ce ND Demo - [OrC/\0 3-1 (active)]

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FIG. 3.36
Ideal collector characteristics for the transistor of Fig. 3.34.

PROBLEMS
*Note: Asterisks indicate more difficult problems.
3.2 Transistor Construdion
•
1. What names are applied to the two types of BJT transistors? Sketch the basic construction of
each and label the various minority and majority carriers in each. Draw the graphic symbol next
to each. Is any of this information altered by changing from a silicon to a germanium base?
2. What is the major difference between a bipolar and a unipolar device?

3.3 Transistor Operation

3. How must the two transistor junctions be biased for proper transistor amplifier operation?
4. What is the source of the leakage current in a transistor?
5. Sketch a figure similar to Fig. 3.4a for the forward-biased junction of an npn transistor.
Describe the resulting carrier motion.
6. Sketch a figure similar to Fig. 3 .4b for the reverse-biased junction of an npn transistor. Describe
the resulting carrier motion.
7. Sketch a figure similar to Fig. 3.5 for the majority- and minority-carrier flow of an npn transis-
tor. Describe the resulting carrier motion.
8. Which of the transistor currents is always the largest? Which is always the smallest? Which
two currents are relatively close in magnitude?
9. If the emitter current ofa transistor is 8 mA and/8 is 1/100 of le, determine the levels of le and I 8.

3.4 Common-Base Configuration

10. From memory, sketch the transistor symbol for a pnp and an npn transistor, and then insert the
conventional flow direction for each current.
11. Using the characteristics of Fig. 3.7, determine V8 e atle = 5 mA for Vcs = 1, 10, and 20 V. Is
it reasonable to assume on an approximate basis that Vcs has only a slight effect on the rela-
tionship between V8 e and fe?
BIPOLAR JUNCTION 12. a. Determine the average ac resistance for the characteristics of Fig. 3.10b.
TRANSISTORS b. For networks in which the magnitude of the resistive elements is typically in kilohms, is the
approximation of Fig. 3.10c a valid one [based on the results of part (a)]?
13. a. Using the characteristics of Fig. 3.8, determine the resulting collector current if le= 3.5 mA
and VCB = lOV.
b. Repeat part (a) for le = 3.5 mA and VCB = 20 V.
c. How have the changes in V CB affected the resulting level of l c?
d. On an approximate basis, how are le and le related based on the results above?
14. a. Using the characteristics of Figs. 3.7 and 3.8, determine le if Vc8 = 5 V and V8 e = 0.7 V.
b. Determine V8 eif le= 5 mA and VCB = 15 V.
c. Repeat part (b) using the characteristics of Fig. 3.10b.
d. Repeat part (b) using the characteristics of Fig. 3.10c.
e. Compare the solutions for VBE for parts (b) through (d). Can the difference be ignored if
voltage levels greater than a few volts are typically encountered?
15. a. Given an adc of 0.998, determine le if le = 4 mA.
b. Determine adc if le= 2.8 mA, le= 2.75 mA and lcBO = 0.1 µA.
16. From memory only, sketch the common-base BIT transistor configuration (for npn and pnp)
and indicate the polarity of the applied bias and resulting current directions.

3.5 Common-Emitter Configuration

17. Define lcBO and lcEO• How are they different? How are they related? Are they typically close
in magnitude?
18. Using the characteristics of Fig. 3.13:
a. Find the value of le corresponding to V8 e = + 750 m V and VCE = +4 V.
b. Find the value of VCE and V8 e corresponding to le = 3.5 mA and 18 = 30 JLA.
*19. a. For the common-emitter characteristics of Fig. 3.13, find the de beta at an operating point
of Vee= 6V and le= 2mA.
b. Find the value of a corresponding to this operating point.
c. At Vee= +6 V, find the corresponding value of lcEO-
d. Calculate the approximate value of lcBO using the de beta value obtained in part (a).
*20. a. Using the characteristics of Fig. 3.13a, determine lceo at VCE= 10 V.
b. Determine f3dc at l B = 10 f-LA and VCE = 10 V.
c. Using the f3dc determined in part (b), calculate lCBo•
21. a. Using the characteristics of Fig. 3.13a, determine f3dc at/8 = 60 f-LA and Vee= 4 V.
b. Repeat part (a) at/8 = 30 f-LA and VCE = 7 V.
c. Repeat part (a) at 18 = 10 f-LA and VCE= 10 V.
d. Reviewing the results of parts (a) through (c), does the value of f3dc change from point to point
on the characteristics? Where were the higher values found? Can you develop any general con-
clusions about the value of f3dc on a set of characteristics such as those provided in Fig. 3.13a?
*22. a. Using the characteristics of Fig. 3.13a, determine f3ac at/8 = 60 f-LA and Vee= 4 V.
b. Repeat part (a) at/8 = 30 f-LA and VCE = 7 V.
c. Repeat part (a) at 18 = 10 f-LA and VCE= 10 V.
d. Reviewing the results of parts (a) through (c), does the value of f3ac change from point to
point on the characteristics? Where are the high values located? Can you develop any gen-
eral conclusions about the value of f3ac on a set of collector characteristics?
e. The chosen points in this exercise are the same as those employed in Problem 21. If Prob-
lem 21 was performed, compare the levels of f3dc and f3ac for each point and comment on
the trend in magnitude for each quantity.
23. Using the characteristics of Fig. 3.13a, determine f3dc at 18 = 25 f-LA and Vee= 10 V. Then
calculate adc and the resulting level of le- (Use the level of le determined by le = f3dcl8 .)
24. a. Given that adc = 0.980, determine the corresponding value of f3dc·
b. Given f3dc = 120, determine the corresponding value of a.
c. Given that f3dc = 120 and le = 2.0 mA, find le and 18 .
25. From memory only, sketch the common-emitter configuration (for npn and pnp) and insert the
proper biasing arrangement with the resulting current directions for 18 , le, and le-

3.6 Common-Colledor Configuration

26. An input voltage of 2 V rms (measured from base to ground) is applied to the circuit of Fig. 3.21.
Assuming that the emitter voltage follows the base voltage exactly and that Vb, (rms) = 0.1 V,
calculate the circuit voltage amplification (Av = V0 /V;) and emitter current for Re= 1 kil.
 27. For a transistor having the characteristics of Fig. 3.13, sketch the input and output characteris-
tics of the common-collector configuration.

3.7 Limitsof Operation

28. Determine the region of operation for a transistor having the characteristics of Fig. 3.13 if
lcnrnx = 6mA,BVcEo = 15V,andPcnrnx = 35mW.
29. Determine the region of operation for a transistor having the characteristics of Fig. 3.8 if
lcnrnx = 7 mA, BVCBo = 20V, andPcnrnx = 42mW.

3.8 Transistor Specification Sheet

30. Referring to Fig. 3.23, determine the temperature range for the device in degrees Fahrenheit.
31. Using the information provided in Fig. 3.23 regarding Pnmax, VCEmax, lcmax and VcEsat' sketch the
boundaries of operation for the device.
32. Based on the data of Fig. 3.23, what is the expected value of lcEo using the average value of f3dc?
33. How does the range of hFE (Fig. 3.23c, normalized from hFE = 100) compare with the range
of hfe (Fig. 3.23b) for the range of le from 0.1 to 10 mA?
34. Using the characteristics of Fig. 3 .23d, determine whether the input capacitance in the common-
base configuration increases or decreases with increasing levels of reverse-bias potential. Can
you explain why?
*35. Using the characteristics of Fig. 3.23b, determine how much the level of hfe has changed from
its value at 1 mA to its value at 10 mA. Note that the vertical scale is a log scale that may require
reference to Section 11.2. Is the change one that should be considered in a design situation?
*36. Using the characteristics of Fig. 3.23c, determine the level of f3dc at le = 10 mA at the three
levels of temperature appearing in the figure. Is the change significant for the specified tem-
perature range? Is it an element to be concerned about in the design process?

3.9 Transistor Testing

37. a. Using the characteristics of Fig. 3.24, determine f3ac atlc = 14 mA and VcE = 3 V.
b. Determine /3dc at l c = 1 mA and VCE = 8 V.
c. Determine f3ac atlc = 14 mA and VcE = 3 V.
d. Determine f3dc at le= 1 mA and VcE = 8 V.
e. How does the level of f3ac and f3dc compare in each region?
f. Is the approximation f3dc ~ f3ac a valid one for this set of characteristics?
CHAPTER OBJECTIVES
•
• Be able to determine the de levels for the variety of important BJT configurations.
• Understand how to measure the important voltage levels of a BJT transistor configura-
tion and use them to determine whether the network is operating properly.
• Become aware of the saturation and cutoff conditions of a BJT network and the
expected voltage and current levels established by each condition.
• Be able to perform a load-line analysis of the most common BJT configurations.
• Become acquainted with the design process for BJT amplifiers.
• Understand the basic operation of transistor switching networks.
• Begin to understand the troubleshooting process as applied to BJT configurations.
• Develop a sense for the stability factors of a BJT configuration and how they affect its
operation due to changes in specific characteristics and environmental changes.

4.1 INTRODUCTION
•
The analysis or design of a transistor amplifier requires a knowledge of both the de and the
ac response of the system. Too often it is assumed that the transistor is a magical device
that can raise the level of the applied ac input without the assistance of an external energy
source. In actuality,
any increase in ac voltage, current, or power is the result of a transfer of energy from
the applied de supplies.
The analysis or design of any electronic amplifier therefore has two components: a de and
an ac portion. Fortunately, the superposition theorem is applicable, and the investigation of
the de conditions can be totally separated from the ac response. However, one must keep in
mind that during the design or synthesis stage the choice of parameters for the required de
levels will affect the ac response, and vice versa.
The de level of operation of a transistor is controlled by a number of factors, includ-
ing the range of possible operating points on the device characteristics. In Section 4.2
we specify the range for the bipolar junction transistor (BJT) amplifier. Once the desired
de current and voltage levels have been defined, a network must be constructed that will
establish the desired operating point. A number of these networks are analyzed in this
chapter. Each design will also determine the stability of the system, that is, how sensitive
the system is to temperature variations, another topic to be investigated in a later section
of this chapter.
 Although a number of networks are analyzed in this chapter, there is an underlying OPERATING POINT 161
similarity in the analysis of each configuration due to the recurring use of the following
important basic relationships for a transistor:

I VnE ~ 0.7V I (4.1)

Ih = (/3 + 1)/n ~ le I (4.2)

I le= f3In I (4.3)

In fact, once the analysis of the first few networks is clearly understood, the path toward
the solution of the networks to follow will begin to become quite apparent. In most instances
the base current In is the first quantity to be determined. Once In is known, the relationships
of Eqs. (4.1) through (4.3) can be applied to find the remaining quantities of interest. The
similarities in analysis will be immediately obvious as we progress through the chapter.
The equations for In are so similar for a number of configurations that one equation can be
derived from another simply by dropping or adding a term or two. The primary function of
this chapter is to develop a level of familiarity with the BIT transistor that would permit a
de analysis of any system that might employ the BJT amplifier.

4.2 OPERATING POINT

The term biasing appearing in the title of this chapter is an all-inclusive term for the appli-
cation of de voltages to establish a fixed level of current and voltage. For transistor ampli-
fiers the resulting de current and voltage establish an operating point on the characteristics
•
that define the region that will be employed for amplification of the applied signal. Because
the operating point is a fixed point on the characteristics, it is also called the quiescent
point (abbreviated Q-point). By definition, quiescent means quiet, still, inactive. Figure 4.1
shows a general output device characteristic with four operating points indicated. The

80µA

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VcEmax

FIC. 4.1
Various operating points within the limits of operation of a transistor.
162 DC BIASING-BJTs biasing circuit can be designed to set the device operation at any of these points or others
within the active region. The maximum ratings are indicated on the characteristics of Fig.
4.1 by a horizontal line for the maximum collector current Icmax and a vertical line at the
maximum collector-to-emitter voltage VCEmax· The maximum power constraint is defined
by the curve P cmax in the same figure. At the lower end of the scales are the cutoff region,
defined by IB =s Oµ,A, and the saturation region, defined by VcE =s VcE,.,·
The BJT device could be biased to operate outside these maximum limits, but the
result of such operation would be either a considerable shortening of the lifetime of
the device or destruction of the device. Confining ourselves to the active region, we
can select many different operating areas or points. The chosen Q-point often depends
on the intended use of the circuit. Still, we can consider some differences among the
various points shown in Fig. 4.1 to present some basic ideas about the operating point
and, thereby, the bias circuit.
If no bias were used, the device would initially be completely off, resulting in a Q-
point at A-namely, zero current through the device (and zero voltage across it). Because
it is necessary to bias a device so that it can respond to the entire range of an input signal,
point A would not be suitable. For point B, if a signal is applied to the circuit, the device
will vary in current and voltage from the operating point, allowing the device to react to
(and possibly amplify) both the positive and negative excursions of the input signal. If
the input signal is properly chosen, the voltage and current of the device will vary, but not
enough to drive the device into cutoff or saturation. Point C would allow some positive
and negative variation of the output signal, but the peak-to-peak value would be limited
by the proximity of VCE = 0 V and le = 0 mA. Operating at point C also raises some
concern about the nonlinearities introduced by the fact that the spacing between IB curves
is rapidly changing in this region. In general, it is preferable to operate where the gain
of the device is fairly constant (or linear) to ensure that the amplification over the entire
swing of input signal is the same. Point B is a region of more linear spacing and therefore
more linear operation, as shown in Fig. 4.1. Point D sets the device operating point near
the maximum voltage and power level. The output voltage swing in the positive direction
is thus limited if the maximum voltage is not to be exceeded. Point B therefore seems the
best operating point in terms of linear gain and largest possible voltage and current swing.
This is usually the desired condition for small-signal amplifiers (Chapter 5) but not the
case necessarily for power amplifiers, which will be considered in Chapter 12. In this
discussion, we will be concentrating primarily on biasing the transistor for small-signal
amplification operation.
One other very important biasing factor must be considered. Having selected and
biased the BJT at a desired operating point, we must also take the effect of temperature
into account. Temperature causes the device parameters such as the transistor current
gain (J3ac) and the transistor leakage current (/CEo) to change. Higher temperatures result
in increased leakage currents in the device, thereby changing the operating condition set
by the biasing network. The result is that the network design must also provide a degree
of temperature stability so that temperature changes result in minimum changes in the
operating point. This maintenance of the operating point can be specified by a stability
factor S, which indicates the degree of change in operating point due to a temperature
variation. A highly stable circuit is desirable, and the stability of a few basic bias circuits
will be compared.
For the BJT to be biased in its linear or active operating region the following must be true:
1. The base-emitter junction must be forward-biased (p-region voltage more positive),
with a resulting forward-bias voltage of about 0.6 V to 0.7 V.
2. The base-collector junction must be reverse-biased (n-region more positive), with
the reverse-bias voltage being any value within the maximum limits of the device.
[Note that for forward bias the voltage across the p-n junction is p-positive, whereas for
reverse bias it is opposite (reverse) with n-positive.]
Operation in the cutoff, saturation, and linear regions of the BJT characteristic are pro-
vided as follows:
1. Linear-region operation:
Base-emitter junction forward-biased
Base-collector junction reverse-biased
2. Cutoff-region operation: FIXED-BIAS 163
Base-emitter junction reverse-biased CONFIGURATION
Base-collector junction reverse-biased
3. Saturation-region operation:
Base-emitter junction forward-biased
Base-collector junction forward-biased

4.3 FIXED-BIAS CONFIGURATION

•
The fixed-bias circuit of Fig. 4.2 is the simplest transistor de bias configuration. Even
though the network employs an npn transistor, the equations and calculations apply equally
well to a pnp transistor configuration merely by changing all current directions and voltage
polarities. The current directions of Fig. 4.2 are the actual current directions, and the volt-
ages are defined by the standard double-subscript notation. For the de analysis the network
can be isolated from the indicated ac levels by replacing the capacitors with an open-circuit
equivalent because the reactance of a capacitor is a function of the applied frequency. For
de, f = 0 Hz, and Xe = ½nfC = ½7T(0)C = 00 fl. In addition, the de supply Vcc can
be separated into two supplies (for analysis purposes only) as shown in Fig. 4.3 to permit a
separation of input and output circuits. It also reduces the linkage between the two to the
base current In. The separation is certainly valid, as we note in Fig. 4.3 that Vee is con-
nected directly to Rn and Re just as in Fig. 4.2.

Vee
Vee Vee

Re
Re ile

L
ac RB

ac B
input o--------) t----------c>-----1
-----------1(----o output
C
+
C
2
signal
L_ C
+
VCE
........-0

signal c1 n+ B+
VBE - E VBE - E

■::"

FIG. 4.2 FIG. 4.3

Fixed-bias circuit. DC equivalent of Fig. 4.2.

Forward Bias of Base-Emitter

+
Consider first the base-emitter circuit loop of Fig. 4.4. Writing Kirchhoff's voltage equa-
Rn
tion in the clockwise direction for the loop, we obtain
+
+Vee - InRn - VnE = 0
Vee ...:::...
Note the polarity of the voltage drop across Rn as established by the indicated direction of +
In. Solving the equation for the current In results in the following: IH ~
\IBE
\
(4.4) ';" ■::"

FIG. 4.4
Equation (4.4) is certainly not a difficult one to remember if one simply keeps in mind Base-emitter loop.
that the base current is the current through Rn and by Ohm's law that current is the voltage
across Rn divided by the resistance Rn. The voltage across Rn is the applied voltage Vcc
at one end less the drop across the base-to-emitter junction (VnE). In addition, because the
supply voltage Vcc and the base-emitter voltage VnE are constants, the selection of a base
resistor Rn sets the level of base current for the operating point.
164 DC BIASING-BJTs Collector-Emitter Loop
The collector-emitter section of the network appears in Fig. 4.5 with the indicated direc-
tion of current le and the resulting polarity across Re. The magnitude of the collector cur-
rent is related directly to IB through

I le= f3IB I (4.5)

It is interesting to note that because the base current is controlled by the level of RB and
le is related to IB by a constant /3, the magnitude of le is not a function of the resistance
Re. Changing Re to any level will not affect the level of IB or le as long as we remain in
the active region of the device. However, as we shall see, the level of Re will determine the
magnitude of VCE, which is an important parameter.
Applying Krrchhoffs voltage law in the clockwise direction around the indicated closed
loop of Fig. 4.5 results in the following:
FIC. 4.5 VcE + IcRc - Vee= 0
Collector-emitter loop.
and I VcE = Vee - IcRc I (4.6)
which states that the voltage across the collector-emitter region of a transistor in the fixed-
bias configuration is the supply voltage less the drop across Re.
As a brief review of single- and double-subscript notation recall that

I VcE = Ve - VE I (4.7)
where VCE is the voltage from collector to emitter and V c and VE are the voltages from col-
lector and emitter to ground, respectively. In this case, since VE= 0 V, we have

I VcE = Ve I (4.8)
In addition, because

(4.9)
and VE = 0 V, then

(4.10)

Keep in mind that voltage levels such as VCE are determined by placing the positive lead
(normally red) of the voltmeter at the collector terminal with the negative lead (normally
black) at the emitter terminal as shown in Fig. 4.6. Vc is the voltage from collector to ground
and is measured as shown in the same figure. In this case the two readings are identical, but
in the networks to follow the two can be quite different. Clearly understanding the differ-
FIC. 4.6 ence between the two measurements can prove to be quite important in the troubleshooting
Measuring VCE and VC· of transistor networks.

EXAMPLE 4.1 Determine the following for the fixed-bias configuration of Fig. 4.7.
a. IBQ and lcQ•
b. VcEQ·
c. VB and Ve.
d. VBc•

Solution:
Vee - VBE 12V - 0.7V
a. Eq. (4.4): JB = - - - - 240 kll = 47.08 µA
Q RB
Eq. (4.5): lcQ = f3IBQ = (50)(47.08 µ,A) = 2.35 mA
 Vcc=+12V FIXED-BIAS 165
CONFIGURATION

C1
Rs
240kQ

G_
J +\ lOµF
0
ac
output

ac ,..________1,___.___ _ _____
input~ Vrn /3 = 50
lOµF
J
FIG. 4.7
DC.fixed-bias circuit for Example 4.1.

b. Eq. (4.6): VcEQ = Vee - IcRc

= 12 V - (2.35 mA)(2.2 kll)
= 6.83 V
C. Vn = VnE = 0.7V
Vc = VCE = 6.83 V
d. Using double-subscript notation yields
Vnc = Vn - Ve= 0.7V - 6.83V
= -6.13V
with the negative sign revealing that the junction is reversed-biased, as it should be for
linear amplification.

Transistor Saturation
The term saturation is applied to any system where levels have reached their maximum values.
A saturated sponge is one that cannot hold another drop of water. For a transistor operating in
the saturation region, the current is a maximum value for the particular design. Change the
design and the corresponding saturation level may rise or drop. Of course, the highest saturation
level is defined by the maximum collector current as provided by the specification sheet.
Saturation conditions are normally avoided because the base-collector junction is no
longer reverse-biased and the output amplified signal will be distorted. An operating point
in the saturation region is depicted in Fig. 4.8a. Note that it is in a region where the char-
acteristic curves join and the collector-to-emitter voltage is at or below VCEsat" In addition,
the collector current is relatively high on the characteristics.

le

1csat - 1csat -

0 I VcEsat 0

(a) (b)

FIG. 4.8
Saturation regions: (a) actual; (b) approximate.
166 DC BIASING-BJTs If we approximate the curves of Fig. 4.8a by those appearing in Fig. 4.8b, a quick, direct
method for determining the saturation level becomes apparent. In Fig. 4.8b, the current is
relatively high, and the voltage VcE is assumed to be O V. Applying Ohm's law, we can
determine the resistance between collector and emitter terminals as follows:
c ~•1 VCE OV
RcE=-=-=0O
RcE =00 le lc,.1
(VcE = O V, le= Ic,.1 ) Applying the results to the network schematic results in the configuration of Fig. 4.9.
E
For the future, therefore, if there were an immediate need to know the approximate
maximum collector current (saturation level) for a particular design, simply insert a short-
circuit equivalent between collector and emitter of the transistor and calculate the resulting
FICi. 4.9 collector current. In short, set VCE = 0 V. For the fixed-bias configuration of Fig. 4.10, the
Determining Ic,.t" short circuit has been applied, causing the voltage across Re to be the applied voltage Vcc-
The resulting saturation current for the fixed-bias configuration is

Vee
I
c,.1 - -
- Re (4.11)

~-------<>---0Vcc

+
VcE=OV

FICi. 4.10
Determining Ic,.Jor the fixed-bias
configuration.

Once lc,.1 is known, we have some idea of the maximum possible collector current for the
chosen design and the level to stay below if we expect linear amplification.

EXAMPLE 4.2 Determine the saturation level for the network of Fig. 4. 7.
Solution:
Vee 12 V
le =-=--=S.45mA
sat Re 2.2k0

The design of Example 4.1 resulted in lcQ = 2.35 mA, which is far from the saturation
level and about one-half the maximum value for the design.

Load-Line Analysis
Recall that the load-line solution for a diode network was found by superimposing the actual
diode characteristics of the diode on a plot of the network equation involving the same network
variables. The intersection of the two plots defined the actual operating conditions for the net-
work. It is referred to as load-line analysis because the load (network resistors) of the network
defined the slope of the straight line connecting the points defined by the network parameters.
The same approach can be applied to BJT networks. The characteristics of the BJT are
superimposed on a plot of the network equation defined by the same axis parameters. The
load resistor Re for the fixed-bias configuration will define the slope of the network equa-
tion and the resulting intersection between the two plots. The smaller the load resistance, the
 le (mA)

s~ 50µA_

7 ~t 40µA

6
r 30µA
5
'r
.-------o---oVcc 4
+ i le
3
20µA

Re T
lOµA
2 ~
+
1 -
r
I I I
0 5
t 15 VcE (V)
lCEo

(a) (b)

FIG. 4.11
Load-line analysis: (a) the network; (b) the device characteristics.

steeper the slope of the network load line. The network of Fig. 4.1 la establishes an output
equation that relates the variables / c and VCE in the following manner:

(4.12)
The output characteristics of the transistor also relate the same two variables / c and VCE as
shown in Fig. 4.llb.
The device characteristics of le versus VcE are provided in Fig. 4.llb. We must now
superimpose the straight line defined by Eq. (4.12) on the characteristics. The most direct
method of plotting Eq. (4.12) on the output characteristics is to use the fact that a straight line
is defined by two points. If we choose I c to be OmA, we are specifying the horizontal axis as
the line on which one point is located. By substituting le= OmAinto Eq. (4.12), we find that
VcE = Vee - (O)Rc

and I VcE = Vccl1c=OmA I (4.13)

defining one point for the straight line as shown in Fig. 4.12.

/
VCE=ov-

\ 1- -~
-+---------------__:,---]►
0 ~ Vee VCE

FIG. 4.12
Fixed-bias load line. 167
168 DC BIASING-BJTs If we now choose VCE to be O V, which establishes the vertical axis as the line on which
the second point will be defined, we find that le is determined by the following equation:
0 = Vee - IcRc

and Vccl
Ic= - (4.14)
Rc vCE=ov

as appearing on Fig. 4.12.

By joining the two points defined by Eqs. (4.13) and (4.14), we can draw the straight
line established by Eq. (4.12). The resulting line on the graph of Fig. 4.12 is called the load
line because it is defined by the load resistor Re. By solving for the resulting level of IB, we
can establish the actual Q-point as shown in Fig. 4.12.
If the level of IB is changed by varying the value of RB, the Q-point moves up or down
the load line as shown in Fig. 4.13 for increasing values of/B· If Vcc is held fixed and Re
increased, the load line will shift as shown in Fig. 4.14. If IB is held fixed, the Q-point will
move as shown in the same figure. If Re is fixed and Vcc decreased, the load line shifts as
shown in Fig. 4.15.

FICi. 4.13 FICi. 4.14

Movement of the Q-point with increasing level of IB. Effect of an increasing level of Re on the load line
and the Q-point.

FICi. 4.15
Effect of lower values of Vcc on the load line and the Q-point.
 EMITTER-BIAS 169
EXAMPLE 4.3 Given the load line of Fig. 4.16 and the defined Q-point, determine the CONFIGURATION
required values of V cc, Re, and RB for a fixed-bias configuration.
/c(mA)

12
50µA
10

4
_ _ _ _ _ _ _ ___,;~ : - - - - lOµA
2 =0 µA
-------------.~- 18

0 5 15 20 VCE

FIG. 4.16
Example 4.3.
Solution: From Fig. 4.16,
VcE =Vee= 20Vatlc = OmA
Vee
le= -atVCE = ov
Re
Vee 20V
and Re= - = - - = 2k!l
le lOmA
Vee - VBE
RB

and RB = _V:_cc_-_v_B_E = _20_V_-_o_.7_V_ = 772 k!l

lB 25 µ,A

4.4 EMITTER-BIAS CONFIGURATION

•
The de bias network of Fig. 4.17 contains an emitter resistor to improve the stability
level over that of the fixed-bias configuration. The more stable a configuration, the less
its response will change due to undesireable changes in temperature and parameter

~
vi o-------)1---+-------1

C1

FIG. 4.17
BJT bias circuit with emitter resistor.
170 DC BIASING-BJTs variations. The improved stability will be demonstrated through a numerical example
later in the section. The analysis will be performed by first examining the base-emitter
loop and then using the results to investigate the collector-emitter loop. The de equiva-
lent of Fig. 4.17 appears in Fig 4.18 with a separation of the source to create an input
and output section.

Base-Emitter Loop
The base-emitter loop of the network of Fig. 4.18 can be redrawn as shown in Fig. 4.19.
Writing Kirchhoff's voltage law around the indicated loop in the clockwise direction
results in the following equation:
+Vee - lsRs - VsE - JERE= O (4.15)
Recall from Chapter 3 that
IE = ({3 + l)ls (4.16)
FICi. 4.18
DC equivalent of Fig. 4.17. Substituting for IE in Eq. (4.15) results in
Vee - IsRs - VsE - (/3 + l)IsRE = 0
Grouping terms then provides the following:
+
-Is(Rs + (/3 + l)RE) + Vee - VsE = 0
Rn +In
Multiplying through by (-1), we have
-B ls(Rs + (/3 + l)RE) - Vee+ VsE =0
+ +
Vee with

() VsE and solving for ls gives

Vee - VsE
I------- (4.17)
s - Rs + (/3 + l)RE

-=- Note that the only difference between this equation for ls and that obtained for the fixed-
FICi. 4.19 bias configuration is the term (/3 + l)RE.
Base-emitter loop. There is an interesting result that can be derived from Eq. (4.17) if the equation is used to
sketch a series network that would result in the same equation. Such is the case for the net-
work of Fig. 4.20. Solving for the current ls results in the same equation as obtained above.
Note that aside from the base-to-emitter voltage VsE, the resistor RE is reflected back to the
input base circuit by a factor (/3 + 1). In other words, the emitter resistor, which is part of
the collector-emitter loop, "appears as" (/3 + l)RE in the base-emitter loop. Because f3 is
typically 50 or more, the emitter resistor appears to be a great deal larger in the base circuit.
In general, therefore, for the configuration of Fig. 4.21,
(j3 + I)RE
(4.18)

-=- Equation (4.18) will prove useful in the analysis to follow. In fact, it provides a fairly
FICi. 4.20
easy way to remember Eq. (4.17). Using Ohm's law, we know that the current through a
Network derived from Eq. (4.17). system is the voltage divided by the resistance of the circuit. For the base-emitter circuit
the net voltage is Yee - VsE• The resistance levels are Rs plus RE reflected by (/3 + 1).
The result is Eq. (4.17).

B
Colledor-Emitter Loop
The collector-emitter loop appears in Fig. 4.22. Writing Kirchhoffs voltage law for the
R; = (/3+ l)RE
....... indicated loop in the clockwise direction results in
+hRE + VeE + IeRe - Vee= 0
Substituting IE ~ le and grouping terms gives

-=- VCE - Vee+ Ic(Re +RE)= 0

FICi. 4.21
Reflected impedance level of RE.
and I VeE = Vee - Ie(Re + RE) I (4.19)
 The single-subscript voltage VE is the voltage from emitter to ground and is deter- EMITTER-BIAS 171
mined by CONFIGURATION

(4.20)

whereas the voltage from collector to ground can be determined from

VeE = Ve - VE

and Ve= VeE + VE I (4.21)

or I Ye= Vee - leRe I (4.22)

T
The voltage at the base with respect to ground can be determined using Fig. 4.18
FIC. 4.ll
Collector-emitter loop.
I VB= Vee - lsRB I (4.23)

or (4.24)

EXAMPLE4.4 For the emitter-bias network of Fig. 4.23, determine:

a. lB.
b. le.
+20V
C. VeE•
d. Ve.
e. VE·
f. VB· 2 kQ
g. VBe•
430 kQ 10 µF
_____, ( - - - - - - - Vo

10 µF
V; ---)n----------1 /3 = 50

l kQ I40µF

":" -=-
FIC. 4.23
Emitter-stabilized bias circuit for Example 4.4.

Solution:
Vee - VBE
l --- ----
20V-0.7V
a. Eq. (4.17):
B- RB + (/3 + l)RE 430 kll + (51)(1 kD,)
19.3 V
481 kll = 40.l µA
b. le= f31B
= (50)(40.1 µ,A)
~ 2.0lmA
172 DC BIASING-BJTs C. Eq. (4.19): VcE = Vee - lc(Rc + RE)
= 20 V - (2.01 mA)(2 kD + 1 kil) = 20 V - 6.03 V
= 13.97V
d. Ve= Vee - IcRc
= 20 V - (2.01 mA)(2 kD) = 20 V - 4.02 V
= 15.98V
e. VE= Ve - VcE
= 15.98 V - 13.97 V
= 2.01 V
or VE = feRE ~ IcRE
= (2.01 mA)(l kD)
= 2.01 V
f. VB = VBE + VE
= 0.7V + 2.01 V
= 2.71 V
g. VBc = VB - Vc
= 2.71 V - 15.98 V
= - 13.27 V (reverse-biased as required)

Improved Bias Stability

The addition of the emitter resistor to the de bias of the BJT provides improved stability,
that is, the de bias currents and voltages remain closer to where they were set by the circuit
when outside conditions, such as temperature and transistor beta, change. Although a
mathematical analysis is provided in Section 4.12, some comparison of the improvement
can be obtained as demonstrated by Example 4.5.

EXAMPLE 4.5 Prepare a table and compare the bias voltage and currents of the circuits of
Fig. 4.7 and Fig. 4.23 for the given value of /3 = 50 and for a new value of /3 = 100. Com-
pare the changes in / c and VCE for the same increase in {3.
Solution: Using the results calculated in Example 4.1 and then repeating for a value of
f3 = 100 yields the following:
Effect of f3 variation on the response of the
fixed-bias configuration of Fig. 4. 7.

/3 Ic(mA) VCE (V)

50 47.08 2.35 6.83

100 47.08 4.71 1.64

The BJT collector current is seen to change by 100% due to the 100% change in the value
of {3. The value of IB is the same, and VCE decreased by 76%.
Using the results calculated in Example 4.4 and then repeating for a value of f3 = 100,
we have the following:

Effect of f3 variation on the response of the

emitter-bias configuration of Fig. 4.23.

/3 Ic(mA) VCE (V)

50 40.1 2.01 13.97

100 36.3 3.63 9.11
Now the BJT collector current increases by about 81 % due to the 100% increase in /3. EMITTER-BIAS 173
Notice that In decreased, helping maintain the value of le-or at least reducing the overall CONFIGURATION
change in le due to the change in /3. The change in VeE has dropped to about 35%. The
network of Fig. 4.23 is therefore more stable than that of Fig. 4.7 for the same change in /3.

Saturation Level
The collector saturation level or maximum collector current for an emitter-bias design can
be determined using the same approach applied to the fixed-bias configuration: Apply a
short circuit between the collector-emitter terminals as shown in Fig. 4.24 and calculate
the resulting collector current. For Fig. 4.24

I Vee
- -- --
e,., - Re+ RE (4.25)

The addition of the emitter resistor reduces the collector saturation level below that FIC. 4.24
obtained with a fixed-bias configuration using the same collector resistor. Determining Ic,.Jor the emitter-
stabilized bias circuit.

EXAMPLE 4.6 Determine the saturation current for the network of Example 4.4.
Solution:
I Vee
- -- --
e,., - Re+ RE
20V 20V
2kfl + 1 kfl 3 kfl
= 6.67mA
which is about three times the level of IeQ for Example 4.4.

Load-Line Analysis
The load-line analysis of the emitter-bias network is only slightly different from that
encountered for the fixed-bias configuration. The level of In as determined by Eq. (4.17)
defines the level of In on the characteristics of Fig. 4.25 (denoted InQ).
The collector-emitter loop equation that defines the load line is
VCE = Vee - Ic(Re + RE)

FIC. 4.25
Load line for the emitter-bias configuration.
174 DC BIASING-BJTs Choosing le= 0 mA gives

(4.26)

as obtained for the fixed-bias configuration. Choosing VCE = 0 V gives

lc=- -- I
Yee (4.27)
Rc + RE vCE=ov

as shown in Fig. 4.25. Different levels of lnQ will, of course, move the Q-point up or down
the load line.

EXAMPLE4.7
a. Draw the load line for the network of Fig. 4.26a on the characteristics for the transistor
appearing in Fig. 4.26b.
b. For a Q-point at the intersection of the load line with a base current of 15 µ,A, find the
values of lcQ and VCEQ·
c. Determine the de beta at the Q-point.
d. Using the beta for the network determined in part c, calculate the required value of Rn
and suggest a possible standard value.

le (mA)
Vee= 18 V
30µA
6
Re
2.2 kQ
5

vi o------------)1---------1 3
- - - - - - - - - 15 µA
C1
- - - - - - - - - - lOµA
2
RE - - - - - - - - - - - - 5µA
1.1 kQ
18 = 0 µA
--------------
0 5 10 15 20

FIG. 4.l&a FIG. 4.l&b

Networkfor Example 4.7. Example 4. 7.

Solution:
a. Two points on the characteristics are required to draw the load line.
Yee 18 V 18 V
AtVcE=OV: le= Rc+RE 2.2k0+1.lk0 =3.3k0 =5.45mA
Atlc = OmA: VcE =Vee= 18 V
The resulting load line appears in Fig. 4.27.
b. From the characteristics of Fig. 4.27 we find
VcEQ ~ 7.5 V, lcQ ~ 3.3 mA
c. The resulting de beta is:
/3 = lcQ = 3.3 mA = 220
lnQ 15 µ,A
 le (mA) VOLTAGE-DIVIDER BIAS 175
CONFIGURATION

6
5.45mA--
5

0 15 \ 20 VCE
Vee= 18V

FIG. 4.27
Example 4. 7.

d. Applying Eq. 4.17:

Vee - VBE 18V - 0.7V
I-------
B - RB + (/3 + l)RE RB + (220 + 1)(1.1 kil)
17.3 V 17.3 V
and 15 µ,A = RB + (221)(1.1 kil) + 243.1 kil
RB
so that (15 µ,A)(RB) + (15 µ,A)(243.l kil) = 17.3 V
and (15 µ,A)(RB) = 17.3 V - 3.65 V = 13.65 V
. . 13.65 V
resultmg m RB + - - - = 910 kfl
15 µ,A

4.5 VOLTAGE-DIVIDER BIAS CONFIGURATION

In the previous bias configurations the bias current IeQ and voltage VeEQ were a func- •
tion of the current gain /3 of the transistor. However, because f3 is temperature sensi-
tive, especially for silicon transistors, and the actual value of beta is usually not well
defined, it would be desirable to develop a bias circuit that is less dependent on, or in
fact is independent of, the transistor beta. The voltage-divider bias configuration of
Fig. 4.28 is such a network. If analyzed on an exact basis, the sensitivity to changes in
beta is quite small. If the circuit parameters are properly chosen, the resulting levels of
IeQ and VeEQ can be almost totally independent of beta. Recall from previous discus-
sions that a Q-point is defined by a fixed level of IeQ and VeEQ as shown in Fig. 4.29.
The level of IBQ will change with the change in beta, but the operating point on the
characteristics defined by IeQ and VeEQ can remain fixed if the proper circuit parame-
ters are employed.
As noted earlier, there are two methods that can be applied to analyze the voltage-divider
configuration. The reason for the choice of names for this configuration will become obvi-
ous in the analysis to follow. The first to be demonstrated is the exact method, which can be
applied to any voltage-divider configuration. The second is referred to as the approximate
method and can be applied only if specific conditions are satisfied. The approximate ap-
proach permits a more direct analysis with a savings in time and energy. It is also particu-
larly helpful in the design mode to be described in a later section. All in all, the approximate
approach can be applied to the majority of situations and therefore should be examined with
the same interest as the exact method.
176 DC BIASING-BJTs le

_ _ _ _ _ _ _ ... ~ in,!. ___ (resulting IBQ)

R1
(---00
-----1 ..
C2
V0
,.=----+ ---
----- I

V; 0 )
C1

R2

FIG. 4.28 FIG. 4.29

Voltage-divider bias configuration. Defining the Q-point for the voltage-divider bias
configuration.

Exac:t Analysis
For the de analysis the network of Fig. 4.28 can be redrawn as shown in Fig. 4.30. The
input side of the network can then be redrawn as shown in Fig. 4.31 for the de analysis.
The Thevenin equivalent network for the network to the left of the base terminal can then
be found in the following manner:

R1h The voltage source is replaced by a short-circuit equivalent as shown in Fig. 4.32:

(4.28)
':"

FIG. 4.30 E,h The voltage source V cc is returned to the network and the open-circuit Thevenin
DC components of the voltage- voltage of Fig. 4.33 determined as follows:
divider configuration. Applying the voltage-divider rule gives

(4.29)

The Thevenin network is then redrawn as shown in Fig. 4.34, and lsQ can be determined
by first applying Kirchhoff's voltage law in the clockwise direction for the loop indicated:
ETh - lsRTh - VsE - JERE = 0
Substituting IE = (/3 + 1)/s and solving for ls yields

Thevenin
(4.30)
FIG. 4.31
Redrawing the input side of the
network of Fig. 4.28. Although Eq. (4.30) initially appears to be different from those developed earlier, note
that the numerator is again a difference of two voltage levels and the denominator is the base
resistance plus the emitter resistor reflected by (/3 + 1)-certainly very similar to Eq. (4.17).
Once ls is known, the remaining quantities of the network can be found in the same
manner as developed for the emitter-bias configuration. That is,

':" ':"
I VcE = Vee - Ic(Rc + RE) I (4.31)

FIG. 4.32 which is exactly the same as Eq. (4.19). The remaining equations for VE, Ve, and Vs are
Determining RTh. also the same as obtained for the emitter-bias configuration.
 VOLTAGE-DIVIDER BIAS 177
EXAMPLE 4.8 Determine the de bias voltage VCE and the current / e for the voltage- CONFIGURATION
divider configuration of Fig. 4.35.
Solution: Eq. (4.28): RTh = R1IIR2 + +
+
= (39 kfl)(3.9 kfl) = 3 _55 k!l -=- Vee
39 k!l + 3.9 k!l
R2Vee
Eq. (4.29): ETh =- --
R1 + R2
(3.9 kfl)(22 V) = 2V FIC. 4.33
39 k!l + 3.9 k!l Determining ETh.
ETh - VnE
Eq. (4.30): I-------
n - RTh + (/3 + l)RE
2V- 0.7V 1.3V
3.55 k!l + (101)(1.5 kil) 3.55 k!l + 151.5 k!l
= 8.38 µ,A
le= f3ln
= (100)(8.38 µ,A)
= 0.84mA
FIC. 4.34
+22V Inserting the Thevenin equivalent
circuit.

lOkQ
39kQ
let 10 µF
( Va
+
10 µF
V; ) VcE ,8 = 100

3.9kQ
tg I50µF

-=- -=- -=-

FIC. 4.35
Beta-stabilized circuit for Example 4. 8.

Eq. (4.31): VCE = Vee - Ie(Re + RE)

= 22 V - (0.84 rnA)(lO k!l + 1.5 kfl)
= 22 V - 9.66V
= 12.34 V

Approximate Analysis
The input section of the voltage-divider configuration can be represented by the network of
Fig. 4.36. The resistance Ri is the equivalent resistance between base and ground for the
transistor with an emitter resistor RE. Recall from Section 4.4 [Eq. (4.18)] that the reflected
resistance between base and emitter is defined by R; = (/3 + l)RE. If R; is much larger
than the resistance R2 , the current In will be much smaller than Ii (current always seeks the
path of least resistance) and Ii will be approximately equal to Ji. If we accept the approxi-
mation that In is essentially 0 A compared to Ii or Ii, then Ii = Ii, and R 1 and R2 can be
considered series elements. The voltage across R2 , which is actually the base voltage, can be
178 DC BIASING-BJTs

/1 t R1
+ IB
~
Vee-=-
+

1 "II"
lz t
"II"

FIG. 4.36
Rz
t
VB

! "II"
R;
R;»R 2
U1 =12 )

Partial-bias circuit for calculating the approximate base

voltage VB·

determined using the voltage-divider rule (hence the name for the configuration). That is,

(4.32)

Because Ri = (/3 + l)RE ~ f3RE the condition that will define whether the approxi-
mate approach can be applied is

I f3RE 2: lOR2 I (4.33)

In other words, if /3 times the value of RE is at least 10 times the value of R2, the approximate
approach can be applied with a high degree of accuracy.
Once V8 is determined, the level of VE can be calculated from

(4.34)

and the emitter current can be determined from

~ (4.35)
~
and

(4.36)

The collector-to-emitter voltage is determined by

VeE = Vee - IeRe - feRE
but because / E ~ I e,

I VeEQ = Vee - Ic(Re + RE) I (4.37)

Note in the sequence of calculations from Eq. (4.33) through Eq. (4.37) that f3 does not
appear and / 8 was not calculated. The Q-point (as determined by IeQ and VeEQ) is therefore
independent of the value of {3.

EXAMPLE 4.9 Repeat the analysis of Fig. 4.35 using the approximate technique, and
compare solutions for IeQ and VeEQ·
Solution: Testing:
f3RE 2: lOR2
(100)(1.5 kll) 2: 10(3.9 kll)
150 kll 2: 39 kll (satisfied)
 R2Vee VOLTAGE-DIVIDER BIAS 179
Eq. (4.32): VB=--- CONFIGURATION
R1 + R2
(3.9 kil)(22 V)
39 kil + 3.9 kil
=2V
Note that the level of VB is the same as ETh determined in Example 4.7. Essentially,
therefore, the primary difference between the exact and approximate techniques is the
effect of RTh in the exact analysis that separates ETh and VB.
Eq. (4.34): VE = VB - VBE
= 2V - 0.7V
= 1.3V
VE 1.3 V
leQ ~le= - = k" = 0.867mA
RE 1. 5 H

compared to 0.84 mA with the exact analysis. Finally,

VeEQ = Vee - lc(Re + RE)
= 22 V - (0.867 mA)(lO kV + 1.5 kil)
= 22 V - 9.97V
= 12.03V
versus 12.34 V obtained in Example 4.8.
The results for IeQ and VeEQ are certainly close, and considering the actual variation in
parameter values, one can certainly be considered as accurate as the other. The larger the
level of Ri compared to R 2 , the closer is the approximate to the exact solution. Example
4.11 will compare solutions at a level well below the condition established by Eq. (4.33).

EXAMPLE 4.10 Repeat the exact analysis of Example 4.8 if /3 is reduced to 50, and com-
pare solutions for IeQ and VeEQ·
Solution: This example is not a comparison of exact versus approximate methods, but a test-
ing of how much the Q-point will move if the level of f3 is cut in half. RTh and ETh are the same:
RTh = 3.55 kil, ETh = 2 V
ETh - VBE
I--------
B - RTh + (/3 + l)RE
2V- 0.7V 1.3 V
3.55 kil + (51)(1.5 kil) 3.55 kil + 76.5 kil
= 16.24µ,A
leQ = f3/B
= (50)(16.24 µ.,A)
= 0.81mA
VeEQ = Vee - le(Re + RE)
= 22 V - (0.81 mA)(lO kil + 1.5 kil)
= 12.69V
Tabulating the results, we have:
Effect of /3 variation on the response of the
voltage-divider configuration of Fig. 4.35.

p
100 0.84mA 12.34 V
50 0.81 mA 12.69 V

The results clearly show the relative insensitivity of the circuit to the change in /3. Even though
/3 is drastically cut in half, from 100 to 50, the levels of/eQ and VeEQ are essentially the same.
180 DC BIASING-BJTs hnportant Note: Looking back on the results for the fixed-bias configuration, we find the cur-
rent decreased from 4.71 rnA to 2.35 rnA when beta dropped from 100 to 50. For the voltage-
divider configuration, the same change in beta only resulted in a change in current from
0.84 rnA to 0.81 rnA. Even more noticeable is the change in VeEQ for the fixed-bias configuration.
Dropping beta from 100 to 50 resulted in an increase in voltage from 1.64 to 6.83 V (a change of
over 300%). For the voltage-divider configuration, the increase in voltage was only from 12.34 V
to 12.69 V, which is a change of less than 3%. fu summary, therefore, changing beta by 50%
resulted in a change in an important network parameter of over 300% for the fixed-bias configura-
tion and less than 3% for the voltage-divider configuration-a significant difference.

EXAMPLE 4.11 Determine the levels of IeQ and VeEQ for the voltage-divider configura-
tion of Fig. 4.37 using the exact and approximate techniques and compare solutions. In this
case, the conditions of Eq. (4.33) will not be satisfied and the results will reveal the differ-
ence in solution if the criterion of Eq. (4.33) is ignored.

18V

5.6k0
82k0
ilc lOµF
Q ( O Vo

+
V; o ) VcEQ f3 = 50
10 µF

22k0
1.2 kO

FIG. 4.37
Voltage-divider configuration for Example 4.11.

Solution: Exact analysis:

Eq. (4.33):
f3RE ~ lOR2
(50)(1.2 kfl) ~ 10(22 kfl)
60 kfl ;f. 220 kfl (not satisfied)
RTh = R1IIR2 = 82kDll22kfl = 17.35kfl
R2Vee 22kfl(l8V)
ETh = R1 + R2 82kfl + 22kfl = 3 ·81 V

ETh - VnE 3.81 V - 0.7 V 3.11 V

I-------
B - RTh + (/3 + l)RE 17.35 kfl + (51)(1.2 kfl) 78.55 kfl = 39 ·6 µ.,A
IeQ = f3In = (50)(39.6 µ.,A) = 1.98 mA
VeEQ = Vee - Ic(Re + RE)
= 18 V - (1.98 rnA)(5.6 kfl + 1.2 kfl)
= 4.54 V
Approximate analysis:
Vn = ETh = 3.81 V
VE = Vn - VnE = 3.81 V - 0.7 V = 3.11 V
VE 3.11 V
leQ =IE= - = - - = 2.59mA
RE l.2kfl
VeEQ = Vee - Ie(Re + RE)
= 18 V - (2.59 rnA)(5.6 kfl + 1.2 kfl)
= 3.88V
Tabulating the results, we have: COLLECTOR FEEDBACK 181
CONFIGURATION
Comparing the exact and approximate approaches.

Exact 1.98 4.54

Approximate 2.59 3.88

The results reveal the difference between exact and approximate solutions. IeQ is about
30% greater with the approximate solution, whereas VeEQ is about 10% less. The results
are notably different in magnitude, but even though f3RE is only about three times larger
than R2 , the results are still relatively close to each other. For the future, however, our
analysis will be dictated by Eq. (4.33) to ensure a close similarity between exact and
approximate solutions.

Transistor Saturation
The output collector-emitter circuit for the voltage-divider configuration has the same
appearance as the emitter-biased circuit analyzed in Section 4.4. The resulting equation for
the saturation current (when VCE is set to O Von the schematic) is therefore the same as
obtained for the emitter-biased configuration. That is,

I - I - Vee (4.38)
e,at - emax - Re + RE

Load-Line Analysis
The similarities with the output circuit of the emitter-biased configuration result in the
same intersections for the load line of the voltage-divider configuration. The load line will
therefore have the same appearance as that of Fig. 4.25, with

---I
I e =Vee (4.39)
Re + RE vCE=ov

and I VCE = Veclic=OmA I (4.40)

The level of In is of course determined by a different equation for the voltage-divider bias
and the emitter-bias configurations.

4.6 COLLECTOR FEEDBACK CONFIGURATION

•
An improved level of stability can also be obtained by introducing a feedback path from
collector to base as shown in Fig. 4.38. Although the Q-point is not totally independent of
beta (even under approximate conditions), the sensitivity to changes in beta or temperature
variations is normally less than encountered for the fixed-bias or emitter-biased configura-
tions. The analysis will again be performed by first analyzing the base-emitter loop, with
the results then applied to the collector-emitter loop.

Base-Emitter Loop
Figure 4.39 shows the base-emitter loop for the voltage feedback configuration. Writing
Kirchhoff's voltage law around the indicated loop in the clockwise direction will result in
Vee - leRe - lnRF - VnE - feRE = 0
Itis important to note that the current through Reis not/e, but/e (where le = le + In).
However, the level of le and le far exceeds the usual level of In, and the approximation
le ~ le is normally employed. Substituting le ~ le = f3ln and /e ~ le results in
Vee - f3lnRe - InRF - VnE - f3lnRE = 0
182 DC BIASING-BJTs Vee
+
Re

~ le +
(
RF ~ le c2 + I ~F

~
v;o-------------}t-----+-------1
+
VeE
Vee-=-
~
C1
+
_ _ _ VBE _
~ IE +
RE

FIC. 4.38 FIC. 4.39

DC bias circuit with voltage feedback. Base-emitter loop for the network of Fig. 4.38.

Gathering terms, we have

Vee - VsE - f3ls(Re + RE) - lsRF = 0
and solving for ls yields

Vee - VsE
fs=------- (4.41)
RF + {3(Rc + RE)
The result is quite interesting in that the format is very similar to equations for ls ob-
tained for earlier configurations. The numerator is again the difference of available voltage
levels, whereas the denominator is the base resistance plus the collector and emitter resis-
tors reflected by beta. In general, therefore, the feedback path results in a reflection of the
resistance Re back to the input circuit, much like the reflection of RE.
In general, the equation for ls has the following format, which can be compared with the
result for the fixed-bias and emitter-bias configurations.
V'
I-----
s - RF+ f3R'
For the fixed-bias configuration f3R' does not exist. For the emitter-bias setup (with
f3 + 1 = /3), R' = RE.
Because le = f3ls,

I
{3V'
-----
V'
eQ - RF+ {3R' R
_f_ + R'
/3
+ R
Re In general, the larger R' is compared with ; , the more accurate the approximation that

V'
IeQ = R'
The result is an equation absent of {3, which would be very stable for variations in {3.
Because R' is typically larger for the voltage-feedback configuration than for the emitter-
bias configuration, the sensitivity to variations in beta is less. Of course, R' is O n for the
fixed-bias configuration and is therefore quite sensitive to variations in beta.

Collector-Emitter Loop
The collector-emitter loop for the network of Fig. 4.38 is provided in Fig. 4.40. Applying
FIC. 4.40
Collector-emitter loop for the Kirchhoff's voltage law around the indicated loop in the clockwise direction results in
network of Fig. 4.38. JERE+ VCE + 1/::Re - Vee= 0
Because le ~ le and IE ~ le, we have COLLECTOR FEEDBACK 183
CONFIGURATION
Ie(Re + RE) + VeE - Vee= 0

and I VeE = Vee - Ie(Re + RE) I (4.42)

which is exactly as obtained for the emitter-bias and voltage-divider bias configurations.

EXAMPLE 4.12 Determine the quiescent levels of I eQ and VeEQ for the network of Fig.
4.41.

lOV

4.7 kQ

~ - - ~ ~ - + - - - - I ~ Vo
10 µF
v;o--------)1----+-----1 /3 = 90
10 µF

1.2 kQ

FIG. 4.41
Networkfor Example 4.12.

Vee - VsE
Solution: Eq. (4.41): ls=-------
Rp + {3(Re + RE)
10V-0.7V
250 kll + (90)( 4. 7 kll + 1.2 kll)
9.3 V 9.3 V
250 kll + 531 kll 781 kll
= 11.91 µ,A
IeQ = {318 = (90)(11.91 µ,A)
= 1.07mA
VeEQ = Vee - Ic(Re + RE)
= lOV - (l.07mA)(4.7kll + l.2kll)
= lOV - 6.31 V
= 3.69V

EXAMPLE 4.13 Repeat Example 4.12 using a beta of 135 (50% greater than in Example
4.12).
Solution: It is important to note in the solution for 18 in Example 4.12 that the second
term in the denominator of the equation is much larger than the first. Recall in a recent
discussion that the larger this second term is compared to the first, the less is the sensitivity
to changes in beta. In this example, the level of beta is increased by 50%, which will
increase the magnitude of this second term even more compared to the first. It is more
important to note in these examples, however, that once the second term is relatively large
compared to the first, the sensitivity to changes in beta is significantly less.
184 DC BIASING-BJTs Solving for In gives
Vee - VnE
In=-------
Rn + f3(Re + RE)
10V-0.7V
250kO + (135)(4.7kO + l.2kO)
9.3V 9.3V
250 kO + 796.5 kO 1046.5kO
= 8.89 µ,A
and IeQ = f3In
= (135)(8.89 µ,A)
= 1.2mA
and VeEQ = Vee - Ie(Re + RE)
= 10 V - (1.2 mA)(4.7 kO + 1.2 kD,)
= lOV - 7.08V
= 2.92V
Even though the level of f3 increased 50%, the level of IeQ only increased 12.1 %, whereas
the level of VCEQ decreased about 20.9%. If the network were a fixed-bias design, a 50%
increase in /3 would have resulted in a 50% increase in IeQ and a dramatic change in the
location of the Q-point.

EXAMPLE 4.14 Determine the de level of In and Ve for the network of Fig. 4.42.

18V

3.3 kQ
91 kQ 110 kQ 10 µF
~...,,._.~-------~'-------1(--------o Vo

10 µF
v;o------)t-------------1 f3 = 75

510 Q

FIG. 4.42
Networkfor Example 4.14.

Solution: In this case, the base resistance for the de analysis is composed of two resistors
with a capacitor connected from their junction to ground. For the de mode, the capacitor
assumes the open-circuit equivalence, and Rn = RF1 + RF2 -
Solving for In gives
Vee - VnE
In=-------
Rn + f3(Re + RE)
18V- 0.7V
(91 kO + 110 kD,) + (75)(3.3 kO + 0.51 kil)
17.3 V 17.3 V
201 kO + 285.75 kO 486.75 kO
= 35.SµA
 le= {318 COLLECTOR FEEDBACK 185
CONFIGURATION
= (75)(35.5 µ.,A)
= 2.66mA
Ve= Vee - leRe ~ Vee - leRe
= 18 V - (2.66 mA)(3.3 kil)
= 18V- 8.78V
= 9.22V

Saturation Conditions
Using the approximation le = le, we find that the equation for the saturation current is the
same as obtained for the voltage-divider and emitter-bias configurations. That is,

l _ l _ Vee (4.43)
e,at - emax - Re + RE

Load-Line Analysis
Continuing with the approximation le = le results in the same load line defined for the
voltage-divider and emitter-biased configurations. The level of l 8 Q is defined by the chosen
bias configuration.

EXAMPLE 4.15 Given the network of Fig. 4.43 and the BJT characteristics of Fig. 4.44.
a. Draw the load line for the network on the characteristics.
b. Determine the de beta in the center region of the characteristics. Define the chosen
point as the Q-point.
c. Using the de beta calculated in part b, find the de value of 18 .
d. Find leQ and leEQ·

le (mA)
36V

15
2.7 kQ 50µA
150 kQ 360kQ 10 µF
,---#V"'-~0--~"'-------1(--------o Vo

10

10 µF - - - - - - - - 30µA
vio----------)t----+----------1

5 - - - - - - - - - - 20µA
f . - - - - - - - - - - - - lOµA
330 n
i--~---:----:------:--- OµA
0 10 20 30 40 50 VCE(V)

FICi. 4.43 FICi. 4.44

Network for Example 4.15. BIT characteristics.

Solution:
a. The load line is drawn on Fig. 4.45 as determined by the following intersections:
Vee 36V
VeE = OV:le =Re+ RE 2.7kil + 3300 = 11.88mA
le= 0mA: VeE =Vee= 36V
186 DC BIASING-BJTs le (mA)

~----- - ~"""----lOµA
~- - ~- ~- -~ :---:--- 0~
0 30 I 40 50 VcE(V)
36V

FIG. 4.45
Defining the Q-point for the voltage-divider bias configuration of
Fig. 4.43.

b. The de beta was determined using IB = 25 µ,A and VCE about 17 V.

/3 ~ IeQ = 6.2mA = 248
IBQ 25 µ,A
c. Using Eq. 4.41:
Vee - VBE 36V - 0.7V
IB = RB + {3(Re + RE) = 510 kll + 248(2.7 kll + 330 ll)
35.3 V
510 kll + 751.44 kll
d 35.3 V _ 28 A
an IB = 1.261 Mll - µ
d. From Fig. 4.45 the quiescent values are
I eQ ~ 6.9 mA and VCEQ ~ 15 V

4.7 EMITTER-FOLLOWER CONFIGURATION

•
The previous sections introduced configurations in which the output voltage is typically
taken off the collector terminal of the BJT. This section will examine a configuration where
the output is taken off the emitter terminal as shown in Fig. 4.46. The configuration of Fig.
4.46 is not the only one where the output can be taken off the emitter terminal. In fact, any of
the configurations just described can be used so long as there is a resistor in the emitter leg.

C1
Vi 0>-----1) l-------.>--------11 /3 = 90
C2
----1,(,----0V0

-VEE

FIG. 4.46
Common-collecter (emitter-follower) configuration.
The de equivalent of the network of Fig. 4.46 appears in Fig. 4.47 COMMON-BASE 187
Applying Kirchhoff's voltage rule to the input circuit will result in CONFIGURATION

-JnRn - VnE - feRE + VEE = 0

and using le = (/3 + l)Jn
JnRn + (/3 + l)JnRE = VEE - VnE
VEE - VnE
so that Jn=------ (4.44)
Rn + (/3 + l)RE

For the output network, an application of Kirchhoff s voltage law will result in
-VCE - JERE+ VEE= 0

and (4.45) FIG. 4.47

de equivalent of
Fig. 4.46.

EXAMPLE 4.16 Determine VCEQ and JEQ for the network of Fig. 4.48.

C1
V; O>----l)l------+-------11
10 µF
C2
RB 240Hl ------11--(---0V0
JO µF
-=-

FIG. 4.48
Example 4.16.

Solution:
J -------
VEE - VnE
Eq. 4.44:
B - Rn + (/3 + l)RE
20V-0.7V 19.3 V
240 kO + (90 + 1)2 kO 240 kO + 182 kO
19.3 V
422 kO = 45.73 µ,A

and Eq. 4.45: VCEQ = VEE - JERE

= VEE - (/3 + l)JnRE
= 20 V - (90 + 1)(45.73 µ,A)(2 kO)
= 20V - 8.32V
= 11.68V
JEQ = (/3 + l)Jn = (91)(45.73 µ,A)
= 4.16mA

4.8 COMMON-BASE CONFIGURATION

•
The common-base configuration is unique in that the applied signal is connected to the
emitter terminal and the base is at, or just above, ground potential. It is a fairly popular
configuration because in the ac domain it has a very low input impedance, high output
impedance, and good gain.
188 DC BIASING-BJTs A typical common-base configuration appears in Fig. 4.49. Note that two supplies are
used in this configuration and the base is the common terminal between the input emitter
terminal and output collector terminal.
The de equivalent of the input side of Fig. 4.49 appears in Fig. 4.50.

v;o-----------}t----------- ,---+------l~Vo
c, Cz

FICi. 4.49
Common-base configuration.

Applying Kirchhoff s voltage law will result in

-VEE+ /ERE+ VsE = 0

(4.46)

Applying Kirchhoff s voltage law to the entire outside perimeter of the network of Fig.
FICi. 4.50 4.51 will result in
Input de equivalent of -VEE+ /ERE+ VcE + IcRc - Vee= 0
Fig. 4.49.
and solving for VCE: VCE = VEE+ Vee - feRE - IcRc
Because

(4.47)
+ The voltage VCB of Fig. 4.51 can be found by applying Kirchhoff s voltage law to the
+ output loop of Fig 4.51 to obtain:
Vcs + IcRc - Vee= 0
+ or Vcs = Vee - IcRc
VEE-;;;;;-
+ .___ _ _ _ _ _=-., Using le~ IE

-=- we have (4.48)

FICi. 4.51
Determining VCE and VCB·

EXAMPLE 4.17 Determine the currents IE and /8 and the voltages VcE and Vcs for the
common-base configuration of Fig. 4.52.

Vi o-----------} t-----+--------.. r - - - - - - + - - - - - - 1 ~ Vo

-=-
FICi. 4.52
Example 4.17.
 VEE - VnE MISCELLANEOUS BIAS 189
Solution: Eq. 4.46: IE=---- CONFIGURATIONS
RE
4 V - 0.7V = 2_75 mA
l.2k!l
IE 2.75mA 2.75 mA
In = /3 + 1 = 60 + 1 61
= 45.0SµA
Eq. 4.47: VCE =VEE+ Vee - hf.Re+ RE)
= 4 V + 10 V - (2.75 mA)(2.4 k!l + 1.2 k!l)
= 14 V - (2.75 mA)(3.6 k!l)
= 14 V - 9.9V
= 4.1 V
Eq. 4.48: Ven= Vee - IeRe = Vee - f3InRe
= 10 V - (60)(45.08 JLA)(24 k!l)
= lOV - 6.49V
= 3.51 V

4.9 MISCELLANEOUS BIAS CONFIGURATIONS

•
There are a number of BJT bias configurations that do not match the basic mold of those
analyzed in the previous sections. In fact, there are variations in design that would require
many more pages than is possible in a single publication. However, the primary purpose
here is to emphasize those characteristics of the device that permit a de analysis of the
configuration and to establish a general procedure toward the desired solution. For each
configuration discussed thus far, the first step has been the derivation of an expression for
the base current. Once the base current is known, the collector current and voltage levels of
the output circuit can be determined quite directly. This is not to imply that all solutions
will take this path, but it does suggest a possible route to follow if a new configuration is
encountered.
The first example is simply one where the emitter resistor has been dropped from the
voltage-feedback configuration of Fig. 4.38. The analysis is quite similar, but does require
dropping RE from the applied equation.

EXAMPLE 4.18 For the network of Fig. 4.53:

a. Determine IeQ and VeEQ·
b. Find Vn, Ve, VE, and Vne•

Vee =20V

4.7kQ

10 µF
~~~-+----1(-------ovo
C2
IOµF
vio-------)1----..-----1 f3 = 120
C1

FIG. 4.53
Collector feedback with RE = 0 n.
190 DC BIASING-BJTs Solution:
a. The absence of RE reduces the reflection of resistive levels to simply that of Re, and the
equation for / B reduces to
Vee - VBE
/B=----
RB + f3Re
20 V - 0.7 V 19.3 V
680 kll + (120)(4.7 kll) 1.244 MO
= 15.SlµA
leQ = f3/B = (120)(15.51 µ,A)
= 1.86mA
VeEQ = Vee - IeRe
= 20 V - (1.86 mA)(4.7 kll)
= 11.26V
b. VB= VBE = 0.7V
Ve= VeE = 11.26V
VE= OV
VBe = VB - Ve= 0.7V - 11.26V
= -10.56V

In the next example, the applied voltage is connected to the emitter leg and Re is con-
nected directly to ground. Initially, it appears somewhat unorthodox and quite different
from those encountered thus far, but one application of Kirchhoff's voltage law to the base
circuit will result in the desired base current.

EXAMPLE 4.19 Determine Ve and VB for the network of Fig. 4.54.

Re

c,
Vi o,......_---1) l - - - - + - - - - - - - 1 f3 = 45
10 µF

R8 IOOHl VEE =-9V

FIG. 4.54
Example 4.19.

Solution: Applying Kirchhoff's voltage law in the clockwise direction for the base-emitter
loop results in
-IBRB - VBE + VEE = 0
VEE - VBE
and /B=----
RB
Substitution yields
9V-0.7V
I------
B - lO0kll
8.3V
100 kll
= 83 µ,A
 le= f3Is MISCELLANEOUS BIAS 191
CONFIGURATIONS
= (45)(83 JLA)
= 3.735 mA
Ve= -IeRe
= -(3.735 mA)(l.2 kil)
= -4.48 V
Vs= -IsRs
= -(83 JLA)(lOO kil)
= -8.3V

Example 4.20 employs a split supply and will require the application of Thevenin's
theorem to determine the desired unknowns.

EXAMPLE 4.20 Determine Ve and Vs for the network of Fig. 4.55.

vcc=+20V

Re 2.7 kQ
R, 8.2 kQ e2
e
( oVo

c, 10 µF
B
ViO ) {3 = 120
lOµF
E
R2 2.2 kQ
RE 1.8 kQ

VEE=-20V

FIG. 4.55
Example 4.20.

Solution: The Thevenin resistance and voltage are determined for the network to the left
of the base terminal as shown in Figs. 4.56 and 4.57.

RTh = 8.2 kil II 2.2 kil = 1.73 kil

r
8.2 kQ
.---~~""-----.------oB
R, + +
8.2 kQ + R2 2.2 kQ
Erb

FIG. 4.56 FIG. 4.57

Determining RTh. Determining ETh·
192 DC BIASING-BJTs
Vee+ VEE
/=----
20V + 20V 40V
R1 + R2 8.2 kfl + 2.2 kfl 10.4kfl
= 3.85mA
ETh = IR2 - VEE
= (3.85 mA)(2.2 kfl) - 20 V
= -11.53 V
The network can then be redrawn as shown in Fig. 4.58, where the application of
Kirchhoff's voltage law results in
-ETh - IBRTh - VBE - JERE+ VEE= 0

/3 = 120
( : 1.73 kQ +
I LB E
+
1.8 kQ

VEE =-20V

FIG. 4.58
Substituting the Thevenin equivalent circuit.

Substituting IE = (/3 + 1)/B gives

VEE - ETh - VBE - (/3 + 1)/BRE - IBRTh = 0
VEE - ETh - VBE
and I--------
B - RTh + (/3 + l)RE
20 V - 11.53 V - 0. 7 V
1.73 kfl + (121)(1.8 kfl)
7.77V
219.53 kfl
= 35.39 JLA
le= f3/B
= (120)(35.39 µ,A)
= 4.25mA
Ve= Vee - IeRe
= 20 V - (4.25 mA)(2.7 kfl)
= 8.53 V
VB = -ETh - l#Th
= -(11.53 V) - (35.39 µ,A)(l.73 kfl)
= -11.59V

4.1 O SUMMARY TABLE

Table 4.1 is a review of the most common single-stage BJT configurations with their •
respective equations. Note the similarities that exist between the equations for the various
configurations.
 TABLE 4.1
BIT Bias Configurations

Type Configuration Pertinent Equations

Fixed-bias

Vee - Vse
Re ls=----
Rs
le = /3ls, le = (/3 + l)ls
/3 Vee= Vee - lcRc

Emitter-bias

Re l
Vee - Vse
-------
s - Rs + (/3 + l)Re
le = /3ls, le = (/3 + l)ls
R; = (/3 + l)Re
Vee = Vee - le (Re + Re)

Voltage-divider Vee
bias

R2Vcc APPROXIMATE: f3Re 2a: lOR2

EXACT: RTh = R1IIR2, ETh = - - -
Ri + R2 R2Vcc
Vs = - - - , Ve = Vs - Vse
l --- ETh-- -Vse
-- Ri + R2
/3 s - RTh + (/3 + l)Re Ve le
l --1 - - -
le = /3ls, le = (/3 + l)ls e - Re' s - /3 + 1
Vee= Vee - lc(Rc + Re) Vee = Vee - lc(Rc + Re)

Collector-feedback

Vee - Vse
ls=------
RF + /3(Rc + Re)
le = /3ls, le = (/3 + l)ls
Vee = Vee - le (Re + Re)

Emitter-follower

Vee - Vse
ls=------
Rs + (/3 + l)Re
le = /3ls, le = (/3 + l)ls
Vee= Vee - leRe

Common-base Vee - Vse

le=----
Re
le
+ ls = /3 + 1 Jc = /3ls
-=- Vee Vee= Vee+ Vee - le(Rc + Re)
Vcs = Vee - lcRc
193
194 DC BIASING-BJTs 4.11 DESIGN OPERATIONS
•
Discussions thus far have focused on the analysis of existing networks. All the elements
are in place, and it is simply a matter of solving for the current and voltage levels of the
configuration. The design process is one where a current and/or voltage may be specified
and the elements required to establish the designated levels must be determined. This syn-
thesis process requires a clear understanding of the characteristics of the device, the basic
equations for the network, and a firm understanding of the basic laws of circuit analysis,
such as Ohm's law, Kirchhoff's voltage law, and so on. In most situations the thinking
process is challenged to a higher degree in the design process than in the analysis sequence.
The path toward a solution is less defined and in fact may require a number of basic
assumptions that do not have to be made when simply analyzing a network.
The design sequence is obviously sensitive to the components that are already specified
and the elements to be determined. If the transistor and supplies are specified, the design
process will simply determine the required resistors for a particular design. Once the theo-
retical values of the resistors are determined, the nearest standard commercial values are
normally chosen and any variations due to not using the exact resistance values are accepted
as part of the design. This is certainly a valid approximation considering the tolerances
normally associated with resistive elements and the transistor parameters.
If resistive values are to be determined, one of the most powerful equations is simply
Ohm's law in the following form:

(4.49)

In a particular design the voltage across a resistor can often be determined from specified
levels. If additional specifications define the current level, Eq. (4.49) can then be used to
calculate the required resistance level. The first few examples will demonstrate how par-
ticular elements can be determined from the design specifications. A complete design pro-
cedure will then be introduced for two popular configurations.

EXAMPLE 4.21 Given the device characteristics of Fig. 4.59a, determine Vcc, RB, and Re
for the fixed-bias configuration of Fig. 4.59b.

le (mA)

8
Re

0 20 V Ve£

(a) (b)

FIG. 4.59
Example 4.21.

Solution: From the load line

Vee= 20V

le= Vccl
Re vCE=ov
Vee 20V
and Re = - = -- = 2.5 kO
le 8mA
Vee - VBE
lB=----
RB
 Vee - VnE DESIGN OPERATIONS 195
with Rn=----
In
20 V - 0.7 V 19.3 V
40µ,A 40µ,A
= 482.Sk!l
Standard resistor values are
Re= 2.4kll
Rn= 470kll
Using standard resistor values gives
In= 41.1 µ,A
which is well within 5% of the value specified.

EXAMPLE 4.22 Given that IcQ = 2 mA and VcEQ = 10 V, determine R 1 and Re for the
network of Fig. 4.60.

18 V

Re; 10 µF
,.__--,1-(--oVa
10 µF
V; o----1)t----+-------11

18 kQ
1.2 kQ

FIG. 4.60
Example 4.22.

Solution:
VE = hRE ~ IcRE
= (2 mA)(l.2 kll) = 2.4 V
Vn = VnE + VE= 0.1 V + 2.4 V = 3.1 V
R V:
Vn = 2 cc = 3.1 V
R1+ R2
(18 kD,)(18 V) = 3 _1 V
and
R1 + 18 kll
324kll = 3.1R 1 + 55.8kll
3.1R 1 = 268.2 kll
_ 268.2kll _ n
R 1 - - - - - 865 2 k~£
3.1
VRc Vee - Ve
Eq. (4.49): Re== - = ----
le le
with Ve= VcE +VE= lOV + 2.4V = 12.4V
18 V - 12.4 V
and Re = __2_mA_ __

= 2.Sk!l
196 DC BIASING-BJTs The nearest standard commercial values to R 1 are 82 kll and 91 kll. However, using
the series combination of standard values of 82 kll and 4. 7 kll = 86. 7 kll would result in
a value very close to the design level.

~--------.--o 28 V
EXAMPLE 4.23 The emitter-bias configuration of Fig. 4.61 has the following specifica-
tions: lcQ = ½lsat, le,., = 8 mA, Vc = 18 V, and /3 = 110. Determine Re, RE, and RB.
Solution:
Ve= 18 V
leQ = 1
21csat = 4mA
/3 = 110 VRc Vee - Ve
Rc=-=----
lcQ lcQ
= 28V - 18V = 2.Sk!l
4mA

I - -Vee
--
FICi. 4.61 c,., - Re+ RE
Example 4.23. Vee 28 V
and Re + RE = - = - - = 3.5 kll
8 mA
le,.,
RE= 3.5kll - Re
= 3.5 kll - 2.5 kll
= lk!l
lcQ 4mA
lB = - = - - = 36.36 µ.,A
Q /3 110

l --- Vee-- -VBE

--
BQ - RB + ({3 + l)RE
V: - V
and RB + ({3 + l)RE = CC BE
lBQ

with RB= Vee - VBE - (/3 + l)RE

lBQ
= 28V6 -60.7V - (111)(1kll)
3 .3 µ.,A
27.3V
111 kll
36.36 µ.,A
= 639.Sk!l
For standard values,
Re= 2.4kll
RE= l kll
RB= 620kll

The discussion to follow will introduce one technique for designing an entire circuit
to operate at a specified bias point. Often the manufacturer's specification (spec) sheets
provide information on a suggested operating point (or operating region) for a particular
transistor. In addition, other system components connected to the given amplifier stage may
also define the current swing, voltage swing, value of common supply voltage, and so on,
for the design.
In actual practice, many other factors may have to be considered that may affect the
selection of the desired operating point. For the moment we concentrate on determining the
component values to obtain a specified operating point. The discussion will be limited to
the emitter-bias and voltage-divider bias configurations, although the same procedure can
be applied to a variety of other transistor circuits.
Design of a Bias Circuit with an Emitter Feedback Resistor DESIGN OPERATIONS 197

Consider first the design of the de bias components of an amplifier circuit having emitter-
resistor bias stabilization as shown in Fig. 4.62. The supply voltage and operating point
were selected from the manufacturer's information on the transistor used in the amplifier.
The selection of collector and emitter resistors cannot proceed directly from the infor-
mation just specified. The equation that relates the voltages around the collector-emitter
loop has two unknown quantities present-the resistors Re and RE. At this point some en-
gineering judgment must be made, such as the level of the emitter voltage compared to the
applied supply voltage. Recall that the need for including a resistor from emitter to ground
was to provide a means of de bias stabilization so that the change of collector current due
to leakage currents in the transistor and the transistor beta would not cause a large shift in
the operating point. The emitter resistor cannot be unreasonably large because the voltage
across it limits the range of swing of the voltage from collector to emitter (to be noted when
the ac response is discussed). The examples examined in this chapter reveal that the voltage
from emitter to ground is typically around one-fourth to one-tenth of the supply voltage.
Selecting the conservative case of one-tenth will permit calculating the emitter resistor RE
and the resistor Re in a manner similar to the examples just completed. fu the next example
we perform a complete design of the network of Fig. 4.62 using the criteria just introduced
for the emitter voltage.

Vcc=20V

IcQ = 2mA{ Re
RB e2
ac
( •output
+
ac
input
. C1
)
VB 2N4401
lOµF
VceQ = IOV
IOµF
(/3 =
150)

r Ce

SOµF

FIG. 4.62
Emitter-stabilized bias circuit for design consideration.

EXAMPLE 4.24 Determine the resistor values for the network of Fig. 4.62 for the indicated
operating point and supply voltage.
Solution:
VE= foVee = fo(20V) = 2 V
RE= VE~ VE= _}:y_ = lkfl
le le 2mA
VRc Vee - VeE - VE 20V - lOV - 2V 8V
Re= - =- ------
le le 2mA 2mA
= 4kfl
le 2mA
ls = /3 = lS0 = 13.33 µ,A
20V - 0.7V - 2V
13.33 µ,A
198 DC BIASING-BJTs Design of a Current-Gain-Stabilized (Beta-Independent) Circuit
The circuit of Fig. 4.63 provides stabilization both for leakage and current gain (beta)
changes. The four resistor values shown must be obtained for the specified operating point.
Engineering judgment in selecting a value of emitter voltage VE, as in the previous design
consideration, leads to a direct, straightforward solution for all the resistor values. The
design steps are all demonstrated in the next example.

EXAMPLE 4.25 Determine the levels of Re, RE, R1, and R 2 for the network of Fig. 4.63
for the operating point indicated.

IcQ= 10 mA
.,.
I
____ C2
(------ ac
+ output
C1 10 µF
ac ___}1--~------1
input----------, VCEQ= 8 V f:l(min) = 80
lOµF

FIG. 4.63
Current-gain-stabilized circuit for design considerations.

Solution:
VE= foVee = fo(20 V) = 2 V
VE VE 2V
RE=-~-=--=200O
IE le lOrnA
VRc 20V - 8V - 2V lOV
Re=-=
le lOrnA lOrnA
= lk!l
VB = VBE + VE= 0.7 V + 2 V 2.7 V
=
The equations for the calculation of the base resistors R 1 and R 2 will require a little
thought. Using the value of base voltage calculated above and the value of the supply volt-
age will provide one equation-but there are two unknowns, R 1 and R 2 . An additional
equation can be obtained from an understanding of the operation of these two resistors in
providing the necessary base voltage. For the circuit to operate efficiently, it is assumed
that the current through R1 and R2 should be approximately equal to and much larger than
the base current (at least 10:1). This fact and the voltage-divider equation for the base volt-
age provide the two relationships necessary to determine the base resistors. That is,
R2 ~ tof3RE
R2
and VB= - - - V e e
R1 + R2
Substitution yields
R2 ~ fo(80)(0.2 k11)
= 1.6k!l
(1.6 kll)(20 V)
VB= 2.7V = -----
R1 + l.6kll
and 2.7R 1 + 4.32 kf! = 32 kf! MULTIPLE BJT 199
NETWORKS
2.7R1 = 27.68 kf!
R1 = 10.25 kfl (use 10 kf!)

4.12 MULTIPLE BJT NETWORKS

The BJT networks introduced thus far have only been single-stage configurations. This
section will cover some of the most popular networks using multiple transistors. It will
demonstrate how the methods introduced thus far in this chapter can be applied to net-
•
works with any number of components.
The R-C coupling of Fig. 4.64 is probably the most common. The collector output of one
stage is fed directly into the base of the next stage using a coupling capacitor Cc. The capaci-
tor is chosen to ensure that it will block de between the stages and act like a short circuit to
any ac signal. The network of Fig. 4.64 has two voltage-divider stages, but the same coupling
can be used between any combination of networks such as the fixed-bias or emitter-follower
configurations. Substituting an open-circuit equivalent for Cc and the other capacitors of the
network will result in the two bias arrangements shown in Fig. 4.65. The methods of analysis
introduced in this chapter can then be applied to each stage separately since one stage will not
affect the other. Of course, the 20 V de supply must be applied to each isolated component.

Re R1 Re
R1 Cc Cc

re,
Vo

Q1 Q2

r, r,
+ RL
Vs '\, R2 + R2 +
RE RE

l -=- -=-
FIG. 4.64
-=- -=- -=-

R-C coupled BJT amplifiers.

Vee Vee

-=- -=- -=-

FIG. 4.65
DC equivalent of Fig. 4.64.
200 DC BIASING-BJTs The Darlington configuration of Fig. 4.66 feeds the output of one stage directly into
the input of the succeeding stage. Since the output of Fig. 4.66 is taken directly off the
emitter terminal, you will find in the next chapter that the ac gain is very close to 1 but
the input impedance is very high, making it attractive for use in amplifiers operating off
sources that have a relatively high internal resistance. If a load resistor were added to
the collector leg and the output taken off the collector terminal, the configuration would
provide a very high gain.

+Vee

c,

R,
Qz
Cc

+
v, '\,

t
FIC. 4.66
-=-
RE
T'' L

Darlington amplifier.

For the de analysis of Fig. 4.67 assuming a beta {3 1 for the first transistor and {3 2 for the
second, the base current for the second transistor is
IB2 = IE1 = (/31 + l)/B1
and the emitter current for the second transistor is
/e2 = (/32 + l)/B2 = (/32 + 1)(/31 + l)/B 1
Assuming f3 >> I for each transistor, we find the net beta for the configuration is

I f3v = /31/32 1 (4.so)

which compares directly with a single-stage amplifier having a gain of /3v-
Applying an analysis similar to that of Section 4.4 will result in the following equation
for the base current:

FIC. 4.67 Defining

DC equivalent of Fig. 4.66.
(4.51)

we have

Vee - VBEv
IB =------- (4.52)
[ RB + (f3v + l)RE

The currents

(4.53)
and the de voltage at the emitter terminal is MULTIPLE BJT 201
NETWORKS
I VE2 = fe2RE I (4.54)

The collector voltage for this configuration is obviously equal to that of the source V.

(4.55)

and the voltage across the output of the transistor is

VcE2 = Vc2 - VE2

and I VcE2 = Vee - VE2 I (4.56)

The Cascode configuration of Fig. 4.68 ties the collector of one transistor to the emitter
of the other. In essence it is a voltage-divider network with a common-base configuration at
the collector. The result is a network with a high gain and a reduced Miller capacitance-a
topic to be examined in Section 9.9.

Vee

Vee Vee

Re
R1 R1
Vo Vc2
Ic2

r
C1 Cc
Qz VB2 Qz
RL +
R2 R2 Vcl = VE2

r· Q1
-=- Vnl
+
VBE1
R, VEI
R3 R3 /El
+
RE
v, 'v ICE
~ -=- -=- -=- -=-
FIC. 4.68 FIC. 4.69
Cascade amplifier. DC equivalent of Fig. 4.68.

The de analysis is initiated by assuming the current through the bias resistors R 1, R2 , and
R 3 of Fig. 4.69 is much larger than the base current of each transistor. That is,
IR1 ~ IR2 ~ IR3 >> IB1 or /B2
The result is that the voltage at the base of the transistor Q 1 is simply determined by an
application of the voltage-divider rule:

(4.57)

The voltage at the base of the transistor Q2 is found in the same manner:

(4.58)
202 DC BIASING-BJTs The emitter voltages are then determined by

I VE1 = VB1 - VBE1 I (4.59)

and (4.60)

with the emitter and collector currents determined by:

(4.61)

The collector voltage Vc 1:

(4.62)

and the collector voltage Vc 2 :

(4.63)

The current through the biasing resistors is

(4.64)

and each base current is determined by

~ (4.65)
~
with
~ (4.66)
~
The next multistage configuration to be introduced is the Feedback Pair of Fig. 4.70,
which employs both an npn and pnp transistor. The result is a configuration that provides
high gain with increased stability.
The de version with all the currents labeled appears in Fig. 4.71.

Rs cs

+
vs
I
'\, Rs
'-------1 Qz

1
"II" "II" "II" "II"

FICi. 4.70 FICi. 4.71

Feedback Pair amplifier. DC equivalent of Fig. 4. 70.
 The base current MULTIPLE BJT 203
NETWORKS
IB 2 = le, = f31IB,
and Ie2 = f32IB2
so that (4.67)

The collector current

le= IE,+ IE2
~ + f31f32IB,
f31IB,
= /31(1 + f32)IB,
so that I le ~ f31f32IB, I (4.68)

Applying Kirchhoff's voltage law down from the source to ground will result in
Vee - IeRe - VEB 1 - IB,RB = 0
or Vee - VEB, - f31f32IB,Re - IBlB = 0

Vee - VEB 1
and JB = - - - - - (4.69)
1 RB + f31f32Re

The base voltage VB, is

(4.70)

and (4.71)

The collector voltage Ve 2 = VE, is

I Ve2 = Vee - IeRe I (4.72)

and (4.73)

In this case

(4.74)

and

so that (4.75)

The last multistage configuration to be introduced is the Direct Coupled amplifier such
as appearing in Example 4.26. Note the absence of a coupling capacitor to isolate the de
levels of each stage. The de levels in one stage will directly affect the de levels in succeed-
ing stages. The benefit is that the coupling capacitor typically limits the low-frequency
response of the amplifier. Without coupling capacitors, the amplifier can amplify signals of
very low frequency-in fact down to de. The disadvantage is that any variation in de levels
due to a variety of reasons in one stage can affect the de levels in the succeeding stages of
the amplifier.

EXAMPLE 4.26 Determine the de levels for the currents and voltages of the direct-coupled
amplifier of Fig. 4.72. Note that it is a voltage-divider bias configuration followed by a
common-collector configuration; one that is excellent in cases wherein the input imped-
ance of the next stage is quite low. The common-collector amplifier is acting like a buffer
between stages.
204 DC BIASING-BJTs Vee
14 V

Re 6.8 kfl
R1 33 kfl

R, c, Q1 ------111-------0 V0

+
~~
v, '\J R2 lOkfl
REI 2.2 ill

l -=-
FIG. 4.72
-=-

Direct-coupled amplifier.

Solution: The de equivalent of Fig. 4.72 appears as Fig. 4.73. Note that the load and
source are no longer part of the picture. For the voltage-divider configuration, the follow-
ing equations for the base current were developed in Section 4.5.
ETh - VnE
I --------
Bi - RTh + (/3 + l)RE1

with RTh = R1 II R2

and

14 V 14 V

t1c
Re 6.8 kfl
R1 33 kfl
Vc2
VB2
/32 =50

~ lei
VE2
~
/31 = 100
/Bl
RE2 1.2 kfl
R2 lOkfl REI 2.2 kfl
t/Ez
-=-
-=- -=-
FIG. 4.73
DC equivalent of Fig. 4. 72.

In this case,
RTh = 33 kfl II 10 kfl = 7.67 kfl
10 kfl(14 V)
and ETh = 10 kfl + 33 kfl = 3·26 V
 3.26V-0.7V CURRENT MIRRORS 205
so that I------------
ni - 7.67 kO + (100 + 1) 2.2 kO
2.56V
229.2kO
= 11.17 µA
with lei = f3ln1
= 100 (11.17 µ,A)
= 1.12mA
In Fig. 4.73 we find that

I Vn2 = Vee - IeRe I (4.76)

= 14 V - (1.12 mA)(6.8 kn)

= 14 V - 7.62V
= 6.38V
and VE2 = Vn2 - VnE2
= 6.38V - 0.7V
= 5.68V
resulting in

(4.77)

5.68V
l.2kO
= 4.73mA
Obviously,
(4.78)

= 14 V
and VeE2 = Ve2 - VE2

VeE2 = Vee - VE2 I (4.79)

= 14 V - 5.68V
= 8.32V

4.1 J CURRENT MIRRORS

•
The current mirror is a de network in which the current through a load is controlled by a
current at another point in the network. That is, if the controlling current is raised or low-
ered the current through the load will change to the same level. The discussion to follow
will demonstrate that the effectiveness of the design is dependent on the fact that the two
transistors employed have identical characteristics. The basic configuration appears in
Fig. 4.74. Note that the two transistors are back to back and the collector of one is con-
nected to the base of the two transistors.
Assume identical transistors will result in VnE1 = VnE2 and ln 1 = ln2 as defined by the
base-to-emitter characteristics of Fig. 4.75. Raise the base to emitter voltage, and the current
of each will rise to the same value.
Since the base to emitter voltages of the two transistors in Fig. 4. 74 are in parallel, they
must have the same voltage. The result is that ln1 = ln2at every set base to emitter voltage.
It is clear from Fig. 4.74 that In = ln1 + ln2
and if ln1 = ln2
then
206 DC BIASING-BJTs Vee lOV

/control~

R
In
___._ In

In, 1n2
~ ~
In,

0 VnE
FICi. 4.74 FICi. 4.75
Current mirror using back-to-back BJTs. Base characteristics
for transistor Q1
(and Qz).

In addition, lcontrol = lc1 + IB = lc 1 + 2IB1

but Ic1 = f31IB1
so lcontrol = f31IB1 + 2JB1 = (/31 + 2)JB1
and since /31 is typically >> 2, Icontrol ~ f31IB 1

or (4.80)

If the control current is raised, the resulting IBi will increase as determined by Eq. 4.80. If IBi
increases, the voltage VBEi must increase as dictated by the response curve of Fig. 4.75. If VBEi
increases, then VBE2 must increase by the same amount and IB2 will also increase. The result is
that h = I c2 = {31B 2 will also increase to the level established by the control current.
Referring to Fig. 4.74 we find the control current is determined by

Vee - VBE
Icontrol = R (4.81)

revealing that for a fixed V cc, the resistor R can be used to set the control current.
The network also has a measure of built-in control that will try to ensure that any varia-
tion in load current will be corrected by the configuration itself. For instance, if h should
try to increase for whatever reason, the base current of Q2 will also increase due to the
relationship IB2 = lcz/ {3 2 = h/{3 2 . Returning to Fig. 4.101, we find that an increase in IB 2
will cause voltage VBE2 to increase also. Because the base of Q2 is connected directly to the
collector of Q 1, the voltage VCEi will increase also. This action causes the voltage across
the control resistor R to decrease, causing IR to drop. But if IR drops, the base current IB will
drop, causing both IBi and IB2 to drop also. A drop in IB2 will cause the collector current and
therefore the load current to drop also. The result, therefore, is a sensitivity to unwanted
changes that the network will make every effort to correct.
The entire sequence of events just described can be presented on a single line as shown
below. Note that at one end the load current is trying to increase, and at the end of the se-
quence the load current is forced to return to its original level.
hi Ic i IB i VBE i VcE 1 t, IR t, IB t, IB t Ic th t
2 2 2 2 2

~Note~

EXAMPLE 4.27 Calculate the mirrored current I in the circuit of Fig. 4. 76.
Solution: Eq. (4.75):
Vee - VBE 12V - 0.7V = 10_27 mA
I== lcontrol == - - - -
R l.lkO
 +12V CURRENT MIRRORS 207

1.1 kil

-=-
FIG. 4.76
Current mirror circuit for Example 4.27.

EXAMPLE 4.28 Calculate the current I through each of the transistor Q2 and Q3 in the
circuit of Fig. 4.77.
Solution: Since VsE, = VsE2 = VsE3 then Is, = ls2 = ls3
I _ !control I I
Substituting B1 - f3 and / 82 = {3 with ls3 = {3
Icontrol I I
we have
/3 /3 /3
so I must equal /control
Vee - VsE 6V-0.7V
and !control = R ---- = 4.08 mA
1.3 kll

+6V

~ 1control
1.3 kQ

FIG. 4.77
Current mirror circuit for Example 4.28.

Figure 4.78 shows another form of current mirror to provide higher output impedance
than that of Fig. 4.74. The control current through R is
Vee - 2VsE le /3 + 1
!control= R = le+ /3 = -
13- I e = le
Assuming that Q1 and Q2 are well matched, we find that the output current I is held
constant at
I = le = /control
208 DC BIASING-BJTs Again we see that the output current I is a mirrored value of the current set by the fixed
current through R.
Figure 4. 79 shows still another form of current mirror. The junction field effect transistor
(see Chapter 6) provides a constant current set at the value of lvss• This current is mirrored,
resulting in a current through Q2 of the same value:
I= lvss

+Vi lvss
lcontroli R

-=-
FIG. 4.78 FIG. 4.79
Current mirror circuit with higher output Current mirror connection.
impedance.

4.14 CURRENT SOURCE CIRCUITS

•
The concept of a power supply provides the starting point in our consideration of current
source circuits. A practical voltage source (Fig. 4.80a) is a voltage supply in series with a
resistance. An ideal voltage source has R = 0, whereas a practical source includes some
small resistance. A practical current source (Fig. 4.80b) is a current supply in parallel with
a resistance. An ideal current source has R = oo D, whereas a practical current source
includes some very large resistance.

~ ~
1 Practical
° El--
Ideal
R

Practical Ideal
voltage source voltage source current source current source

(a) (b)

FIG. 4.80
Voltage and current sources.

+ An ideal current source provides a constant current regardless of the load connected to
it. There are many uses in electronics for a circuit providing a constant current at a very
high impedance. Constant-current circuits can be built using bipolar devices, FET devices,
and a combination of these components. There are circuits used in discrete form and others
more suitable for operation in integrated circuits.

-=- Bipolar Transistor Constant-Current Source

FIG. 4.81 Bipolar transistors can be connected in a circuit that acts as a constant-current source in a
Discrete constant-current source. number of ways. Figure 4.81 shows a circuit using a few resistors and an npn transistor for
operation as a constant-current circuit. The current through le can be determined as fol- CURRENT SOURCE 209
lows. Assuming that the base input impedance is much larger than R 1 or R2 , we have CIRCUITS
R1
VB= ---(-VEE)
R1 + R2
and VE= VB - 0.7V
VE - (-VEE)
with IE = - - - - - = le (4.82)
RE
where I c is the constant current provided by the circuit of Fig. 4.81.

EXAMPLE 4.29 Calculate the constant current I in the circuit of Fig. 4.82.
Solution:
R1 5.1 kO
VB= R1 + R2 (-VEE)= 5.1 kO + 5.1 kO (-20V) = -lOV
5.1 ill
5.1 lill 2 ill
VE= VB - 0.7V = lOV - 0.7V = -10.7V
VE - (-VEE) -10.7V - (-20V)
I= IE= - - - - - = -------
RE 2k0
9.3V -20V
= lkO = 4.65mA
FIG. 4.82
Constant-current source for
Example 4.29.
Transistor/Zener Constant-Current Source
Replacing resistor R 2 with a Zener diode, as shown in Fig. 4.83, provides an improved
constant-current source over that of Fig. 4.81. The Zener diode results in a constant current
calculated using the base-emitter KVL (Kirchhoff voltage loop) equation. The value of I
can be calculated using

(4.83) +
VBE(on)

A major point to consider is that the constant current depends on the Zener diode voltage,
which remains quite constant, and the emitter resistor RE. The voltage supply VEE has no
effect on the value of I.

EXAMPLE 4.30 Calculate the constant current I in the circuit of Fig. 4.84.

FIG. 4.83
Constant-current circuit using Zener
diode.

+
6.2V
2.2 ill 1.8 ill

-18 V
FIG. 4.84
Constant-current circuit for Example 4.30.

Solution:
. : I = Vz - VBE = 6.2V - 0.7V = 3 _06 mA
Eq. (483) = 3 mA
RE l.8k0
210 DC BIASING-BJTs 4.15 pnp TRANSISTORS
•
The analysis thus far has been limited totally to npn transistors to ensure that the initial
analysis of the basic configurations was as clear as possible and uncomplicated by switch-
ing between types of transistors. Fortunately, the analysis of pnp transistors follows the
same pattern established for npn transistors. The level of IB is first determined, followed by
Re the application of the appropriate transistor relationships to determine the list of unknown
Ra + quantities. In fact, the only difference between the resulting equations for a network in
+ t le which an npn transistor has been replaced by a pnp transistor is the sign associated with

U: Ia
+
+
VCE
particular quantities.
As noted in Fig. 4.85, the double-subscript notation continues as normally defined. The
current directions, however, have been reversed to reflect the actual conduction directions.
VaE
Using the defined polarities of Fig. 4.85, both VBE and VCE will be negative quantities.
RE Applying Kirchhoff's voltage law to the base-emitter loop results in the following equa-
+ tion for the network of Fig. 4.85:
"=" tIE -JERE+ VBE - IsRB + Vee= 0
FICi. 4.85 Substituting IE = (/3 + 1)/B and solving for IB yields
pnp transistor in an emitter-
stabilized configuration.
Vee+ VBE
I-------- (4.84)
B - RB + ({3 + l)RE

The resulting equation is the same as Eq. (4.17) except for the sign for VBE· However, in
this case VBE = -0. 7 V and the substitution of values results in the same sign for each term
ofEq. (4.84) as Eq. (4.17). Keep in mind that the direction of IB is now defined opposite of
that for a pnp transistor as shown in Fig. 4.85.
For VCE Kirchhoff s voltage law is applied to the collector-emitter loop, resulting in the
following equation:
-JERE+ VCE - IcRc + Vee= 0
Substituting IE == le gives

(4.85)

The resulting equation has the same format as Eq. (4.19), but the sign in front of each
term on the right of the equal sign has changed. Because Vcc will be larger than the mag-
nitude of the succeeding term, the voltage VCE will have a negative sign, as noted in an
earlier paragraph.

EXAMPLE 4.31 Determine VCE for the voltage-divider bias configuration of Fig. 4.86.

~------<l--0 -18 V

2.4kQ
47kQ 10 µF
( 0 V0

lOµF B
C +
Vi 0>------,)11-----+---0----0 VCE f3 = 120

E
lOkQ
1.1 kQ

FICi. 4.86
pnp transistor in a voltage-divider bias configuration.
Solution: Testing the condition TRANSISTOR SWITCHING 211
NETWORKS
f3RE 2='. lOR2
results in
(120)(1.1 kll) 2:: 10(10 kll)
132 kll 2:: 100 kll (satisfied)
Solving for VB, we have
V = R2Vcc =(10kll)(-18V)=_ 3 _16 V
B R1 + R2 47kll + lOkll
Note the similarity in format of the equation with the resulting negative voltage for VB.
Applying Kirchhoff's voltage law around the base-emitter loop yields
+VB - VBE - VE= 0
and
Substituting values, we obtain
VE = -3.16 V - (-0.7 V)
= -3.16V + 0.7V
= -2.46V
Note in the equation above that the standard single- and double-subscript notation is
employed. For an npn transistor the equation VE = VB - VBE would be exactly the same.
The only difference surfaces when the values are substituted.
The current is
VE 2.46V
IE=-=--=2.24mA
RE 1.1 kll
For the collector-emitter loop,
-JERE+ VCE - IcRc + Vee= 0
Substituting le == le and gathering terms, we have
VcE =-Vee+ Ic(Rc + RE)
Substituting values gives
VCE = -18 V + (2.24 mA)(2.4 kll + 1.1 kD,)
= -18 V + 7.84 V
= -10.16V

4.16 TRANSISTOR SWITCHING NETWORKS

•
The application of transistors is not limited solely to the amplification of signals. Through
proper design, transistors can be used as switches for computer and control applications.
The network of Fig. 4.87a can be employed as an inverter in computer logic circuitry. Note
that the output voltage Vc is opposite to that applied to the base or input terminal. In addi-
tion, note the absence of a de supply connected to the base circuit. The only de source is
connected to the collector or output side, and for computer applications is typically equal
to the magnitude of the "high" side of the applied signal-in this case 5 V. The resistor RB
will ensure that the full applied voltage of 5 V will not appear across the base-to-emitter
junction. It will also set the IB level for the "on" condition.
Proper design for the inversion process requires that the operating point switch from
cutoff to saturation along the load line depicted in Fig. 4.87b. For our purposes we will
assume that le = IcEo == 0 mA when IB = 0 µ,A (an excellent approximation in light of
improving construction techniques), as shown in Fig. 4.87b. In addition, we will assume
that VCE = VCE,., == 0 V rather than the typical 0.1-V to 0.3-V level.
When V; = 5 V, the transistor will be "on" and the design must ensure that the network
is heavily saturated by a level of IB greater than that associated with the IB curve appearing
212 DC BIASING-BJTs Vee =5V

V; 0.82 kQ

5V 5V

..,.._ hFE = 125 ..,.._

ov

(a)

le (mA)

7
Ie,.,=6.lmA"---- L - - - - - - - - - - - - - - - - -
6

0 2 3 4 5
Vee=5V

(b)

FICi. 4.87
Transistor inverter.

near the saturation level. In Fig. 4.87b, this requires that lB > 50 µ,A. The saturation level
for the collector current for the circuit of Fig. 4.87a is defined by

I Vee
e,., - -
- Re (4.86)

The level of lB in the active region just before saturation results can be approximated by
the following equation:
le,.,
l =-
Bmax - /3 de
For the saturation level we must therefore ensure that the following condition is satisfied:

le
lB > ~ (4.87)
f3ctc

For the network of Fig. 4.87b, when Vi = 5 V, the resulting level of lB is

Vi-0.7V 5V-0.7V
lB = - - - - - - - - - = 63 µ,A
RB 68 k!1
Vee 5V
and le,a, = -Re = 082
. k"'
,l.L
= 6.lmA
 Testing Eq. (4.87) gives TRANSISTOR SWITCHING 213
NETWORKS
lesat 6.1 mA
IB = 63µ,A > - = -- = 48.8µ,A
/3ac 125
which is satisfied. Certainly, any level of IB greater than 60 µ,A will pass through a Q-point
on the load line that is very close to the vertical axis.
For V; = 0 V, IB = 0 µ,A, and because we are assuming that le = IeEo = 0 mA, the
voltage drop across Re as determined by VRe = IeRe = 0 V, resulting in Ve= +5 V for
the response indicated in Fig. 4.87a.
In addition to its contribution to computer logic, the transistor can also be employed as a
switch using the same extremities of the load line. At saturation, the current I e is quite high
and the voltage VCE very low. The result is a resistance level between the two terminals
determined by
VeEsat
Rsat = --
lesat
and is depicted in Fig. 4.88.

e
==:>-
E

FIC. 4.88 FIC. 4.89

Saturation conditions and the resulting Cutoff conditions and the resulting terminal
terminal resistance. resistance.

Using a typical average value of VCEsat such as 0.15 V gives

_ VeEsat _ 0.15 V _ n
R t - - - - - - - 24.6 u
sa / Csat 6.1 mA
which is a relatively low value and can be considered as approximately O O when placed
in series with resistors in the kilohm range.
For V; = 0 V, as shown in Fig. 4.89, the cutoff condition results in a resistance level of
the following magnitude:

R - Vee
- 5V - oofi
cutoff - IeEO - 0 mA -

resulting in the open-circuit equivalence. For a typical value of ICEo = 10 µ,A, the magni-
tude of the cutoff resistance is
Vee 5V
R t ff = - - = - - = 500 kfi
cu o IeEo 10 µ,A
which certainly approaches an open-circuit equivalence for many situations.

EXAMPLE4.32 DetermineRBandReforthetransistorinverterofFig.4.90iflesat = 10 mA.

Vee= lOV

v;
lOV lOV lOV

v; o---"'~---<I hFE = 250

ov ov ov

FIC. 4.90
Inverter for Example 4.32.
214 DC BIASING-BJTs Solution: At saturation,
I
Vee
e,., - -
- Re
lOV
and lOmA =- -
Re
lOV
so that Re= - - = lkll
lOmA
At saturation,
le,., 10 mA
ls == - = -- = 40 µ,A
250 f3ctc
Choosing ls = 60 µ,A to ensure saturation and using
V; - 0.7V
ls=----
Rs
we obtain
V; - 0.7V 10V-0.7V
Rs=---- 155 kll
ls 60µ,A
Choose Rs= 150 kll, which is a standard value. Then
V; - 0.7V lOV - 0.7V
ls = Rs 150 kll = 62 µ,A

le,.,
and ls= 62µ,A > - = 40µ,A
f3ctc
Therefore, use Rs = 150 kfl and Re = 1 kfl.

There are transistors that are referred to as switching transistors due to the speed with
which they can switch from one voltage level to the other. fu Fig. 3 .23c the periods of time
defined as ts, td, tr, and t_rare provided versus collector current. Their impact on the speed of
response of the collector output is defined by the collector current response of Fig. 4.91. The
total time required for the transistor to switch from the "off" to the "on" state is designated
as t0 n and is defined by

(4.88)

Transistor "on" Transistor "off"

100% I
90% - ----------,---

10% -L-------------~--
1 I I

0 I I I I
l-+-1---- I I
1 I -----..i I
I I I I
~ I 4------
I I tr I 1
I I I
--------..:I ~ ----.i ~
ton

FIG. 4.91
Defining the time intervals of a pulse waveform.
with td the delay time between the changing state of the input and the beginning of a response TROUBLESHOOTING 215
at the output. The time element tr is the rise time from 10% to 90% of the final value. TECHNIQUES
The total time required for a transistor to switch from the "on" to the "off" state is re-
ferred to as t0 ff and is defined by

toff = ts + tr (4.89)

where ts is the storage time and trthe fall time from 90% to 10% of the initial value.
For the general-purpose transistor of Fig. 3.23c at/c = 10 mA, we find that
ts= 120ns
td = 25 ns
tr = 13 ns
and tr= 12ns
so that ton = tr + td= 13 ns + 25 ns = 38 ns
and t 0 ff =ts+ tr= 120ns + 12ns = 132ns
Comparing the values above with the following parameters of a BSV52L switching tran-
sistor reveals one of the reasons for choosing a switching transistor when the need arises:
t0 n = 12 ns and t 0 ff = 18 ns

4.17 TROUBLESHOOTING TECHNIQUES

The art of troubleshooting is such a broad topic that a full range of possibilities and tech-
niques cannot be covered in a few sections of a book. However, the practitioner should be
aware of a few basic maneuvers and measurements that can isolate the problem area and
•
possibly identify a solution.
Quite obviously, the first step in being able to troubleshoot a network is to fully under-
stand the behavior of the network and to have some idea of the expected voltage and current
levels. For the transistor in the active region, the most important measurable de level is the
base-to-emitter voltage.
For an "on" transistor, the voltage VBE should be in the neighborhood of 0.7 V.
The proper connections for measuring VsE appear in Fig. 4.92. Note that the positive
(red) lead is connected to the base terminal for an npn transistor and the negative (black)
lead to the emitter terminal. Any reading totally different from the expected level of about = 0.7 V Si
= 0.3 VGe
0.7 V, such as 0, 4, or 12 Vora negative value, would be suspect and the device or network = 1.2 V GaAs
connections should be checked. For a pnp transistor, the same connections can be used, but
a negative reading should be expected.
A voltage level of equal importance is the collector-to-emitter voltage. Recall from the
general characteristics of a BIT that levels of VCE in the neighborhood of 0.3 V suggest a
saturated device-a condition that should not exist unless it is being employed in a switch-
ing mode. However:
For the typical transistor amplifier in the active region, VCE is usually about FICi. 4.92
25% to 75% of V cc- Checking the de level of VsE•

For Vcc = 20 V, a reading of VCE of 1 V to 2 V or from 18 V to 20 V as measured in Fig.

4.93 is certainly an uncommon result, and unless the device was knowingly designed for this
response, the design and operation should be investigated. If VCE= 20 V (with Vcc = 20 V)
at least two possibilities exist-either the device (BIT) is damaged and has the characteristics

0.3 V = saturation
0 V = hort-circuit state
or poor connection
'ormally a few volt
or more

FICi. 4.93
Checking the de level ofVCE·
 216 DC BIASING-BJTs of an open circuit between collector and emitter terminals or a connection in the collector-
emitter or base-emitter circuit loop is open as shown in Fig. 4.94, establishing le at OmA and
VRc = 0 V. In Fig. 4.94, the black lead of the voltmeter is connected to the common ground
of the supply and the red lead to the bottom terminal of the resistor. The absence of a collector
current and a consequent zero voltage drop across Re will result in a reading of 20 V. If the
+ meter is connected between the collector terminal and ground of the BJT, the reading will be
½le= 0 V 0 V because Vcc is blocked from the active device by the open circuit. One of the most
common errors in the laboratory is the use of the wrong resistance value for a given design.
Imagine the impact of using a 680-0 resistor for RB rather than the design value of 680 kll.
For Vcc = 20 V and a fixed-bias configuration, the resulting base current would be
20V - 0.7V
lB = 680 ,n = 28.4 mA

rather than the desired 28.4 µ,A-a significant difference!

A base current of 28.4 mA would certainly place the design in a saturation region and pos-
sibly damage the device. Because actual resistor values are often different from the nominal
color-code value (recall the common tolerance levels for resistive elements), it is time well
FICi. 4.94 spent to measure a resistor before inserting it in the network. The result is measurements closer
Effect of a poor connection or to theoretical levels and some insurance that the correct resistance value is being employed.
damaged device. There are times when frustration will develop. You check the device on a curve tracer
or other BJT testing instrumentation and it looks good. All resistor levels seem correct,
the connections appear solid, and the proper supply voltage has been applied-what next?
Vee Now the troubleshooter must strive to attain a higher level of sophistication. Could it be
... that the internal connection of a lead is faulty? How often has simply touching a lead at the
'
'\\ proper point created a "make or break" situation between connections? Perhaps the supply
Re \
was turned on and set at the proper voltage but the current-limiting knob was left in the
I zero position, preventing the proper level of current as demanded by the network design.
Ve .... '- I CJ
\ \ Obviously, the more sophisticated the system, the broader is the range of possibilities. In
I \ (!) any case, one of the most effective methods of checking the operation of a network is to
\ \
'- + - check various voltage levels with respect to ground by hooking up the black (negative)
' '-?
VE ... ___ /
/
lead of a voltmeter to ground and "touching" the important terminals with the red (posi-
tive) lead. In Fig. 4.95, if the red lead is connected directly to Vcc, it should read Vcc volts
because the network has one common ground for the supply and network parameters. At
V c the reading should be less, as determined by the drop across Re, and VE should be less
than Vc by the collector-emitter voltage VCE· The failure of any of these points to register
what would appear to be a reasonable level may be sufficient in itself to define the faulty
FICi. 4.95
connection or element. If VRc and VRE are reasonable values but V CE is O V, the possibility
Checking voltage levels with respect
exists that the BJT is damaged and displays a short-circuit equivalence between collector
to ground.
and emitter terminals. As noted earlier, if VCE registers a level of about 0.3 Vas defined by
V CE = V c - VE ( the difference of the two levels as measured above), the network may be
in saturation with a device that may or may not be defective.
It should be somewhat obvious from the discussion above that the voltmeter section of the
VOM or DMM is quite important in the troubleshooting process. Current levels are usually
20V calculated from the voltage levels across resistors rather than "breaking" the network to insert
the milliammeter section of a multimeter. On large schematics, specific voltage levels are pro-
vided with respect to ground for easy checking and identification of possible problem areas.
Of course, for the networks covered in this chapter, one must simply be aware of typical levels
+ 3.3 kQ within the system as defined by the applied potential and general operation of the network.
19.85 V 250k0 All in all, the troubleshooting process is a true test of your clear understanding of the
20V
proper behavior of a network and the ability to isolate problem areas using a few basic
measurements with the appropriate instruments. Experience is the key, and that will come
f3 = 100
only with continued exposure to practical circuits.

2 kQ EXAMPLE 4.33 Based on the readings provided in Fig. 4.96, determine whether the net-
work is operating properly and, if not, the probable cause.
Solution: The 20 Vat the collector immediately reveals that le = 0 mA, due to an open
FICi. 4.96 circuit or a nonoperating transistor. The level of VR8 = 19.85 V also reveals that the
Network for Example 4.33. transistor is "off" because the difference of V cc - VR8 = 0.15 Vis less than that required
to tum "on" the transistor and provide some voltage for VE· In fact, if we assume a short- BIAS STABILIZATION 217
circuit condition from base to emitter, we obtain the following current through RB:

I = -V:cc
-- -20V
-- = 79.4 A
Rs RB + RE 252 kO µ,
which matches that obtained from
VRB 19.85 V
/RB = RB = 250 kO = 79 .4 µ,A

If the network were operating properly, the base current should be

Vee - VBE 20V - 0.7V 19.3V
I = ----- ------- -- = 42.7 µ,A
B RB + (/3 + l)RE 250 kO + (101)(2 kO) 452 kO
The result, therefore, is that the transistor is in a damaged state, with a short-circuit condi-
tion between base and emitter.

EXAMPLE 4.J4 Based on the readings appearing in Fig. 4.97, determine whether the tran- 20V
sistor is "on" and the network is operating properly.
Solution: Based on the resistor values of R 1 and R2 and the magnitude of Vcc, the volt-
age VB = 4 V seems appropriate (and in fact it is). The 3.3 V at the emitter results in a 4.7kO
0.7-V drop across the base-to-emitter junction of the transistor, suggesting an "on" transis- 80kO
20V
tor. However, the 20 Vat the collector reveals that le = 0 mA, although the connection to
the supply must be "solid" or the 20 V would not appear at the collector of the device. Two
4V
possibilities exist-there can be a poor connection between Re and the collector terminal
of the transistor or the transistor has an open base-to-collector junction. First, check the 3.3 V
continuity at the collector junction using an ohm-meter, and if it is okay, check the transis- 20kO
tor using one of the methods described in Chapter 3. lkO

4.18 BIAS STABILIZATION

The stability of a system is a measure of the sensitivity of a network to variations in its
parameters. In any amplifier employing a transistor the collector current le is sensitive to
each of the following parameters:
• FIC. 4.97
Networkfor Example 4.34.

/3: increases with increase in temperature

IVBEi: decreases about 2.5 m V per degree Celsius (°C) increase in temperature
Ico (reverse saturation current): doubles in value for every 10°C increase in temperature
Any or all of these factors can cause the bias point to drift from the designed point of
operation. Table 4.2 reveals how the levels of Ico and VBE change with increase in tempera-
ture for a particular transistor. At room temperature (about 25°C) Ico = 0.1 nA, whereas
at 100°C (boiling point of water) Ico is about 200 times larger, at 20 nA. For the same tem-
perature variation, /3 increases from 50 to 80 and VBE drops from 0.65 V to 0.48 V. Recall
that IB is quite sensitive to the level of VBE, especially for levels beyond the threshold value.

TABLE 4.2
Variation of Silicon Transistor Parameters
with Temperature

T(°C) Ico (nA) p VBE (V)

-65 0.2 X 10- 3 20 0.85

25 0.1 50 0.65
100 20 80 0.48
175 3.3 X 103 120 0.3

The effect of changes in leakage current (/co) and current gain (/3) on the de bias point
is demonstrated by the common-emitter collector characteristics of Fig. 4.98a and b. Fig-
ure 4.98 shows how the transistor collector characteristics change from a temperature of
 le (mA) le (mA)

50 µ,A
6 70µA 6

60µA 40 µ,A
5 5

50µA
30 µ,A
4

Increase (/3)

10 µ,A
2 11 . - - - - - -~ - - - - - - -

I CEO'= /3 I CBO Increase (/co)

0 5 15 20
leEo'= /3 leBO

(a) (b)

FIG. 4.98
Shift in de bias point (Q-point) due to change in temperature: (a) 25°C; (b) 100°C.

25°C to a temperature of 100°C. Note that the significant increase in leakage current not
only causes the curves to rise, but also causes an increase in beta, as revealed by the larger
spacing between curves.
An operating point may be specified by drawing the circuit de load line on the graph
of the collector characteristic and noting the intersection of the load line and the de base
current set by the input circuit. An arbitrary point is marked in Fig. 4.98a at IB = 30 µ,A.
Because the fixed-bias circuit provides a base current whose value depends approximately
on the supply voltage and base resistor, neither of which is affected by temperature or the
change in leakage current or beta, the same base current magnitude will exist at high tem-
peratures as indicated on the graph of Fig. 4.98b. As the figure shows, this will result in the
de bias point's shifting to a higher collector current and a lower collector-emitter voltage
operating point. In the extreme, the transistor could be driven into saturation. In any case,
the new operating point may not be at all satisfactory, and considerable distortion may result
because of the bias-point shift. A better bias circuit is one that will stabilize or maintain the
de bias initially set, so that the amplifier can be used in a changing-temperature environment.

Stability Fadors S(lco), S(V8 £), and S(/J)

A stability factor Sis defined for each of the parameters affecting bias stability as follows:

Mc
SUco) = Meo (4.90)

(4.91)

M
S(/3) = 11; (4.92)

In each case, the delta symbol (/1) signifies change in that quantity. The numerator of each
equation is the change in collector current as established by the change in the quantity
218
in the denominator. For a particular configuration, if a change in lco fails to produce a BIAS STABILIZATION 219
significant change in le, the stability factor defined by S(lco) = t::..lc/!::..lco will be quite
small. In other words:
Networks that are quite stable and relatively insensitive to temperature variations have
low stability factors.
In some ways it would seem more appropriate to consider the quantities defined by
Eqs. (4.90) through (4.92) to be sensitivity factors because:
The higher the stability factor, the more sensitive is the network to variations in that
parameter.
The study of stability factors requires the knowledge of differential calculus. Our pur-
pose here, however, is to review the results of the mathematical analysis and to form an
overall assessment of the stability factors for a few of the most popular bias configurations.
A great deal of literature is available on this subject, and if time permits, you are encour-
aged to read more on the subject. Our analysis will begin with the S(lco) level for each
configuration.

S(lco)
Fixed-Bias Configuration
For the fixed-bias configuration, the following equation results:

I S(/co) ~ /31 (4.93)

Emitter-Bias Configuration
For the emitter-bias configuration of Section 4.4, an analysis of the network results in

/3(1 + Rn/RE)
S(l ) = ---- (4.94)
co - /3 + Rn/BE

For Rn/RE >> {3, Eq. (4.94) reduces to the following:

I S(lco) ~ /3 I Rs/RE»/3 (4.95)

as shown on the graph of SUco) versus Rn/RE in Fig. 4.99.

For Rn/RE << 1, Eq. (4.94) will approach the following level (as shown in Fig. 4.99):

I S(/co) ~ 1 I R8/RE«l (4.96)

revealing that the stability factor will approach its lowest level as RE becomes sufficiently
large. Keep in mind, however, that good bias control normally requires that Rn be greater
than RE. The result therefore is a situation where the best stability levels are associated with
poor design criteria. Obviously, a trade-off must occur that will satisfy both the stability and
bias specifications. It is interesting to note in Fig. 4.99 that the lowest value of S(lco) is 1,
revealing that le will always increase at a rate equal to or greater than lco•
For the range where Rn/RE ranges between 1 and (/3 + 1), the stability factor will be
determined by

(4.97)

The results reveal that the emitter-bias configuration is quite stable when the ratio Rn/RE is
as small as possible and the least stable when the same ratio approaches {3.
Note that the equation for the fixed-bias configuration matches the maximum value for
the emitter-bias configuration. The result clearly reveals that the fixed-bias configuration
has a poor stability factor and a high sensitivity to variations in lco-
220 DC BIASING-BJTs SUco),
Stability factor
/3 t------ - - - - - -

FIG. 4.99
Variation of stability factor S(Ico) with the resistor ratio RB/RE
for the emitter-bias configuration.

Voltage-Divider Bias Configuration

Recall from Section 4.5 the development of the Thevenin equivalent network appearing in
Fig. 4.100, for the voltage-divider bias configuration. For the network of Fig. 4.100, the
equation for S(/co) is the following:

/3(1 + RTo/RE)
S(/co) - (4.98)
/3 + RTo/RE
Note the similarities with Eq. (4.94), where it was determined that S(/co) had its low-
est level and the network had its greatest stability when RE > RB- For Eq. (4.98), the
FIG. 4.100 corresponding condition is RE > RTh, or RTh/RE should be as small as possible. For the
Equivalent circuit for the voltage- voltage-divider bias configuration, RTh can be much less than the corresponding RTh of
divider configuration. the emitter-bias configuration and still have an effective design.

Feedbac:k-Bias Configuration (RE = o 0)

In this case,

/3(1 + RB/Re)
S(/ ) = (4.99)
co - /3 + RB/Re

Because the equation is similar in format to that obtained for the emitter-bias and voltage-divider
bias configurations, the same conclusions regarding the ratio RB/Re can be applied here also.

Physic:al lmpac:t
Equations of the type developed above often fail to provide a physical sense for why the
networks perform as they do. We are now aware of the relative levels of stability and how
the choice of parameters can affect the sensitivity of the network, but without the equations
it may be difficult for us to explain in words why one network is more stable than another.
The next few paragraphs attempt to fill this void through the use of some of the very basic
relationships associated with each configuration.
For the fixed-bias configuration of Fig. 4.101a, the equation for the base current is
Vee - VBE
IB=----
RB
with the collector current determined by

I le = PIB + (/3 + 1)/co I (4.100)

 ~-----ovee ~-----oVee ~-----+--o Vee BIAS STABILIZATION 221

+
+
Va +
+ VE
VE

-=- -=-
(a) (b) (c) (d)

FIC. 4.101
Review of biasing managements and the stability factor S(Ic0 ).

If le as defined by Eq. (4.93) should increase due to an increase in Ico, there is noth-
ing in the equation for IB that would attempt to offset this undesirable increase in current
level (assuming VBE remains constant). In other words, the level of le would continue
to rise with temperature, with IB maintaining a fairly constant value-a very unstable
situation.
For the emitter-bias configuration of Fig. 4.101b, however, an increase in le due to an
increase in Ico will cause the voltage VE = JERE == IcRE to increase. The result is a drop
in the level of IB as determined by the following equation:

(4.101)

A drop in IB will have the effect of reducing the level of le through transistor action
and thereby offset the tendency of le to increase due to an increase in temperature. In total,
therefore, the configuration is such that there is a reaction to an increase in le that will tend
to oppose the change in bias conditions.
The feedback configuration of Fig. 4.101c operates in much the same way as the emitter-
bias configuration when it comes to levels of stability. If le should increase due to an
increase in temperature, the level of VRe will increase in the equation

VCC - VBE - VRe j

¼~1 =-------- (4.102)
RB

and the level of IB will decrease. The result is a stabilizing effect as described for the
emitter-bias configuration. One must be aware that the action described above does not
happen in a step-by-step sequence. Rather, it is a simultaneous action to maintain the
established bias conditions. In other words, the very instant le begins to rise, the network
will sense the change and the balancing effect described above will take place.
The most stable of the configurations is the voltage-divider bias network of Fig. 4.101d.
If the condition f3RE >> 10R2 is satisfied, the voltage VB will remain fairly constant for
changing levels of le. The base-to-emitter voltage of the configuration is determined by
VBE = VB - VE· If le should increase, VE will increase as described above, and for a con-
stant VB the voltage VBE will drop. A drop in VBE will establish a lower level of IB, which
will try to offset the increased level of le.

EXAMPLE 4.35 Calculate the stability factor and the change in le from 25°C to 100°C for
the transistor defined by Table 4.2 for the following emitter-bias arrangements:
a. RB/RE = 250 (RB = 250RE).
b. RB/RE = 10 (RB = lORE).
c. RB/RE = 0.0l(RE = lOORB).
222 DC BIASING-BJTs Solution:
/3(1 + Rn/RE)
a. S(/co) = P + Rn/RE
50(1 + 250)
50 + 250
=
41.83
which begins to approach the level defined by P = 50.
The change in le is given by
Mc= [S(lco)] (Meo) = (41.83)(19.9 nA)
= 0.83µA
/3(1 + Rn/RE)
b. S(/co) = p + Rn/RE
50(1 + 10)
50 + 10
= 9.17
Mc= [S(Ico)](Mco) = (9.17)(19.9nA)
= 0.18µA
/3(1 + Rn/RE)
C. S(lco) = p + Rn/RE
50(1 + 0.01)
50 + 0.01
=1.01
which is certainly very close to the level of 1 forecast if Rn/RE << 1.
We have
Mc = [S(/co)] (Meo) = 1.01(19.9 nA)
= 20.lnA

Example 4.35 reveals how lower and lower levels of lco for the modem-day BJT
transistor have improved the stability level of the basic bias configurations. Even though
the change in le is considerably different in a circuit having ideal stability (S = 1) from
one having a stability factor of 41.83, the change in le is not that significant. For example,
the amount of change in l c from a de bias current set at, say, 2 mA, would be from 2 mA
to 2.00083 mA in the worst case, which is obviously small enough to be ignored for most
applications. Some power transistors exhibit larger leakage currents, but for most amplifier
circuits the lower levels of lco have had a very positive impact on the stability question.

The stability factor S(VnE) is defined by

Mc
S(VnE) = d VnE
Fixed-Bias Configuration
For the fixed-bias configuration:

(4.103)

Emitter-Bias Configuration
For the emitter-bias configuration:

(4.104)
 Substituting the condition f3 >> R8 /RE results in the following equation for S(YsE): BIAS STABILIZATION 223

-{3/RE 1
S(VsE) == - - = - - (4.105)
/3 RE

which shows that the larger the resistance RE, the lower is the stability factor and the more
stable is the system.

Voltage-Divider Configuration
For the voltage-divider configuration:

-{3/RE
(4.106)

Feedback-Bias Configuration
For the feedback-bias configuration:

-{3/Re
S(V ) - - - - (4.107)
BE - /3 + Rs/Re

EXAMPLE 4.36 Determine the stability factor S(VsE) and the change in le from 25°C to
100°C for the transistor defined by Table 4.2 for the following bias arrangements.
a. Fixed-bias with R8 = 240 kll and /3 = 100.
b. Emitter-bias with R8 = 240 kll, RE = 1 kll, and /3 = 100.
c. Emitter-bias with R8 = 47 kll, RE = 4.7 kll, and /3 = 100.

Solution:
a. Eq. (4.103): S(VsE) = _}!_
Rs
100
240 kll
-0.417 X 10- 3
and = [S(VsE)](aVsE)
t::..le
= (-0.417 X 10-3)(0.48V - 0.65V)
= (-0.417 X 10-3)(-0.17 V)
= 70.9 µA
b. In this case, f3 = 100 and R8 /RE = 240. The condition f3 >> R8 /RE is not satisfied,
negating the use of Eq. (4.105) and requiring the use of Eq. (4.104 ).
-{3/RE
Eq. (4.104): S(VsE) = /3 + Rs/RE
-(100)/(1 kll) -0.1
100 + (240 kll / 1 kll) 100 + 240
- 0.294 X 10- 3
which is about 30% less than the fixed-bias value due to the additional RE term in the
denominator of the S(VsE) equation. We have
t::..le = [ S(VsE)] (a VsE)
= (-0.294 X 10-3)(-0.17 V)
== 50 µA
c. In this case,
Rs 47 kll
f3 = 100 >> - = - - = 10 (satisfied)
RE 4.7kll
224 DC BIASING-BJTs 1
Eq. (4.105): S(VsE) = --
RE
1
4.7kll
= - 0.212 X 10- 3
and Mc = [S(VsE)](Li VsE)
= (-0.212 X 10-3)(-0.17V)
= 36.04µA

In Example 4.36, the increase of 70.9 µ,A will have some impact on the level of IcQ• For
a situation where IcQ = 2 mA, the resulting collector current increases to a 3.5% increase.
lcQ = 2 mA + 70.9 µ,A
= 2.0709mA
For the voltage-divider configuration, the level of Rs will be changed to RTh in Eq.
(4.104) (as defined by Fig. 4.100). In Example 4.36, the use of Rs = 47 kll is a question-
able design. However, RTh for the voltage-divider configuration can be this level or lower
and still maintain good design characteristics. The resulting equation for S(VsE) for the
feedback network will be similar to that of Eq. (4.104) with RE replaced by Re.

S(/J)
The last stability factor to be investigated is that of S(/3). The mathematical development is
more complex than that encountered for S(/co) and S(VsE), as suggested by some of the
following equations.

Fixed-Bias Configuration
For the fixed-bias configuration

le
S(/3) = /3: (4.108)

Emitter-Bias Configuration
For the emitter-bias configuration

Mc lei(l + Rs/RE)
S(/3) = - = ---- (4.109)
Li/3 /31 (/32 + Rs/RE)

The notation/c 1 and {3 1 is used to define their values under one set of network conditions,
whereas the notation {3 2 is used to define the new value of beta as established by such causes
as temperature change, variation in /3 for the same transistor, or a change in transistors.

EXAMPLE 4.37 Determine IcQ at a temperature of 100°C if IcQ = 2 mA at 25°C for the
emitter-bias configuration. Use the transistor described by Table 4.2, where {3 1 = 50 and
{32 = 80, and a resistance ratio Rs/RE of 20.
Solution:
IcP + Rs/RE)
Eq. (4.109): S(/3) = -/3-1(-1_+_f3_2_+_R_s_/R_E_)

(2 X 10-3)(1+ 20) 42 X 10-3

(50)(1 + 80 + 20) 5050
= 8.32 X 10- 6
and Mc = [S(,8) ][ Ll,B] BIAS STABILIZATION 225
= (8.32 X 10-6)(30)
~ 0.25mA
In conclusion, therefore, the collector current changed from 2 mA at room temperature to
2.25 mA at 100°C, representing a change of 12.5%.

Voltage-Divider Bias Configuration

For the voltage-divider bias configuration

IcP + RTo/RE)
S(,B) = -,8-1(,8_2_+_R_To_/R_E_) (4.110)

Feedback-bias Configuration
For the collector feedback-bias configuration

(4.111)

Summary
Now that the three stability factors of importance have been introduced, the total effect on
the collector current can be determined using the following equation for each configuration

I Mc = SUco)Mco + S(VsE)Ll VsE + S(,B)Ll,B I (4.112)

The equation may initially appear quite complex, but note that each component is simply
a stability factor for the configuration multiplied by the resulting change in a parameter
between the temperature limits of interest. In addition, the Mc to be determined is simply
the change in I c from the level at room temperature.
For instance, if we examine the fixed-bias configuration, Eq. (4. 78) becomes

(4.113)

after substituting the stability factors as derived in this section. Let us now use Table 4.2 to
find the change in collector current for a temperature change from 25°C (room temperature)
to 100°C (the boiling point of water). For this range the table reveals that
Meo= 20nA - 0.1 nA = 19.9nA
Ll VsE = 0.48 V - 0.65 V = -0.17 V (note the sign)
and Ll,B = 80 - 50 = 30
Starting with a collector current of 2 mA with an R8 of 240 kD, we obtain the resulting
change in le due to an increase in temperature of 75°C as follows:
50 2mA
Mc = (50)(19.9 nA) - 240 kD (-0.17 V) + 50 (30)
= 1 µ,A + 35.42 µ,A + 1200 µ,A
= l.236mA
which is a significant change due primarily to the change in ,8. The collector current has
increased from 2 mA to 3.236 mA, but this was expected in the sense that we recognize
from the content of this section that the fixed-bias configuration is the least stable.
If the more stable voltage-divider configuration is employed with a ratio RTo/RE = 2
and RE= 4.7 k!l, then
S(Ico) = 2.89, S(VsE) = -0.2 X 10-3, S(,8) = 1.445 X 10-6
226 DC BIASING-BJTs and Mc = (2.89)(19.9 nA) - 0.2 X 10-3(-0.17 V) + 1.445 X 10-6(30)
= 57.51 nA + 34 µ.,A + 43.4 µ.,A
= 0.077mA
The resulting collector current is 2.077 mA, or essentially 2.1 mA, compared to the 2.0 mA
at 25°C. The network is obviously a great deal more stable than the fixed-bias configuration,
as mentioned in earlier discussions. In this case, S(/3) did not override the other two factors,
and the effects of S(V8 E) and S(/co) were equally important. In fact, at higher temperatures,
the effects of S(/co) and S(VsE) will be greater than S(/3) for the device of Table 4.2. For
temperatures below 25°C, le will decrease with increasingly negative temperature levels.
The effect of S(/co) in the design process is becoming a lesser concern because of
improved manufacturing techniques, which continue to lower the level of Ico = lcso• It
should also be mentioned that for a particular transistor the variation in levels of/cso and VBE
from one transistor to another in a lot is almost negligible compared to the variation in beta.
In addition, the results of the analysis above support the fact that for a good stabilized design:

General Conclusion:
The ratio RsfRE or Rn/RE shouul be as small as possible with due consideration to
all aspects of the design, including the ac response.

Although the analysis above may have been clouded by some of the complex equations
for some of the sensitivities, the purpose here was to develop a higher level of awareness of
the factors that go into a good design and to be more intimate with the transistor parameters
and their impact on the network's performance. The analysis of the earlier sections was for
idealized situations with nonvarying parameter values. We are now more aware of how the
de response of the design can vary with the parameter variations of a transistor.

4.19 PRAOICAL APPLICATIONS

•
As with the diodes in Chapter 2, it would be virtually impossible to provide even a surface
treatment of the broad areas of application of BJTs. However, a few applications are cho-
sen here to demonstrate how different facets of the characteristics of BJTs are used to
perform various functions.

BJT Diode Usage and Protective Capabilities

As you begin to scan complex networks you will often find transistors being used where
all three terminals are not connected in the network-particularly the collector lead. In
such cases it is most likely being used as a diode rather than a transistor. There are a num-
ber of reasons for such use, including the fact that it is cheaper to buy a large number of
transistors rather than a smaller bundle and then pay separately for specific diodes. Also, in
ICs the manufacturing process may be more direct to make additional transistors that
introduce the diode construction sequence. Two examples of its use as a diode appear in
Fig. 4.102. In Fig. 4.102a it is being used in a simple diode network. In Fig. 4.102b it is
being used to establish a reference level.
Often times you will see a diode connected directly across a device as shown in Fig.
4.103 to simply ensure that the voltage across a device or system with the polarity shown
cannot exceed the forward bias voltage of 0. 7 V. In the reverse direction if the breakdown
strength is sufficiently high it will simply appear as an open circuit. Again, however, only
two terminals of the BJT are being employed.
The point to be made is that one should not assume that every BJT transistor in a network
is being used for amplification or as a buffer between stages. The number of areas of
application for BJTs beyond these areas is quite extensive.

Relay Driver
This application is in some ways a continuation of the discussion introduced for diodes
about how the effects of inductive kick can be minimized through proper design. In Fig.
4.104a, a transistor is used to establish the current necessary to energize the relay in the
 6V PRACTICAL 227
APPLICATIONS

2.2k0 +v

IE R
-~►----0 VL = 8 V - VBE =7.3 V
tIL=lE=ls

System

-=- -=-
(a) (b)

FICi. 4.102 -=-

BJT applications as a diode: (a) simple series diode circuit; (b) setting a reference level.
FICi. 4.103
Acting as a protective device.
collector circuit. With no input at the base of the transistor, the base current, collector cur-
rent, and coil current are essentially O A, and the relay sits in the unenergized state (nor-
mally open, NO). However, when a positive pulse is applied to the base, the transistor
turns on, establishing sufficient current through the coil of the electromagnet to close the
relay. Problems can now develop when the signal is removed from the base to tum off the
transistor and deenergize the relay. Ideally, the current through the coil and the transistor
will quickly drop to zero, the arm of the relay will be released, and the relay will simply
remain dormant until the next "on" signal. However, we know from our basic circuit
courses that the current through a coil cannot change instantaneously, and, in fact, the
more quickly it changes, the greater the induced voltage across the coil as defined by
vL = L(diLfdt). In this case, the rapidly changing current through the coil will develop a
large voltage across the coil with the polarity shown in Fig. 4.104a, which will appear
directly across the output of the transistor. The chances are likely that its magnitude will
exceed the maximum ratings of the transistor, and the semiconductor device will be per-
manently damaged. The voltage across the coil will not remain at its highest switching
level but will oscillate as shown until its level drops to zero as the system settles down.

At tum-off Vee

~ - ~NO NO
v------a
V; ~~-
§' VL ~ NC VL

11~NC
"'"o II - High-voltage spike
Von
lt/ V;
+ V;
When transistor
0 v~ E~vL /
voff R tumedoff

At tum-off
-=- -=-
(a) (b)

FICi. 4.104
Relay driver: (a) absence of protective device; (b) with a diode across the relay coil.

This destructive action can be subdued by placing a diode across the coil as shown in
Fig. 4.104b. During the "on" state of the transistor, the diode is back-biased; it sits like an
open circuit and doesn't affect a thing. However, when the transistor turns off, the voltage
across the coil will reverse and will forward-bias the diode, placing the diode in its "on"
state. The current through the inductor established during the "on" state of the transistor
can then continue to flow through the diode, eliminating the severe change in current level.
Because the inductive current is switched to the diode almost instantaneously after the
"off' state is established, the diode must have a current rating to match the current through
the inductor and the transistor when in the "on" state. Eventually, because of the resistive
228 DC BIASING-BJTs elements in the loop, including the resistance of the coil windings and the diode, the high-
frequency (quickly oscillating) variation in voltage level across the coil will decay to zero,
and the system will settle down.

Light Control
In Fig. 4.105a, a transistor is used as a switch to control the "on" and "off' states of the light-
bulb in the collector branch of the network. When the switch is in the "on" position, we have
a fixed-bias situation where the base-to-emitter voltage is at its 0.7-V level, and the base cur-
rent is controlled by the resistor R 1 and the input impedance of the transistor. The current
through the bulb will then be beta times the base current, and the bulb will light up. A prob-
lem can develop, however, if the bulb has not been on for a while. When a lightbulb is first
turned on, its resistance is quite low, even though the resistance will increase rapidly the
longer the bulb is on. This can cause a momentary high level of collector current, which
could damage the bulb and the transistor over time. In Fig. 4.105b, for instance, the load line
for the same network with a cold and a hot resistance for the bulb is included. Note that even
though the base current is set by the base circuit, the intersection with the load line results in
a higher current for the cold lightbulb. Any concern about the turn-on level can easily be cor-
rected by inserting an additional small resistor in series with the lightbulb, as shown in Fig.
43.105c, just to ensure a limit on the initial surge in current when the bulb is first turned on.

____. . ,__~-- V;

(a) (b) (c)

FICi. 4.105
Using the transistor as a switch to control the on-off states of a bulb: (a) network; (b) effect of low bulb resistance
on collector current; (c) limiting resistor.

Maintaining a Fixed Load Current

If we assume that the characteristics of a transistor have the ideal appearance of Fig. 4.106a
(constant beta throughout) a source, fairly independent of the applied load, can be constructed
using the simple transistor configuration of Fig. 4.106b. The base current is fixed so no matter
where the load line is, the load or collector current remains the same. In other words, the
collector current is independent of the load in the collector circuit. However, because the
actual characteristics are more like those in Fig. 4.106b, where beta will vary from point to
point, and even though the base current may be fixed by the configuration, the beta will vary
from point to point with the load intersection, and / c = h will vary-not characteristic of
a good current source. Recall, however, that the voltage-divider configuration resulted in a
low level of sensitivity to beta, so perhaps if that biasing arrangement is used, the current
source equivalent is closer to reality. In fact, that is the case. If a biasing arrangement such
as shown in Fig. 4.107 is employed, the sensitivity to changes in operating point due to
varying loads is much less, and the collector current will remain fairly constant for changes
in load resistance in the collector branch. In fact, the emitter voltage is determined by
VE= Vn - 0.7V
with the collector or load current determined by
VE Vn - 0.7V
le == IE = - = -----
RE RE
 Vee PRACTICAL 229
APPLICATIONS

Vee No variation
le Rs R1oad
inle
fs4
fs3
fs2 ~ IeQ
lsQ
Is,
0 VeE
-=-
(a) (b) (c)

FIC. 4.106
Building a constant-current source assuming ideal BJT characteristics: (a) ideal characteristics;
(b) network; (c) demonstrating why le remains constant.

Using Fig. 4.107, we can describe the improved stability by examining the case where
le may be trying to rise for any number of reasons. The result is that IE = le will also rise
and the voltage VRE = JERE will increase. However, if we assume Vn to be fixed (a good
assumption because its level is determined by two fixed resistors and a voltage source), the
base-to-emitter voltage VnE = Vn - VRE will drop. A drop in VnE will cause In and there-
fore le(= f3In) to drop. The result is a situation where any tendency for Ieto increase will
be met with a network reaction that will work against the change to stabilize the system.

Vee

Vs
R,

+
0.7V
⇒

VE
,,
0 ~

R2

-=-
FIC. 4.107
Network establishing a fairly constant current source
due to its reduced sensitivity to changes in beta.

Alarm System with a CCS

An alarm system with a constant-current source of the type just introduced appears in Fig.
4.108. Because f3RE = (100)(1 kll) = 100 kll is much greater than R 1, we can use the
approximate approach and find the voltage VR 1,
2 kll(l6 V) = 4 _78 V
VR, = 2 kll + 4.7 kll
and then the voltage across RE,
VRE = VR1 - 0.7V = 4.78V - 0.7V = 4.08V
and finally the emitter and collector current,
VRE 4.08 V
IE = - = - - = 4.08 mA = 4 mA = le
RE I kll
no DC BIASING-BJTs .-----------,---0 +16 V

+16V
Door
switch
Rret t
4mA
~=HJO
current
source R 2 =4.7 k!l
Window
foil
Out ut
4 To alarm
4mA bell circuit

FICi. 4.108
An alarm system with a constant-current source and an op-amp comparator.

Because the collector current is the current through the circuit, the 4-mA current will
remain fairly constant for slight variations in network loading. Note that the current passes
through a series of sensor elements and finally into an op-amp designed to compare the
4-mA level with the set level of 2 mA. (Although the op-amp may be a new device to you,
it will be discussed in detail in Chapter 10--you will not need to know the details of its
behavior for this application.)
The LM2900 operational amplifier of Fig. 4.108 is one of four found in the dual-in-
line integrated circuit package appearing in Fig. 4.109a. Pins 2, 3, 4, 7, and 14 were used

Dual-in-line package
y+ INPUT 3+ INPUT 4+ INPUT 4- OUTPUT 4 OUTPUT 3 INPUT 3-

14

On package
to identify
pin numbers
4

INPUT I+ INPUT z+ INPUT Z- OUTPUT 2 OUTPUT 1 INPUT 1- GND

TOPVIEW

(a)
.__ __,__--o Output
4

+
3
-Inputo------+----11 Rseries
+ V

2
+lnputo-----1

v,ow
0

"II"
7
l
"II" "II"

(b) (c)

FICi. 4.109
LM2900 operational amplifier: (a) dual-in-line package (DIP); (b) components; (c) impact of low-input impedance.
for the design of Fig. 4.108. For the sake of interest only, note in Fig. 4.109b the number PRACTICAL 2J1
of elements required to establish the desired terminal characteristics for the op-amp-as APPLICATIONS
mentioned earlier, the details of its internal operation are left for another time. The 2 rnA at
terminal 3 of the op-amp is a reference current established by the 16-V source and Rref at
the negative side of the op-amp input. The 2-mA current level is required as a level against
which the 4-rnA current of the network is to be compared. As long as the 4-mA current on
the positive input to the op-amp remains constant, the op-amp will provide a "high" output
voltage, exceeding 13.5 V, with a typical level of 14.2 V (according to the specification
sheets for the op-amp). However, if the sensor current drops from 4 rnA to a level below
2 rnA, the op-amp will respond with a "low" output voltage, typically about 0.1 V. The
output of the op-amp will then signal the alarm circuit about the disturbance. Note from
the above that it is not necessary for the sensor current to drop all the way down to O rnA
to signal the alarm circuit. Only a variation around the reference level that appears unusual
is required-a good alarm feature.
One very important characteristic of this particular op-amp is the low-input impedance
as shown in Fig. 4.109c. This feature is important because one does not want alarm circuits
reacting to every voltage spike or turbulence that comes down the line because of some
external switching action or outside forces such as lightning. In Fig. 4.109c, for instance,
if a high-voltage spike should appear at the input to the series configuration, most of the
voltage will appear across the series resistor rather than the op-amp-thus preventing a false
output and an activation of the alarm.

Logic: Gates
In this application we will expand on the coverage of transistor switching networks in Sec-
tion 4.15. To review, the collector-to-emitter impedance of a transistor is quite low near or
at saturation and large near or at cutoff. For instance, the load line defines saturation as the
point where the current is quite high and the collector-to-emitter voltage quite low as shown
. F"1g. 4110
m . . The resu1· .
tmg resistance, de f.medb y R sat = . low an d"1s of ten
VcE,.,(low).1s qmte
,
lc,.,<high)
approximated as a short circuit. At cutoff, the current is relatively low and the voltage near
its maximum value as shown in Fig. 4.110, resulting in a very high impedance between the
collector and emitter terminal, which is often approximated by an open circuit.

- - - - - - - - - - " ~ - - - - : : . . - - -Cutoff
1ccukltr- - ln=OµA

O : VeEsatumtion 1 Yee VCE

' VeE,utoff

FICi. 4.110
Points of operation for a BJT logic gate.

The above impedance levels established by "on" and "off' transistors make it relatively
easy to understand the operation of the logic gates of Fig. 4.111. Because there are two
inputs to each gate, there are four possible combinations of voltages at the input to the
transistors. A 1, or "on," state is defined by a high voltage at the base terminal to tum the
transistor on. A 0, or "off," state is defined by OVat the base, ensuring that transistor is off.
If both A and B of the OR gate of Fig. 4.111 a have a low or 0-V input, both transistors are
off (cutoff), and the impedance between the collector and the emitter of each transistor can
be approximated by an open circuit. Mentally replacing both transistors by open circuits
2J2 DC BIASING-BJTs between the collector and the emitter will remove any connection between the applied
bias of 5 V and the output. The result is zero current through each transistor and through
the 3.3-k!1 resistor. The output voltage is therefore 0 V, or "low"-a 0 state. On the other
hand, if transistor Q1 is on and Q2 is off due to a positive voltage at the base of Q1 and
0 V at the base of Q2, then the short-circuit equivalent between the collector and emitter
for transistor Q 1 can be applied, and the voltage at the output is 5 V, or "high"-a 1 state.
Finally, if both transistors are turned on by a positive voltage applied to the base of each,
they will both ensure that the output voltage is 5 V, or "high"-a 1 state. The operation
of the OR gate is properly defined: an output if either input terminal has applied turn-on
voltage or if both are in the "on" state. A 0 state exists only if both do not have a 1 state
at the input terminals.
The AND gate of Fig. 4.111 b requires that the output be high only if both inputs have a
tum-on voltage applied. If both are in the "on" state, a short-circuit equivalent can be used
for the connection between the collector and the emitter of each transistor, providing a di-
rect path from the applied 5-V source to the output-thereby establishing a high, or 1, state
at the output terminal. If one or both transistors are off due to 0 V at the input terminal, an
open circuit is placed in series with the path from the 5-V supply voltage to the output, and
the output voltage is 0 V, or an "off' state.

Vee 5V

Vee 5V
A o---"".,....,__....,,
lOkO

B R2 R2
B
lOkO lOkO lOkO
Qz Q2

C=A+B C=A•B
AND Gate
OR Gate RE 3.3k0 RE 3.3k0

"II" "II"

A B C A B C
0 0 0 0 0 0
0 1 1 0 1 0
1 0 1 1 0 0
1 1 1 1 1 1

1 = high (b)

0=low
(a)

FIC. 4.111
BJT logic gates: (a) OR; (b) AND.

Voltage Level Indicator

The last application to be introduced in this section, the voltage level indicator, includes
three of the elements introduced thus far: the transistor, the Zener diode, and the LED. The
voltage level indicator is a relatively simple network using a green LED to indicate when
the source voltage is close to its monitoring level of 9 V. In Fig. 4.112 the potentiometer is
set to establish 5.4 V at the point indicated. The result is sufficient voltage to tum on both
the 4.7-V Zener and the transistor and establish a collector current through the LED suffi- SUMMARY 2J3
cient in magnitude to turn on the green LED.
Once the potentiometer is set, the LED will emit its green light as long as the supply
voltage is near 9 V. However, if the terminal voltage of the 9-V battery should decrease,
the voltage set up by the voltage-divider network may drop to 5 V from 5.4 V. At 5 V there
is insufficient voltage to turn on both the Zener and the transistor, and the transistor will be
in the "off' state. The LED will immediately tum off, revealing that the supply voltage has
dropped below 9 V or that the power source has been disconnected.

9V - - - - - - - - - - - ~

l0kO.,,...f-----tla--~
+- ~
+ 0.7V
5.4V

-=-
FIC. 4.112
Voltage level indicator.

4.20 SUMMARY
Important Conclusions and Concepts •
1. No matter what type of configuration a transistor is used in, the basic relationships
between the currents are always the same, and the base-to-emitter voltage is the
threshold value if the transistor is in the "on" state.
2. The operating point defines where the transistor will operate on its characteristic
curves under de conditions. For linear (minimum distortion) amplification, the de
operating point should not be too close to the maximum power, voltage, or current
rating and should avoid the regions of saturation and cutoff.
3. For most configurations the de analysis begins with a determination of the base current.
4. For the de analysis of a transistor network, all capacitors are replaced by an open-
circuit equivalent.
5. The fixed-bias configuration is the simplest of transistor biasing arrangements, but it
is also quite unstable due its sensitivity to beta at the operating point.
6. Determining the saturation (maximum) collector current for any configuration can
usually be done quite easily if an imaginary short circuit is superimposed between
the collector and emitter terminals of the transistor. The resulting current through the
short is then the saturation current.
7. The equation for the load line of a transistor network can be found by applying
Kirchhoff's voltage law to the output or collector network. The Q-point is then deter-
mined by finding the intersection between the base current and the load line drawn on
the device characteristics.
8. The emitter-stabilized biasing arrangement is less sensitive to changes in beta-
providing more stability for the network. Keep in mind, however, that any resistance
in the emitter leg is "seen" at the base of the transistor as a much larger resistor, a
fact that will reduce the base current of the configuration.
9. The voltage-divider bias configuration is probably the most common of all the con-
figurations. Its popularity is due primarily to its low sensitivity to changes in beta
from one transistor to another of the same lot (with the same transistor label). The
exact analysis can be applied to any configuration, but the approximate one can be
applied only if the reflected emitter resistance as seen at the base is much larger than
the lower resistor of the voltage-divider bias arrangement connected to the base of the
transistor.
2J4 DC BIASING-BJTs 10. When analyzing the de bias with a voltage feedback configuration, be sure to
remember that both the emitter resistor and the collector resistor are reflected
back to the base circuit by beta. The least sensitivity to beta is obtained when the
reflected resistance is much larger than the feedback resistor between the base and
the collector.
11. For the common-base configuration the emitter current is normally determined
first due to the presence of the base-to-emitter junction in the same loop. Then the fact
that the emitter and the collector currents are essentially of the same magnitude is
employed.
12. A clear understanding of the procedure employed to analyze a de transistor network
will usually permit a design of the same configuration with a minimum of difficulty
and confusion. Simply start with those relationships that minimize the number of
unknowns and then proceed to make some decisions about the unknown elements of
the network.
13. In a switching configuration, a transistor quickly moves between saturation and cut-
off, or vice versa. Essentially, the impedance between collector and emitter can be
approximated as a short circuit for saturation and an open circuit for cutoff.
14. When checking the operation of a de transistor network, first check that the base-to-
emitter voltage is very close to 0.7 V and that the collector-to-emitter voltage is
between 25% and 75% of the applied voltage V cc-
15. The analysis of pnp configurations is exactly the same as that applied to npn transis-
tors with the exception that current directions will reverse and voltages will have the
opposite polarities.
16. Beta is very sensitive to temperature, and VnE decreases about 2.5 mV (0.0025 V)
for each 1° increase in temperature on a Celsius scale. The reverse saturation current
typically doubles for every 10° increase in Celsius temperature.
17. Keep in mind that networks that are the most stable and least sensitive to temperature
changes have the smallest stability factors.

Equations
VnE == 0.7V, IE = (/3 + l)In - le, le= f3ln
Fixed bias:

le= f3ln

Emitter stabilized:
Vee - VnE
I--------
B - Rn + (/3 + l)RE'
R; = (/3 + l)RE
Voltage-divider bias:

Approximate: Test

V -- R2Vcc
--
B - R1 + R2'

DC bias with voltage feedback:

Vee - VnE
In=-------
Rn + {3(Rc + RE),
Common base:

Transistor switching networks:

Vee le VcE,at
I
c,., - -
- Re, IB > ~{3 , Rsat = -1--, toff = ts + t_r
de Csat
Stability factors: COMPUTER ANALYSIS ns
Mc Mc
S(/co) = ~, S(/3) = Ll/3
u.lco
S(/co):
Fixed bias: S(Ico) ~ /3
/3(1 + Rs/RE)*
Emitter bias:
S(/co) = /3 + Rs/RE
*Voltage-divider bias: Change Rs to RTh in above equation.
*Feedback bias: Change RE to Re in above equation.
S(VsE):

Fixed bias:

Emitter bias:

tv oltage-divider bias: Change Rs to RTh in above equation.

tFeedback bias: Change RE to Re in above equation.
S(/3):

Fixed bias:

Emitter bias:

+Voltage-divider bias: Change Rs to RTh in above equation.

+Feedback bias: Change RE to Re in above equation.

4.21 COMPUTER ANALYSIS

Cadenc:e OrCAD
Voltage-Divider Configuration The results of Example 4.8 will now be verified using
•
Cadence OrCAD. Using methods described in detail in the previous chapters, we can con-
struct the network of Fig. 4.113. Recall from the previous chapter that the transistor is
found under the EVAL library, the de source under the SOURCE library, and the resistors
under the ANALOG library. The capacitor has not been called up earlier but can also be
found in the ANALOG library. For the transistor, the list of available transistors can be
found in the EVAL library.
The value of beta is changed to 140 to match Example 4.8 by first clicking on the
transistor symbol on the screen. It will then appear boxed in red to reveal it is in an active
status. Then proceed with Edit-PSpice Model, and the PSpice Model Editor Demo dialog
box will appear in which Bf can be changed to 140. As you try to leave the dialog box the
Model Editor/16.3 dialog box will appear asking if you want to save the changes in the
network library. Once they are saved, the screen will automatically return with beta set at
its new value.
The analysis can then proceed by selecting the New simulation profile key (looks like
a printout with an asterisk in the top left comer) to obtain the New Simulation dialog box.
Insert Fig. 4.113 and select Create. The Simulation Settings dialog box will appear in
which Bias Point is selected under the Analysis Type heading. An OK, and the system is
ready for simulation.
Proceed by selecting the Run PSpice key (white arrow in green background) or these-
quence PSpice-Run. The bias voltages will appear as shown in Fig. 4.113 if the V option
selected. The collector-to-emitter voltage is 13.19 V - 1.333 V = 11.857 V versus 12.22 V
of Example 4.8. The difference is primarily due to the fact that we are using an actual
transistor whose parameters are very sensitive to the operating conditions. Also recall the
difference in beta from the specification value and the value obtained from the plot of the
previous chapter.
2J6 DC BIASING-BJTs
Macro P~lce Aoces,orles
cadence

'"
FIG. 4.113 FIG. 4.114
Applying PSpice Windows to the voltage- Response obtained after changing /3 from 140
divider configuration of Example 4.8. to 255.9 for the network of Fig. 4.113.

Because the voltage-divider network has a low sensitivity to changes in beta, let us return
to the transistor specifications and replace beta by the default value of 255.9 and see how
the results change. The result is the printout of Fig. 4.114, with voltage levels very close to
those obtained in Fig. 4.113.
Note the distinct advantage of having the network set up in memory. Any parameter
can now be changed and a new solution obtained almost instantaneously-a wonderful
advantage in the design process.

Fixed-Bias Configuration Although the voltage-divider bias network is relatively

insensitive to changes in the beta value, the fixed-bias configuration is very sensitive to
beta variations. This can be demonstrated by setting up the fixed-bias configuration of
Example 4.1 using a beta of 50 for the first run. The results of Fig. 4.115 demonstrate
that the design is a fairly good one. The collector or collector-to-emitter voltage is
appropriate for the applied source. The resulting base and collector currents are fairly
common for a good design.
However, if we now go back to the transistor specifications and change beta back to the
default value of255.9, we obtain the results of Fig. 4.116. The collector voltage is now only
0.113 Vat a current of 5.4 rnA-a terrible operating point. Any applied ac signal would be
severely truncated due to the low collector voltage.

~ file fdit ~.,., _ • !!lace Macro PSpiee

I\Gcessories Qptions '!Yindow a
elp C d I! n CI! O!CAD C.,,tur•CIS - l>omo EdtlKln - V (SCt!EM..
flit fd;1 ~;ew lOOIS et,.Q 11!,>ao PSJ)lc, - • • •

': SCI-IEWATIC1-0Plcad .-
Ill>"""' w,.- l:lelP cadence

~r-. . 1f"'"
!.. f=a
-
""""'
-
'""'

FIG. 4.115 FIG. 4.116

Fixed-bias configuration with a /3 of 50. Network of Fig. 4.115 with a /3 of 255.9.
 Clearly, therefore, from the preceding analysis, the voltage-divider configuration is the COMPUTER ANALYSIS 2J7
preferred design if there is any concern about beta variations.

Multisim
Multisim will now be applied to the fixed-bias network of Example 4.4 to provide an
opportunity to review the transistor options internal to the software package and to com-
pare results with the handwritten approximate solution.
All the components of Fig. 4.117 except the transistor can be entered using the procedure
described in Chapter 2. Transistors are available through the Transistor key pad, which
is the fourth option down on the Component toolbar. When it is selected, the Select a
Component dialog box will appear, from which BJT_NPN is chosen. The result is a Com-
ponent list, from which 2N2222A can be selected. An OK, and the transistor will appear
on the screen with the labels Ql and 2N2222A. The label Bf = 50 can be added by first
selecting Place in the top toolbar followed by the Text option. Place the resulting marker
in the area you want to place the text and click once more. The result is a blank space with
a blinking marker for where the text will appear when entered. When finished, a second
double-click, and the label is set. To move the label to the position shown in Fig. 4.117,
simply click on the label to place the four small squares around the device. Then click it
once more and drag it to the desired position. Release the clicker, and it is in place. Another
click, and the four small markers will disappear.

ms 4 I Mulrisim {ms 4 l •)

IE'J fi le fd it l,'.iew f lace MCU ~mulate Tra.!J•ler Ioo15

Reports Qp110ns W,ndow Jielp

Cl '' JOl!ll~liiil~Eil'ica~ • mn n □ " GI"

~-~~------<

E -=- 20v

FICi. 4.117
Verifying the results of Example 4.4 using Multisim.

Even though the label may say Bf = 50, the transistor will still have the default param-
eters stored in memory. To change the parameters, the first step is to click on the device
to establish the device boundaries. Then select Edit, followed by Properties, to obtain
the BJT_NPN dialog box. If it is not already present, select Value and then Edit Model.
The result will be the Edit Model dialog box in which f3 and ls can be set to 50 and 1 nA,
respectively. Then choose Change Part Model to obtain the BJT_NPN dialog box again
and select OK. The transistor symbol on the screen will now have an asterisk to indicate
that the default parameters have been modified. One more click to remove the four markers,
and the transistor is set with its new parameters.
The indicators appearing in Fig. 4.117 were set as described in the previous chapter.
Finally, the network must be simulated using one of the methods described in Chapter 2.
For this example the switch was set to the 1 position and then back to the Oposition after the
Indicator values stabilized. The relatively low levels of current were partially responsible
for the low level of this voltage.
DC BIASING-BJTs The results are a close match with those of Example 4.4 with le = 2.217 mA, Vn =
2.636 V, Ve= 15.557 V, and VE= 2.26 V.
The relatively few comments required here to permit the analysis of transistor networks
is a clear indication that the breadth of analysis using Multisim can be expanded dramati-
cally without having to learn a whole new set of rules-a very welcome characteristic of
most technology software packages.

PROBLEMS
*Note: Asterisks indicate more difficult problems.
4.3 Fixed-Bias Configuration
•
1. For the fixed-bias configuration of Fig. 4.118, determine:
a. lsQ•
b. IeQ•
c. VeEQ·
d. Ve,
e. Vs,
f. VE,
16 V

1.8 kQ
510kQ

FIG. 4.118
Problems I, 4, 6, 7, 14, 65, 69,
71, and 75.

2. Given the information appearing in Fig. 4.119, determine:

a. le,
b. Re,
c. Rs,
d. VCE,
3. Given the information appearing in Fig. 4.120, determine:
a. le,
b. Vee•
c. {3.
d. Rs,

12 V

i le

2.2kQ
Rs
Ve =6V
+ +

~
l 8 =40µA
VeE /3= 80 ~ VCE =7.2 V J3

i IE=4mA
"II" "II"

FIG. 4.119 FIG. 4.120

Problem 2. Problem 3.
 4. Find the saturation current (le,.,) for the fixed-bias configuration of Fig. 4.118.
*5. Given the BJT transistor characteristics of Fig. 4.121:
a. Draw a load line on the characteristics determined by E = 21 V and Re = 3 kil for a
fixed-bias configuration.
b. Choose an operating point midway between cutoff and saturation. Determine the value of
RB to establish the resulting operating point.
c. What are the resulting values of I cQ and VcEQ?
d. What is the value of /3 at the operating point?
e. What is the value of a defined by the operating point?
f. What is the saturation current (/c,.,) for the design?
g. Sketch the resulting fixed-bias configuration.
h. What is the de power dissipated by the device at the operating point?
i. What is the power supplied by Vcc?
j. Determine the power dissipated by the resistive elements by taking the difference between
the results of parts (h) and (i).

lc(mA)

110,, A

100 I
10 90µ

80_.
9
70µ
8
60µ
7
50
6
40
5
3011
4

3 20µ

2
16'
1
ls 0 µ~

0 5 10 15 20 25 30 VcE (V)

FIC. 4.121
Problems 5, 6, 9, 13, 24, 44, and 57.

6. a. Ignoring the provided value of /3(120) draw the load line for the network of Fig. 4.118 on the
characteristics of Fig. 4.121.
b. Find the Q-point and the resulting IcQ and VCEQ·
c. What is the beta value at this Q-point?
7. If the base resistor of Fig. 4.118 is increased to 910 kil, find the new Q-point and resulting
values of IcQ and VCEQ·
4.4 Emitter-Bias Configuration
8. For the emitter-stabilized bias circuit of Fig. 4.122, determine:
a. IBQ·
b. IcQ•
c. VcEQ·
d. Ve,
e. VB,
r. vE.
DC BIASING-BJTs 20V

470Q
270kQ
i IcQ
Ve
+
VB
lJ.
JBQ
VcEQ /3= 125

VE

2.2kQ

FIG. 4.122
Problems 8, 9, 12, 14, 66, 69, 72, and 76.

9. a. Draw the load line for the network of Fig. 4.122 on the characteristics of Fig. 4.121 using /3
from problem 8 to find lBQ·
b. Find the Q-point and resulting values leQ and VCEQ·
c. Find the value of /3 at the Q-point.
d. How does the value of part (c) compare with /3 = 125 in problem 8?
e. Why are the results for problem 9 different from those of problem 8?
10. Given the information provided in Fig. 4.123, determine:
a. Re.
b. RE.
c. RB.
d. VCE.
e. VB.
11. Given the information provided in Fig. 4.124, determine:
a. /3.
b. Vee-
c. RB.

12 V

7)2mA
Re
20 µAi
2.7 kQ
7.6V
+
VB + /3
VcE /3 = 80
Vrn=7.3V

2.4 V 0------02.1 V

RE 0.68 kn

-=- T

FIG. 4.123 FIG. 4.124

Problem 10. Problem 11.

12. Determine the saturation current (les.,) for the network of Fig. 4.122.
*13. Using the characteristics of Fig. 4.121, determine the following for an emitter-bias configura-
tion if a Q-point is defined atleQ = 4 mA and VeEQ = 10 V.
a. Reif Vee= 24 V and RE= 1.2kil.
b. /3 at the operating point.
c. RB.
d. Power dissipated by the transistor.
e. Power dissipated by the resistor Re.
*14. a. Determine le and VCE for the network of Fig. 4.118.
b. Change f3 to 180 and determine the new value of le and VCE for the network of Fig. 4.118.
c. Determine the magnitude of the percentage change in le and VCE using the following
equations:

%Mc= llc(part;- lc(parta)I X 100%, %-<lVcE = IVcE(part;) - VcE(parta)I X 100%

C(parta) CE(parta)
d. Determine le and VcE for the network of Fig. 4.122.
e. Change /3 to 187.5 and determine the new value of le and VcE for the network of Fig. 4.122.
f. Determine the magnitude of the percentage change in le and VCE using the following
equations:
1c(part ) - 1c(partd) I V CE(partc) - VCE(partd) I
%Mc= I ' X 100%, %.<lVCE = I- - - - - - - X 100%
1c(partdJ VcEcpartd)
g. In each of the above, the magnitude of f3 was increased 50%. Compare the percentage
change in le and VCE for each configuration, and comment on which seems to be less sensi-
tive to changes in {3.

4.5 Voltage-Divider Bias Configuration

15. For the voltage-divider bias configuration of Fig. 4.125, determine:
a. lsQ•
b. lcQ•
c. VcEQ·
d. Ve.
e. VE.
f. Vs.
16. a. Repeat problem 15 for /3 = 140 using the general approach (not the approximate).
b. What levels are affected the most? Why?
17. Given the information provided in Fig. 4.126, determine:
a. le.
b. VE.
c. Vs.
d. R 1.

16V

~--------018 V
3.9 kn
62 kil ~ IcQ
Ve
Vs +
VCEQ f3 = 80
~

lsQ
VE
9.1 kn
0.68 kn
1.2 kn

"II" "II" "II"

FICi. 4.125 FICi. 4.126

Problems l 5, 16, 20, 23, 25, 67, Problems 17 and 19.
69, 70, 73, and 77.

18. Given the information appearing in Fig. 4.127, determine:

a. le.
b. VE.
c. Vcc-
d. VcE•
e. Vs.
f. R1.
DC BIASING-BJTs Vee

2.7 kQ
R1
pc
--
10.6 V
ZOµA +
VcE /3 = 100
Vs
VE
8.2 kQ
1.2 kQ

FIG. 4.127
Problem 18.

19. Determine the saturation current (le,.,) for the network of Fig. 4.125.
20. a. Repeat problem 16 with {3 = 140 using the approximate approach and compare results.
b. Is the approximate approach valid?
*21. Determine the following for the voltage-divider configuration of Fig. 4.128 using the approxi-
mate approach if the condition established by Eq. (4.33) is satisfied.
a. le.
b. VCE.
c. IB.
d. VE.
e. VB·

18 V

3.3 kQ
39 kQ
i le

--
ls

Vs
+
VCE

VE
f3 = 120

8.2 kQ
1 kQ

-=- -=-
FIG. 4.128
Problems 21, 22, and 26.

*22. Repeat Problem 21 using the exact (Thevenin) approach and compare solutions. Based on the
results, is the approximate approach a valid analysis technique if Eq. (4.33) is satisfied?
23. a. Determine IeQ, VCEQ, and IBQ for the network of Problem 15 (Fig. 4.125) using the approxi-
mate approach even though the condition established by Eq. (4.33) is not satisfied.
b. Determine I eQ, VCEQ, and / BQ using the exact approach.
c. Compare solutions and comment on whether the difference is sufficiently large to require
standing by Eq. (4.33) when determining which approach to employ.
*24. a. Using the characteristics of Fig. 4.121, determine Re and RE for a voltage-divider network
having a Q-point ofleQ = 5 rnA and VCEQ = 8 V. Use Vee= 24 V and Re = 3RE.
b. Find VE.
c. Determine VB.
d. Find R2 if R 1 = 24 kil assuming that f3RE > 10R2 .
e. Calculate {3 at the Q-point.
f. Test Eq. (4.33), and note whether the assumption of part (d) is correct.
*25. a. Determine le and VcE for the network of Fig. 4.125.
b. Change /3 to 120 (50% increase), and determine the new values of le and VcE for the net-
work of Fig. 4.125.
c. Determine the magnitude of the percentage change in le and VCE using the following
equations:

%Mc = Ilc(partb) - lc(parta) IX 100%, %~ VCE = IVCE(partb)- VCE(parta) I X 100%

lc(parta) VCE(parta)
d. Compare the solution to part (c) with the solutions obtained for parts (c) and (t) of Problem 14.
e. Based on the results of part (d), which configuration is least sensitive to variations in /3?
*26. a. Repeat parts (a) through (e) of Problem 25 for the network of Fig. 4.128. Change /3 to 180
in part (b).
b. What general conclusions can be made about networks in which the condition f3RE > 10R2
is satisfied and the quantities l c and V CE are to be determined in response to a change in /3?

4.6 Collector-Feedback Configuration

27. For the collector-feedback configuration of Fig. 4.129, determine:
a. lB.
b. le.
c. Ve. +16 V

3.6 kQ
270kQ
Ve
Jic
/3 = 120

l.2kQ

FICi. 4.129
Problems 27, 28, 74, and 78.

28. For the network of problem 27

. . h . V'
a. Determme lcQ usmg t e equation lcQ = R'

b. Compare with the results of problem 27 for lcQ•

c. Compare R' to RF//3·
d. Is the statement valid that the larger R' is compared with RF//3• the more accurate the
equation lcQ = ;: ? Prove using a short derivation for the exact current lcQ•
e. Repeat parts (a) and (b) for /3 = 240 and comment on the new level of lcQ•
29. For the voltage feedback network of Fig. 4.130, determine:
a. le.
b. Ve.
c. VE.
d. VcE•
30. a. Compare levels of R' = Re + RE to RF//3 for the network of Fig. 4.131.
b. Is the approximation lcQ = V'/R' valid?
*31. a. Determine the levels of le and VcE for the network of Fig. 4.131.
b. Change /3 to 135 (50% increase), and calculate the new levels of le and VcE•
c. Determine the magnitude of the percentage change in le and VCE using the following equations:

%Mc = Ilc(partb) - lc(parta) IX 100%, %~ VCE = IVCE(partb) - VCE(parta) I X 100%

l C(part a) VCE(part a)
d. Compare the results of part (c) with those of Problems 14(c), 14(f ), and 25(c). How does
the collector-feedback network stack up against the other configurations in sensitivity to
changes in /3?
DC BIASING-BJTs 30V
+22V

8.2kQ

V l0µF 9.1 kQ
~"""""".....--.----"'""'f'u--+--c---t{-------o Vo 470 kQ

IOµF
)Ic +

V; o------)1-----+---------t VCE /3= 180

1.8 kQ 9.1 kQ
r5µF

FIG. 4.130 FIG. 4.131

Problems 29 and 30. Problems 30 and 31.

32. Determine the range of possible values for Ve for the network of Fig. 4.132 using the 1-Mf!
potentiometer.
*33. Given VB = 4 V for the network of Fig. 4.133, determine:
a. VE.
b. le.
c. Ve.
d. VCE·
e. JB.
f. /3. +12 V
18 V

4.7 kQ 2.2Hl

330 kQ
~ - - - " ' ~ - - + - - - - o Ve
lMQ

f3 = 180 V8 =4V --
ls 'jlc +
VCE {3

3.3 kQ 1.2 kQ

'II' 'II'

FIG. 4.132 FIG. 4.133

Problem 32. Problem 33.

4.7 Emitter-Follower Configuration

*34. Determine the level of VE and h for the network of Fig. 4.134.

6V

330kQ

f3 = 120

1.2 kQ

-6 V

FIG. 4.134
Problem 34.
 35. For the emitter follower network of Fig. 4.135
a. Find IB, le, and h-
b. Determine VB, Ve, and VE.
e. Calculate VBc and VCE·

12V

(----o V0

l.2k0

FIC. 4.135
Problem 35.

4.8 Common-Base Configuration

*36. For the network of Fig. 4.136, determine:
a. IB.
b. Ic-
e. VCE.
d. Ve.
*37. For the network of Fig. 4.137, determine:
a. h-
b. Ve-
e. VCE-
38. For the common-base network of Fig. 4.138
a. Using the information provided determine the value of Re-
h. Find the currents IB and h-
e. Determine the voltages VBc and VCE.

+16 V 14 V

12 kn

Ve ~le -8V
Is + (---<>V 0

Ii VCE /3 = 80 2.2 kO
- VCE + Ve 4V

9.1 kn t1 IE
f-----o V;
15 kn
1.8 kn
-=- -=-
-12 V lOV
-=-
FIC. 4.136 FIC. 4.137 FIC. 4.138
Problem 36. Problem 37. Problem 38.

4.9 Miscellaneous Bias Configurations

*39. For the network of Fig. 4.139, determine:
a. IB.
b. le.
e. VE.
d. VCE-
DC BIASING-BJTs 18 V

510 kil 3.9 kil

Ve
560 kQ
/3 = 130 1 Ve= 8 V
'f le
+
VE
510 kil 11<.'l fJ

"II"

FIG. 4.139 FIG. 4.140

Problem 39. Problems 40 and 68.

40. Given Ve = 8 V for the network of Fig. 4.140, determine:

a. lB.
b. le,
c. /3.
d. VCE,

4.11 Design Operations

41. DetermineReandRBforafixed-biasconfigurationifVee = 12V,/3 = 80,andleQ = 2.5mA
with VeEQ = 6 V. Use standard values.
42. Design an emitter-stabilized network at leQ = ½lesat and VeEQ = ½Vee• Use Vee= 20 V,
lesat = 10 mA, /3 = 120, and Re = 4RE. Use standard values.
43. Design a voltage-divider bias network using a supply of 24 V, a transistor with a beta of 110,
and an operating point of leQ = 4 mA and VeEQ = 8 V. Choose VE= !Vee- Use standard
values.
*44. Using the characteristics of Fig. 4.121, design a voltage-divider configuration to have a satura-
tion level of 10 mA and a Q-point one-half the distance between cutoff and saturation. The
available supply is 28 V, and VE is to be one-fifth of Vee• The condition established by Eq.
(4.33) should also be rnet to provide a high stability factor. Use standard values.

4.12 Multiple BJT Networks

45. For the R-C-coupled amplifier of Fig. 4.141 determine
a. the voltages VB, Ve, and VE for each transistor.
b. the currents lB, le, and le for each transistor

+20V

2.2 kn 22kn 2.2kn

18kn
----1(t------<O V0
lOµF lOµF
V; o )
lOµF QI /3 = 160 /3= 90

4.7kn 3.3 kn
lkn l.2kn

FIG. 4.141
Problem 45.
46. For the Darlington amplifier of Fig. 4.142 determine
a. the level of f3v,
b. the base current of each transistor.
c. the collector current of each transistor.
d. the voltages Vel' Ve2 , VE!' and VE2 -
 18 V

2.2MQ

0.1 µF
V; o----1)1--------11 /31 = 50, /32 = 75
VBE, = VBE, = 0.7 V

----1 11 -------<o V0
+ {
20µF
470Q

FIC. 4.142
Problem 46.

47. For the cascode amplifier of Fig. 4.143 determine

a. the base and collector currents of each transistor.
b. the voltages V81 , V82 , VE,, Ve,, VE2 , and Vc2 -

Vee= 22 V

Re
2.2kQ

-----11-(---0 V 0

C=5µF
Q2 /3 2 = 120

V, o>----_,),1----+-----
C8 = 5 µF

-=-
FIC. 4.143
Problem 47.

48. For the feedback amplifier of Fig. 4.144 determine

a. the base and collector current of each transistor.
b. the base, emitter, and collector voltages of each transistor.

4.13 Current Mirror Circuits

49. Calculate the mirrored current I in the circuit of Fig. 4.145.
DC BIASING-BJTs 12 V

2200 +18V

. ~---------1(---o Vo 2kO
V; o-------) o----+----o
/Ji =80
/32 = 160
1.8 MO /3 = 200

FIC. 4.144 FIC. 4.145

Problem 48. Problem 49.

*50. Calculate collector currents for Q1 and Q2 in Fig. 4.146.

+12 V

I
~ 2mA 3 ill

FIC. 4.146
Problem 50.

4.14 Current Source Circuits

51. Calculate the current through the 2.2-kil load in the circuit of Fig. 4.147.
52. For the circuit of Fig. 4.148, calculate the current/.

28V

2.2 ill

+6V Rn /3 = 100
/3 = 120
lO0ill

4.3kO
1.2 ill

-18 V

FIC. 4.147 FIC. 4.148

Problem 51. Problem 52.
*53. Calculate the current/ in the circuit of Fig. 4.149.

f3 = 200

+
1.5 ill 5.1 V

-12V

FICi. 4.149
Problem 53.

4.15 pnp Transistors

54. Determine Ve, VCE, and/c for the network of Fig. 4.150.
55. Deterruine Ve and IB for the network of Fig. 4.151.

-22V

2.2kQ
-12 V
82kQ
Ve
.....
IB
f3 = 220
510kQ

16kQ
0.75 kQ
VCE f3 = 100

FICi. 4.150 FICi. 4.151

Problem 54. Problem 55.

56. Determine h and Vc for the network of Fig. 4.152.

+8V 3.3 kQ

f3 = I 10

3.9kQ
~---""-"'---o-12 V

Ve
FICi. 4.152
Problem 56.

4.16 Transistor Switching Networks

*57. Using the characteristics of Fig. 4.121, deterruine the appearance of the output waveform for
the network of Fig. 4.153. Include the effects of VCE,.,, and determine IB,/Bmax' and le,., when
V; = 10 V. Determine the collector-to-emitter resistance at saturation and cutoff.
DC BIASING-BJTs lOV

v;
lOV
180 kQ
v; o---""~---
ov
"II"

FIG. 4.153
Problem 57.

*58. Design the transistor inverter of Fig. 4.154 to operate with a saturation current of 8 mA using a
transistor with a beta of 100. Use a level of IB equal to 120% of /Brnax and standard resistor values.

sv

v;

sv
v; o--~,..,'v---1 /3 = 100
ov

FIG. 4.154
Problem 58.

59. a. Using the characteristics of Fig. 3.23e, determine ton and t0 ff at a current of 2 mA. Note the
use of log scales and the possible need to refer to Section 9.2.
b. Repeat part (a) at a current of 10 mA. How have t0 n and t0 ff changed with increase in col-
lector current?
c. For parts (a) and (b), sketch the pulse waveform of Fig. 4.91 and compare results.
4.17 Troubleshooting Techniques
*60. The measurements of Fig. 4.155 all reveal that the network is not functioning correctly. List as
many reasons as you can for the measurements obtained.

20V 20V 20V

4.7kQ 4.7Hl 4.7kQ

470kQ 470kQ 470kQ
20V
+
20V

1.2kQ

"II"

(a) (b) (c)

FIG. 4.155
Problem 60.

*61. The measurements appearing in Fig. 4.156 reveal that the networks are not operating properly.
Be specific in describing why the levels obtained reflect a problem with the expected network
behavior. In other words, the levels obtained reflect a very specific problem in each case.
 16V 16V

3.6kQ 3.6 kQ
91kQ 91kQ

Vn= 9.4 V
/3 = 100 /3 = 100

4V
18 kQ 18 kQ
1.2 kQ 1.2 kQ

"II"

(a) (b)

FICi. 4.156
Problem 61.

62. For the circuit of Fig. 4.157:

a. Does Vc increase or decrease if RB is increased?
b. Does le increase or decrease if /3 is reduced?
c. What happens to the saturation current if /3 is increased?
d. Does the collector current increase or decrease if Vcc is reduced?
e. What happens to VCE if the transistor is replaced by one with smaller /3?
63. Answer the following questions about the circuit of Fig. 4.158:
a. What happens to the voltage Vc if the transistor is replaced by one having a larger value of /3?
b. What happens to the voltage VCE if the ground leg of resistor RB2 opens (does not connect
to ground)?
c. What happens to le if the supply voltage is low?
d. What voltage VCE would occur if the transistor base-emitter junction fails by becoming
open?
e. What voltage VCE would result if the transistor base-emitter junction fails by becoming a
short?
*64. Answer the following questions about the circuit of Fig. 4.159:
a. What happens to the voltage Vc if the resistor RB is open?
b. What should happen to VCE if /3 increases due to temperature?
c. How will VE be affected when replacing the collector resistor with one whose resistance is
at the lower end of the tolerance range?
d. If the transistor collector connection becomes open, what will happen to VE?
e. What might cause VCE to become nearly 18 V?

Yee=+l8V
+Yee=20V

Re
2.2kQ
+Vee= 16V Re
Rr lOkQ
Re
Rn 3.6 kQ 75 kQ
Ve
240 kQ
/3 = 90
/3 = 80
/3 = 120

VE
R2
lOkQ RE
RE= 1.5 kQ
l.2kQ

"II" "II" "II" 'II'

FICi. 4.157 FICi. 4.158 FICi. 4.159

Problem 62. Problem 63. Problem 64.
DC BIASING-BJTs 4.18 Bias Stabilization
65. Determine the following for the network of Fig. 4.118:
a. S(lea)-
b. S(VBE)-
c. S(/3), using Ti as the temperature at which the parameter values are specified and /3(T2) as
25% more than /3(Ti)-
d. Determine the net change in le if a change in operating conditions results in lea increasing
from 0.2 µ,A to 10 µ,A, VBE drops from 0.7 V to 0.5 V, and /3 increases 25%.
*66. For the network of Fig. 4.122, determine:
a. S(/ea)-
b. S(VBE)-
c. S(/3), using Ti as the temperature at which the parameter values are specified and /3(T2) as
25% more than /3(Ti)-
d. Determine the net change in le if a change in operating conditions results in lea increasing
from 0.2 µ,A to 10 µ,A, VBE drops from 0.7 V to 0.5 V, and /3 increases 25%.
*67. For the network of Fig. 4.125, determine:
a. S(lea)-
b. S(VBE)-
c. S(/3), using Ti as the temperature at which the parameter values are specified and /3(T2) as
25% more than /3(Ti)-
d. Determine the net change in le if a change in operating conditions results in lea increasing
from 0.2 µ,A to 10 µ,A, VBE drops from 0.7 V to 0.5 V, and /3 increases 25%.
*68. For the network of Fig. 4.140, determine:
a. S(/ea)-
b. S(VBE)-
c. S(/3), using Ti as the temperature at which the parameter values are specified and /3(T2) as
25% more than /3(Ti)-
d. Determine the net change in le if a change in operating conditions results in lea increasing
from 0.2 µ,A to 10 µ,A, VBE drops from 0.7 V to 0.5 V, and /3 increases 25%.
*69. Compare the relative values of stability for Problems 65 through 68. The results for Exercises
65 and 67 can be found in Appendix E. Can any general conclusions be derived from the
results?
*70. a. Compare the levels of stability for the fixed-bias configuration of Problem 65.
b. Compare the levels of stability for the voltage-divider configuration of Problem 67.
c. Which factors of parts (a) and (b) seem to have the most influence on the stability of the
system, or is there no general pattern to the results?

4.21 Computer Analysis

71. Perform a PSpice analysis of the network of Fig. 4.118. That is, determine le, VCE, and lB.
72. Repeat Problem 71 for the network of Fig. 4.122.
73. Repeat Problem 71 for the network of Fig. 4.125.
74. Repeat Problem 71 for the network of Fig. 4.129.
75. Repeat Problem 71 using Multisim.
76. Repeat Problem 72 using Multisim.
77. Repeat Problem 73 using Multisim.
78. Repeat Problem 74 using Multisim.
CHAPTER OBJECTIVES
•
• Become familiar with the re, hybrid, and hybrid 7T models for the BJT transistor.
• Learn to use the equivalent model to find the important ac parameters for an amplifier.
• Understand the effects of a source resistance and load resistor on the overall gain and
characteristics of an amplifier.
• Become aware of the general ac characteristics of a variety of important BJT
configurations.
• Begin to understand the advantages associated with the two-port systems approach to
single- and multistage amplifiers.
• Develop some skill in troubleshooting ac amplifier networks.

5.1 INTRODUCTION
•
The basic construction, appearance, and characteristics of the transistor were introduced in
Chapter 3. The de biasing of the device was then examined in detail in Chapter 4. We now
begin to examine the ac response of the BJT amplifier by reviewing the models most fre-
quently used to represent the transistor in the sinusoidal ac domain.
One of our first concerns in the sinusoidal ac analysis of transistor networks is the mag-
nitude of the input signal. It will determine whether small-signal or large-signal techniques
should be applied. There is no set dividing line between the two, but the application-and
the magnitude of the variables of interest relative to the scales of the device characteristics-
will usually make it quite clear which method is appropriate. The small-signal technique is
introduced in this chapter, and large-signal applications are examined in Chapter 12.
There are three models commonly used in the small-signal ac analysis of transistor
networks: the re model, the hybrid 7T model, and the hybrid equivalent model. This chapter
introduces all three but emphasizes the re model.

5.2 AMPLIFICATION IN THE AC DOMAIN

•
It was demonstrated in Chapter 3 that the transistor can be employed as an amplifying device.
That is, the output sinusoidal signal is greater than the input sinusoidal signal, or, stated
another way, the output ac power is greater than the input ac power. The question then arises
as to how the ac power output can be greater than the input ac power. Conservation of energy
dictates that over time the total power output, P 0 , of a system cannot be greater than its power
254 BJT AC ANALYSIS input, P;, and that the efficiency defined by 77 = P0 /P; cannot be greater than 1. The factor
missing from the discussion above that permits an ac power output greater than the input ac
power is the applied de power. It is the principal contributor to the total output power even
though part of it is dissipated by the device and resistive elements. In other words, there is an
"exchange" of de power to the ac domain that permits establishing a higher output ac power.
In fact, a conversion efficiency is defined by 77 = Pa(ac)/P;(ctc), where Pa(ac) is the ac power
to the load and P;(ctc) is the de power supplied.
Perhaps the role of the de supply can best be described by first considering the simple
de network of Fig. 5.1. The resulting direction of flow is indicated in the figure with a plot
of the current i versus time. Let us now insert a control mechanism such as that shown in
+ Fig. 5.2. The control mechanism is such that the application of a relatively small signal to
-=- E the control mechanism can result in a substantial oscillation in the output circuit.

"II"

FIC. 5.1 FIC. 5.2

Steady current established by a Effect of a control element on the steady-state flow of the electrical
de supply. system of Fig. 5.1.

That is, for this example,

iac(p-p) >> ic(p-p)
and amplification in the ac domain has been established. The peak-to-peak value of the
output current far exceeds that of the control current.
For the system of Fig. 5.2, the peak value of the oscillation in the output circuit is con-
trolled by the established de level. Any attempt to exceed the limit set by the de level will
result in a "clipping" (flattening) of the peak region at the high and low end of the output
signal. In general, therefore, proper amplification design requires that the de and ac com-
ponents be sensitive to each other's requirements and limitations.
However, it is extremely helpful to realize that:
The superposition theorem is applicable for the analysis and design of the de and ac
components of a BJT network, permitting the separation of the analysis of the de and
ac responses of the system.
In other words, one can make a complete de analysis of a system before considering the
ac response. Once the de analysis is complete, the ac response can be determined using a
completely ac analysis. It happens, however, that one of the components appearing in the
ac analysis of BJT networks will be determined by the de conditions, so there is still an
important link between the two types of analysis.

5.3 BIT TRANSISTOR MODELING

•
The key to transistor small-signal analysis is the use of the equivalent circuits (models) to
be introduced in this chapter.
A model is a combination of circuit elements, properly chosen, that best approximates
the actual behavior of a semiconductor device under specific operating conditions.
Once the ac equivalent circuit is determined, the schematic symbol for the device can
be replaced by this equivalent circuit and the basic methods of circuit analysis applied to
determine the desired quantities of the network.
In the formative years of transistor network analysis the hybrid equivalent network was
employed the most frequently. Specification sheets included the parameters in their listing,
and analysis was simply a matter of inserting the equivalent circuit with the listed values.
The drawback to using this equivalent circuit, however, is that it is defined for a set of oper- BJT TRANSISTOR 255
ating conditions that might not match the actual operating conditions. In most cases, this is MODELING
not a serious flaw because the actual operating conditions are relatively close to the chosen
operating conditions on the data sheets. In addition, there is always a variation in actual
resistor values and given transistor beta values, so as an approximate approach it was quite
reliable. Manufacturers continue to specify the hybrid parameter values for a particular
operating point on their specification sheets. They really have no choice. They want to give
the user some idea of the value of each important parameter so comparisons can be made
between transistors, but they really do not know the user's actual operating conditions.
In time the use of the re model became the more desirable approach because an impor-
tant parameter of the equivalent circuit was determined by the actual operating conditions
rather than using a data sheet value that in some cases could be quite different. Unfortu-
nately, however, one must still tum to the data sheets for some of the other parameters of
the equivalent circuit. The re model also failed to include a feedback term, which in some
cases can be important if not simply troublesome.
The re model is really a reduced version of the hybrid 7T model used almost exclusively
for high-frequency analysis. This model also includes a connection between output and
input to include the feedback effect of the output voltage and the input quantities. The full
hybrid model is introduced in Chapter 9.
Throughout the text the re model is the model of choice unless the discussion centers
on the description of each model or a region of examination that predetermines the model
that should be used. Whenever possible, however, a comparison between models will be
discussed to show how closely related they really are. It is also important that once you gain
a proficiency with one model it will carry over to an investigation using a different model,
so moving from one to another will not be a dramatic undertaking.
In an effort to demonstrate the effect that the ac equivalent circuit will have on the
analysis to follow, consider the circuit of Fig. 5.3. Let us assume for the moment that the
small-signal ac equivalent circuit for the transistor has already been determined. Because
we are interested only in the ac response of the circuit, all the de supplies can be replaced
by a zero-potential equivalent (short circuit) because they determine only the de (quiescent
level) of the output voltage and not the magnitude of the swing of the ac output. This is
clearly demonstrated by Fig. 5.4. The de levels were simply important for determining the
proper Q-point of operation. Once determined, the de levels can be ignored in the ac analy-
sis of the network. In addition, the coupling capacitors C1 and C2 and bypass capacitor C3
were chosen to have a very small reactance at the frequency of application. Therefore, they,
too, may for all practical purposes be replaced by a low-resistance path or a short circuit.
Note that this will result in the "shorting out" of the de biasing resistor RE. Recall that ca-
pacitors assume an "open-circuit" equivalent under de steady-state conditions, permitting
an isolation between stages for the de levels and quiescent conditions.

Vee

ilo
Re
R1 ( 0

C C2 +
.....
r,
I,
B
zD
Vo
R, + E

+ R2
V;
v, '\, Rt:
r C3
-i ':' ':' l
FIG. 5.3
Transistor circuit under examination in this introductory discussion.
256 BJT AC ANALYSIS .:. .:.
~1
0

Re

R1
C +
I;
~ B ~

~
zo
+ Vo
Z;
Rs E

+ V; R2

V, '\,
-1 T T T
1
T

FIG. 5.4
The network of Fig. 5.3 following removal of the de
supply and insertion of the short-circuit equivalent
for the capacitors.

It is important as you progress through the modifications of the network to define the ac
equivalent that the parameters of interest such as Z;, Za, I;, and Ia as defined by Fig. 5.5 be
carried through properly. Even though the network appearance may change, you want to be
sure the quantities you find in the reduced network are the same as defined by the original
network. In both networks the input impedance is defined from base to ground, the input
current as the base current of the transistor, the output voltage as the voltage from collector
to ground, and the output current as the current through the load resistor Re-

I; lo
~ ~

+ + ~
[.

V; ~ System ~ Vo
Z; zo + ---+- + ~'
V; R;
- ---+- -
I
I

T
..L.
T

FIG. 5.5 FIG. 5.6

Defining the important parameters Demonstrating the reason for the defined
of any system. directions and polarities.

The parameters of Fig. 5.5 can be applied to any system whether it has one or a thou-
sand components. For all the analysis to follow in this text, the directions of the currents,
the polarities of the voltages, and the direction of interest for the impedance levels are as
appearing in Fig. 5.5. In other words, the input current!; and output current/a are, by defini-
tion, defined to enter the system. If, in a particular example, the output current is leaving the
system rather than entering the system as shown in Fig. 5.5, a minus sign must be applied.
The defined polarities for the input and output voltages are also as appearing in Fig. 5.5. If
Va has the opposite polarity, the minus sign must be applied. Note that Z; is the impedance
"looking into" the system, whereas Za is the impedance "looking back into" the system
from the output side. By choosing the defined directions for the currents and voltages as
appearing in Fig. 5.5, both the input impedance and output impedance are defined as having
positive values. For example, in Fig. 5.6 the input and output impedances for a particular
system are both resistive. For the direction of I; and Ia the resulting voltage across the resis-
tive elements will have the same polarity as V; and Va, respectively. If Ia had been defined
as the opposite direction in Fig. 5.5 a minus sign would have to be applied. For each case
Z; = V;/ I; and Za = Va/ Ia with positive results if they all have the defined directions and
polarity of Fig. 5.5. If the output current of an actual system has a direction opposite to that
of Fig. 5 .5 a minus sign must be applied to the result because V0 must be defined as appear- THE re TRANSISTOR 257
ing in Fig. 5.5. Keep Fig. 5.5 in mind as you analyze the BJT networks in this chapter. It is MODEL
an important introduction to "System Analysis," which is becoming so important with the
expanded use of packaged IC systems.
Ifwe establish a common ground and rearrange the elements of Fig. 5.4, R 1 andR 2 will
be in parallel, and Re will appear from collector to emitter as shown in Fig. 5.7. Because
the components of the transistor equivalent circuit appearing in Fig. 5.7 employ familiar
components such as resistors and independent controlled sources, analysis techniques
such as superposition, Thevenin's theorem, and so on, can be applied to determine the
desired quantities.

Transistor small-signal
ac equivalent circuit
1--------------<>-------o
I; ,........
I
;---,, C +
~ I :0," \
------{;-: 1
~+ B ,....... _. ,;'
Z;
Rs Re
V;
+ E

vs '\,
-i -l- "II" "II"

FIG. 5.7
Circuit of Fig. 5.4 redrawn for small-signal ac analysis.

Let us further examine Fig. 5.7 and identify the important quantities to be determined
for the system. Because we know that the transistor is an amplifying device, we would
expect some indication of how the output voltage V0 is related to the input voltage V,-
the voltage gain. Note in Fig. 5.7 for this configuration that the current gain is defined
by Ai= Io/h
In summary, therefore, the ac equivalent of a transistor network is obtained by:
1. Setting all de sources to zero and replacing them by a short-circuit equivalent
2. Replacing all capacitors by a short-circuit equivalent
3. Removing all elements bypassed by the short-circuit equivalents introduced by steps
1 and2
4. Redrawing the network in a more convenient and logical form
In the sections to follow, a transistor equivalent model will be introduced to complete
the ac analysis of the network of Fig. 5. 7.

5.4 THE re TRANSISTOR MODEL

The re model for the CE, CB, and CC BJT transistor configurations will now be introduced
with a short description of why each is a good approximation to the actual behavior of a
BJT transistor.
•
C
lb
Common-Emitter Configuration ___._ B

The equivalent circuit for the common-emitter configuration will be constructed using the + +
device characteristics and a number of approximations. Starting with the input side, we find V; Vbe
E
the applied voltage Vi is equal to the voltage Vbe with the input current being the base cur- {I,
rent lb as shown in Fig. 5.8.
"II"
Recall from Chapter 3 that because the current through the forward-biased junction of
the transistor is fe, the characteristics for the input side appear as shown in Fig. 5.9a for FIG. 5.8
various levels of VBE· Taking the average value for the curves of Fig. 5.9a will result in the Finding the input equivalent circuit
single curve of Fig. 5 .9b, which is simply that of a forward-biased diode. for a BJT transistor.
258 BJT AC ANALYSIS

Various Average
values value
ofVc8 ofVc8

0 0.7V 0 0.7V
(a) (b)

FIG. 5.9
Defining the average curve for the characteristics of Fig. 5.9a.

For the equivalent circuit, therefore, the input side is simply a single diode with a current
le, as shown in Fig. 5.10. However, we must now add a component to the network that will
+ establish the current le of Fig. 5.10 using the output characteristics.
If we redraw the collector characteristics to have a constant /3 as shown in Fig. 5.11
(another approximation), the entire characteristics at the output section can be replaced by
a controlled source whose magnitude is beta times the base current as shown in Fig. 5.11.
Because all the input and output parameters of the original configuration are now present, the
FIG. 5.10 equivalent network for the common-emitter configuration has been established in Fig. 5.12.
Equivalent circuit for the input side
of a BJT transistor.

IC
~
0
I +
Constant 13 { f3lb
lb
1----------------[B2 ~

1------------------fsl +
Vbe
re Vee

FIG. 5.11 FIG. 5.12

Constant f3 characteristics. BJT equivalent circuit.

The equivalent model of Fig. 5.12 can be awkward to work with due to the direct con-
nection between input and output networks. It can be improved by first replacing the diode
by its equivalent resistance as determined by the level of IE, as shown in Fig. 5.13. Recall
from Section 1.8 that the diode resistance is determined by rv = 26 mV/Iv. Using the sub-
script e because the determining current is the emitter current will result in re = 26 m V / / E·
V; Vbe
{ f3Ib Now, for the input side: Z;=-=-
lb h h
~

+ +
~
Z;
re
re
Solving for Vbe: Vbe = lere =
= (/3 +
Uc + h)re
l)lbre
= (/3h + h)re

V; Vbe _ Vbe _ (/3 + l)hre

and Z·- -------
' lb h
FIG. 5.13
Defining the level of Z;.
(5.1)
The result is that the impedance seen "looking into" the base of the network is a resistor THE re TRANSISTOR 259
equal to beta times the value of re, as shown in Fig. 5.14. The collector output current is MODEL
still linked to the input current by beta as shown in the same figure.

!
lb IC

~~
-cc
I
p,, l f3Ib

l oe
-¥
FIG. 5.14
Improved BJT equivalent circuit.

The equivalent circuit has therefore been defined for the ideal characteristics of Fig. 5.11,
but now the input and output circuits are isolated and only linked by the controlled source-a
form much easier to work with when analyzing networks.

Early Voltage
We now have a good representation for the input circuit, but aside from the collector out-
put current being defined by the level of beta and / B, we do not have a good representation
for the output impedance of the device. In reality the characteristics do not have the ideal
appearance of Fig. 5.11. Rather, they have a slope as shown In Fig. 5.15 that defines the
output impedance of the device. The steeper the slope, the less the output impedance and
the less ideal the transistor. In general, it is desirable to have large output impedances to
avoid loading down the next stage of a design. If the slope of the curves is extended until
they reach the horizontal axis, it is interesting to note in Fig. 5.15 that they will all intersect
at a voltage called the Early voltage. This intersection was first discovered by James M.
Early in 1952. As the base current increases the slope of the line increases, resulting in an
increase in output impedance with increase in base and collector current. For a particular
collector and base current as shown in Fig. 5.15, the output impedance can be found using
the following equation:

(5.2)

FIG. 5.15
Defining the Early voltage and the output impedance of a transistor.
260 BJT AC ANALYSIS Typically, however, the Early voltage is sufficiently large compared with the applied
collector-to-emitter voltage to permit the following approximation.

(5.3)

Clearly, since VA is a fixed voltage, the larger the collector current, the less the output
impedance.
For situations where the Early voltage is not available the output impedance can be found
from the characteristics at any base or collector current using the following equation:
Liy Lile 1
Slope =- = -- = -
Lix Li VeE r0

and (5.4)

For the same change in voltage in Fig. 5.15 the resulting change in current Lile is signifi-
cantly less for r02 than r01 , resulting in r02 being much larger than r01 •
In situations where the specification sheets of a transistor do not include the Early volt-
age or the output characteristics, the output impedance can be determined from the hybrid
parameter hoe that is normally plotted on every specification sheet. It is a quantity that will
be described in detail in Section 5 .19.
In any event, an output impedance can now be defined that will appear as a resistor in
parallel with the output as shown in the equivalent circuit of Fig. 5.16.

bo------ ---------<>-------o C

f3r,

eo------------<,___ __._ _ _ _ _ _ _ _ _~e

FIG. 5.16
r, model for the common-emitter transistor configuration
including effects of r0 •

The equivalent circuit of Fig. 5.16 will be used throughout the analysis to follow for the
common-emitter configuration. Typical values of beta run from 50 to 200, with values of
f3re typically running from a few hundred ohms to a maximum of 6 kfl to 7 kfl. The output
resistance r is typically in the range of 40 kfl to 50 kfl.

Common-Base Configuration
The common-base equivalent circuit will be developed in much the same manner as
applied to the common-emitter configuration. The general characteristics of the input and
output circuit will generate an equivalent circuit that will approximate the actual behavior
of the device. Recall for the common-emitter configuration the use of a diode to represent
the connection from base to emitter. For the common-base configuration of Fig. 5.17 a the
pnp transistor employed will present the same possibility at the input circuit. The result is
the use of a diode in the equivalent circuit as shown in Fig. 5.17b. For the output circuit, if
we return to Chapter 3 and review Fig. 3.8, we find that the collector current is related to
the emitter current by alpha a. In this case, however, the controlled source defining the
collector current as inserted in Fig. 5 .17b is opposite in direction to that of the controlled
source of the common-emitter configuration. The direction of the collector current in the
output circuit is now opposite that of the defined output current.
 IC Io I; le IC Io
~ ~ ~ ~ ~ ~
C E+ +c
+

~
zo
V0 V; ~
Z; t lc=al, ~
zo
Vo

B0----------------------0 B B B

(a) (b)

FIG. 5.17
(a) Common-base BJT transistor; (b) equivalent circuit for configuration of (a).

For the ac response, the diode can be replaced by its equivalent ac resistance determined
by re = 26 mV / IE as shown in Fig. 5.18. Take note of the fact that the emitter current
continues to determine the equivalent resistance. An additional output resistance can be
determined from the characteristics of Fig. 5.19 in much the same manner as applied to the
common-emitter configuration. The almost horizontal lines clearly reveal that the output
resistance rO as appearing in Fig. 5 .18 will be quite high and certainly much higher than that
for the typical common-emitter configuration.
The network of Fig. 5.18 is therefore an excellent equivalent circuit for the analysis of
most common-base configurations. It is similar in many ways to that of the common-emitter
configuration. In general, common-base configurations have very low input impedance
because it is essentially simply re. Typical values extend from a few ohms to perhaps 50 !1.
The output impedance rO will typically extend into the megohm range. Because the output
current is opposite to the defined I0 direction, you will find in the analysis to follow that
there is no phase shift between the input and output voltages. For the common-emitter
configuration there is a 180° phase shift.

I; le IC Io
~ ~ ~ ~
E
+ +

V;
~
Z;
r, t l c=a.l, ro ~
zo Vo

FIG. 5.18
Common base r, equivalent circuit.

le (mA) 1
/ Slope=,:;;-

41--- -------------
31 -- -------------
2 Y..,,--------------
IE= 1 mA

FIG. 5.19
Defining Z 0 •
261
262 BJT AC ANALYSIS Common-Collector Configuration
For the common-collector configuration, the model defined for the common-emitter configu-
ration of Fig. 5.16 is normally applied rather than defining a model for the common-collector
configuration. In subsequent chapters, a number of common-collector configurations will be
investigated, and the effect of using the same model will become quite apparent.

npn versus pnp

The de analysis of npn and pnp configurations is quite different in the sense that the currents
will have opposite directions and the voltages opposite polarities. However, for an ac analy-
sis where the signal will progress between positive and negative values, the ac equivalent
circuit will be the same.

5.5 COMMON-EMITTER FIXED-BIAS

CONFIGURATION
•
The transistor models just introduced will now be used to perform a small-signal ac analy-
sis of a number of standard transistor network configurations. The networks analyzed rep-
resent the majority of those appearing in practice. Modifications of the standard
configurations will be relatively easy to examine once the content of this chapter is reviewed
and understood. For each configuration, the effect of an output impedance is examined for
completeness.
The computer analysis section includes a brief description of the transistor model em-
ployed in the PSpice and Multisim software packages. It demonstrates the range and depth
of the available computer analysis systems and how relatively easy it is to enter a complex
network and print out the desired results. The first configuration to be analyzed in detail is
the common-emitter.fixed-bias network of Fig. 5.20. Note that the input signal V; is applied
to the base of the transistor, whereas the output V0 is off the collector. In addition, recognize
that the input current I; is not the base current, but the source current, and the output current
I0 is the collector current. The small-signal ac analysis begins by removing the de effects
of Vcc and replacing the de blocking capacitors C 1 and C2 by short-circuit equivalents,
resulting in the network of Fig. 5.21.

Vee

Vo
I;
___._ B
I;
___._ B
tlo
V;
V; o-------)-----0----1 Re ~
C1
___._ ___._ Rs
zo
E
E Z;
Z;

-=- -=- -=-

FICi. 5.20 FICi. 5.21
Common-emitter fixed-bias configuration. Network of Fig. 5.20 following the removal
of the effects of Vee, C1, and C2.

Note in Fig. 5.21 that the common ground of the de supply and the transistor emitter
terminal permits the relocation of Rs and Re in parallel with the input and output sections
of the transistor, respectively. In addition, note the placement of the important network
parameters Z;, Z 0 , I;, and I0 on the redrawn network. Substituting the re model for the
common-emitter configuration of Fig. 5.21 results in the network of Fig. 5.22.
The next step is to determine /3, re, and r0 • The magnitude of f3 is typically obtained
from a specification sheet or by direct measurement using a curve tracer or transistor
 COMMON-EMITTER 263
lb
___._
I; IC FIXED-BIAS

l
~
CONFIGURATION
+___._
Z;
b C
t
lo
+

~
V; f3r, Vo
RB f3/b ro Re
~-
J. -=- -=-
FIG. 5.22
t -=- -=-
zo i
Substituting the r, model into the network of Fig. 5.21.

testing instrument. The value of re must be determined from a de analysis of the system,
and the magnitude of rO is typically obtained from the specification sheet or characteristics.
Assuming that {3, re, and r0 have been determined will result in the following equations for
the important two-port characteristics of the system.

Z1 Figure 5.22 clearly shows that

I Zi = Rnllf3re I ohms (5.5)

For the majority of situations Rn is greater than f3re by more than a factor of 10 (recall
from the analysis of parallel elements that the total resistance of two parallel resistors is
always less than the smallest and very close to the smallest if one is much larger than the
other), permitting the following approximation:

ohms (5.6)

Z0 Recall that the output impedance of any system is defined as the impedance Z 0
determined when Vi = 0. For Fig. 5.22, when Vi = 0, Ii = lb = 0, resulting in an open-
circuit equivalence for the current source. The result is the configuration of Fig. 5.23.
We have

If rO ~ lORe, the approximation Rell rO

I Z = Rcllr I
0 0 ohms

== Re is frequently applied, and

(5.7) l -=-
FIG. 5.23
-=-

Determining Z 0 for the network

(5.8) of Fig. 5.22.

Av The resistors r0 and Re are in parallel, and

V0 = -{3Ib(Rdr0 )
Vi
but h=-
f3re

so that V0 = -{3(~)cRcllr0 )

(Rellro)
and (5.9)

If r0 ~ lORe, so that the effect of r0 can be ignored,

(5.10)

Note the explicit absence of {3 in Eqs. (5.9) and (5.10), although we recognize that {3 must
be utilized to determine re·
264 BJT AC ANALYSIS Phase Relationship The negative sign in the resulting equation for Av reveals that a 180°
phase shift occurs between the input and output signals, as shown in Fig. 5.24. The is a
result of the fact that f3h establishes a current through Re that will result in a voltage across
Re, the opposite of that defined by V0 •

FIG. 5.24
Demonstrating the 180° phase shift between input and output waveforms.

EXAMPLE 5.1 For the network of Fig. 5 .25:

a. Determine re.
b. FindZ; (with r0 = 00 0).
c. Calculate Z 0 (with r0 = oc 0).
d. Determine Av (with rO = 00 0).
e. Repeat parts (c) and (d) including r0 = 50 kO in all calculations and compare results.

~------<t----o 12 V

3kQ
470 ill

I;
~ 10 µF
V; o--------) t - - - + - - - - - 1
10 µF /3 = 100 Z0
r = 50 ill
0

FIG. 5.25
Example 5.1.

Solution:
a. DC analysis:
Vee - VsE 12V - 0.7V
ls = Rs 470 kO = 24.04 µ.,A
IE = (/3 + l)Is = (101)(24.04 µ.,A) = 2.428 mA
re= 26mV = 26mV = l0.7l O
le 2.428mA
b. f3re = (100)(10.71 0) = 1.071 kO
Z; = Rsllf3re = 470kOIIL071kO = 1.07kO
C. Zo = Re = 3 kO
Re 3k0
d. Av= - - -280.11
re 10.71 0
e. Z0 = r0 IIRc = 50 kil 113 kil = 2.83 kO vs. 3 kil VOLTAGE-DIVIDER BIAS 265

Av = --- =
rollRc 2.83 kil
"' = - 264.24 vs. -280.11
re 10.71 H

5.6 VOLTAGE-DIVIDER BIAS

The next configuration to be analyzed is the voltage-divider bias network of Fig. 5.26.
Recall that the name of the configuration is a result of the voltage-divider bias at the input
side to determine the de level of V8 .
•
Substituting the re equivalent circuit results in the network of Fig. 5.27. Note the absence
of RE due to the low-impedance shorting effect of the bypass capacitor, CE. That is, at the
frequency (or frequencies) of operation, the reactance of the capacitor is so small compared
to RE that it is treated as a short circuit across RE. When Vcc is set to zero, it places one
end of R 1 and Re at ground potential as shown in Fig. 5.27. In addition, note that R 1 and
R2 remain part of the input circuit, whereas Re is part of the output circuit. The parallel
combination of R 1 and R2 is defined by

(5.11)

Z1 From Fig. 5.27

(5.12)

f I0
Re
R1
C (------o V 0
I; Cz

V;
~

o----J
Cl
B
.....Zo
~ E
Z; Rz
CE

"II"

FIG. 5.26
Voltage-divider bias configuration.

I;
~ b C
=i+Ib
+ ~
Z;
t I0 +
V; R1 Rz f3r, ~ /3/b ro Re
..... Vo

e e zo

"II"
'----y-----J "II"
R'

FIG. 5.27
Substituting the r, equivalent circuit into the ac equivalent network of Fig. 5.26.
266 BJT AC ANALYSIS Z0 From Fig. 5.27 with V; set to 0 V, resulting in lb = 0 µ,A and f3lb = 0 mA,

(5.13)

If ra 2: lORc,

(5.14)

Av Because Re and r a are in parallel,

Va = -({3/b)(Rcllra)

and

so that

and (5.15)

which you will note is an exact duplicate of the equation obtained for the fixed-bias con-
figuration.
For ra 2: lORc,

(5.16)

Phase Relationship The negative sign of Eq. (5.15) reveals a 180° phase shift between
Va and V;.

EXAMPLE 5.2 For the network of Fig. 5.28, determine:

a. re.
b. Z;.
c. Za(ra =
oof!).
d. Av (ra =
00 fl).

e. The parameters of parts (b) through (d) if ra = 50 kf! and compare results.

22V

t 10
6.8kQ
56kQ 10 µF
(----a V
0

10 µF
V; o-------} t - - - + - - - - - - 1
___._
I;

8.2 kQ

FIG. 5.28
Example 5.2.
Solution: CE EMITTER-BIAS 267
CONFIGURATION
a. DC: Testing f3RE > lOR2,
(90)(1.5 kO) > 10(8.2 kO)
135 kO > 82 kO (satisfied)
Using the approximate approach, we obtain
R2 (8.2 kO)(22 V)
VB= R1 + R2 Vee= 56kO + 8.2kO = 2·81 V
VE= VB - VBE = 2.81 V - 0.7V = 2.11 V
VE 2.11 V
IE= RE= l.5kO = l.41mA
re=26mV = 26mV =l 8.440
IE 1.41 mA
b. R' = R1IIR2 = (56kO)ll(8.2kO) = 7.15kO
zi = R'llf3re = 7.15kOll(90)(18.44O) = 7.15kOIIL66kO
= 1.35kfi
c. Z 0 =Re= 6.8kfi
Re 6.8kO
d. Av=-----;:;= 18.44 0 -368.76
e. zi = 1.35 kfi
Zo = Rcllro = 6.8 kO llso kO = 5.98 kfi vs. 6.8 kO
5.98kO
18.44 O = -324.3 vs. -368.76

There was a measurable difference in the results for Z 0 and Av, because the condition
rO ~ I ORe was not satisfied.

5.7 CE EMITTER-BIAS CONFIGURATION

The networks examined in this section include an emitter resistor that may or may not be
bypassed in the ac domain. We first consider the unbypassed situation and then modify the
resulting equations for the bypassed configuration.
•
Unbypassed
The most fundamental of unbypassed configurations appears in Fig. 5.29. The re equiva-
lent model is substituted in Fig. 5.30, but note the absence of the resistance r0 • The effect
of rO is to make the analysis a great deal more complicated, and considering the fact that in

I;
__._ b C
71/b
+ +
Re
Z;
__._ __._
f3r,
+ f3/b
t
Rs {10 10 ~

(--o va zb zo
__._
I; e2 V; Rs Re Vo
V; o-------) e
e1
~
zo
__._ RE
Z;
-=-
FICi. 5.29 FICi. 5.30
CE emitter-bias configuration. Substituting the r, equivalent circuit into the ac equivalent network of Fig. 5.29.
268 BJT AC ANALYSIS most situations its effect can be ignored, it will not be included in the present analysis.
However, the effect of r0 will be discussed later in this section.
Applying Kirchhoffs voltage law to the input side of Fig. 5.30 results in
vi = hf3re + leRE
or vi = hf3re + (/3 + I)lifiE
and the input impedance looking into the network to the right of RB is

Zb = -Vi = f3re + ({3 + l)RE

lb
The result as displayed in Fig. 5.31 reveals that the input impedance of a transistor with
an unbypassed resistor RE is determined by

(5.17)

Because {3 is normally much greater than 1, the approximate equation is

zb = f3re + f3RE
and (5.18)
FIC. 5.31
Defining the input impedance of a Because RE is usually greater than re, Eq. (5.18) can be further reduced to
transistor with an unbypassed
emitter resistor. (5.19)

Z1 Returning to Fig. 5.30, we have

(5.20)

Z0 With Vi set to zero, lb = 0, and f3lb can be replaced by an open-circuit equivalent.

The result is

(5.21)

h=-
vi
zb
and V0 = -l0 Rc = -{31,fic

= -{3(;JRc

with (5.22)

Substituting Zb = {3(re + RE) gives

(5.23)

and for the approximation Zb = f3RE,

I A,~~~-~ I (5.24)

Note the absence of /3 from the equation for Av demonstrating an independence in variation
of {3.

Phase Relationship The negative sign in Eq. (5.22) again reveals a 180° phase shift
between V0 and Vi.
Effed of r0 The equations appearing below will clearly reveal the additional complexity CE EMITTER-BIAS 269
resulting from including r0 in the analysis. Note in each case, however, that when certain CONFIGURATION
conditions are met, the equations return to the form just derived. The derivation of each
equation is beyond the needs of this text and is left as an exercise for the reader. Each
equation can be derived through careful application of the basic laws of circuit analysis
such as Kirchhoff's voltage and current laws, source conversions, Thevenin's theorem,
and so on. The equations were included to remove the nagging question of the effect of r0
on the important parameters of a transistor configuration.

(/3-+-1)-+-Rc/r
= r + [- -- 0 ]
Z R (5.25)
b {3 e 1 + (Re + RE)/ro E

Because the ratio Rc/r0 is always much less than (/3 + 1),
zb ~ f3re + - -(/3-+-l)RE
---
1 + (Re + RE)/ro
For rO 2='. lO(Re + RE),
Zb ~ f3re + ({3 + l)RE
which compares directly with Eq. (5.17).
In other words, if r0 2='. lO(Re + RE), all the equations derived earlier result. Because
{3 + 1 ~ /3, the following equation is an excellent one for most applications:

(5.26)

(5.27)

However, r0 >> re, and

which can be written as

Typically 1//3 and re/RE are less than one with a sum usually less than one. The result
is a multiplying factor for r0 greater than one. For {3 = 100, re = 10 !1, and RE = 1 kll,
1
- - -1- - - = -1- = 50
1 10 n 0.02
-+---
100 1000 n
and Z0 = Rcll51r0
which is certainly simply Re. Therefore,

(5.28)
Any level of r0

which was obtained earlier.

270 BJT AC ANALYSIS Av

(5.29)

re
The ratio - << 1, and
ro
f3Re Re
--+-
zb ro
1 + Re
ro
For r O ~ lORe,

(5.30)

as obtained earlier.

Bypassed
If RE of Fig. 5.29 is bypassed by an emitter capacitor CE, the complete re equivalent model
can be substituted, resulting in the same equivalent network as Fig. 5.22. Equations (5.5)
to (5.10) are therefore applicable.

EXAMPLE 5.3 For the network of Fig. 5.32, without CE (unbypassed), determine:
a. re.
b. Z;. 20V
C. Z 0 •
d. Av.

2.2kQ
10 µF
470 kQ
( o V0
e2 ~
10 µF zo
V; 0 ~ )1---+-----11 /3 = 120, r0 = 40 ill
I; e1

0.56kQ lieE 10 µF

-=-
FIG. 5.32
Example 5.3.

Solution:
a. DC:

I = -Vee-- - - - -20V
- VBE
- -- - 0.7V
- - - = 35.89 µ.,A
B RB + (/3 + l)RE 470 kfl + (121)0.56 kfl
IE = (/3 + 1)/B = (121)(35.89 µ.,A) = 4.34 mA

and r = 26 m V = 26 m V = S O
e fe 4.34 mA •99
b. Testing the condition rO 2: lO(Re + RE), we obtain CE EMITTER-BIAS 271
CONFIGURATION
40 kO 2: 10(2.2 kO + 0.56 kO)
40 kO 2: 10(2.76 kO) = 27.6 kO (satisfied)
Therefore,
Zb ~ f3(re+ RE) = 120(5.99 0 + 560 0)
= 67.92kO
and zi = Rsllzh = 470kDll67.92kO
= 59.34kfl
c. Z0 =Re= 2.2kfl
d. r0 2: lORe is satisfied. Therefore,
V0 /3Re (120)(2.2 kO)
Av=-~--
Vi Zb 67.92kO
= -3.89
compared to -3.93 using Eq. (5.20): Av ~ -Re/RE.

EXAMPLE 5.4 Repeat the analysis of Example 5 .3 with CE in place.

Solution:
a. The de analysis is the same, and re = 5.99 0.
b. RE is "shorted out" by CE for the ac analysis. Therefore,
zi = Rsllzb = Rsllf3re = 470 kDll(120)(5.99 0)
= 470 kO 11718.8 0 ~ 717.70 fl
c. Z0 =Re= 2.2kfl
Re
d. Av= - -
re
2.2kO
--- - 367.28 (a significant increase)
5.99 n

EXAMPLE 5.5 For the network of Fig. 5.33 (with CE unconnected), determine (using
appropriate approximations):
a. re.
b. Zi.
C. Z 0 •
d. Av.

16V

i 10
2.2 kQ
90kQ

( 0

e2 +
v~ /3 =210,r =50kQ
,....
-
0

l
I -
I; C1
zo
Z; lOkQ
0.68kQ
_]CE
-=-
FIG. 5.33
Example 5.5.
272 BJT AC ANALYSIS Solution:
a. Testing f3RE > lOR2,
(210)(0.68 k!l) > 10(10 kn)
142.8 kD > 100 kD (satisfied)
we have
R2 lOkD
VB= R1 + R 2 Vee= 90k!l + lOkD (l 6 V) = 1. 6 V
VE= VB - VBE = 1.6V - 0.7V = 0.9V
VE 0.9V
IE = RE = 0.68 k!l = 1.3 24 mA

re= 26mV = 26mV = 19 _640

l.324mA
IE
b. The ac equivalent circuit is provided in Fig. 5.34. The resulting configuration is differ-
ent from Fig. 5.30 only by the fact that now
RB= R' = R1IIR2 = 9kD

+
-+-
Z; 2.2kQ
V; lOkQ 90kQ
0.68kQ

'---y---J
R'

FIG. 5.34
The ac equivalent circuit of Fig. 5.33.

The testing conditions of r0 2: lO(Re + RE) and r0 2: lORe are both satisfied. Using
the appropriate approximations yields
Zb ~ f3RE = 142.8 k!l
Z; = RBllzb = 9 kn I 142.8 kn
= 8.47k!l
c. Z 0 =Re= 2.2k!l
Re 2.2kD
d. Av= - RE= 0.68kD -3.24

EXAMPLE 5.6 Repeat Example 5.5 with CE in place.

Solution:
a. The de analysis is the same, and re = 19.64 fl.
b. Zb = f3re = (210)(19.64 D) ~ 4.12k!l
Z; = RBllzb = 9kDll4.12kD
= 2.83 k!l
c. Z 0 =Re= 2.2k!l
Re 2.2kD
d. Av= - - -112.02 (a significant increase)
re 19.64 D

Another variation of an emitter-bias configuration is shown in Fig. 5.35. For the de

analysis, the emitter resistance is RE1 + RE2, whereas for the ac analysis, the resistor RE in
the equations above is simply RE1 with RE2 bypassed by CE.
 EMITTER-FOLLOWER 273
CONFIGURATION

c,
V; o--------J
~
-------
I;

FIG. 5.35
An emitter-bias configuration with a
portion of the emitter-bias resistance
bypassed in the ac domain.

5.8 EMITTER-FOLLOWER CONFIGURATION

•
When the output is taken from the emitter terminal of the transistor as shown in Fig. 5.36,
the network is referred to as an emitter-follower. The output voltage is always slightly less
than the input signal due to the drop from base to emitter, but the approximation Av ~ 1
is usually a good one. Unlike the collector voltage, the emitter voltage is in phase with the
signal Vi. That is, both V0 and Vi attain their positive and negative peak values at the same
time. The fact that V0 "follows" the magnitude of Vi with an in-phase relationship accounts
for the terminology emitter-follower.

C
I;
..,._ B
V; o--)1---+-0----0
C1
E

FIG. 5.36
Emitter-follower configuration.

The most common emitter-follower configuration appears in Fig. 5.36. In fact, because
the collector is grounded for ac analysis, it is actually a common-collector configuration.
Other variations of Fig. 5.36 that draw the output off the emitter with V0 ~ Vi will appear
later in this section.
The emitter-follower configuration is frequently used for impedance-matching pur-
poses. It presents a high impedance at the input and a low impedance at the output, which
is the direct opposite of the standard fixed-bias configuration. The resulting effect is much
the same as that obtained with a transformer, where a load is matched to the source imped-
ance for maximum power transfer through the system.
Substituting the re equivalent circuit into the network of Fig. 5.36 results in the network
of Fig. 5.37. The effect of r0 will be examined later in the section.
274 BJT AC ANALYSIS I;
___._ b C

-=- -=-
FIG. 5.37
Substituting the r, equivalent circuit into the ac
equivalent network of Fig. 5.36.

Z1 The input impedance is determined in the same manner as described in the preceding
section:

(5.31)

with (5.32)

or (5.33)

and (5.34)

Z0 The output impedance is best described by first writing the equation for the current h,
V;
lb= -
Zb
and then multiplying by (J3 + 1) to establish le. That is,
V
le = (/3 + l)lb = (/3 + 1)~
Zb
Substituting for Zb gives

l =------
(/3 + l)Vi
e f3re + (/3 + l)RE
y.
or l = '
e [f3re/(f3 + l)] + RE
but (/3 + 1) = /3
f3re f3re
and --=-=r
{3+1-{3 e
r,
~4""'.---+-1-1--0 Vo
so that le= - - -
vi (5.35)
t' e
re+ RE

If we now construct the network defined by Eq. (5.35), the configuration of Fig. 5.38
results.
-=-
To determine Za, V; is set to zero and
FIG. 5.38
Defining the output impedance for
(5.36)
the emitter-follower configuration.
Because RE is typically much greater than re, the following approximation is often applied: EMITTER-FOLLOWER 275
CONFIGURATION
(5.37)

Av Figure 5.38 can be used to determine the voltage gain through an application of the
voltage-divider rule:
REY.
V = l
o RE+ re

and (5.38)

Because RE is usually much greater than re, RE + re ~ RE and

(5.39)

Phase Relationship As revealed by Eq. (5.38) and earlier discussions of this section, V0
and V; are in phase for the emitter-follower configuration.

Effed of r0
z,
(5.40)

If the condition rO 2='. lORE is satisfied,

Zb = f3re + (/3 + l)RE
which matches earlier conclusions with

(5.41)

(5.42)

Using f3 + I ~ /3, we obtain

and because r0 >> re,

Any r0
(5.43)

(5.44)

If the condition r0 2='. lORE is satisfied and we use the approximation f3 +I ~ /3, we find
f3RE
Av~ -
zb
276 BJT AC ANALYSIS But Zb ~ {3(re + RE)
f3RE
so that Av~----
{3(re + RE)

and (5.45)

EXAMPLE 5.7 For the emitter-follower network of Fig. 5.39, determine:

a. re.
b. Z;.
C. Z 0 •
d. Av.
e. Repeat parts (b) through (d) with r0 = 25 kll and compare results.
12 V

RB 220 k.Q
10 µF
V; o-------) t - - - - - - - - 1 /3=100,r0 = 00 Q
___._
I;

RE 3.3 k.Q

FIG. 5.39
Example 5. 7.
Solution:
Vee - VBE
a. I B = - - - - -
RB + (/3 + l)RE

= - -12V-- -0.7V- - - = 20.42 µ,A

220 kll + (101)3.3 kll
IE= ({3 + l)IB
= (101)(20.42 µ,A) = 2.062 mA
re= 26mV = 26mV = 12_61 O
IE 2.062mA
b. Zb = f3re + ({3 + l)RE
= (100)(12.61 0) + (101)(3.3 kll)
= 1.261 kll + 333.3 kll
= 334.56 kll ~ f3RE
Z; = RBllzb = 22oknll334.56kll
= 132.72 kfi
C. Zo = REIi re= 3.3 kllll 12.61 !1
= 12.56 0 ~ re
V0 RE 3.3 kll
d. Av= -
V; RE + re 3.3 kll + 12.61 !1
= 0.996 ~ 1
e. Checking the condition r0 2: lORE, we have COMMON-BASE 277
CONFIGURATION
25 kll 2: 10(3.3 kll) = 33 kll
which is not satisfied. Therefore,
z = /3 + (/3 + l)RE = (100)(l 2 61 !1) + (100 +
1)3.3 kll
b re RE · 3.3 kll
1+- I+--
r0 25 kll
= 1.261 kll + 294.43 kll
= 295.7 kll
with zi = Rsllzb = 220k!!ll295.7k!1
= 126.15 kfi vs. 132. 72 kll obtained earlier
Z0 = REllre = 12.56 fi as obtained earlier
(/3 + l)RE/Zb (100 + 1)(3.3 kD,)/295.7 kll
Av= [1 + RE] = [1 + 3.3k!1]
r0 25 kll
= 0.996 5:! 1
matching the earlier result.

In general, therefore, even though the condition r0 2: lORE is not satisfied, the results
for Z 0 and Av are the same, with Zi only slightly less. The results suggest that for most ap-
plications a good approximation for the actual results can be obtained by simply ignoring
the effects of r0 for this configuration.
The network of Fig. 5.40 is a variation of the network of Fig. 5.36, which employs
a voltage-divider input section to set the bias conditions. Equations (5.31) to (5.34) are
changed only by replacing Rs by R' = R1 IIRz.
The network of Fig. 5.41 also provides the input/output characteristics of an emitter-
follower, but includes a collector resistor Re. In this case Rs is again replaced by the parallel
combination of R 1 and R2 . The input impedance Zi and output impedance Z 0 are unaffected
by Re because it is not reflected into the base or emitter equivalent networks. In fact, the
only effect of Re is to determine the Q-point of operation.

Re
R1 R1

~ e1
V; ~ V; o---}
e1

---+- R2 ---+- R2
Z;
,.,_ Z;
,.,_
Zo Zo
-=- -=- -=- -=-
FIG. 5.40 FIG. 5.41
Emitter-follower configuration with a Emitter-follower configuration with
voltage-divider biasing arrangement. a collector resistor Re.

5.9 COMMON-BASE CONFIGURATION

The common-base configuration is characterized as having a relatively low input and a high
output impedance and a current gain less than 1. The voltage gain, however, can be quite
large. The standard configuration appears in Fig. 5.42, with the common-base re equivalent
•
model substituted in Fig. 5.43. The transistor output impedance r0 is not included for the
 ~
I;
~
I,
~
le e
lt/, ~
IC
C

+ E C
t lo
+ + ~
I; tlo
+
..,_
~ Re
..,_ V; ~
RE re
t al, Re Vo zo
V; Z; Vo zo Z;
B
-=- VEE --;;;-vcc
+
FICi. 5.42 FICi. 5.43
Common-base configuration. Substituting the r, equivalent circuit into the ac equivalent network
of Fig. 5.44.

common-base configuration because it is typically in the megohm range and can be ignored
in parallel with the resistor Re.

(5.46)

(5.47)

with

so that

and (5.48)

A1 Assuming that RE >> re yields

le= I;
and 10 = -ale = -al;

with (5.49)

Phase Relationship The fact that Av is a positive number shows that V0 and V; are in
phase for the common-base configuration.

Effed of r0 For the common-base configuration, r0 = I/hob is typically in the megohm

range and sufficiently larger than the parallel resistance Re to permit the approximation
rollRe ~ Re.

EXAMPLE 5.8 For the network of Fig. 5.44, determine:

a. re.
10 µF 10 µF
b. Z;.
C. Z 0 •
d. Av.
+~)
I; t ~ lo
a= 0.98 Re 5 ill
e. A;. V; r0 = 1 MQ
--;;;- 8 V
+
FICi. 5.44
Example 5.8.
278
Solution: COLLECTOR FEEDBACK 279
CONFIGURATION
VEE - VnE 2V - 0.7V 1.3V
a. IE= RE 1 kfl = 1 kfl = 1.3 mA
r = 26mV = 26mV = 200
e fe l.3mA
b. Z; = REIi re= 1 kflll20 fl
= 19.61 fl "" re
C. Zo = Re = 5 kfl
d. Av "" Re = 5 kfl = 250
re 20 fl
e. A; = - 0.98 "" -1

5.1 O COLLECTOR FEEDBACK CONFIGURATION

The collector feedback network of Fig. 5.45 employs a feedback path from collector to •
base to increase the stability of the system as discussed in Section 4.6. However, the sim-
ple maneuver of connecting a resistor from base to collector rather than base to de supply
has a significant effect on the level of difficulty encountered when analyzing the network.
Some of the steps to be performed below are the result of experience working with
such configurations. It is not expected that a new student of the subject would choose
the sequence of steps described below without taking a wrong step or two. Substituting the
equivalent circuit and redrawing the network results in the configuration of Fig. 5.46. The
effects of a transistor output resistance rO will be discussed later in the section.

Vee

B -RF+ e
I;
---+-
V; o------} - - -
B + ---+-
I; tib j ' tle
e, V; ---+- f3r, f3/b
E
Z; +
-=- i -=-
FICi. 5.45 FICi. 5.46
Collector feedback configuration. Substituting the re equivalent circuit into the ac
equivalent network of Fig. 5.45.

I0 = I' + f3Ib
V0 - V;
and I'= - - -
RF
but V0 = -I0 Re = -(I' + f3Ib)Re
with V; = Ibf3re
(I' + /3h)Re - Ibf3re I'Re f3IbRe hf3re
so that I' = ---------
RF RF RF
which when rearranged in the following:
r(i + Re) = -{3I/Re + re)
RF RF
280 BJT AC ANALYSIS
and finally,

and

or

Substituting for Vi in the above equation for Zi leaves

z. -_ -Vi_- - - -lbf3re
----
Ii ( (Re + re))
lb 1 + { 3 - - -
l

Re + Rp

Since Re>> re
f3re
Zi=-----
f3Re
I+---
Re + Rp

re
or Z-=----- (5.50)
1 1 Re
-+---
/3 Re+ Rp

Z0 Ifwe set Vi to zero as required to define Z0 , the network will appear as shown in Fig. 5.47.
The effect of f3re is removed, and Rp appears in parallel with Re and

(5.51)

/3r,

-=- -=-
FIG. 5.47
Defining Z 0 for the collector feedback configuration.

V0 = -I0 Re = -(I' + /3h)Re

(Re+ re)
= - ( -f3Ib--- + /3h ) Re
Re+ Rp

or V
o
= -{31
b
(1 - (Re+ re)R
Re+ Rp e
Then

For Re>> re
Re )Re
Av = - ( 1 - Re + Rp -;:;
 (Rf: + RF - /Yc)Re COLLECTOR FEEDBACK 281
or AV = CONFIGURATION
Re+ RF re

and (5.52)

For RF>> Re

~ (5.53)
~
Phase Relationship The negative sign ofEq. (5.52) indicates a 180° phase shift between
V0 and V;.

Effed of r0
Z1 A complete analysis without applying approximations results in

(5.54)

Applying the condition r 0 2='. lORe, we obtain

l + Re
RF
Z-= - - - - - - - - - -
1 1 1 Re Re
-+-+--+-- -1 + -1 [ re+ -Re + Re ]
f3re RF f3reRF RFre /3 RF /3
Re
Applying Re >> re and /3'
re [
RF+
Rf;,
Re]
RF+ f3Re 1( RF ) Re
f3Rf;, /3 RF + Re + Re + RF
R
but, since RF typically >> Re, RF + Re ~ RF and F = I
RF+ Re

re
Z;~----- (5.55)
1 Re
-+---
/3 Re+ RF
as obtained earlier.

Z0 Including r 0 in parallel with Re in Fig. 5.47 results in

(5.56)

For r O 2=: lORe,

(5.57)

as obtained earlier. For the common condition of RF >> Re,

(5.58)
282 BJT AC ANALYSIS Av

(5.59)

For r 0 ~ lORe,

(5.60)

and for RF >> Re

~ (5.61)
~
as obtained earlier.

EXAMPLE 5.9 For the network of Fig. 5.48. determine:

a. re.
b. Zi.
C. Z0 •
d. Av.
e. Repeat parts (b) through (d) with rO = 20 kll and compare results.

9V

2.7 kQ

I;
V; o ~ ) 1-------------0 /3 =200,r
0 = 00 Q
lOµF ~
zo

FIG. 5.48
Example 5.9.

Solution:
Vee - VBE 9V-0.7V
a. J B = - - - -
RF + f3Re 180 kll + (200)2.7 kll
= 11.53 µA
le = (/3 + l)IB = (201)(11.53 JLA) = 2.32 mA
re= 26mV = 26mV = 11 _21 !l
le 2.32mA
re 11.21 0 11.21 0
b. zi = 1 Re 1 2.7 kll 0.005 + 0.0148
-+--- -+---
/3 Re+ RF 200 182.7 kll

= 11. 21 n= 566.16 n
0.0198
C. Z 0 = RdRF = 2.7 kll II 180 kll = 2.66 k!l
2.7kll
-240.86
11.21 n
e. Z;: The condition r 0 2: lORc is not satisfied. Therefore, COLLECTOR FEEDBACK 283
CONFIGURATION
Rcllr0 2.7kll ll20 kll
1+-- 1+-----
RF 180kll
z. = ----------
! 1 1 Rcllr Rcllr 1 1 2.7k!!ll20kll 2.7k!!ll20kll
- + - + - -0 + - -0 ---- + + -------- + ------
f3re RF {3reRF RFre (200)(11.21) 180 kll (200)(11.21 !1)(180 kll) (180 kll)(l l.21 ll)
2.38 kll
l + 180kll 1 + 0.013
3 3 6
0.45 X 10- + 0.006 X 10- + 5.91 X 10- + 1.18 X 10- 3 1.64 X 10-3
= 617.7 fl vs. 566.16 !1 above
Za:
Z0 = r 0 IIRcllRF = 20k!!ll2.7k!!ll180k!1
= 2.35 k!l vs. 2.66 kll above
A.,:
( RF )Rell ro [ 180 kll ] 2.38 kll
= - Rell r + RF
0 -----;:;- = - 2.38 kll + 180 kll 11.21
= - [ 0.987] 212.3
= -209.54

For the configuration of Fig. 5.49, Eqs. (5.61) through (5.63) determine the variables of
interest. The derivations are left as an exercise at the end of the chapter.

Vee

Re

Fo (------o V 0

I; Cz
V; o _._ )

_._ C1 ....zo
Z;
RE

FIG. 5.49
Collector feedback configuration with an emitter resistor RE.

Z; == - - -RE- - - (5.62)
[ _!_+(RE + Re)]
f3 RF

(5.63)

(5.64)
284 BJT AC ANALYSIS 5.11 COLLECTOR DC FEEDBACK CONFIGURATION
The network of Fig. 5.50 has a de feedback resistor for increased stability, yet the capacitor
C3 will shift portions of the feedback resistance to the input and output sections of the net-
work in the ac domain. The portion of RF shifted to the input or output side will be deter-
•
mined by the desired ac input and output resistance levels.

Re
RF1 RF2 ~lo
,---4~"6---.--~""-,,__....._--1(----o Vo
C2

-=-
FIG. 5.50
Collector de feedback configuration.

At the frequency or frequencies of operation, the capacitor will assume a short-circuit

equivalent to ground due to its low impedance level compared to the other elements of the
network. The small-signal ac equivalent circuit will then appear as shown in Fig. 5.51.

_._
I;

+
ltlb +
_._
V;
Z;
RF, f3re ~

-=- -=- i -=- -=-

'------y-------
R'
-=-

FIG. 5.51
Substituting the r, equivalent circuit into the ac equivalent network of Fig. 5.50.

(5.65)

(5.66)

For r O 2: lORc,

(5.67)

R' = rollRFJRc
and Vo = -{3/lfl'
but
 COLLECTOR 285
and DC FEEDBACK
CONFIGURATION
so that

Vo
A=- (5.68)
v V;

For rO 2=: lORc,

(5.69)

Phase Relationship The negative sign in Eq. (5.68) clearly reveals a 180° phase shift
between input and output voltages.

EXAMPLE 5.10 For the network of Fig. 5 .52, determine:

a. re.
b. Z;.
C. Z 0 •
d. Av.
e. V0 if V;=2mV

12V

3k.Q

120 kf.l 68 kQ Fo
,.........~~--~,v,.,-~,(-------<, ~
lOµF
I; l0.01 µF ~
Zo
V; o...,._ )1------------11 /3 =l40,r0 =30kQ
lOµF

FIG. 5.52
Example 5.10.

Solution:
·I
a. Dc . B
= Vcc-VBE
Rp + f3Rc
12V- 0.7V
(120 kll + 68 kll) + (140)3 kll
11.3 V
608 kll = 18 ·6 µ,A

le = (/3 + l)Is = (141)(18.6 µ,A)

= 2.62mA
re= 26mV = 26mV = 9_920
le 2.62mA
b. f3re = (140)(9.92 ll) = 1.39 kll
The ac equivalent network appears in Fig. 5.53.
Z; = RpJf3re = 120 kll II 1.39 kll
~ 1.37kfi
286 BJT AC ANALYSIS
+
120 ill f3re 68ill 3ill
1.395 ill

-=- -=- -=- -=-

FIG. 5.53
Substituting the re equivalent circuit into the ac equivalent network of Fig. 5.52.

c. Testing the condition r O ~ lORc, we find

30 kO ~ 10(3 kD) = 30 kO
which is satisfied through the equals sign in the condition. Therefore,
Z0 ::c Rc11Rp2 = 3 kO I 68 kO
= 2.87k0
d. ro ~ lORc; therefore,

Av ::,
RF2IIRc 68kOll3kO
re 9.920
2.87kO
-
9.920
= -289.3
Vo
e. IAvl = 289.3 =-
V;
V0 = 289.3V; = 289.3(2 mV) = 0.579 V

5.12 EFFEO OF Rt AND Rs

All the parameters determined in the last few sections have been for an unloaded amplifier
with the input voltage connected directly to a terminal of the transistor. In this section the
effect of applying a load to the output terminal and the effect of using a source with an
•
internal resistance will be investigated. The network of Fig. 5 .54a is typical of those inves-
tigated in the previous section. Because a resistive load was not attached to the output ter-
minal, the gain is commonly referred to as the no-load gain and given the following
notation:

~ (5.70)
~
In Fig. 5.54b a load has been added in the form of a resistor RL, which will change the
overall gain of the system. This loaded gain is typically given the following notation:

(5.71)

In Fig. 5.54c both a load and a source resistance have been introduced, which will have
an additional effect on the gain of the system. The resulting gain is typically given the fol-
lowing notation:

(5.72)
with RL and R,

The analysis to follow will show that:

The loaded voltage gain of an amplifier is always less than the no-load gain.
 Vee Vee Vee

Re
Rs Rs

~ + +

+ ~
Vo R Vo R Vo
L L
V; V;

A VNL =
-=-
Vo * -=- -=-

vi
(a) (b) (c)

FIG. 5.54
Amplifier configurations: (a) unloaded; (b) loaded; (c) loaded with a source resistance.

In other words, the addition of a load resistor RL to the configuration of Fig. 5.54a will
always have the effect ofreducing the gain below the no-load level.
Furthermore:
The gain obtained with a source resistance in p'lace will always be less than that
obtained under loaded or unloaded conditions due to the drop in applied voltage across
the source resistance.
In total, therefore, the highest gain is obtained under no-load conditions and the lowest
gain with a source impedance and load in place. That is:
For the same configuration A.NL > AvL > Av,

It will also be interesting to verify that:

For a particu'lar design, the 'larger the level of Ru the greater is the level of ac gain.
In other words, the larger the load resistance, the closer it is to an open-circuit approxi-
mation that would result in the higher no-load gain.
In addition:
For a particu'lar amplifier, the smaller the internal resistance of the signal source, the
greater is the overall gain.
In other words, the closer the source resistance is to a short-circuit approximation, the
greater is the gain because the effect of Rs will essentially be eliminated.
For any network, such as those shown in Fig. 5.54 that have coupling capacitors, the
source and load resistance do not affect the de biasing levels.
The conclusions listed above are all quite important in the amplifier design process.
When one purchases a packaged amplifier, the listed gain and all the other parameters are
for the unloaded situation. The gain that results due to the application of a load or source
resistance can have a dramatic effect on all the amplifier parameters, as will be demon-
strated in the examples to follow.
In general, there are two directions one can take to analyze networks with an applied
load and/or source resistance. One approach is to simply insert the equivalent circuit, as
was demonstrated in Section 5.11, and use methods of analysis to determine the quantities
of interest. The second is to define a two-port equivalent model and use the parameters
determined for the no-load situation. The analysis to follow in this section will use the first
approach, leaving the second method for Section 5.14.
For the fixed-bias transistor amplifier of Fig. 5.54c, substituting the re equivalent circuit
for the transistor and removing the de parameters results in the configuration of Fig. 5.55.
287
288 BJT AC ANALYSIS
+

V, " '
Rs

Z;
+
---+- V; Rs
lt[b

f3re ~ f3Jb ro -
Re
zo
RL
+
Vo

1
"II' -!- "II' "II' l "II' "II'
'----------...,
"II'
-
-!-
R~ = r 0 IIRcllRL'= RcllRL

FIG. 5.55
The ac equivalent networkfor the network of Fig. 5.54c.

It is particularly interesting that Fig. 5.55 is exactly the same in appearance as Fig. 5.22
except that now there is a load resistance in parallel with Re and a source resistance has
been introduced in series with a source Vs.
The parallel combination of
Rl = r0 IIRcllRL ~ RcllRL
and Ya = -{3/ifil = -{3/b(RcllRL)
V;
with h=-
f3re

gives V0 = -{3(£:-)<RcllRL)

RellRL
so that (5.73)

The only difference in the gain equation using V; as the input voltage is the fact that Re
of Eq. (5.10) has been replaced by the parallel combination of Re andRL. This makes good
sense because the output voltage of Fig. 5.55 is now across the parallel combination of the
two resistors.
The input impedance is

(5.74)

as before, and the output impedance is

(5.75)

as before.
If the overall gain from signal source Vs to output voltage V0 is desired, it is only neces-
sary to apply the voltage-divider rule as follows:
Z;Vs
V;=---
Z; + Rs
and
Vs Z; + Rs
V0 V0 V; Z;
or A ----•--A
Vs - Vs - V; Vs - vL Z; + Rs

so that (5.76)

Because the factor Z;/ (Z; + Rs) must always be less than one, Eq. (5. 76) clearly supports
the fact that the signal gain Avs is always less than the loaded gain Ave
 EFFECT OF RL AND Rs 289
EXAMPLE 5.11 Using the parameter values for the fixed-bias configuration of Example 5.1
with an applied load of 4. 7 kll and a source resistance of 0.3 kll, determine the following
and compare to the no-load values:
a. Ave
b. Av,
C. Zi.
d. Z 0 •

Solution:
RcllRL 3 kDll4.7kll 1.831 kll
a. Eq. (5.73): AvL = ----------;;- = 10.71 D, -170.98
10.11 n
which is significantly less than the no-load gain of -280.11.
zi
b. Eq. (5.76): Av,= ---AvL
zi + Rs
With Zi = 1.07 kll from Example 5.1, we have
l.07kll
Av,= l.0 7 kll + 0 _3 kll (-170.98) = -133.54
which again is significantly less than AvNL or Ave
c. Zi = 1.07 kfi as obtained for the no-load situation.
d. Z0 = Re = 3 kfi as obtained for the no-load situation.
The example clearly demonstrates that AVNL > AvL > Av,

For the voltage-divider configuration of Fig. 5.56 with an applied load and series source
resistor the ac equivalent network is as shown in Fig. 5.57.

Cz
R,
+

r-t.
___._
lb

+
+ ~ RL Vo
zo
V, '\J ___._ V;
R2
Z; CE

FIG. 5.56
Voltage-divider bias configuration with Rs and RL.

I;
R, ~

+ ~+
b
=irb C

+
Z;
v, '\J f3r,
V; R1 Rz
+ f3/b ro Re
~
Vo

e e zo

"II' ~
R'
"II' -!
FIG. 5.57
Substituting the r, equivalent circuit into the ac equivalent network of Fig. 5.56.
290 BJT AC ANALYSIS First note the strong similarities with Fig. 5.55, with the only difference being the par-
allel connection of R1 and R2 instead of just R8 . Everything else is exactly the same. The
following equations result for the important parameters of the configuration:

(5.77)

I Zi = R1IIR2llf3re I (5.78)

I Z = Rcllr I
0 0 (5.79)

For the emitter-follower configuration of Fig. 5.58 the small-signal ac equivalent net-
work is as shown in Fig. 5.59. The only difference between Fig. 5.59 and the unloaded
configuration of Fig. 5.37 is the parallel combination of RE and RL and the addition of the
source resistor Rs. The equations for the quantities of interest can therefore be determined
by simply replacing RE by REIIRL wherever RE appears. If RE does not appear in an equation,
the load resistor RL does not affect that parameter. That is,

V0 REIIRL
(5.80)
AvL = vi REIIRL + re

B
+
10
CT
z Vo

--- z

FIG. 5.58
Emitter-follower configuration with Rs and RL.

I;

Rs
---+-
+
b
71/b C

---+-
Z;
f3r,
~ f3/b

+
v,I\, V; RB
lo
~ +
---
e

1 "II" -+- "II"

FIG. 5.59
RE

"II"
Zo
RL

Substituting the r, equivalent circuit into the ac equivalent network of Fig. 5.58.
"II"
Vo

"II"
 DETERMINING THE 291
I Z; = RBllzb I (5.81) CURRENT GAIN

I zb ~ ,B(REIIRL) I (5.82)

I Za ~ re I (5.83)

The effect of a load resistor and a source impedance on the remaining BJT configura-
tions will not be examined in detail here, although Table 5.1 in Section 5.14 will review
the results for each configuration.

5.1 l DETERMINING THE CURRENT GAIN

•
You may have noticed in the previous sections that the current gain was not determined for
each configuration. Earlier editions of this text did have the details of finding that gain, but
in reality the voltage gain is usually the gain of most importance. The absence of the deri-
vations should not cause concern because:
For each transistor configuration, the current gain can be determined directly from the
voltage gain, the defined load, and the input impedance.
The derivation of the equation linking the voltage and current gains can be derived using
the two-port configuration of Fig. 5.60.

System

FIG. 5.60
Determining the current gain using the voltage gain.

The current gain is defined by

A- = Ia (5.84)
I; l

Applying Ohm's law to the input and output circuits results in

V; Va
I- = - and I = --
1 Z; a RL
The minus sign associated with the output equation is simply there to indicate that the polar-
ity of the output voltage is determined by an output current having the opposite direction. By
definition, the input and output currents have a direction entering the two-port configuration.
Substituting into Eq. (5.84) then results in
Va
Ia RL Va Z;
A;L - - = -- = -- . -
I; V; V; RL
Z;
and the following important equation:

(5.85)

The value of RL is defined by the location of Va and Ia.

292 BJT AC ANALYSIS To demonstrate the validity of Eq. (5.82), consider the voltage-divider bias configura-
tion of Fig. 5.28.
Using the results of Example 5.2, we find

----
Vo
6.8 kfl

so that A;, - lo - ( - ~ ) - -(Yo)(l.35kfl)

l; V; V; 6.8 kfl
1.35 kfl

= -(-368.76) ( l.35kfl)
6 _8 kfl = 73.2

Using Eq. 5.82: Z; ( 1.35 kfl)

A;L = -AVLRL = -(-368.76) 6.8 kfl = 73.2
which has the same format as the resulting equation above and the same result.
The solution to the current gain in terms of the network parameters will be more com-
plicated for some configurations if a solution is desired in terms of the network parameters.
However, if a numerical solution is all that is desired, it is simply a matter of substituting
the value of the three parameters from an analysis of the voltage gain.
As a second example, consider the common-base bias configuration of Section 5.9. In
this case the voltage gain is

and the input impedance is

Z; == REllre - re
with RL defined as Re due to the location of 10 .
The result is the following:

A·
'L
= -A Z;
VLRL
= (-Re)(
Ye ReYe) == -1

which agrees with the solution of that section because le == le. Note, in this case, that the
output current has the opposite direction to that appearing in the networks of that section
due to the minus sign.

5.14 SUMMARY TABLES

•
The last few sections have included a number of derivations for unloaded and loaded BJT
configurations. The material is so extensive that it seemed appropriate to review most of
the conclusions for the various configurations in summary tables for quick comparisons.
Although the equations using the hybrid parameters have not been discussed in detail at
this point, they are included to make the tables complete. The use of hybrid parameters
will be considered in a later section of this chapter. In each case the waveforms included
demonstrate the phase relationship between input and output voltages. They also reveal the
relative magnitude of the voltages at the input and output terminals.
Table 5.1 is for the unloaded situation, whereas Table 5.2 includes the effect of Rs and Rv

5.15 TWO-PORT SYSTEMS APPROACH

•
In the design process, it is often necessary to work with the terminal characteristics of a
device rather then the individual components of the system. In other words, the designer is
handed a packaged product with a list of data regarding its characteristics but has no access
to the internal construction. This section will relate the important parameters determined
for a number of configurations in the previous sections to the important parameters of this
packaged system. The result will be an understanding of how each parameter of the
 TABLE 5.1
Unloaded BJT Transistor Amplifiers

Configuration
Fixed-bias: Medium (1 kil) Medium (2 kn) High (-200) High (100)

= I RBll/3r, I (Rcllr0 ) --
f3RBro
r, (r0 + Rc)(RB + /3r,)
=~
(RB ~ l0f3r,)
(r0 ~ lORc,
RB~ l0f3r,)

Voltage-divider Medium (1 kn) Medium (2 kn) High (-200) High (50)

bias:
= I R1IIR2ll/3r, I = I Rdr I
=~
0
~I~ Relic"
r,
~ I '-----------'

~~ /3(R1 IIR2)
R1 IIR2 + /3r,

Unbypassed High (100 kn) Medium (2 kn) Low (-5) High (50)
emitter bias:

Zb
=

=
I RBllzb I
f3(r, + Re)
= ~
(any level of r0 )
~ 1,. :CR, I
= I RBll/3Re I
(Re>> r,)
(Re>> r,)

Emitter- High (100 kil) Low (20 D) Low (=1) High (-50)
follower:
= I RBllzb I =I Rellr, I
Zb = /3(r, + Re)

V; ---+- = I RBll/3Re I (Re>> r,)

Z;
(Re>> r,)

Low (20 D) Medium (2 kn) High(200) Low (-1)

= I Rellr, I =~
=G] ~
(Re>> r,)

Collector Medium (1 kil) Medium (2 kil) High (-200) High (50)

feedback:
= I RcllRF I
~~
r, f3RF

(r0 ~ lORc)
(r0 ~ lORc)
(r0 ~ lORc) (RF>> Re)

293
 TABLE 5.2
BIT Transistor Amplifiers Including the Effect of Rs and RL

Configuration AvL =V 0 /Vi zi Zo

Vee
-(RLIIRc) Rsllf:lr, Re
Re r,

Vo

+
~

v,. '\,
Rs V

-
I

z,
- zn Including r0 :

(RLi!Reliro)
r,
Rsllf:lr, Rcllro
- .1....

Vee

-(RLi!Rc)
R1 IIR2llf:lr, Re
r,

r4 v, v,,

Including r0 :

-1 -
+
v, '\, -(RLi!Reliro)
R1 IIR2llf:lr, Rcllro
z,
r,
...
11ee

RE= RLIIRE R; = RsllR1IIR2

Re

r4v, R1
=1 R1 IIR2llf:l(r, + RE) Rd(1;} + r,)

-1 - tf'
+
v, '\, R2 .;_ n Including r0 :
z, RE
=1 R1IIR2llf:l(r, + RE) /3 + r, )
REIi (R;
....

-(RLi!Rc) REllr, Re
-
r,
v.
+
v, '\,
- 1... -
1.,
-=- VEE
I ':'
RE

' ... -.
Vee -=- -- z.
RL
Including r0 :

-
-(RLi!Rellr
r,
0)
REllr, Rcllro

Vee
-(RLi!Rc)
R1IIR2llf:l(r, + RE) Re
RE
Vo

+
r4V;-
v, '\, Z;
Including r0 :

-(RLi!Rc)
RE
R1 IIR2llf:l(r, + R,) =Re
-l ...

294
 TABLE 5.2 (Continued)
BIT Transistor Amplifiers Including the Effect of Rs and RL

Configuration Z;

Vee

Re

Including r0 :

Vee

Re
r,

Rs V- Including r0 :
+ · ',.,........__--1
~I-<>-

vs '\,
- l... r,

Vee

Including r0 :

packaged system relates to the actual amplifier or network. The system of Fig. 5.61 is
called a two-port system because there are two sets of terminals-one at the input and the
other at the output. At this point it is particularly important to realize that
the data surrounding a packaged system is the no-load data.
This should be fairly obvious because the load has not been applied, nor does it come with
the load attached to the package.

I;
---+-
0----
.._
lo
----0

+ ~ ..,_+
Z; zo
V; AvNL Vo

0----

+Thevenin

FIG. 5.61
Two-port system.
295
296 BJT AC ANALYSIS For the two-port system of Fig. 5.61 the polarity of the voltages and the direction of
the currents are as defined. If the currents have a different direction or the voltages have
a different polarity from that appearing in Fig. 5.61, a negative sign must be applied.
Note again the use of the label AVNL to indicate that the provided voltage gain will be the
no-load value.
For amplifiers the parameters of importance have been sketched within the boundaries
of the two-port system as shown in Fig. 5.62. The input and output resistance of a packaged
amplifier are normally provided along with the no-load gain. They can then be inserted as
shown in Fig. 5.62 to represent the seated package.

FIG. 5.62
Substituting the internal elements for the two-port system of Fig. 5.61.

For the no-load situation the output voltage is

(5.86)

due to the fact that I = OA, resulting in I0 R0 = OV.

The output resistance is defined by V; = OV. Under such conditions the quantity AvNLV;
is zero volts also and can be replaced by a short-circuit equivalent. The result is

(5.87)

Finally, the input impedance Z; simply relates the applied voltage to the resulting input
current and

(5.88)

For the no-load situation, the current gain is undefined because the load current is zero.
There is, however, a no-load voltage gain equal to AvNL·
The effect of applying a load to a two-port system will result in the configuration of
Fig. 5.63. Ideally, all the parameters of the model are unaffected by changing loads and
levels of source resistance. However, for some transistor configurations the applied load
can affect the input resistance, whereas for others the output resistance can be affected by
the source resistance. In all cases, however, by simple definition, the no-load gain is unaf-
fected by the application of any load. In any case, once AvNL' R;, and R 0 are defined for a
particular configuration, the equations about to be derived can be employed.

___..
I;
_.,_
Io

+ + Ro +

V; R; '\J AvmYi RL Vo

FIG. 5.63
Applying a load to the two-port system of Fig. 5.62.
 Applying the voltage-divider rule to the output circuit results in TWO-PORT SYSTEMS 297
APPROACH
RLAvNLvi
V =---
o RL + Ro

and (5.89)

Because the ratio RL/(RL + R 0 ) is always less than 1, we have further evidence that the
loaded voltage gain of an amplifier is always less than the no-load level.
The current gain is then determined by
I0 -V0 /RL
A·= -
lL Ii

and (5.90)

as obtained earlier. In general, therefore, the current gain can be obtained from the voltage
gain and impedance parameters Zi and RL. The next example will demonstrate the useful-
ness and validity of Eqs. (5.89) and (5.90).
Our attention will now turn to the input side of the two-port system and the effect of an
internal source resistance on the gain of an amplifier. In Fig. 5.64, a source with an internal
resistance has been applied to the basic two-port system. The definitions of Zi and AvNL are
such that:
The parameters Zi and AvNL of a two-port system are unaffected by the internal resis-
tance of the applied source.

Is I; Io
~ ~ ~
VV\, 0

+ R, + + I Ro +
V, '\, ~
V; R; '\, ¾V·
LI ~Vo
zo
Z;

FIG. 5.64
Including the effects of the source resistance Rs.

However:
The output impedance may be affected by the magnitude of R8•
The fraction of the applied signal reaching the input terminals of the amplifier of Fig. 5 .64
is determined by the voltage-divider rule. That is,

R;Vs
Vi=--- (5.91)
Ri + Rs

Equation (5.91) clearly shows that the larger the magnitude of Rs, the lower is the voltage
at the input terminals of the amplifier. In general, therefore, as mentioned earlier, for a
particular amplifier, the larger the internal resistance of a signal source, the lower is the
overall gain of the system.
For the two-port system of Fig. 5.64,
Vo= AvNLV;
RiVs
and Vi=---
R; + Rs
298 BJT AC ANALYSIS
so that

and (5.92)

The effects of Rs and RL have now been demonstrated on an individual basis. The next
natural question is how the presence of both factors in the same network will affect the
total gain. In Fig. 5.65, a source with an internal resistance Rs and a load RL have been
applied to a two-port system for which the parameters Z;, AvNL' and Z 0 have been specified.
For the moment, let us assume that Z; and Z 0 are unaffected by RL and Rs, respectively.

ls I; Io
~
I, ,.,_
~

+
+ R, + Ro
+ zo
vs '\, ~
V; R; '\, AuNLV;
RL Vo
z,

FIC. 5.65
Considering the effects of Rs and RL on the gain of an amplifier.

At the input side we find

Eq. (5.91): v- = -R;Vs

--
' R; + Rs

or (5.93)

and at the output side,

or (5.94)

For the total gain Avs = V0 /Vs, the following mathematical steps can be performed:

(5.95)

and substituting Eqs. (5.93) and (5.94) results in

(5.96)

Because I; = V;/R;, as before,

(5.97)

(5.98)
However, Ii = ls, so Eqs. (5.97) and (5.98) generate the same result. Equation (5.96) TWO-PORT SYSTEMS 299
clearly reveals that both the source and the load resistance will reduce the overall gain of APPROACH
the system.
The two reduction factors of Eq. (5.96) form a product that has to be carefully consid-
ered in any design procedure. It is not sufficient to ensure that Rs is relatively small if the
effect of the magnitude of RL is ignored. For instance, in Eq. (5.96), if the first factor is 0.9
and the second factor is 0.2, the product of the two results in an overall reduction factor
equal to (0.9)(0.2) = 0.18, which is close to the lower factor. The effect of the excellent
0.9 level was completely wiped out by the significantly lower second multiplier. If both
were 0.9-level factors, the net result would be (0.9)(0.9) = 0.81, which is still quite high.
Even if the first were 0.9 and the second 0.7, the net result of 0.63 would still be respect-
able. In general, therefore, for good overall gain the effects of Rs and RL must be evaluated
individually and as a product.

EXAMPLE 5.12 Determine AvL and Av, for the network of Example 5.11 and compare
solutions. Example 5.1 showed that AvNL = -280, Zi = 1.07 kll, and Z 0 = 3 kll. In
Example 5.11, RL = 4.7 kll and Rs = 0.3 kll.
Solution:
RL
a. Eq. (5.89): AvL = ---AVNL
RL + Ro
4.7kll
= 4.7 kll + 3 kll (- 280.ll)
= -170.98
as in Example 5.11.
Ri RL
b. Eq. (5.96): Av = - - - ---Av
s Ri + Rs RL + Ro NL
l.07kll 4.7kll
l.07kll + 0.3kll. 4.7kll + 3kll (- 28 0.ll)
= (0.781)(0.610)(-280. l 1)
= -133.45
as in Example 5.11.

EXAMPLE 5.13 Given the packaged (no-entry-possible) amplifier of Fig. 5.66:

a. Determine the gain AvL and compare it to the no-load value with RL = 1.2 kll.
b. Repeat part (a) with RL = 5.6 kll and compare solutions.
c. Determine Av with RL = 1.2 kll.
' I I
d. Find the current gain Ai = ~ = ~ with RL = 5.6 kll.
Ii ls

FIG. 5.66
Amplifier for Example 5.13.
300 BJT AC ANALYSIS Solution:
RL
a. Eq. (5.89): AvL = ---AVNL
RL + Ra

k~ kll n (-480) = (0.375)(-480)

1.2 + 2k
-180
which is a dramatic drop from the no-load value.
RL
b. Eq. (5.89): AvL = ---AVNL
RL + Ra

= 5 _6 :~6 !ll2 kll (-480) = (0.737)(-480)

= -353.76
which clearly reveals that the larger the load resistor, the better is the gain.
Ri RL
C. Eq. (5.96): Av, = - - - ----AvNL
Ri + Rs RL + Ra
4kll l.2kll
= 4 kll + 0.2 kll • 1.2 kll + 2 kll (- 480)
= (0.952)(0.375)(-480)
= -171.36
which is fairly close to the loaded gain Av because the input impedance is considerably
more than the source resistance. In other words, the source resistance is relatively
small compared to the input impedance of the amplifier.
Ia Ia Z;
d. AiL = - = - = -AvL
I; ls RL

= -(-353.76) ( 54kll
_6 kll ) = -(-353.76)(0.714)
= 252.6

It is important to realize that when using the two-port equations in some configurations
the input impedance is sensitive to the applied load (such as the emitter-follower and collec-
tor feedback) and in some the output impedance is sensitive to the applied source resistance
(such as the emitter-follower). In such cases the no-load parameters for Zi and Za have to
first be calculated before substituting into the two-port equations. For most packaged sys-
tems such as op-amps this sensitivity of the input and output parameters to the applied load
or source resistance is minimized to eliminate the need to be concerned about changes from
the no-load levels when using the two-port equations.

5.16 CASCADED SYSTEMS

•
The two-port systems approach is particularly useful for cascaded systems such as that
appearing in Fig. 5.67, where Av 1, Av2 , Av3 , and so on, are the voltage gains of each stage
under loaded conditions. That is, Av 1 is determined with the input impedance to Av2 acting
as the load on Av 1• For Av2, Av1 will determine the signal strength and source impedance at
the input to Av2 • The total gain of the system is then determined by the product of the indi-
vidual gains as follows:

(5.99)

and the total current gain is given by

(5.100)
 No matter how perfect the system design, the application of a succeeding stage or load CASCADED SYSTEMS 301
to a two-port system will affect the voltage gain. Therefore, there is no possibility of a
situation where Av,, Av2 , and so on, of Fig. 5.67 are simply the no-load values. The no-load
parameters can be used to determine the loaded gains of each stage, but Eq. (5.99) requires
the loaded values. The load on stage 1 is Zi2 , on stage 2 Zi 3 , on stage 3 Zin' and so on.

1--------- I
+ o------1
Av 1 Av 2 ----! Avn r----JRL ~
z,
'2

FIG. 5.67
z'3
--

8
Z;n
----------• 9 i-- --

Zan =Zo
-

Cascaded system.

EXAMPLE 5.14 The two-stage system of Fig. 5.68 employs a transistor emitter-follower
configuration prior to a common-base configuration to ensure that the maximum percentage
of the applied signal appears at the input terminals of the common-base amplifier. In Fig.
5.68, the no-load values are provided for each system, with the exception of Zi and Z 0 for the
emitter-follower, which are the loaded values. For the configuration of Fig. 5.68, determine:
a. The loaded gain for each stage.
b. The total gain for the system, Av and Av,
c. The total current gain for the system.
d. The total gain for the system if the emitter-follower configuration were removed.

+
v,
I;

~ 1 kil
Rs
-"' +
Au
1
Emitter-follower
Z;= !OkQ
Common-base
Z;= 26Q
+

"'
V RL 8.2 kil
'1 Z0 =12Q Z0 =5.l kil
AvNL= 1 A,NL =240

FIG. 5.68
Example 5.14.

Solution:
a. For the emitter-follower configuration, the loaded gain is (by Eq. (5.94))
zi2 26 n
Vo, = z. + Z AvNLV;, = 26 D, + 12 D, (1) Vi, = 0.684 V;,
Zz OJ

Vo,
and Av.= - = 0.684
l V;,
For the common-base configuration,
RL 8.2kll
Vo 2 = _R_+_R-AvNL V; 2 = -8 _-2-k_0_+_5__l_k_D,_(240) V; 2 = 147.97 Vi2
L 02

Vo2
and Av2 =-= 147.97
V;2
b. Eq. (5.99): Avr = Av,Av2
= (0.684)(147.97)
= 101.20
302 BJT AC ANALYSIS Zi 1 (10 kll)(lOl.20)
Eq. (5.91): Av, = z. + R Avr - 10 kll + 1 kll
lJ S

= 92

d. Eq. (5.91):

V- V
and ___!_ = 0.025 with _!!_ = 147.97 from above
Vs Vi
Vo Vi Vo
and Av = - = - • - = (0.025)(147.97) = 3.7
s Vs Vs Vi
In total, therefore, the gain is about 25 times greater with the emitter-follower configuration
to draw the signal to the amplifier stages. Note, however, that it is also important that the
output impedance of the first stage is relatively close to the input impedance of the second
stage, otherwise the signal would have been "lost" again by the voltage-divider action.

RC-Coupled BJT Amplifiers

One popular connection of amplifier stages is the RC-coupled variety shown in Fig. 5.69 in
the next example. The name is derived from the capacitive coupling capacitor Cc and the
fact that the load on the first stage is an RC combination. The coupling capacitor isolates
the two stages from a de viewpoint but acts as a short-circuit equivalent for the ac response.
The input impedance of the second stage acts as a load on the first stage, permitting the
same approach to the analysis as described in the last two sections.

EXAMPLE 5.15
a. Calculate the no-load voltage gain and output voltage of the RC-coupled transistor
amplifiers of Fig. 5.69.
b. Calculate the overall gain and output voltage if a 4.7 kll load is applied to the second
stage, and compare to the results of part (a).
c. Calculate the input impedance of the first stage and the output impedance of the second
stage.
+20V

2.2kQ 15kQ 2.2kQ

15kQ Cc
( 0V0
lOµF lOµF
V; =25 µYo )
Q1
/3= 200 Q2
/3= 200
lOµF

Iw~ Iw~
4.7kQ 4.7kQ
+ +
1 kQ 1 kQ

"=' "=' "=' "='

FIG. 5.69
RC-coupled BJT amplifier for Example 5.15.

Solution:
a. The de bias analysis results in the following for each transistor:
Vs= 4.8V, VE= 4.1 V, Ve= 11 V, /e = 4.lmA
 At the bias point, CASCADED SYSTEMS 303
re=26mV = 26mV = 6 _340
IE 4.1 mA
The loading of the second stage is
zi2 = R1 IIR2II.Bre
which results in the following gain for the first stage:
Rell (R1IIR211.Bre)
AV[=

(2.2 kO) I [15k0114.7 kO I (200)(6.34 0)]

6.34 o
659.2 o = _ 104
6.340
For the unloaded second stage the gain is
Re 2.2 kO
Av2(NL) = ----;:- =
e
6.34 o = - 347
resulting in an overall gain of
AvT(NL) = Av 1Av 2(NL) = (-104)(-347) ~ 36.1 X 103
The output voltage is then
Va = AVT(NL) vi = (36.1 X 103)(25 JL V) ~ 902.5 mV
b. The overall gain with the 10-kO load applied is

A = Va RL A = ~7k0 0 (36.1 X 103) ~ 24.6 X 103

VT vi RL + Za VT(NL) 4.7 k + 2.2 k
which is considerably less than the unloaded gain because RL is relatively close to Re.
Va = AvrVi
= (24.6 X 103)(25 JL V)
= 615mV
c. The input impedance of the first stage is
zit = R1 IIR2II.Bre = 4.7 kO I 15 kO 11(200)(6.34 D,) = 0.94 k!l
whereas the output impedance for the second stage is
Za2 = Re = 2.2 k!l

Cascode Connection
The cascode configuration has one of two configurations. In each case the collector of the
leading transistor is connected to the emitter of the following transistor. One possible
arrangement appears in Fig. 5.70; the second is shown in Fig. 5.71 in the following example.

Vee

Re Qz
Rat
Vo

V; QI

RL
Ra2
RE
ICE Ra
lea
':" ':" ':" ':" ':" ':"

FIG. 5.70
Cascade configuration.
304 BJT AC ANALYSIS The arrangements provide a relatively high-input impedance with low voltage gain for the
first stage to ensure the input Miller capacitance (to be discussed in Section 9.9) is at a
minimum, whereas the following CB stage provides an excellent high-frequency response.

EXAMPLE 5.16 Calculate the no-load voltage gain for the cascade configuration of Fig. 5.71.

Vee= 18 V

Re
1.8 ill

---u(------<O V02
e=SµF
Qz

V;
1
a------),_________,,
es=5 µF

FIG. 5.71
Practical cascade circuit for Example 5.16.

Solution: The de analysis results in

VB1 = 4.9 V, VB2 = 10.8 V, lei ~ Ie2 = 3.8 mA
because fe 1 ~ fe 2 the dynamic resistance for each transistor is
re= 26mV ~ 26mV = 6 _8 ll
IE 3.8mA
The loading on the transistor Q1 is the input impedance of the Q2 transistor in the CB
configuration as shown by re in Fig 5.72.
The result is the replacement of Re in the basic no-load equation for the gain of the CB
configuration, with the input impedance of a CB configuration as follows:
Re re
= - - = - - = -1
Av1
re re
with the voltage gain for the second stage (common base) of
Re l.8kll
A =-=--=265
Vz re 6.8 fl

FIG. 5.72
Defining the load of Qi.
The overall no-load gain is
Avr = Av 1Av2 = (-1 )(265) = - 265
As expected, in Example 5.16, the CE stage provides a higher input impedance than can
be expected from the CB stage. With a voltage gain of about 1 for the first stage, the
Miller-effect input capacitance is kept quite low to support a good high-frequency response.
A large voltage gain of 265 was provided by the CB stage to give the overall design a good
input impedance level with desirable gain levels.

5.17 DARLINGTON CONNECTION

•
A very popular connection of two bipolar junction transistors for operation as one "super-
beta" transistor is the Darlington connection shown in Fig. 5.73. The main feature of the
Darlington connection is that the composite transistor acts as a single unit with a current
gain that is the product of the current gains of the individual transistors. If the connection
is made using two separate transistors having current gains of {3 1 and /32, the Darlington American (Pittsburgh, PA; Exeter, NH)
(1906-1997)
connection provides a current gain of
Department Head at Bell Laboratories
I f3v = /31/32 I (5.101) Professor, Department of Electrical and
Computer Engineering, University of
New Hampshire
C
Dr. Sidney Darlington earned his B.S. in
physics at Harvard, his B.S. in electrical
communication at MIT, and his Ph.D. at
B Columbia University. In 1929 he joined
Bell Laboratories, where he was head of
the Circuits and Control Department. Dur-
Q2 ing that period he became good friends
with other important contributors such as
Edward Norton and Hendrik Bode. A
holder of 24 U.S. patents, he was awarded
E
the Presidential Medal of Freedom, the
highest civilian honor in the United States,
FIG. 5.73
in 1945 for his contributions to network
Darlington combination.
design during World War IL An elected
The configuration was first introduced by Dr. Sidney Darlington in 1953. A short biog- member of the National Academy of
Engineering, he also received the IEEE
raphy appears as Fig 5.74.
Edison Medal in 1975 and the IEEE
Medal of Honor in 1981. His U.S. patent
Emitter-Follower Configuration 2 663 806 titled "Semiconductor Signal
Translating Device" was issued on Decem-
A Darlington amplifier used in an emitter-follower configuration appears in Fig. 5.75. The ber 22, 1953, describing how two transis-
primary impact of using the Darlington configuration is an input impedance much larger than tors could be constructed in the Darlington
configuration on the same substrate--
often looked upon as the beginnings of
compound IC construction. Dr. Darlington
was also responsible for the introduction
and development of the Chirp technique,
used throughout the world in waveguide
transmission and radar systems. He is a
C primary contributor to the Bell Laborato-
ries Command Guidance System that
guides most of the rockets used today to
V; o----1)t------8---il place satellites in orbit. It uses a combina-
C1 tion of radar tracking on the ground with
/32 inertial control in the rocket itself. Dr.
+ Darlington was an avid outdoorsman as a
VBE - ( o V0
hiker and member of the Appalachian
E
IE,'i Cz Mountain Club. One of his proudest
accomplishments was being able to climb
RE Mt. Washington at the age of 80.

FIG. 5.74
Sidney Darlington (Courtesy of
AT&T Archives and History Center.)
FIG. 5.75
Emitter-follower configuration with a Darlington amplifier.
305
306 BJT AC ANALYSIS that obtained with a single-transistor network. The current gain is also larger, but the voltage
gain for a single-transistor or Darlington configuration remains slightly less than one.

DC Bias The case current is determined using a modified version of Eq. 4.44. There are
now two base-to-emitter voltage drops to include and the beta of a single transistor is
replaced by the Darlington combination ofEq. 5.101.

Vcc - VBE1 - VBE2

/B =------- (5.102)
1 RB+ f3vRE

The emitter current of Q 1 is equal to the base current of Q2 so that

h2 = f32IB2 = /32h1 = /32(/31/£1) = f31f32IB1
resulting in

I 1c2 ~ 1E2 = f3vlB1 I (5.103)

The collector voltage of both transistors is

(5.104)

the emitter voltage of Q2

(5.105)

the base voltage of Q 1

I VB1 = Vee - IB1RB = VE2 + VBE1 + VBE2 I (5.106)

the collector-emitter voltage of Q

I VcE2 = Vc2 - VE2 = Vee - VE2 (5.107)

EXAMPLE 5.17 Calculate the de bias voltages and currents for the Darlington configura-
tion of Fig. 5.76.

+18 V

3.3MQ

V; o>----)-1t----+----1
C1
/32 = 100
-------11-(----<O V 0

Cz

390Q

"II"

FIG. 5.76
Circuit for Example 5.17.
Solution: DARLINGTON 307
CONNECTION
f3v = /31/32 = (50)(100) = 5000
V cc - VBE1 - VBE2 18V - 0.7V - 0.7V
RB+ f3vRE 3.3 Mfl + (5000)(390 fl)
18 V - 1.4 V 16.6V
3.3 Mfl + 1.95 Mfl 5.25 Mfl = 3"16 µA
/c2 ::, /E2 = f3vlB1 = (5000)(3.16 mA) = 15.80 mA
Vc 1 = Vc2 = 18 V
VE2 = h2RE = (15.80 mA)(390 fl) = 6.16 V
VB 1 = VE2 + VBE 1 + VBEz = 6.16V + 0.7V + 0.7V = 7.56V
VcE2 = Vee - VE2 = 18 V - 6.16 V = 11.84 V

AC Input Impedance The ac input impedance can be determined using the ac equivalent
network of Fig. 5.77.

.:.

_ ___,, Qz

Z; R
___._ B

-=-
FIG. 5.77
Finding Z;.

As defined in Fig. 5.77:

= f32(re2 + RE)
Z;z
Z;1 = /31(re1 + Z;2)
so that Z;1 = /31(re1 + f32(re2 + RE))
Assuming RE>> rez
and Z;1 = /31(re 1 + f32RE)
Since f32RE >> re1
Z;l ::, f31f32RE
and since Z; = RBIIZ;
(5.108)

For the network of Fig. 5.76

Z; = RBllf3vRE
= 3.3 Mfl I (5000)(390 fl) = 3.3 Mfl I 1.95 Mfl
= 1.38MO
Note in the preceding analysis that the values of re were not compared but dropped com-
pared to much larger quantites. In a Darlington configuration the values of re will be differ-
ent because the emitter current through each transistor will be different. Also, keep in mind
that chances are the beta values for each transistor will be different because thay deal with
different current levels. The fact remains, however, that the product of the two beta values
will equal f3v, as indicated on the specification sheet.
308 BJT AC ANALYSIS AC Current Gain The current gain can be determined from the equivalent network of
Fig. 5.78. The output impedance of each transistor is ignored and the parameters for each
transistor are employed.

I;
B, f31re1 E, B2 J32re2 E2
---+-
---+-
lb1
---+-
lbz
q
Rn
t /3ifb1
t f32lb2 RE

"II"
lei
"II"
lc2
"II" "II"

FIG. 5.78
Determining A;Jor the network of Fig. 5.75.

Solving for the output current: /0 = lb2 + f32lb 2 = (/32 + l)/b2

with lb2 = f31lb, + lb, = (/31 + 1)/b,
Then / 0 = (/32 + 1)(/31 + l)Ib,
Using the current-divider rule on the input circuit:
Rn
lb = ---[. = - - -Rn- - / -
' Rn + zi I
Rn + f31f32RE 1

and /0 = (/32 + 1)(/3 1 + 1)( Rn

Rn+ f31f32RE
)ii
A- = lo = _(/3_1_+_1_)(/3_2_+_l)_R_s
so that
' Ii Rn + f31f32RE
Using /31, /32 >> 1

(5.109)

or (5.110)

For Fig. 5.76:

_ 10 _ f3vRn (5000)(3.3 MO)
A--------
1 Ii Rn + f3vRE 3.3 MO + 1.95 MO
= 3.14 X 103

AC Voltage Gain The voltage gain can be determined using Fig. 5.77 and the following
derivation:
V0 = laRE
vi = I/RnllzD

and

and (5.111)

an expected result for the emitter-follower configuration.

AC Output Impedance The output impedance will be determined by going back to Fig. 5.78 DARLINGTON 309
and setting Vi to zero volts as shown in Fig. 5.79. The resistor Rn is "shorted out," resulting CONNECTION
in the configuration of Fig. 5.80. Note in Figs. 5.82 and 5.83 that the output current has
been redefined to match standard nomenclature and properly defined Z 0 •

V;=OV

FIG. 5.79
Determining Z 0 •

lbz (/32 + 1)/bz Io

+ ---➔- - ---➔- a ~

f3zr,2 I, i ~
zo
+

[bit f31re1
t /3/b1
t f32lb2 RE Vo

+
"II"
1 "II"
1"II" "II"

FIG. 5.80
Redrawn of network of Fig. 5. 79.

At point a Kirchhoff's current law will result in 10 + (/32 + l)h2 = le:

10 = le - (/32 + l)/b2
Applying Kirchhoff's voltage law around the entire outside loop will result in
-hil31re 1 - h 2 f32re 2 - Vo = 0
and =V0 lb 1/31re 1 + h 2/32re 2
Substituting h 2 = (/31 + l)h 1
Vo = -h 1f31re 1 - + l)Ib 1f32re 2
(/31
= -hJ/31re 1 + (/31 + l)f32re2]
Vo
and h1 = f31re 1 + (/31 + 1)/32re2

with h2 = (/31 + l)hl = (/31 + l)[ f31re 1 + (/31 + l)f32re2

so that h2 = - [ - - -/31-+-1- - - ] Vo
f31re 1 + (/31 + l)f32re2

Going back 10 = le - (/32 + l)/b2 = le - (/32 + 1)(

or I = -V0 + -(/31-+-1)(/32
--+ - l)V
-- 0

0 RE f31re 1 + (/31 + l)f32re 2

310 BJT AC ANALYSIS Because /31, {3 2 >> 1

I = Va + __V_o_
o RE rel
- + re2
/32
Io which defines the parallel resistance network of Fig. 5.81.
~

+ In general, RE >> (~~ + re 2) so the output impedance is defined by

~
rel Zo
-+re RE Vo
/32 2
(5.112)

Using the de results, the value of re2 and re 1 can be determined as follows.
FIC. 5.81
Resulting network defined by Z0 • re2 = 26mV = 26mV = 1. 65 O
IE2 15.80mA
/E2 15.80mA
and IE = / 8 = - = --- = 0 158 mA
I 2 /32 100 .
26mV
SO that re 1 = 0 _158 mA = 164.5 Q
The output impedance for the network of Fig. 5.78 is therefore:
rel 164.5 Q
Zo == /3 2 + re 2 = lOO + 1.65 0 = 1.645 0 + 1.65 0 = 3.30 !l

In general, the output impedance for the configuration of Fig. 5.78 is very low-in the
order of a few ohms at most.

Voltage-Divider Amplifier
DC Bias Let us now investigate the effect of the Darlington configuration in a basic
amplifier configuration as shown in Fig. 5.82. Note that now there is a collector resistor
Re, and the emitter terminal of the Darlington circuit is connected to ground for ac condi-
tions. As noted on Fig. 5.82, the beta of each transistor is provided along with the resulting
voltage from base to emitter.

Vcc=27V

Re 1.2kQ

~---+-----1(--------o Vo
Darlingt C2

-- --
Pair
V; o )1---+-----0
I; C1
/31 = /32 = 110.
Z.'l
VBE= l.SV
z,
l R2
220kQ

680Ql CE

FIC. 5.82
Amplifier configuration using a Darlington pair.
 The de analysis can proceed as follows: DARLINGTON 311
CONNECTION
/3v = /31/32 = (110 X 110) = 12,100
R2 220 kfl(27 V)
Vs= R2 + R/cc = 220kfl + 470kfl = 8•61 V
VE = Vs - VsE = 8.61 V - 1.5 V = 7.11 V
_ VE _ 7.11 V _ 10 46 A
fe - RE - 680 fl - • m
10.46mA
12, 100 = 0.864 µA
Using the preceding results the values of re 2 and re 1 can be determined:
rez = 26 mV = 26 mV = 2_49 O
lE2 10.46 mA
h2 10.46mA
Ie1 = Is2 = 132 = 110 = 0.095 mA

and 'e1 = 26 mV = 26 mV = 273.7 0

Ie1 o.095mA

AC Input Impedance The ac equivalent of Fig. 5.82 appears as Fig. 5.83. The resistors R1
and R 2 are in parallel with the input impedance to the Darlington pair, assuming the second
transistor found by assuming the second transistor acts like an RE load on the first as
shown in Fig. 5.83.
That is, z; = f31re 1 + /31(f32re 2)

Vo

V;
I;
---+- 17
Z;
---+-
Re
R1 Rz

-=- -=-
FIG. 5.83
Defining Z'; and Z;.

and (5.113)

For the network of Fig. 5.82:

Z/ = 110[273.7 fl + (110)(2.49 fl)]
= 110[273.7 fl + 273.9 fl]
= 110[547.6 fl]
= 60.24kO
and zi = R1IIR2llz/
= 470 kfl I 220 kfl I 60.24 kfl
= 149.86kflll60.24kfl
= 42.97kO
312 BJT AC ANALYSIS AC Current Gain The complete ac equivalent of Fig. 5.82 appears as Fig. 5.84.

/.' lb1 /3/b1

BI --+-- ___._ E1,B2
......, C1C2

/31re1 i [bz
~
Vo

---.- R1 R2 ---.-
z.' f32re2 { f32lb2 Re
Z; I

E2 lE2
'II" 'II" 'II" 'II" 'II"

FIG. 5.84
ac equivalent network for Fig. 5.82.

The output current Io = /31h1 + /32h2

with h2 = (/31 + l)hl
so that lo = /31h1 + /32(/31 + l)h1
and with h1 = I[
we find lo = /311[ + /32(/31 + 1)/[
Io
and At = /. = /31 + /32(/3 + 1)
l

= /31 + /32/31 = /31 (1 + /32)

= /31/32

and finally (5.114)

For the original structure:

but

so that
A- = /3v(R1 IIR2) (5.115)
1
R1IIR2 + Z[
A- = (12,100)(149.86 kfl)
For Fig. 5.82 1 149.86kO + 60.24kO
= 8630.7
Note the significant drop in current gain due to R 1 and R 2.

AC Voltage Gain The input voltage is the same across R1 and R2 and at the base of the
first transistor as shown in Fig. 5.84.
The result is

v -
0 Rc
A - -V0 - -I- - - -A -
v- - rzi - i
(Re)
zi
l l l l

(5.116)

For the network of Fig. 5.82,

A = _/3vRc = (12,000)(1.2 kfl)
-241.04
v Z[ 60.24kO
AC Output lmpedence Because the output impedance in Re is parallel with the collector DARLINGTON 313
to emitter terminals of the transistor, we can look back on similar situations and find that CONNECTION
the output impedance is defined by

(5.117)

where r02 is the output resistance of the transistor Q2.

Packaged Darlington Amplifier

Because the Darlington connection is so popular, a number of manufacturers provide
packaged units such as shown in Fig. 5.85. Typically, the two BJTs are constructed on a
single chip rather than separate BJT units. Note that only one set of collector, base, and
emitter terminals is provided for each configuration. These, of course, are the base of the
transistor Q1, the collector of Q1 and Q2 , and the emitter of Q2 .

(a)

FIG. 5.85
Packaged Darlington amplifiers: (a) T0-92 package;
(b) Super SOTJ:M-3 package.

In Fig. 5.86 some of the ratings for an MPSA28 Fairchild Semiconductor Darlington
amplifier are provided. In particular, note that the maximum collector-to-emitter voltage of
80 V is also the breakdown voltage. The same is true for the collector-to-base and emitter-
to-base voltages, although notice how much lower the maximum ratings are for the base-
to-emitter junction. Because of the Darlington configuration, the maximum current rating
for the collector current has jumped to 800 mA-far exceeding levels we have encountered

Absolute Maximum Ratings

VcES Collector-Emitter Voltage 80V

VCBo Collector-Base Voltage 80V
VEBO Emitter-Base Voltage 12V
le Collector Current-Continuous 800mA

Electrical Characteristics

V(BR)CES Collector-Emitter Breakdown Voltage 80V

V(BR)CBO Collector-Base Breakdown Voltage 80V
V(BR)EBO Emitter-Base Breakdown Voltage 12V
lcso Collector Cutoff Current l00mA
leso Emitter Cutoff Current lO0mA

On Characteristics

hFE DC Current Gain 10,000

VcE(sat) Collector-Emitter Saturation Voltage 1.2 V
VBE(on) Base-Emitter on Voltage 2.0V

FIG. 5.86
MPSA 28 Fairchild Semiconductor Darlington amplifier ratings.
314 BJT AC ANALYSIS for single-transistor networks. The de current gain is rated at the high level of 10,000 and
the base-to-emitter potential in the "on" state is 2 V, which certainly exceeds the 1.4 V we
have used for individual transistors. Finally, it is interesting to note that the level of IeEo is
much higher at 500 nA than for a typical single-transistor unit.
In the packaged format the network of Fig. 5.75 would appear as shown in Fig. 5.87.
Using f3v and the provided value of VBE(=VBEi + VBE2), all the equations appearing in
this section can be applied.

+Vee (+18 V)

Rs
3.3MQ C
MPSA 28 Darlin on Amplifier
C1
V o
' ___..
\ 1-----+--___.
1 B f3v = 10,000
VBE =2.0V
I;

......
Z; E
Cz
( o V0
po
RE
390Q

"II"

FIC. 5.87
Darlington emitterjollower circuit.

C
0
5.18 FEEDBACK PAIR
The feedback pair connection (see Fig. 5.88) is a two-transistor circuit that operates like
the Darlington circuit. Notice that the feedback pair uses a pnp transistor driving an npn
transistor, the two devices acting effectively much like one pnp transistor. As with a Dar-
•
B lington connection, the feedback pair provides very high current gain (the product of the
transistor current gains), high input impedance, low output impedance, and a voltage gain
slightly less than one. Initially, it may appear that it would have a high voltage gain because
the output is taken off the collector with a resistor Re in place. However, the pnp-npn
E combination results in terminal characteristics very similar to that of the emitter-follower
FIC. 5.88 configuration. A typical application (see Chapter 12) uses a Darlington and a feedback-pair
Feedback pair connection. connection to provide complementary transistor operation. A practical network employing a
feedback pair is provided in Fig. 5.89 for investigation.

DC Bias
The de bias calculations that follow use practical simplifications wherever possible to pro-
vide simpler results. From the Q 1 base-emitter loop, one obtains
Vee - IeRe - VEB 1 - IBlB = 0
Vee - (/31/32/B)Re - VEB1 - IB1RB = 0
The base current is then

Vee - VBE1
/B = ----- (5.118)
1 RB + /31f32Re

The collector current of Q 1 is

lei = f31IB1 = ls2
which is also the base Q2 current. The transistor Q2 collector current is
Ie2 = f32ls2 = h2
 Vee FEEDBACK PAIR 315
(+18 V)

Re
lei
75 n

(---------<, V0

/E1
~

V; o------) ..,_ i/e2 /31 = 140

/B1 /32 = 180
RB
lh.
lei= /82
2Mn

i1e2
T T

FIG. 5.89
Operation of a feedback pair.

so that the current through Re is

I le = IE1 + Ie2 = IB2 + Ie2 I (5.119)

The voltages (5.120)

and (5.121)

with (5.122)

EXAMPLE 5.18 Calculate the de bias currents and voltages for the circuit of Fig. 5.89 to
provide V0 at one-half the supply voltage (9 V).
Solution:
18 V - 0.7 V 17.3 V
/Bl = 2 MO + (140)(180)(75 0) = 3.89 x 106 = 4.45 µA
The base Q2 current is then

resulting in a Q2 collector current of

le2 = f32IB 2 = 180(0.623 mA) = 112.1 mA

and the current through Re is then

Eq. (5.119): le= /e1 + le2 = 0.623 mA + 112.1 mA = le =
2 112.1 mA
Ve2 = VE 1 = 18 V - (112.1 mA)(75 0)
= 18 V - 8.41 V
= 9.59V
VB1 = IBlB = (4.45 µ,A)(2 MO)
= 8.9V
VBe 1 = VB 1 - 0.7V = 8.9V - 0.7V
= 8.2V
316 BJT AC ANALYSIS AC Operation
The ac equivalent circuit for that of Fig. 5 .89 is drawn in Fig. 5 .90.

I; I'
~ /3 ,re, a
~

+ +
~
~

~ ~ /bl
Z; Z-'I
t f3lb,

v; Rn
~ f32lb2 Re
Vo

~i t,'a
-=-
1

FIG. 5.90
l -=-

ac equivalent for the network of Fig. 5.89.

Input Impedance, Z1 The ac input impedance seen looking into the base of transistor Q 1
is determined as follows:
V-
z! = _..!.
l I[
Applying Kirchhoffs current law at node a and defining le =/ 0:

h, + /31h 1 - /32h2 + Io =0
with /b2 = -f31Ib, as noted in Fig. 5.90.
The result is h,+ /31h, - /32(-f31Ib,) + 1 = 0 0

and Io = -h, - /31h 1 - /31/32h 1

or = -Ibp + /31) - /31f32Ib,
/ 0

but /31 >> 1

and lo = -f31h, - /31/32h, = -lb1(/31 + /31/32)
= -hi/31O + /32)

resulting in: I lo ~ -f31/32Ib, (5.123)

and Vo = -IoRc = -(-f31f32lb 1)Rc = /31f32Ib,Rc

vi - /31/32h,Rc
so lb=-----
' f31re,
Rearranging: Ib,f31re, = Vi - f31f32Ib,Rc
and lb 1(f31re, + f31f32Rc) = Vi
_ / _ vi
so lb - I- - - - - - - -
1 ' f31re, + f31/32Rc
v. vi
vi=_!_=-----
and
' I[ Vi
f31re + f31/32Rc

so that (5.124)

In general,

and (5.125)
 FEEDBACK PAIR 317
with (5.126)

For the network of Fig. 5 .89: re1 = 26 mV =

1
E1
o.!::: = 41. 73 D,

and Z[ = f31re 1 + f31f32Rc

= (140)(41.73 D,) + (140)(180)(75 D,)
= 5842.2 D, + 1.89 Mll
= 1.895M!l
where Eq. (5.125) results in Z[ = {3 1{32Rc = (140)(180)(75 ll) = 1.89 Mil, validating
the above approximations.

Current Gain
Defining Ib 1 = I[
as shown in Fig. 5.90 will permit finding the current gain A[ = Ia/ I;'.
Looking back on the derivation of Zi we found Ia = -f31/32h 1 = -/31/321[

resulting in (5.127)

The current gain Ai = Ia/ Ii can be determined using the fact that
Ia Ia I[
A-=-=-•-
1 Ii I[ Ii

For the input side:

J! = RBJi RBJi
1
RB + z! RB + f31f32Rc
Substituting: la I[ ( RB )
Ai = I[ • Ii = (-/3 1/3 2) RB + f31f32Rc

So that (5.128)

The negative sign appears because both Ii and Ia are defined as entering the network.
Ia
For the network of Fig. 5 .89: A/ = I' = -/31/32
l

= -(140)(180)
= -25.2 X 103
A_ = -f31f32RB (140)(180)(2 Mll)
1
RB+ f31f32Rc 2 Mll + 1.89 Mll
50,400Mll
3.89Mll
= -12.96 X 103 ( = halfof Ai)

Voltage Gain
The voltage gain can quickly be determined using the results obtained above.
Va -IaRc
That is, A=-
v vi I[Z[

(5.129)
318 BJT AC ANALYSIS which is simply the following if we apply the approximation: {3 2Rc >> re 1

Av == /32Rc = 1
/32Rc
/32Rc (180)(75 ll)
For the network of Fig. 5.89: Av = - - - -
re1 + /32Rc 41.73 ll + (180)(75 ll)
13.5 X 103 ll
41.73 n+ 13.5 x 103 n
= 0.997 == 1 (as indicated above)

Output Impedance
The output impedance z; is defined in Fig. 5.91 when V; is set to zero volts.

t /3iJb1
~
Z'
0

'½! t''o
-=-
FIG. 5.91
Determining Z~ and Z 0 •

Using the fact that / 0 = -{3 1{3 2Ib 1 from calculations above, we find that

but

and

so that (5.130)

with (5.131)

re
However, R >> ____!.
C /32

leaving (5.132)

which will be a very low value.

For the network of Fig. 5.89:
ll "'
Z0 == 41.73
180 = 0.23 u
The preceding analysis shows that the feedback pair connection of Fig. 5.89 provides
operation with voltage gain very near 1 (just as with a Darlington emitter-follower), a very
high current gain, a very low output impedance, and a high input impedance.
5.19 THE HYBRID EQUIVALENT MODEL
The hybrid equivalent model was mentioned in the earlier sections of this chapter as one
that was used in the early years before the popularity of the re model developed. Today
there is a mix of usage depending on the level and direction of the investigation.
• THE HYBRID
EQUIVALENT MODEL
319

The re model has the advantage that the parameters are defined by the actual operating
conditions,
whereas
the parameters of the hybrid equivalent circuit are defined in general terms for any
operating conditions.
In other words, the hybrid parameters may not reflect the actual operating conditions
but simply provide an indication of the level of each parameter to expect for general use.
The re model suffers from the fact that parameters such as the output impedance and the
feedback elements are not available, whereas the hybrid parameters provide the entire set
on the specification sheet. In most cases, if the re model is employed, the investigator will
simply examine the specification sheet to have some idea of what the additional elements
might be. This section will show how one can go from one model to the other and how the
parameters are related. Because all specification sheets provide the hybrid parameters and
the model is still extensively used, it is important to be aware of both models. The hybrid
parameters as shown in Fig. 5.92 are derived from the specification sheet for the 2N4400
transistor described in Chapter 3. The values are provided at a de collector current of 1 mA
and a collector-to-emitter voltage of 10 V. In addition, a range of values is provided for
each parameter for guidance in the initial design or analysis of a system. One obvious ad-
vantage of the specification sheet listing is the immediate knowledge of typical levels for
the parameters of the device as compared to other transistors.

Min. Max.
Input impedance
h;e 0.5 7.5 kO
Uc= I mA de, Ve£= IO V dc,.f= I kHz)
Voltage feedback ratio XI0-4
h,,e 0.1 8.0
Uc= I mA de, Ve£= 10 V dc,.f= I kHz)
Small-signal current gain
Uc= I mA de, VCE = 10 V de,.f= I kHz) hre 20 250 -
Output admittance
hoe 1.0 30 lµ.S
Uc= I mA de, VCE = 10 V dc,.f= I kHz)

FIG. 5.92
Hybrid parameters for the 2N4400 transistor.

The description of the hybrid equivalent model will begin with the general two-port
system of Fig. 5.93. The following set of equations (5.131) and (5.132) is only one of
a number of ways in which the four variables of Fig. 5.93 can be related. It is the most
frequently employed in transistor circuit analysis, however, and therefore is discussed in
detail in this chapter.

I; Ia
~
1 o----·--0 0- - - -
~ -----<> 2
+ +
V;
'i'
I
l ' o - - - - - - - - - - - - + - - - - - - - - - - - - - 0 2'

FIG. 5.93
Two-port system.
320 BJT AC ANALYSIS
(5.133)

(5.134)

The parameters relating the four variables are called h-parameters, from the word
"hybrid." The term hybrid was chosen because the mixture of variables (V and/) in each
equation results in a "hybrid" set of units of measurement for the h-parameters. A clearer
understanding of what the various h-parameters represent and how we can determine their
magnitude can be developed by isolating each and examining the resulting relationship.

h11 If we arbitrarily set V0 = 0 (short circuit the output terminals) and solve for h 11 in
Eq. (5.133), we find

ohms (5.135)

The ratio indicates that the parameter h 11 is an impedance parameter with the units of ohms.
Because it is the ratio of the input voltage to the input current with the output terminals
shorted, it is called the short-circuit input-impedance parameter. The subscript 11 of h 11
refers to the fact that the parameter is determined by a ratio of quantities measured at the
input terminals.

h12 If I; is set equal to zero by opening the input leads, the following results for h 12:

unitless (5.136)

The parameter h 12 , therefore, is the ratio of the input voltage to the output voltage with
the input current equal to zero. It has no units because it is a ratio of voltage levels and is
called the open-circuit reverse transfer voltage ratio parameter. The subscript 12 of h 12
indicates that the parameter is a transfer quantity determined by a ratio of input (1) to out-
put (2) measurements. The first integer of the subscript defines the measured quantity to
appear in the numerator; the second integer defines the source of the quantity to appear in
the denominator. The term reverse is included because the ratio is an input voltage over an
output voltage rather than the reverse ratio typically of interest.

h21 If in Eq. (5.134) V0 is set equal to zero by again shorting the output terminals, the
following results for h21:

unitless (5.137)

Note that we now have the ratio of an output quantity to an input quantity. The termforward
will now be used rather than reverse as indicated for h 12. The parameter h21 is the ratio of
the output current to the input current with the output terminals shorted. This parameter,
like h 12, has no units because it is the ratio of current levels. It is formally called the short-
circuit forward transfer current ratio parameter. The subscript 21 again indicates that it
is a transfer parameter with the output quantity (2) in the numerator and the input quantity
(1) in the denominator.

h22 The last parameter, h22 , can be found by again opening the input leads to set 11 = 0
and solving for h22 in Eq. (5.134):

s1emens (5.138)

Because it is the ratio of the output current to the output voltage, it is the output conductance
parameter, and it is measured in siemens (S). It is called the open-circuit output admittance
parameter. The subscript 22 indicates that it is determined by a ratio of output quantities.
 Because each term ofEq. (5.133) has the unit volt, let us apply Kirchhoff's voltage law THE HYBRID 321
"in reverse" to find a circuit that "fits" the equation. Performing this operation results in EQUIVALENT MODEL
the circuit of Fig. 5.94. Because the parameter h 11 has the unit ohm, it is represented by a
resistor in Fig. 5.94. The quantity h 12 is dimensionless and therefore simply appears as a
multiplying factor of the "feedback" term in the input circuit.
Because each term of Eq. (5.134) has the units of current, let us now apply Kirchhoff's
current law "in reverse" to obtain the circuit of Fig. 5.95. Because h22 has the units of
admittance, which for the transistor model is conductance, it is represented by the resistor
symbol. Keep in mind, however, that the resistance in ohms of this resistor is equal to the
O>--------
reciprocal of conductance (l/h22 ).
The complete "ac" equivalent circuit for the basic three-terminal linear device is indi-
cated in Fig. 5.96 with a new set of subscripts for the h-parameters. The notation of Fig. FIC. 5.94
5.96 is of a more practical nature because it relates the h-parameters to the resulting ratio ob- Hybrid input equivalent circuit.
tained in the last few paragraphs. The choice of letters is obvious from the following listing:
h 11 - input resistance - h;
h 12 - reverse transfer voltage ratio - hr +

h21 - forward transfer current ratio - h1 v;,

h22 - output conductance - h 0

FIC. 5.95
__..
I;
Hybrid output equivalent circuit.

~ +
+
V; h,V0 '\,

O>---------~

FIC. 5.96
Complete hybrid equivalent circuit.

The circuit of Fig. 5 .96 is applicable to any linear three-terminal electronic device or system
with no internal independent sources. For the transistor, therefore, even though it has
three basic configurations, they are all three-terminal configurations, so that the resulting
equivalent circuit will have the same format as shown in Fig. 5.96. In each case, the bottom
of the input and output sections of the network of Fig. 5.96 can be connected as shown in
Fig. 5 .97 because the potential level is the same. Essentially, therefore, the transistor model
is a three-terminal two-port system. The h-parameters, however, will change with each
configuration. To distinguish which parameter has been used or which is available, a second

..,_
IC __..
lb h,, ..,_
IC

__..
lb
C ~ C

B hf, lb

+ +
h,, Ve, '\, ~ hoe

+ +
Vb, Vee
I,~
Vi,, Yee
E I,~
e

(a) (b)

FIC. 5.97
Common-emitter configuration: (a) graphical symbol; (b) hybrid equivalent circuit.
322 BJT AC ANALYSIS subscript has been added to the h-parameter notation. For the common-base configuration,
the lowercase letter b was added, whereas for the common-emitter and common-collector
configurations, the letters e and c were added, respectively. The hybrid equivalent network
for the common-emitter configuration appears with the standard notation in Fig. 5.97. Note
that Ii = h, /0 = le, and, through an application of Kirchhoff's current law, le = h + le.
The input voltage is now Vbe, with the output voltage Vee· For the common-base configura-
tion of Fig. 5.98, Ii = le, / 0 = le with Veb = Vi and Veb = V0 • The networks of Figs. 5.97
and 5.98 are applicable for pnp or npn transistors.

I, Jc
I, IC -.. h,b
-+-
-.. -+-
~ C

E C hfble

+ +
h rb vcb '\, ~ hob

V,b
+ +
½:b
- lb~
V,b ½:h
B lb~

(a) (b)

FIG. 5.98
Common-base configuration: (a) graphical symbol; (b) hybrid equivalent circuit.

The fact that both a Thevenin and a Norton circuit appear in the circuit of Fig. 5.96 was
further impetus for calling the resultant circuit a hybrid equivalent circuit. Two additional
transistor equivalent circuits, not to be discussed in this text, called the z-parameter and
y-parameter equivalent circuits, use either the voltage source or the current source, but not
both, in the same equivalent circuit. In Appendix A the magnitudes of the various param-
eters will be found from the transistor characteristics in the region of operation resulting in
the desired small-signal equivalent network for the transistor.
For the common-emitter and common-base configurations, the magnitude of hr and h0
is often such that the results obtained for the important parameters such as Zi, Z0 , Av, and
Ai are only slightly affected if hr and h 0 are not included in the model.
Because hr is normally a relatively small quantity, its removal is approximated by
hr ~ 0 and hrVo = 0, resulting in a short-circuit equivalent for the feedback element as
shown in Fig. 5.99. The resistance determined by 1/ h0 is often large enough to be ignored
in comparison to a parallel load, permitting its replacement by an open-circuit equivalent
for the CE and CB models, as shown in Fig. 5.99.
The resulting equivalent of Fig. 5.100 is quite similar to the general structure of the
common-base and common-emitter equivalent circuits obtained with the re model. In fact,

1,
I;
-.. Io

!
-.. -+-
lo
o-
-+-

~
I

~
0----

I I + +

J FIG. 5.99
~
!
hfli

I
Effect of removing h,, and h0 , from the hybird
v,,

0----
h,

FIG. 5.100
~
!
h1 1;

Approximate hybrid equivalent model.

Vo

0-

equivalent circuit.
the hybrid equivalent and the re models for each configuration are repeated in Fig. 5.101 THE HYBRID 323
for comparison. It should be reasonably clear from Fig. 5.101a that EQUIVALENT MODEL

I hie = f3re (5.139)

and hte = f3ac (5.140)

From Fig. 5.101b,

I hib = re I (5.141)

and I hfb = -a ~ -1 I (5.142)

In particular, note that the minus sign in Eq. (5.142) accounts for the fact that the current
source of the standard hybrid equivalent circuit is pointing down rather than in the actual
direction as shown in the re model of Fig. 5.101b.

h
~
,....le oc lb
~
,....le oc
b b
I I
~ ,.... ~
~
h;e hfelb f3re f3lb

e e e e

(a)

le le le le
~ ~ ~ ~
e oc e oc

I I
h;b
~ htbh ,....
~
re
t ale

b I ob b I oe

(b)

FICi. 5.101
Hybrid versus r, model: (a) common-emitter configuration; (b) common-base configuration.

EXAMPLES.19 Given/£= 2.5 mA, hte = 140, h 0 e = 20 µ,S (µ,mho), andh 0 b = 0.5 µ,S,
determine:
a. The common-emitter hybrid equivalent circuit.
b. The common-base re model.

Solution:
a. re= 26mV = 26mV = 10_40
le 2.5mA
hie = f3re = (140)(10.4 0) = 1.456 kO
1 1
r = - = - - = 50 kO
0
hoe 20 µ,S
324 BJT AC ANALYSIS Note Fig. 5.102.

bo------~
___._ ~-----..-----------oc
h
h;e 1.456k0

eo-------+-------------------------oe

FIG. 5.102
Common-emitter hybrid equivalent circuit for the parameters of Example 5.19.

b. re= 10.4 0
1 1
a ::c 1, r0 = - = - - = 2 M O
hob 0.5 µ,S
Note Fig. 5.103.

___._
eo-----~ r---------------oc
I,

r, 10.4 n I, T0 = 2 Mil

bo------+-------------------'--~b

FIG. 5.103
Common-base r, model for the parameters of Example 5.19.

A series of equations relating the parameters of each configuration for the hybrid
equivalent circuit is provided in Appendix B. In Section 5.23 it is demonstrated that the
hybrid parameter hfe (/3ac) is the least sensitive of the hybrid parameters to a change in col-
lector current. Assuming, therefore, that hfe = f3 is a constant for the range of interest, is
a fairly good approximation. It is hie = f3re that will vary significantly with le and should
be determined at operating levels because it can have a real effect on the gain levels of a
transistor amplifier.

5.20 APPROXIMATE HYBRID EQUIVALENT CIRCUIT

The analysis using the approximate hybrid equivalent circuit of Fig. 5.104 for the common-
emitter configuration and of Fig. 5.105 for the common-base configuration is very similar
to that just performed using the re model. A brief overview of some of the most important
•
configurations will be included in this section to demonstrate the similarities in approach
and the resulting equations.

C E C
~
le
h;,
~ hfe lb 1/ho, h;b
~ hfble 1/hob

£0-----------+----~-----o E
B B

FIG. 5.104 FIG. 5.105

Approximate common-emitter hybrid equivalent circuit. Approximate common-base hybrid equivalent circuit.
 Because the various parameters of the hybrid model are specified by a data sheet or APPROXIMATE HYBRID 325
experimental analysis, the de analysis associated with use of the re model is not an integral EQUIVALENT CIRCUIT
part of the use of the hybrid parameters. In other words, when the problem is presented, the
parameters such as hie, hfe, hib, and so on, are specified. Keep in mind, however, that the
hybrid parameters and components of the re model are related by the following equations, as
discussedearlierinthischapter:hie = f3re,hfe = /3,hoe = I/r0 ,hfb = -a,andhib = re.

Fixed-Bias Configuration
For the fixed-bias configuration of Fig. 5.106, the small-signal ac equivalent network will
appear as shown in Fig. 5.107 using the approximate common-emitter hybrid equivalent
model. Compare the similarities in appearance with Fig. 5.22 and the re model analysis.
The similarities suggest that the analyses will be quite similar, and the results of one can be
directly related to the other.

Vee

Re i 10

( 0
I; C2 + +
~
0 ~ ) l----+--------.11 h;e
zo
hfe
+ c, Vo
V; ~

FIG. 5.106 FIG. 5.107

Fixed-bias configuration. Substituting the approximate hybrid equivalent circuit into the ac
equivalent network of Fig. 5.106.

Z1 From Fig. 5.107,

(5.143)

Z0 From Fig. 5.107,

(5.144)

Av Using R' I/h 0 ellRc, we obtain

V0 = -I0 R 1 = -IcR'
= -hfehR'

and lb= -
vi
h;e
y.
with Vo= -hfe____!_R,
hie

Vo
so that A=- (5.145)
v vi
Ai Assuming that RB >> h;e and 1/ hoe 2: lORc, we find h ~ Ii and 10 = le =
h1eh = h1eh and so

(5.146)
326 BJT AC ANALYSIS
EXAMPLE 5.20 For the network of Fig. 5 .108, determine:
a. Z;.
b. Z 0 •
C. Av.
d. A;.

330kQ
(------o V
0

~
hfe = 120
V; __._ )-----II Z0
h;e= l.175kQ
I; h0 e= 20 µNV

FIG. 5.108
Example 5.20.

Solution:
a. Z; = Rsllh;e = 330 kO I 1.175 kO
~ h;e = 1.171 kfi

b. r0 = h:e = 20 µ,~/V = 50 kO

Z0 = -1 IIRc = 50kOll2.7kO = 2.56kfi ~ Re

hoe
h1eCRc 111 / hoe) (120)(2.7 kO II 50 kn)
C. A= ----= 1.1 7 l kO = - 262.34
v h;e
d. A; ~ hfe = 120

Voltage-Divider Configuration
For the voltage-divider bias configuration of Fig. 5.109, the resulting small-signal ac
equivalent network will have the same appearance as Fig. 5.107, with R8 replaced by
R' = R1IIR2,

I;
V; 0 ► )t---+---
c,

FIG. 5.109
Voltage-divider bias configuration.
Z1 FromFig.5.107withRB = R', APPROXIMATE HYBRID 327
EQUIVALENT CIRCUIT
(5.147)

Z0 From Fig. 5.107,

(5.148)

AV = (5.149)

h1e<R1 IIR2)
A-=---- (5.150)
1 RdR2 + hie

Unbypassed Emitter-Bias Configuration

For the CE unbypassed emitter-bias configuration of Fig. 5.110, the small-signal ac model
will be the same as Fig. 5.30, with f3re replaced by hie and f3Ib by hJeh• The analysis will
proceed in the same manner.

Vee

~1 0

Re
Rs
(--------o V
0

vI___..
~
I;

---+-
Z; RE
h;,
hf,
--Zo

FIC. 5.110
CE unbypassed emitter-bias configuration.

I zb ~ h1eRE I (5.151)

and I zi = RBllzb I (5.152)

(5.153)

and (5.154)
328 BJT AC ANALYSIS

(5.155)

or (5.156)

Emitter-Follower Configuration
For the emitter-follower of Fig. 5.38, the small-signal ac model will match that of Fig.
5.111, with f3re = h;e and /3 = hfe· The resulting equations will therefore be quite similar.

I;
V; 0 )--------1
---+-
Z;
(

FIG. 5.111
Emitter-follower configuration.

I zb ~ h1eRE I (5.157)

I Z; = RBllzb I (5.158)

Z0 For Z 0 , the output network defined by the resulting equations will appear as shown in
Fig. 5.112. Review the development of the equations in Section 5.8 and
II h;e
Zo = RE 1 + h
'fe
or, because 1 + hfe ~ hfe,

(5.159)

+ +

FIG. 5.112
Defining Z 0 for the emitter-follower configuration.
Av For the voltage gain, the voltage-divider rule can be applied to Fig. 5.112 as follows: APPROXIMATE HYBRID 329
EQUIVALENT CIRCUIT
RE(VD
V =-------
0 RE + hie/(1 + hte)
but, since 1 + hte ~ hfe,

(5.160)

(5.161)

or
8 E
l
(5.162)

Common-Base Configuration
The last configuration to be examined with the approximate hybrid equivalent circuit will be
the common-base amplifier of Fig. 5 .113. Substituting the approximate common-base hybrid
equivalent model results in the network of Fig. 5.114, which is very similar to Fig. 5.44.

+ +

FIC. 5.113
Common-base configuration.

+ ---+-
I; ~ +
---+-
Z;
V; RE h;b

FIC. 5.114
Substituting the approximate hybrid equivalent circuit into the ac equivalent network
of Fig. 5.113.

We have the following results from Fig. 5.114.

(5.163)

(5.164)
HO BJT AC ANALYSIS

y.
with I =-l and
e h;b

so that (5.165)

(5.166)

EXAMPLE 5.21 For the network of Fig. 5.115, determine:

a. Z;.
b. Z 0 •
C. Av.
d. A;.

I;

~
---+- ) - - - - - - - ~
+
hfb = -0.99 3.3 kQ
v; h;b = 14.3 0
-=-4v hob= 0.5µAN -;;;- lOV
+

FIG. 5.115
Example 5.21.

Solution:
a. Z; = REllh;b = 2.2killl14.3 il = 14.21 fi ~ h;b
1 1
b. r = - = - - - = 2 M f i
0
hob 0.5 µ,A/V
1
Z0 = -IIRe ~ Re = 3.3 kfi
hob
hfb Re (-0.99)(3.3 kil)
C. Av = -----,;;;- 14 _21 = 229.91
d. A; ~ hfb = - 1

The remaining configurations that were not analyzed in this section are left as an exercise
in the problem section of this chapter. It is assumed that the analysis above clearly reveals the
similarities in approach using the re or approximate hybrid equivalent models, thereby
removing any real difficulty with analyzing the remaining networks of the earlier sections.

5.21 COMPLETE HYBRID EQUIVALENT MODEL

•
The analysis of Section 5.20 was limited to the approximate hybrid equivalent circuit with
some discussion about the output impedance. In this section, we employ the complete
equivalent circuit to show the effect of hr and define in more specific terms the effect of h0 •
It is important to realize that because the hybrid equivalent model has the same appearance
for the common-base, common-emitter, and common-collector configurations, the equa-
tions developed in this section can be applied to each configuration. It is only necessary to
insert the parameters defined for each configuration. That is, for a common-base configu- COMPLETE HYBRID 3J1
ration, htb, hib, and so on, are employed, whereas for a common-emitter configuration, hte, EQUIVALENT MODEL
h;e, and so on, are used. Recall that Appendix A permits a conversion from one set to the
other if one set is provided and the other is required.
Consider the general configuration of Fig. 5.116 with the two-port parameters of particu-
lar interest. The complete hybrid equivalent model is then substituted in Fig. 5.117 using
parameters that do not specify the type of configuration. In other words, the solutions will
be in terms of hi, hr, ht, and h0 • Unlike the analysis of previous sections of this chapter,
here the current gain Ai will be determined first because the equations developed will prove
useful in the determination of the other parameters.

Io
~

---+-
I;
+ +
R, .._
zo
+ ---+- v; Transistor Vo RL
Z;

FIC. 5.116
Two-port system.

I;
---+-
.-------<+>----'V"h;"'""71 lb .-----+-i-I---o+- -71 /o
+
+ ---+- v; I/ho
Z;
v, '\,
-1:_
...- - - - - < 0 > - - - - - ~

FIC. 5.117
Substituting the complete hybrid equivalent circuit into the two-port system of Fig. 5.116.

Current Gain, A; = l 0 / I;
Applying Kirchhoff s current law to the output circuit yields
Vo
I0 - = htii + h0 V0
= hth +I= htii + - 1
1 h 0

Substituting V0 = -I0 RL gives

Io = htii - hoRdo
Rewriting the equation above, we have
Io + hoRdo = htii
and I0 (1 + h0 RL) = htf;

so that A-= Io= ht (5.167)

l ~ 1 + hoRL
Note that the current gain reduces to the familiar result of Ai = ht if the factor hoRL is suf-
ficiently small compared to 1.

Voltage Gain, Av = V0 /V;

Applying Kirchhoff s voltage law to the input circuit results in
Vi = I;h; + hrVo
3J2 BJT AC ANALYSIS Substituting Ii= (1 + hifiL)I0 /h1 from Eq. (5.167) and I0 = -V0 /RL from above
results in

Solving for the ratio V0 /Vi yields

(5.168)

In this case, the familiar form of Av = -h1Rdhi returns if the factor (hiho - hfhr)RL is
sufficiently small compared to hi.

Input Impedance, Z; V;/ I;

For the input circuit,
Vi = hJi + hrVo
Substituting Va= -IaRL
we have Vi = hJi - hrRda
Ia
Because A-=-
1 Ii

I0 = A/i
so that the equation above becomes
Vi = hJi - hrRLAJi
Solving for the ratio Vi/ h we obtain
vi
Zi = J. = hi - hrRLAi
z
and substituting

yields (5.169)

The familiar form of Zi = hi is obtained if the second factor in the denominator (hifiL) is
sufficiently smaller than one.

Output Impedance, Z0 = V0 / / 0
The output impedance of an amplifier is defined to be the ratio of the output voltage to the
output current with the signal Vs set to zero. For the input circuit with Vs = 0,

Ii=

Substituting this relationship into the equation from the output circuit yields
I0 = hJii + h0 V0
hfhrVo
- - - + hoVo
Rs+ hi

and z - -V0 - - - - -1- - - - (5.170)

0 - I0 - h0 - [hfhr/(hi + Rs)]

In this case, the output impedance is reduced to the familiar form Z 0 = 1/ h 0 for the transis-
tor when the second factor in the denominator is sufficiently smaller than the first.
 COMPLETE HYBRID 3J3
EXAMPLE 5.22 For the network of Fig. 5.118, determine the following parameters using EQUIVALENT MODEL
the complete hybrid equivalent model and compare to the results obtained using the
approximate model.
a. Z;andz;.
b. Av.
c. A;= I0 /f;.
d. Z~ (within Re) and Z0 (including Re).

8V

4.7 k!l
470 k!l

~
-- [.'
I

R+sFkn ~+ V; --
z.· I
Q

vs ' \ ,

FICi. 5.118
Example 5.22.

Solution: Now that the basic equations for each quantity have been derived, the order in
which they are calculated is arbitrary. However, the input impedance is often a useful quan-
tity to know, and therefore will be calculated first. The complete common-emitter hybrid
equivalent circuit has been substituted and the network redrawn as shown in Fig. 5.119. A
Thevenin equivalent circuit for the input section of Fig. 5.119 results in the input equivalent
of Fig. 5.120 because ETh ~ Vs and RTh ~ Rs = I kil (a result of RB = 470 kil being
much greater than Rs = I kil). In this example, RL = Re, and I0 is defined as the current
through Re as in previous examples of this chapter. The output impedance Z0 as defined
by Eq. (5.170) is for the output transistor terminals only. It does not include the effects
of Re. Z0 is simply the parallel combination of Z0 and RL- The resulting configuration of

[. [!
..,_
Io

-- --
~ ~
7flb ..,_ ..,_ +
+ 1.6 ill
Z; z:I Z'0 zo
Rs 1 ill +
+ V; 470 ill '\, 2x10- 4 vo
~ 110/b 50k.Q 4.7 k.Q Vo
vs '\,

-=-
Thevenin

FICi. 5.119
Substituting the complete hybrid equivalent circuit into the ac equivalent network of Fig. 5.118.
 + ---+-
z: l.6kQ +
R, 1 lli ' +
- 1- =50kQ
hoe
+ ½ '\, h0 e =20µS

½ '\, -1
-1-.- D > - - - - ~ -
FIC. 5.120
Replacing the input section of Fig. 5.119 with a Thevenin equivalent circuit.

Fig. 5 .120 is then an exact duplicate of the defining network of Fig. 5 .117, and the equa-
tions derived above can be applied.
a. Eq. (5.169):
Vi hfehreRL
Z-=-=h- - - - -
' Ii ,e 1 + hoeRL

= l.6k0 - (lt~(~2~~~~;~~~-:~~)
= 1.6 kO - 94.52 0
= 1.51 kfi
versus 1.6 kO using simply hie; and
z; = 470 kO llzi ~ zi = 1.51 kfi
b. Eq. (5.168):
V0 -hJeRL
A=-
v Vi hie + (hiehoe - hfehre)RL
-(110)(4.7 kO)
1.6 kO + [(1.6 k0)(20 JLS) - (110)(2 X 10-4)]4.7 kO
-517 X 103 0
1.6 kO + (0.032 - 0.022)4.7 kO
-517 X 103 0
l.6kO + 47 0
= -313.9
versus -323.125 using Av ~ -hfeRL/hie·
C. Eq. (5.167):
I0 hfe 110
A'.=-=----
' I/ 1 + h 0 eRL I + (20 JLS)(4.7 kO)
110
- - - = 100.55
1 + 0.094
versus 110 using simply hfe· Because 470 kO >> z;, I; ~ I/ and A; ~ 100.55 also.
d. Eq. (5.170):
Z' = Vo I
0
I0 hoe - [hfehre/(hie + Rs)]
1
20 JLS - [(110)(2 X 10-4)/(1.6 kO + 1 kO)]
1
20 JLS - 8.46 JLS
1
11.54 JLS
= 86.66kfi
3]4
 which is greater than the value determined from 1 / h0 e, 50 kO; and COMPLETE HYBRID 3J5
EQUIVALENT MODEL
Zo = Rcllz; = 4.7kOll86.66kO = 4.46kfi
versus 4. 7 kO using only Re.

Note from the results above that the approximate solutions for Av and Z; were very close
to those calculated with the complete equivalent model. In fact, even A; was off by less than
10%. The higher value of Z~ only contributed to our earlier conclusion that Z~ is often so
high that it can be ignored compared to the applied load. However, keep in mind that when
there is a need to determine the effect of hre and h 0 e, the complete hybrid equivalent model
must be used, as described earlier.
The specification sheet for a particular transistor typically provides the common-emitter
parameters as noted in Fig. 5.92. The next example will employ the same transistor pa-
rameters appearing in Fig. 5.118 in a pnp common-base configuration to introduce the
parameter conversion procedure and emphasize the fact that the hybrid equivalent model
maintains the same layout.

EXAMPLE 5.23 For the common-base amplifier of Fig. 5.121, determine the following
parameters using the complete hybrid equivalent model and compare the results to those
obtained using the approximate model.
a. Z;
b. A;
C. Av.
d. Zo

h;e= 1.6 kQ hfe = 110

hre=2X 10-4 h0 e=20µS

·J~
0
+ +

vs '\,
V;

-=-6v
___._
Z-'I
- - Z'0
-=-12V
+
zo
Vo

-=-
FIC. 5.121
Example 5.23.

Solution: The common-base hybrid parameters are derived from the common-emitter
parameters using the approximate equations of Appendix B:
h;e l.6kO
h·b ::c - - = - - - = 14.41 0
1
1 + hte I + 110
Note how closely the magnitude compares with the value determined from
h;e l.6kO
h;b =re=~= 11() = 14.55 0

h;ehoe (1.6 k0)(20 JLS) _4

Also, hrb ::c - - - - hre = ------ - 2 X 10
1 + hte I+ 110
= 0.883 X 10- 4
-hfe -110
htb ::c --- = - 0.991
1 + hte I + 110
hoe 20 jLS
hob ::c _l_+_hfi-e I + 110 = 0.18 µS
 C

---+- ---+- + ---+- 14.41 Q ,.,_

Z; Z' le Z'0
1 ill ' +
-0.991/e
+ 3kQ
hfb le
vs '\J
b
Thevenin b

FIC. 5.122
Small-signal equivalent for the network of Fig. 5.121.

Substituting the common-base hybrid equivalent circuit into the network of Fig. 5.121
results in the small-signal equivalent network of Fig. 5.122. The Thevenin network for the
input circuit results in RTh = 3 kll 111 kll = 0. 75 kll for Rs in the equation for Z 0 •
a. Eq. (5.169):
V; htbhrbRL
z; = I[ = h;b - 1 + h 0 bRL
(-1.991)(0.883 X 10- )(2.2 kll) 4
= 14.41 n-- ---------
1 + (0.18 µ,S)(2.2 kll)
= 14.41 n + 0.19 n
= 14.600
versus 14.41 n using Z; == h;b; and
Z; = 3 kD llz; == z; = 14.60 n
b. Eq. (5.167):
Ia hfb
Ai=-=----
' I/ 1 + h 0 bRL
-0.991
1 + (0.18µ,S)(2.2kll)
= -0.991
Because 3 kll >> z;, I; == 11 and A; = 10 / I; == -1.
c. Eq. (5.168):

14.41 0 + [(14.410)(0.18 µ,S) - (-0.991)(0.883 X 10-4)]2.2 kll

= 149.25
versus 151.3 using Av == -htbRdhib·
d. Eq. (5.170):
1
Z' =---------
0 hob - [htbhrb/(h;b + Rs)]
1
0.18 µ,S - [(-0.991)(0.883 X 10-4)/(14.41 D, + 0.75 kll)]
1
0.295 µ,S
= 3.39M!l
versus 5.56 MO using Z~ == I/hob· For Z 0 as defined by Fig. 5.122,
Z 0 = Rcllz~ = 2.2 kll II 3.39 MO = 2.199 k!l
versus 2.2 kll using Z0 == Re.

3]6
5.22 HYBRID TT MODEL
•
The last transistor model to be introduced is the hybrid 7T model of Fig. 5.123 which
includes parameters that do not appear in the other two models primarily to provide a more
accurate model for high-frequency effects.
HYBRID 7T MODEL 3J7

Tu

B b' C
---+-
lb
rb
+ t lb
Cu

v,, r,,
c"

E E

FICi. 5.123
Giacoletto (or hybrid 7T) highjrequency transistor small-signal ac equivalent circuit.

The resistors r"" r0 , rb, and ru are the resistances between the indicated terminals of the
device when the device is in the active region. The resistance r71' (using the symbol TT to agree
with the hybrid 7T terminology) is simply f3re as introduced for the common-emitter re model.
That is,

(5.171)

The output resistance r0 is the output resistance normally appearing across an applied
load. Its value, which typically lies between 5 kO and 40 kO, is determined from the hybrid
parameter h0 e, the Early voltage, or the output characteristics.
The resistance rb includes the base contact, base bulk, and base spreading resistance levels.
The first is due to the actual connection to the base. The second includes the resistance from
the external terminal to the active region of the transistor, and the last is the actual resistance
within the active base region. It is typically a few ohms to tens of ohms.
The resistance ru (the subscript u refers to the union it provides between collector and
base terminals) is a very large resistance and provides a feedback path from output to
input circuits in the equivalent model. It is typically larger than {3r0 , which places it in the
megohm range.

All the capacitors that appear in Fig. 5.123 are stray parasitic capacitors between the vari-
ous junctions of the device. They are all capacitive effects that really only come into play
at high frequencies. For low to mid-frequencies their reactance is very large, and they can
be considered open circuits. The capacitor C71' across the input terminals can range from a
few pF to tens of pF. The capacitor Cu from base to collector is usually limited to a few pF
but is magnified at the input and output by an effect called the Miller effect, to be intro-
duced in Chapter 9.

{JI',, or gm Vrr
It is important to note in Fig. 5.123 that the controlled source can be a voltage-controlled
current source (VCCS) or a current-controlled current source (CCCS), depending on the
parameters employed.
Note the following parameter equivalence in Fig. 5.123:

(5.172)
3J8 BJT AC ANALYSIS
and (5.173)

r,,,.
with ---= (5.174)
r,,,. + ru

Take particular note of the fact that the equivalent sources f3Ib and gm V,,,. are both con-
trolled current sources. One is controlled by a current at another place in the network and
the other by a voltage at the input side of the network. The equivalence between the two
is defined by

For the broad range of low- to mid-frequency analysis, the effect of the stray capaci-
tive effects can be ignored due to the very high reactance levels associated with each. The
resistance rb is usually small enough with other series elements to be ignored while the
resistance ru is usually large enough compared to parallel elements to be ignored. The result
is an equivalent network similar to the re model introduced and applied in this chapter.
In Chapter 9, when high-frequency effects are considered, the hybrid 7T model will be
the model of choice.

5.2:S VARIATIONS OF TRANSISTOR PARAMETERS

•
A variety of curves can be drawn to show the variations of the transistor parameters with tem-
perature, frequency, voltage, and current. The most interesting and useful at this stage of the
development include the variations with junction temperature and collector voltage and current.
The effect of the collector current on the re model and hybrid equivalent model is shown
in Fig. 5.124. Take careful note of the logarithmic scale on the vertical and horizontal axes.
Logarithmic scales will be examined in detail in Chapter 9. The parameters have all been
normalized (a process described in detail in Section 9.5) to unity so that the relative change
in magnitude with collector current can easily be determined. On each set of curves, such
as in Figs. 5.124 to 5.126, the operating point at which the parameters were determined
is always indicated. For this particular situation, the quiescent point is at the fairly typical
values of VCE = 5.0 V and le = 1.0 mA. Because the frequency and temperature of operation

Relative magnitude of parameters

50

le= I mA
VC£=5V
T= 25°C
f = I kHz 0.5
..
0.2 .
• ,• I
0.1 hoe (rc,)

0.05

0.02
0.01
0.1 0.2 0.5 2 5 20 50 I c (mA)

FIG. 5.124
Hybrid parameter variations with collector current.
also affect the parameters, these quantities are also indicated on the curves. Figure 5.124 VARIATIONS OF 3J9
shows the variation of the parameters with collector current. Note that at le = 1 mA the TRANSISTOR
PARAMETERS
value of all the parameters has been normalized to 1 on the vertical axis. The result is that
the magnitude of each parameter is compared to the values at the defined operating point.
Because manufacturers typically use the hybrid parameters for plots of this type, they are
the curves of choice in Fig. 5.124. However, to broaden the use of the curves the re and
hybrid 7T equivalent parameters have also been added.
At first glance it is particularly interesting to note that:
The parameter hJe(/3) varies the least of all the parameters of a transistor equivalent
circuit when plotted against variations in collector current.
Figure 5.124 clearly reveals that for the full range of collector current the parameter hieC/3)
varies from 0.5 of its Q-point value to a peak of about 1.5 times that value at a current of
about 6 mA. For a transistor with a /3 of 100, it therefore varies from about 50 to 150. This
seems like quite a bit, but look at h0 e, which jumps to almost 40 times its Q-point value at
a collector current of 50 mA.
Figure 5.124 also shows that h 0 e(l/r0 ) and hie(f3re) vary the most for the chosen current
range. The parameter hie varies from about 10 times its Q-point value down to about one
tenth the Q point value at 50 mA. This variation, however, should be expected because we
know that the value of re is directly related to the emitter current by re = 26 mV/IE. As
IE(=:E./c) increases, the value of re and therefore f3re will decrease, as shown in Fig. 5.124.
Keep in mind as you review the curve of hoe versus current that the actual output resis-
tance rO is I/hoe· Therefore, as the curve increases with current, the value of rO becomes
less and less. Because r0 is a parameter that normally appears in parallel with the applied
load, decreasing values of r0 can become a critical problem. The fact that r0 has dropped to
almost 1/40 of its value at the Q-point could spell a real reduction in gain at 50 mA.
The parameter hre varies quite a bit, but because its Q-point value is usually small enough
to permit ignoring its effect, it is a parameter that is only of concern for collector currents
that are much less, or quite a bit more, than the Q-point level.
This may seem like an extensive description of a set of characteristic curves. However,
experience has revealed that graphs of this nature are too often reviewed without taking the
time to fully appreciate the broad impact of what they are providing. These plots reveal a
lot of information that could be extremely useful in the design process.
Figure 5.125 shows the variation in magnitude of the parameters due to changes in
collector-to-emitter voltage. This set of curves is normalized at the same operating point
as the curves of Fig. 5.124 to permit comparisons between the two. In this case, however,
the vertical scale is in percent rather than whole numbers. The 200% level defines a set of
parameters twice that at the 100% level. A level of 1000% would reflect a 10: 1 change.
Note that hte and hie are relatively steady in magnitude with variations in collector-to-
emitter voltage, whereas for changes in collector current the variation is a great deal more

(o/c of VCE = 5 V value of each quantity)

3000
2000

1000
700
500

300
IE= I mA 200
VcE =5 V
T= 25°C 100
f= I kHz 70
50
30
0.2 0.5 2 5 10 20 50 100

FICi. 5.125
Hybrid parameter variations with collector-emitter potential.
340 BJT AC ANALYSIS significant. In other words, if you want a parameter such as hie(/3re) to remain fairly steady,
keep the variation of le to a minimum while worrying less about variations in the collector-
to-emitter voltage. The variation of h 0 e and hie remains significant for the indicated range
of collector-to-emitter voltage.
In Fig. 5.126, the variation in parameters is plotted for changes injunction temperature.
The normalization value is taken to be room temperature, T = 25°C. The horizontal scale
is now a linear scale rather than the logarithmic scale employed in the two previous figures.
In general:
All the parameters of a hybrid transistor equivalent circuit increase with temperature.

Relative magnitude of parameters

.. ..
3.0 (Freezing H 2O) (Boiling H 2 O)
- h;, (~r,)

.. ... ..
2.0
.. .. ...
•"".,. ~
,,, ~
~ hr,(;;;)

le= I mA 1.5 . ....... hf,(~)

VcE =5 V .................... ho, c+a)

T= 25° C -1.0
I ••••
f= I kHz ho, (ro) .
0.7
hr,(;;;)
0.5 h1,(r,)/
0.4 #

0.3 L__h-=;,_(~-=
~'_.)_ __ _ 1 . _ + - . . . J __ _..___ _..___ _..___ __ . .
-100 -50 0 25° 50 100 150 200 T ( C) 0

I
Room temperature

FIG. 5.126
Hybrid parameter variations with temperature.

However, again keep in mind that the actual output resistance r0 is inversely related
to hoe, so its value drops with an increase in hoe· The greatest change is in hie, although
note that the range of the vertical scale is considerably less than in the other plots. At a
temperature of 200°C the value of hie is almost 3 times its Q-point value, but in Fig. 5.124
parameters jumped to almost 40 times the Q-point value.
Of the three parameters, therefore, the variation in collector current has by far the great-
est effect on the parameters of a transistor equivalent circuit. Temperature is always a factor,
but the effect of the collector current can be significant.

5.24 TROUBLESHOOTING
Although the terminology troubleshooting suggests that the procedures to be described are
designed simply to isolate a malfunction, it is important to realize that the same techniques
can be applied to ensure that a system is operating properly. In any case, the testing, check-
•
ing, and isolating procedures require an understanding of what to expect at various points
in the network in both the de and ac domains. In most cases, a network operating correctly
in the de mode will also behave properly in the ac domain.
In general, therefore, if a system is not working properly, first disconnect the ac source
and check the de biasing levels.
In Fig. 5.127 we have four transistor configurations with specific voltage levels provided
as measured by a DMM in the de mode. The first test of any transistor network is to simply
measure the base-to-emitter voltage of the transistor. The fact that it is only 0.3 V in this
case suggests that the transistor is not "on" and perhaps sitting in its saturation mode. If this
is a switching design then the result is expected, but if in the amplifier mode there is an open
connection preventing the base voltage from reaching an operating level.
 Yee 20V 18V 12V

RB Re R, RB Re

20V
+
3V
+
0.3V
R2
RE

(a) -=- -=- (b) -=- (c) -=- -=- (d)

FIC. 5.127
Checking the de levels to determine if a network is properly biased.

In Fig. 5.127b the fact that the voltage at the collector equals the supply voltage reveals
that there is no drop across the resistor Re and the collector current is zero. The resistor Re
is connected properly because it made the connection from the de source to the collector.
However, any one of the other elements may not have been connected properly, resulting
in the absence of a base or collector current. In Fig. 5.127c the voltage drop across the
collector-to-emitter voltage is too small compared with the applied de voltage. Normally
the voltage VeE is in the mid-range of perhaps 6 V to 14 V. A reading of 18 V would cause
the same concern as the reading of 3 V. The fact that the voltage levels exist at all suggests
that all the elements are connected but the value of one or more of the resistive elements
may be wrong. In Fig. 5.127d we find that the voltage at the base is exactly half the supply
voltage. We know from this chapter that the resistance RE will reflect back to the base by
a factor of beta and appear in parallel with R2 . The result would be a base voltage less than
half the supply voltage. The measurement suggests that the base lead is not connected to
the voltage divider, causing an even split of the 20-V source.
In a typical laboratory setting, the ac response at various points in the network is checked
with an oscilloscope as shown in Fig. 5.128. Note that the black (gnd) lead of the oscillo-
scope is connected directly to ground and the red lead is moved from point to point in the

v 0 (V)

e2 Vo
------(1-------<0

'
0

EB
Oscilloscope

•••
.:.• •
(AC-GND-DC switch on AC)
Ground strap
-=-
FIC. 5.128
Using the oscilloscope to measure and display various voltages of a BJT amplifier.
341
342 BJT AC ANALYSIS network, providing the patterns appearing in Fig. 5.128. The vertical channels are set in
the ac mode to remove any de component associated with the voltage at a particular point.
The small ac signal applied to the base is amplified to the level appearing from collector to
ground. Note the difference in vertical scales for the two voltages. There is no ac response
at the emitter terminal due to the short-circuit characteristics of the capacitor at the applied
frequency. The fact that Va is measured in volts and v; in millivolts suggests a sizable gain
for the amplifier. In general, the network appears to be operating properly. If desired, the
de mode of the multimeter could be used to check VBE and the levels of V8 , VCE, and VE to
review whether they lie in the expected range. Of course, the oscilloscope can also be used
to compare de levels simply by switching to the de mode for each channel.
A poor ac response can be due to a variety of reasons. In fact, there may be more than
one problem area in the same system. Fortunately, however, with time and experience, the
probability of malfunctions in some areas can be predicted, and an experienced person can
isolate problem areas fairly quickly.
In general, there is nothing mysterious about the general troubleshooting process. If you
decide to follow the ac response, it is good procedure to start with the applied signal and
progress through the system toward the load, checking critical points along the way. An
unexpected response at some point suggests that the network is fine up to that area, thereby
defining the region that must be investigated further. The waveform obtained on the oscil-
loscope will certainly help in defining the possible problems with the system.
If the response for the network of Fig. 5.128 is as appears in Fig. 5.129, the network has
a malfunction that is probably in the emitter area. An ac response across the emitter is unex-
pected, and the gain of the system as revealed by Va is much lower. Recall for this configuration
that the gain is much greater if RE is bypassed. The response obtained suggests that RE is not
bypassed by the capacitor, and the terminal connections of the capacitor and the capacitor itself
should be checked. In this case, a checking of the de levels will probably not isolate the problem
area because the capacitor has an "open-circuit" equivalent for de. In general, prior knowledge
of what to expect, familiarity with the instrumentation, and, most important, experience are all
factors that contribute to the development of an effective approach to the art of troubleshooting.

Vee

/
~
Rs

+ 01 "'-/ '\J t

vs '\,

...
FIG. 5.129
The waveforms resulting from a malfunction in the emitter area.

5.25 PRAOICAL APPLICATIONS

Audio Mixer •
When two or more signals are to be combined into a single audio output, mixers such as
shown in Fig. 5. 130 are employed. The potentiometers at the input are the volume controls
for each channel, with potentiometer R 3 included to provide additional balance between
 12 V PRACTICAL 343
APPLICATIONS

R1
R6 470 k!l 3.3 k!l
C3
(---ov 0
68 µ.F

R3 l:l = 120
1 Mil 56 µ.F
~
Z;

R21-----~""-~
Vz 7 _ 4 7 0 k ! l

33 k!l
Rs
1.2 k!l
"II"

r,= 11.71 !l
Z; =l:lr, = 1.4 k!l
FICi. 5.130
Audio mixer.

the two signals. Resistors R4 and Rs are there to ensure that one channel does not load
down the other, that is, to ensure that one signal does not appear as a load to the other,
draw power, and affect the desired balance on the mixed signal.
The effect of resistors R4 and Rs is an important one that should be discussed in some
detail. A de analysis of the transistor configuration results in re = 11.71 !1, which will
establish an input impedance to the transistor of about 1.4 kll. The parallel combination of
R6 llzi is also approximately 1.4 kll. Setting both volume controls to their maximum value
and the balance control R3 to its midpoint result in the equivalent network of Fig. 5.131a.
The signal at v1 is assumed to be a low-impedance microphone with an internal resistance
of 1 kll. The signal at v2 is assumed to be a guitar amplifier with a higher internal imped-
ance of 10 kll. Because the 470-kll and 500-kll resistors are in parallel for the above
conditions, they can be combined and replaced with a single resistor of about 242 kll. Each
source will then have an equivalent such as shown in Fig. 5.131b for the microphone. Ap-
plying Thevenin's theorem shows that it is an excellent approximation to simply drop the
242 kll and assume that the equivalent network is as shown for each channel. The result
is the equivalent network of Fig. 5.131c for the input section of the mixer. Applying the
superposition theorem results in the following equation for the ac voltage at the base of the
transistor:
(1.4 kO II 43 k!1)v 81 (1.4 kO II 34 k!1)v82
V =----------+----------
b 34 kll + (1.4 kll II 43 kn) 43 kll + (1.4 kll II 34 kn)
= 38 X 10- v81 + 30 X 10- V82
3 3

With re= 11.71 !1, thegainoftheamplifieris-Rc/re = 3.3k!1/11.71 !1 = -281.8,

and the output voltage is
V 0 = -10.7V81 - 8.45v 82

which provides a pretty good balance between the two signals, even though they have a
10: 1 ratio in internal impedance. In general, the system will respond quite well. However,
ifwe now remove the 33-kll resistors from the diagram of Fig. 5.131c, the equivalent net-
work of Fig. 5 .132 results, and the following equation for vb is obtained using the superpo-
sition theorem:
(1.4 kO II 10 k!1)v81 (1.4k0111 k!1)V 82
Vb = --------- + ---------
1 kll + 1.4 kO II 10 kll 10 kll + (1.4k0111 kll)
= 0.55v81 + 0.055v82
Using the same gain as before, we obtain the output voltage as
v0 = 155v81 + 15.5v82 ~ 155v81
which indicates that the microphone will be quite loud and clear and the guitar input essen-
tially lost.
 33k!1
Microphone lkil
470 kO 500k!1
+
VS!
'\J
-*
Amplifier

-=- -=-
Z; 1.4 ill

33 kO
Guitar lOkO
470 kil 500k!1
242 kO
+
Vs2 '\J
-i -=- -=- -=-
-1:_--
(a) (b)

Amplifier

lOk!l Z; 1.4 kO

-=-
(c)

FICi. 5.131
(a) Equivalent network with R3 set at the midpoint and the volume controls on their maximum settings;
(b) finding the Thevenin equivalent for channel 1; (c) substituting the Thevenin equivalent networks into Fig. 5.131 a.

Amplifier

Z; 1.4 kO

-=-
FICi. 5.132
Redrawing the network of Fig. 5.131 c with the 33-k[!
resistors removed.

The importance of the 33-kll resistors is therefore defined. It makes each applied signal
appear to have a similar impedance level so that there is good balance at the output. One
might suggest that the larger resistor improves the balance. However, even though the bal-
ance at the base of the transistor may be better, the strength of the signal at the base of the
transistor will be less, and the output level reduced accordingly. In other words, the choice
of resistors R4 and R5 is a give-and-take situation between the input level at the base of the
transistor and the balance of the output signal.
To demonstrate that the capacitors are truly short-circuit equivalents in the audio range,
substitute a very low audio frequency of 100 Hz into the reactance equation of a 56-JLF
capacitor:
1 1
Xe = - - = - - - - - - = 28.42 D,
27TfC 27r(l00 Hz)(56 JLF)
344
A level of 28.42 D compared to any of the neighboring impedances is certainly small PRACTICAL 345
enough to be ignored. Higher frequencies will have even less effect. APPLICATIONS
A similar mixer will be discussed in connection with the junction field effect transistor
(JFET) in the following chapter. The major difference will be the fact that the input imped-
ance of the JFET can be approximated by an open circuit rather than the rather low-level
input impedance of the BJT configuration. The result will be a higher signal level at the
input to the JFET amplifier. However, the gain of the FET is much less than that of the BJT
transistor, resulting in output levels that are actually quite similar.

Preamplifier
The primary function of a preamplifier is as its name implies: an amplifier used to pick up
the signal from its primary source and then operate on it in preparation for its passage
into the amplifier section. Typically, a preamplifier will amplify the signal, control its vol-
ume, perhaps change its input impedance characteristics, and if necessary determine its route
through the stages to follow-in total, a stage of any system with a multitude of functions.
A preamplifier such as shown in Fig. 5.133 is often used with dynamic microphones
to bring the signal level up to levels that are suitable for further amplification or power
amplifiers. Typically, dynamic microphones are low-impedance microphones because
their internal resistance is determined primarily by the winding of the voice coil. The basic
construction consists of a voice coil attached to a small diaphragm that is free to move
within a permanent magnet. When one speaks into the microphone, the diaphragm moves
accordingly and causes the voice coil to move in the same manner within the magnetic
field. In accord with Faraday's law, a voltage will be induced across the coil that will carry
the audio signal.

12 V

3.3 k!l

47 k!l
.-----.--"IV"""°----1-----lt-(- - - - < 0 V 0

J lOµF
20µF

82k!l

[3 = 140
Av= -319.7
Dynamic
microphone
(Rint = 50 !l)
r Z; =1.33 k!l

FIC. 5.133
Preamplifier for a dynamic microphone.

Because it is a low-impedance microphone, the input impedance of the transistor ampli-

fier does not have to be that high to pick up most of the signal. Because the internal imped-
ance of a dynamic microphone may be as low as 20 D to 100 D, most of the signal would
be picked up with an amplifier having an input impedance as low as 1 to 2 kD. This, in
fact, is the case for the preamplifier of Fig. 5.133. For de biasing conditions, the collector
de feedback configuration was chosen because of its high stablity characteristics.
In the ac domain, the 10-µ,F capacitor will assume a short-circuit state (on an approxi-
mate basis), placing the 82-kD resistor across the input impedance of the transistor and the
47 kD across the output of the transistor. A de analysis of the transistor configuration results
in re = 9.64 D, giving an ac gain determined by
(47 kD II 3.3 kD)
Av = 9.64 fl = -319.7
which is excellent for this application. Of course, the gain will drop when this pickup stage
of the design is connected to the input of the amplifier section. That is, the input resistance
346 BJT AC ANALYSIS of the next stage will appear in parallel with the 47-kO and 3.3-kO resistors and will drop
the gain below the unloaded level of 319. 7.
The input impedance of the preamplifier is determined by
zi = 82kOII.Bre = 82kOll(140)(9.64 0) = 82kOIIL34kO = 1.33k!l
which is also fine for most low-impedance dynamic microphones. In fact, for a micro-
phone with an internal impedance of 50 0, the signal at the base would be over 98% of that
available. This discussion is important because if the impedance of the microphone is a
great deal more, say, 1 kO, the preamplifier would have to be designed differently to
ensure that the input impedance was at least 10 kO or more.

Random-Noise Generator
There is often a need for a random-noise generator to test the response of a speaker, micro-
phone, filter, and, in fact, any system designed to work over a wide range of frequencies.
A random-noise generator is just as its name implies: a generator that generates sig-
nals of random amplitude and frequency. The fact that these signals are usually totally
unintelligible and unpredictable is the reason that they are simply referred to as noise.
Thermal noise is noise generated due to thermal effects resulting from the interaction
between free electrons and the vibrating ions of a material in conduction. The result is an
uneven flow of electrons through the medium, which will result in a varying potential
across the medium. In most cases, these randomly generated signals are in the microvolt
range, but with sufficient amplification they can wreak havoc on a system's response. This
thermal noise is also called Johnson noise (named after the original researcher in the area)
or white noise (because in optics, white light contains all frequencies). This type of noise
has a fairly flat frequency response such as shown in Fig. 5. l 34a, that is, a plot of its power
versus frequency from the very low to the very high end is fairly uniform. A second type
of noise is called shot noise, a name derived from the fact that its noise sounds like a
shower of lead shot hitting a solid surface or like heavy rain on a window. Its source is
pockets of carriers passing through a medium at uneven rates. A third is pink, flicker, or
1/fnoise, which is due to the variation in transit times for carriers crossing various junc-
tions of semiconductor devices. It is called 1/ f noise because its magnitude drops off with
increase in frequency. Its effect is usually the most dramatic for frequencies below 1
kHz, as shown in Fig. 5.134b.

S0µV

20µV
20 µV White (Johnson) noise
0 '--------v----' ---------------
Pink or 1 kHz Shot and thermal
0 5 Hz S00kHz 1/Jnoise (Johnson) noise

(a) (b)

FIG. 5.134
Typical noise frequency spectra: (a) white or Johnson; (b) pink, thermal, and shot.

The network of Fig. 5.135 is designed to generate both a white noise and a pink noise.
Rather than a separate source for each, first white noise is developed (level across the entire
frequency spectrum), and then a filter is applied to remove the mid- and high-frequency
components, leaving only the low-frequency noise response. The filter is further designed
to modify the flat response of the white noise in the low-frequency region (to create a 1/f
drop-off) by having sections of the filter "drop in" as the frequency increases. The white
noise is created by leaving the collector terminal of transistor Q 1 open and reverse-biasing
the base-to-emitter junction. In essence, the transistor is being used as a diode biased in
the Zener avalanche region. Biasing a transistor in this region creates a very unstable situ-
ation that is conducive to the generation of random white noise. The combination of the
avalanche region with its rapidly changing charge levels, sensitivity of the current level to
 15-30V

R2 5.6 ldl
5.6 ldl
Cz
L____,, White
~ Noise
1 µF C7
Pink
--------------+-----+------1(---------o Noise
1 µF

C3 25 µF R4

25 µF C4
R3
Qz
----.-
39kfl
Z;

FICi. 5.135
White- and pink-noise generator.

temperature, and quickly changing impedance levels contributes to the level of noise volt-
age and current generated by the transistor. Germanium transistors are often used because
the avalanche region is less defined and less stable than in silicon transistors. In addition,
there are diodes and transistors designed specifically for random-noise generation.
The source of the noise is not some specially designed generator. It is simply due to the
fact that current flow is not an ideal phenomenon but actually varies with time at a level that
generates unwanted variations in the terminal voltage across elements. In fact, that variation
in flow is so broad that it can generate frequencies that extend across a wide spectrum-a
very interesting phenomenon.
The generated noise current of Q1 will then be the base current for Q2, which will be
amplified to generate a white noise of perhaps 100 mV, which for this design would suggest
an input noise voltage of about 170 µ, V. Capacitor C1 will have a low impedance throughout
the frequency range of interest to provide a "shorting effect" on any spurious signals in the
air from contributing to the signal at the base of Q 1. The capacitor C2 is there to isolate the de
biasing of the white-noise generator from the de levels of the filter network to follow. The
39 kll and the input impedance of the next stage create the simple voltage-divider network
of Fig. 5.136. If the 39 kll were not present, the parallel combination of R2 and Z; would
load down the first stage and reduce the gain of Q 1 considerably. In the gain equation, R2
and Z; would appear in parallel (discussed in Chapter 9).

' R3
~ - - c '.........~"---~---
+ 25 µF 39kfl +

FICi. 5.136
Input circuit for the second stage.

The filter network is actually part of the feedback loop from collector to base appear-
ing in the collector feedback network of Section 5.10. To describe its behavior, let us first
consider the extremes of the frequency spectrum. For very low frequencies all the capaci-
tors can be approximated by an open circuit, and the only resistance from collector to base
is the 1-MO resistor. Using a beta of 100, we find that the gain of the section is about 280
and the input impedance about 1.28 kll. At a sufficiently high frequency all the capacitors
347
348 BJT AC ANALYSIS could be replaced by short circuits, and the total resistance combination between collector
and base would be reduced to about 14.5 kil, which would result in a very high unloaded
gain of about 731, more than twice that just obtained with RF = l Mil. Because the 1/f
filter is supposed to reduce the gain at high frequencies, it initially appears as though there
is an error in design. However, the input impedance has dropped to about 19.33 il, which
is a 66-fold drop from the level obtained with RF = l Mil. This would have a significant
impact on the input voltage appearing at the second stage when we consider the voltage-
divider action of Fig. 5.136. In fact, when compared to the series 39-kil resistor, the signal
at the second stage can be assumed to be negligible or at a level where even a gain in excess
of 700 cannot raise it to a level of any consequence. In total, therefore, the effect of dou-
bling the gain is totally lost due to the tremendous drop in Z;, and the output at very high
frequencies can be ignored entirely.
For the range of frequencies between the very low and the very high, the three capacitors
of the filter will cause the gain to drop off with increase in frequency. First, capacitor C4
will be dropped in and cause a reduction in gain (around 100 Hz). Then capacitor C5 will be
included and will place the three branches in parallel (around 500 Hz). Finally, capacitor C6
will result in four parallel branches and the minimum feedback resistance (around 6 kHz).
The result is a network with an excellent random-noise signal for the full frequency
spectrum (white) and the low-frequency spectrum (pink).

Sound-Modulated Light Sourc:e

The light from the 12-V bulb of Fig. 5 .13 7 will vary at a frequency and an intensity that are
sensitive to the applied signal. The applied signal may be the output of an acoustical ampli-
fier, a musical instrument, or even a microphone. Of particular interest is the fact that the
applied voltage is 12 V ac rather than the typical de biasing supply. The immediate ques-
tion, in the absence of a de supply, is how the de biasing levels for the transistor will be
established. In actuality, the de level is obtained through the use of diode Di, which recti-
fies the ac signal, and capacitor C2, which acts as a power supply filter to generate a de
level across the output branch of the transistor. The peak value of a 12-V rms supply is
about 17 V, resulting in a de level after the capacitive filtering in the neighborhood of 16 V.
If the potentiometer is set so that R 1 is about 320 il, the voltage from base to emitter of the
transistor will be about 0.5 V, and the transistor will be in the "off' state. In this state the
collector and emitter currents are essentially O mA, and the voltage across resistor R 3 is
approximately O V. The voltage at the junction of the collector terminal and the diode is
therefore O V, resulting in D 2 being in the "off' state and O Vat the gate terminal of the
silicon-controlled rectifier (SCR). The SCR (see Section 17 .3) is fundamentally a diode
whose state is controlled by an applied voltage at the gate terminal. The absence of a volt-
age at the gate means that the SCR and bulb are off.

=
16 V de '- - - - - - - - - - - - - - - '
r------9--__JL--..-------li.---:1----.----0 12 V ac
60Hz

l0kO-'◄}~R-1-+------11 D
} R2 SCR
10 µ.F

+o----+--~
I
Amplifier R3 1 kO ,c2 12-V bulb
output I

I_ - - - - - ~
"II"
ac~dc conversion

FIG. 5.137
Sound-modulated light source. SCR, Silicon-controlled rectifier.

If a signal is now applied to the gate terminal, the combination of the established bias-
ing level and the applied signal can establish the required 0.7-V tum-on voltage, and the
transistor will be turned on for periods of time dependent on the applied signal. When the
transistor turns on, it will establish a collector current through resistor R 3 that will establish a SUMMARY 349
voltage from collector to ground. If the voltage is more than the required 0. 7 V for diode D2,
a voltage will appear at the gate of the SCR that may be sufficient to tum it on and establish
conduction from the drain to the source of the SCR. However, we must now examine one of
the most interesting aspects of this design. Because the applied voltage across the SCR is ac,
which will vary in magnitude with time as shown in Fig. 5.138, the conduction strength of
the SCR will vary with time also. As shown in the figure, if the SCR is turned on when the
sinusoidal voltage is a maximum, the resulting current through the SCR will be a maximum
also, and the bulb will be its brightest. If the SCR should tum on when the sinusoidal voltage
is near its minimum, the bulb may tum on, but the lower current will result in considerably
less illumination. The result is that the lightbulb turns on in sync with when the input signal
is peaking, but the strength of tum-on will be determined by where one is on the applied 12-V
signal. One can imagine the interesting and varied responses of such a system. Each time one
applies the same audio signal, the response will have a different character.

12-V nns ac signal

Maximum voltage
across bulb--brightest
17V

Bulb will turn

on but with low
light intensity.
-H---+----++-- -
0

FICi. 5.138
Demonstrating the effect of an ac voltage on
the operation of the SCR of Fig. 5.137.

In the above action, the potentiometer was set below the tum-on voltage of the transis-
tor. The potentiometer can also be adjusted so that the transistor is "just on," resulting in a
low-level base current. The result is a low-level collector current and insufficient voltage
to forward-bias diode D 2 and tum on the SCR at the gate. However, when the system is
set up in this manner, the resultant light output will be more sensitive to lower amplitude
components of the applied signal. In the first case, the system acts more like a peak detector,
whereas in the latter case it is sensitive to more components of the signal.
Diode D 2 was included to be sure that there is sufficient voltage to tum on both the diode
and the SCR, in other words, to eliminate the possibility of noise or some other low-level
unexpected voltage on the line turning the SCR on. Capacitor C3 can be inserted to slow
down the response by ensuring the voltage charge across the capacitor before the gate will
reach sufficient voltage to tum on the SCR.

5.26 SUMMARY
Important Conclusions and Concepts •
1. Amplification in the ac domain cannot be obtained without the application of de
biasing level.
2. For most applications the BJT amplifier can be considered linear, permitting the use
of the superposition theorem to separate the de and ac analyses and designs.
3. When introducing the ac model for a BJT:
a. All de sources are set to zero and replaced by a short-circuit connection to
ground.
b. All capacitors are replaced by a short-circuit equivalent.
c. All elements in parallel with an introduced short-circuit equivalent should be
removed from the network.
d. The network should be redrawn as often as possible.
4. The input impedance of an ac network cannot be measured with an ohmmeter.
350 BJT AC ANALYSIS 5. The output impedance of an amplifier is measured with the applied signal set to
zero. It cannot be measured with an ohmmeter.
6. The output impedance for the re model can be included only if obtained from a data
sheet or from a graphical measurement from the characteristic curves.
7. Elements that were isolated by capacitors for the de analysis will appear in the ac
analysis due to the short-circuit equivalent for the capacitive elements.
8. The amplification factor (beta, {3, or hfe) is the least sensitive to changes in collector
current, whereas the output impedance parameter is the most sensitive. The output
impedance is also quite sensitive to changes in VCE, whereas the amplification factor
is the least sensitive. However, the output impedance is the least sensitive to
changes in temperature, whereas the amplification factor is somewhat sensitive.
9. The re model for a BJT in the ac domain is sensitive to the actual de operating con-
ditions of the network. This parameter is normally not provided on a specification
sheet, although h;e of the normally provided hybrid parameters is equal to f3re, but
only under specific operating conditions.
10. Most specification sheets for BJTs include a list of hybrid parameters to establish
an ac model for the transistor. One must be aware, however, that they are provided for
a particular set of de operating conditions.
11. The CE fixed-bias configuration can have a significant voltage gain characteristic,
although its input impedance can be relatively low. The approximate current gain
is given by simply beta, and the output impedance is normally assumed to be Re.
12. The voltage-divider bias configuration has a higher stability than the fixed-bias
configuration, but it has about the same voltage gain, current gain, and output
impedance. Due to the biasing resistors, its input impedance may be lower than that
of the fixed-bias configuration.
13. The CE emitter-bias configuration with an unbypassed emitter resistor has a larger
input resistance than the bypassed configuration, but it will have a much smaller
voltage gain than the bypassed configuration. For the unbypassed or bypassed situa-
tion, the output impedance is normally assumed to be simply Re.
14. The emitter-follower configuration will always have an output voltage slightly less
than the input signal. However, the input impedance can be very large, making it
very useful for situations where a high-input first stage is needed to "pick up" as much
of the applied signal as possible. Its output impedance is extremely low, making it
an excellent signal source for the second stage of a multistage amplifier.
15. The common-base configuration has a very low input impedance, but it can have a
significant voltage gain. The current gain is just less than 1, and the output imped-
ance is simply Re.
16. The collector feedback configuration has an input impedance that is sensitive to
beta and that can be quite low depending on the parameters of the configuration.
However, the voltage gain can be significant and the current gain of some magni-
tude if the parameters are chosen properly. The output impedance is most often
simply the collector resistance Re.
17. The collector de feedback configuration uses the de feedback to increase its stabil-
ity and the changing state of a capacitor from de to ac to establish a higher voltage
gain than obtained with a straight feedback connection. The output impedance is
usually close to Re and the input impedance relatively close to that obtained with the
basic common-emitter configuration.
18. The approximate hybrid equivalent network is very similar in composition to that
used with the re model. In fact, the same methods of analysis can be applied to both
models. For the hybrid model the results will be in terms of the network parameters
and the hybrid parameters, whereas for the re model they will be in terms of the net-
work parameters and /3, re, and r0 •
19. The hybrid model for common-emitter, common-base, and common-collector con-
figurations is the same. The only difference will be the magnitude of the parameters
of the equivalent network.
20. The total gain of a cascaded system is determined by the product of the gains of each
stage. The gain of each stage, however, must be determined under loaded conditions.
21. Because the total gain is the product of the individual gains of a cascaded system, the
weakest link can have a major effect on the total gain.
Equations SUMMARY 351

26mV
r =--
e IE
Hybrid parameters:
h;e = f3re, hfe = f3ac,
CE fixed bias:
Z; ~ f3re, Za ~ Re
Re Z;
Av=--, A;= -Av-~ /3
re Re
Voltage-divider bias:
Z; = R1 IIR2ll/3re, ReZa ~

Re Z;
A
v
= - -' A;= -Av-~ /3
re Re
CE emitter-bias:
Z; ~ Rsllf3RE, Za ~ Re

A=--
Re A;~----
f3Rs
v - RE' Rs + f3RE
Emitter-follower:
Z; ~ Rsllf3RE,

Common-base:
Z; ~ REIi re, Za ~ Re
Re
Av~-, A;~ -1
re
Collector feedback:

Re
A
v
= - -'
re
Collector de feedback:
Z; ~ Rp1 llf3re, Za ~ Rc11Rp2
Rp2 IIRe Z;
Av = re ' A; = -AvRe
Effect of load impedance:
Va RL
AvL = -V; R
L
+ RaAvNL'
Effect of source impedance:
R;Vs Va
V;=--- A =-
R; + Rs v, Vs
Vs
I=---
s Rs+ R;
Combined effect of load and source impedance:
352 BJT AC ANALYSIS Cascade connection:
Av= Av,Av2
Darlington connection (with RE):
f3v = /31/h
A. = f31f32RB
1
(RB + /31/32RE)
Va
A=-= 1
V V;
Darlington connection (without RE):

zi = R1 IIR2ll/31(re, + /31/32re2) A-= 131132(Ri IIR2)

l R1IIR2 + Z[
whereZ[ = /31(re, + f32re 2)

Za ~ Rcllra2 Av = Va
vi
Feedback pair:

5.27 COMPUTER ANALYSIS

PSpice Windows
BJT Voltage-Divider Configuration The last few chapters have been limited to the de anal-
•
ysis of electronic networks using PSpice and Multisim. This section will consider the applica-
tion of an ac source to a BIT network and describe how the results are obtained and interpreted.
Most of the construction of the network of Fig. 5.139 can be accomplished using the
procedures introduced in earlier chapters. The ac source can be found in the SOURCE
library as VSIN. You can scroll down the list of options or simply type in VSIN at the head
of the listing. Once this is selected and placed, a number of labels will appear that define

OrCAD Capture OS - Demo Ed111on

fil e E,dit '{iew Iools f lace Macro P.$pice Accessories
Qptions Window Help Ci denCe
; SCHEMATIC1-01CAO •

im Oul<l 5-l ] ~ PAGU~

G]"""" l_

AC • ok
I' f'-
.
I'"
vcc
de RI

""
,..._.
ab<
l +
1
f-r
C2
o,
02N=

--:-
R<
1.5k
l ~;F

FIG. 5.139
Using PSpice Windows to analyze the network of Fig. 5.28
(Example 5.2).
the parameters of the source. Double-clicking the source symbol or using the sequence COMPUTER ANALYSIS 353
Edit-Properties will result in the Property Editor dialog box, which lists all the param-
eters appearing on the screen and more. By scrolling all the way to the left, you will find a
listing for AC. Select the blank rectangle under the heading and enter the 1 mV value. Be
aware that the entries can use prefixes such as m (milli) and k (kilo). Moving to the right,
the heading FREQ will appear, in which you can enter 10 kHz. Moving again to PHASE,
you will find the default value is 0, so it can be left alone. It represents the initial phase angle
for the sinusoidal signal. Next you will find VAMPL, which is set at 1 mV, also followed
by VOFF at O V. Now that each of the properties has been set, we have to decide what to
display on the screen to define the source. In Fig. 5.139 the only labels are Vs and 1 mV,
so a number of items have to be deleted and the name of the source has to be modified. For
each quantity simply return to the heading and select it for modification. If you choose AC,
select Display to obtain the Display Properties dialog box. Select Value Only because we
prefer not to have the label AC appear. Leave all the other choices blank. An OK, and you
can move to the other parameters within the Property Editor dialog box. We do not want
the FREQ, PHASE, VAMPL and VOFF labels to appear with their values, so in each case
select Do Not Display. To change Vl to Vs, simply go to the Part Reference, and after
selecting it, type in Vs. Then go to Display and select Value Only. Finally, to apply all the
changes, select Apply and exit the dialog box; the source will appear as shown in Fig. 5 .139.
The ac response for the voltage at a point in the network is obtained using the VPRINTl
option found in the SPECIAL library. If the library does not appear, simply select Add
Library followed by special.olb. When VPRINTl is chosen, it will appear on the screen
as a printer with three labels: AC, MAG, and PHASE. Each has to be set to an OK status
to reflect the fact that you desire this type of information about the voltage level. This is
accomplished by simply clicking on the printer symbol to obtain the dialog box and setting
each to OK. For each entry select Display and choose Name and Label. Finally, select
Apply and exit the dialog box. The result appears in Fig. 5.139.
The transistor Q2N2222 can be found under the EVAL library by typing it under the
Part heading or simply scrolling through the possibilities. The levels of ls and f3 can be
set by first selecting the Q2N2222 transistor to make it red and then applying the sequence
Edit-PSpice Model to obtain the PSpice Model Editor Lite dialog box and changing Is to
2E-15A and Bf to 90. The level of Is is the result of numerous runs of the network to find
the value that would result in VsE being closest to 0.7 V.
Now that all the components of the network have been set, it is time to ask the computer
to analyze the network and provide some results. If improper entries were made, the com-
puter will quickly respond with an error listing. First select the New Simulation Profile
key to obtain the New Simulation dialog box. Then, after entering Name as OrCAD 5-1,
select Create and the Simulation Settings dialog box will appear. Under Analysis type,
select AC Sweep/Noise and then under AC Sweep Type choose Linear. The Start Fre-
quency is 10 kHz, the End Frequency is 10 kHz, and the Total Points is 1. An OK, and
the simulation can be initiated by selecting the Run PSpice key (white arrow). A schematic
will result with a graph that extends from 5 kHz to 15 kHz with no vertical scale. Through
the sequence View-Output File the listing of Fig. 5 .140 can be obtained. It starts with a list
of all the elements of the network and their settings followed by all the parameters of the
transistor. In particular, note the level of IS and BF. Next the de levels are provided under
the SMALL SIGNAL BIAS SOLUTION, which match those appearing on the schematic
of Fig. 5.139. The de levels appear on Fig. 5.139 due to the selection of the V option. Also
note that VsE = 2.624 V - 1.924 V = 0. 7 V, as stated above, due to the choice of Is.
The next listing, OPERATING POINT INFORMATION, reveals that even though
beta of the BJT MODEL PARAMETERS listing was set at 90, the operating conditions
of the network resulted in a de beta of 48.3 and an ac beta of 55. Fortunately, however, the
voltage-divider configuration is less sensitive to changes in beta in the de mode, and the
de results are excellent. However, the drop in ac beta had an effect on the resulting level
of V0 : 296.1 mV versus the handwritten solution (with r0 = 50 kil) of 324.3 mV-a 9%
difference. The results are certainly close, but probably not as close as one would like. A
closer result (within 7%) could be obtained by setting all the parameters of the device except
ls and beta to zero. However, for the moment, the impact of the remaining parameters has
been demonstrated, and the results will be accepted as sufficiently close to the handwritten
levels. Later in this chapter, an ac model for the transistor will be introduced with results
354 BJT AC ANALYSIS CIRCUIT DESCRIPTION

*Analysis directives:
.AC LIN 1 lOkHz lOkHz
.OP
.PROBE V(alias(")) !(alias(")) W(alias(")) D(alias(*)) NOISE(alias(*))
.INC " ..\SCHEMATICl.net'
"'source ORCAD 5-1
Q_Ql N00286 N00282 N00319 Q2N2222
RJU N00282 N00254 56k TC=0,0
R_R2 0N00282 8.2k.TC=0,0
R_R3 N00286 N00254 6.Sk TC=0,0
R....ll4 0N00319 1.5kTC=0,0
v_vcc N00254022Vdc
C_Cl 0 N00319 20uF TC=O,0
V_Vs N003420 AC lmV
+SIN0V lmV lOkHz000
.PRINT AC
+ VM ([N00286])
+ VP ([N00286])
C_C2 N00342 N00282 lOuF TC=O,O
.END

BIT MODEL PARAMETERS

Q2N2222
NPN
LEVEL l
IS 2.000000E-15
BF OJ
NF 1
VAF 74.03
1KF .2847
!SE 14.3400008-15
NE 1.307
BR 6.092
NR
ISS
RB 10
RE 0
RC 1
CJE 22.0lOOOOE-12
VJE .75
MJE .377
CJC 7.306000E-12
VJC .75
MJC .3416

XCJC 1
CJS 0
VJS .75
TF 411.lOOOOOE-12
XTF 3
VIF 1.7
ITF .6
TR 46.910000E-09
XTB 1.5
KF 0
AF 1
CN 2.42
D .87

SMALL SIGNAL BIAS SOLUTION TEMPERATURE= 27.000 DEG C

NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE

(N00254) 22.0000 (N(X)282) 2.6239 (N00286) 13.4530 (N00319) 1.9244
(N00342) 0.CXXX>

VOLTAGE SOURCE CURRENTS

NAME CURRENT
V_VCC -l.603E-03
V_Vs 0.OOOE+OO

TOTALPOWERDISSIPATION 3.53E-02 WAITS

*""*"" OPERATING POINT INFORMATION TEMPERATURE = 27 .000 DEG C

*""*"" BIPOLAR JUNCTION TRANSISTORS

NAME Q_Ql
MODEL Q2N2222
IB 2.60E-05
IC 1.26E-03
VBE 6.99E-0l
VBC -l.08E+Ol
VCE l.15E+Ol
BETADC 4.83E+Ol
GM 4.84E-02
RPI l.14E+03
RX l.OOE+Ol
RO 6.75E+04
CBE 5.78E-11
CBC 2.87E-12
CJS 0.OOE+OO
BETAAC 5.50E+Ol
CBX/CBX2 0.OOE+OO
Ff/FT2 l.27E+08

**** AC ANALYSIS TEMPERATURE= 27.000 DEG C

FREQ VM(N00286) VP(N00286)

l.OOOE+04 2.961E-01 -l.780E+02

FIG. 5.140
Output file for the network of Fig. 5.139.

that will be an exact match with the handwritten solution. The phase angle is -178° versus
the ideal of - 180°, a very close match.
A plot of the voltage at the collector of the transistor can be obtained by setting up a new
simulation process to calculate the value of the desired voltage at a number of data points.
The more points, the more accurate is the plot. The process is initiated by returning to the
Simulation Settings dialog box and under Analysis type selecting Time Domain(Transient). COMPUTER ANALYSIS 355
Time domain is chosen because the horizontal axis will be a time axis, requiring that the
collector voltage be determined at a specified time interval to permit the plot. Because the
period of the waveform is 1/10 kHz = 0.1 ms = 100 µ,s, and it would be convenient to
display five cycles of the waveform, the Run to time(TSTOP) is set at 500 µ,s. The Start
saving data after point is left at 0 s and under Transient option, the Maximum step
size is set at 1 µ,s to ensure 100 data points for each cycle of the waveform. An OK, and a
SCHEMATIC window will appear with a horizontal axis broken down in units of time
but with no vertical axis defined. The desired waveform can then be added by first select-
ing Trace followed by Add Trace to obtain the Add Trace dialog box. In the provided
listing V(Ql:c) is selected as the voltage at the collector of the transistor. The instant it is
selected it will appear as the Trace Expression at the bottom of the dialog box. Referring to
Fig. 5.139, we find that because the capacitor CE will essentially be in the short-circuit state
at 10 kHz, the voltage from collector to ground is the same as that across the output terminals
of the transistor. An OK, and the simulation can be initiated by selecting the Run PSpice key.
The result will be the waveform of Fig. 5.141 having an average value of about 13.45 V,
which corresponds exactly with the bias level of the collector voltage in Fig. 5.139.
The range of the vertical axis was chosen automatically by the computer. Five full
cycles of the output voltage are displayed with 100 data points for each cycle. The data
points appear in Fig. 5.139 because the sequence Tools-Options-Mark Data Points
was applied. The data points appear as small dark circles on the plot curve. Using the
scale of the graph, we see that the peak-to-peak value of the curve is approximately
13.76 V - 13.16 V = 0.6 V = 600 mV, resulting in a peak value of 300 mV. Because a
1-mV signal was applied, the gain is 300, or very close to the calculator solution of296.l.

, SCHEMIITICl--OrCAD 5-1 - PSpt<e ND

--------------

13. 6 U+-+-->-<1---+-++~-+--,;----+-->+-+-;-+>-------;-++--+--+-_,,__--+-<..+--+--,......;s--------<>-__,

--

13_41J .i\-.;,_"'l--'_..;.---!,_,.;.._ __.___.__..__ r...

;_.;--,;1-'--'""!'-'-__.,_-ll--'--'-1-,;._;. ---1-

13 .2u+-1;-;.;,'---',_...._--+-i.-+'---'---'--l-l~'---;_.;--+~-l-'---'---'--I-I'-,-+'-
, --'-'.
••--'."!._···-1·

··-r-·1····:···1··· -··r·1···tt·· ····: ···:···tt··· ···r··t"""tt· ···:--··(J"""t""·

···:···r···: ... r .. ···:···:···-r···:··· ····:···:···:···:··· ···:···:···r···:··· ···-r···:···:···:···
1~ . DU + - - ~ - - - + ~ - ~ - + - - - ~ - - + - ~ - ~ - + - ~ - - ' - - - - - <
10Du~ 2oa,,, aoa,,,
°"

-
o U(Q1 :c)
11...
ii 0.CADS-..
For Hel , ress Fl T1me- 500.0E-06 100%

FICi. 5.141
Voltage vcfor the network Fig. 5.139.

If a comparison is to be made between the input and output voltages on the same screen,
the Add Y-Axis option under Plot can be used. After you select it, choose the Add Trace
icon and select V(Vs:+) from the provided list. The result is that both waveforms will ap-
pear on the same screen as shown in Fig. 5.142, each with its own vertical scale.
If two separate graphs are preferred, we can start by selecting Plot followed by Add Plot
to Window after the graph of Fig. 5 .141 is in place. The result will be a second set of axes
waiting for a decision about which curve to plot. Using Trace-Add Trace-V(Vs:+) will
result in the graphs of Fig. 5.143. The SEL >> (from SELECT) appearing next to one of
the plots defines the "active" plot.
356 BJT AC ANALYSIS . SCI l[MAilCI -OrCAll 5-1 - PSp!c<, A/D Demo - (OrCA □ 5-1 (llcllv,e)J

13 • .tiU

13. 2 U

)) . .
... , . u,w+-~ ~....u--1~~-...._-+-_~..._...__f-o_~......._-+-~---'"'-l
100U'" 51:! 0U'i"
Oi
CD n U(Qlcc) rn • U(Usc•) 2 "Du'
liM

Fil, Time- 500.0E-06 100%

FIG. 5.142
The voltages Ve and vsfor the network of Fig. 5.139.

SCHEMATIO -OtCAD S 1 - l'Spr<e A/D Demo - [OrCAD 5-1 (OC!lve)I

11t.nu-..-.• ...
~•...~ ~~~~~-~~~~--------~~~~
,....

:::t:::

·-- ~---
SH~> -- l---
1 3.00 ,
Is 1 Hus 2 00u 5i S alus 'Jtftu s S Oftu !li

-
a U(Q I :1:)
n ...
■ DrCAD5-
C."\ECET11 ORCAD\Orcad 5-l-PSp,ceFi Tlme• 500.0E-06 100%

FIG. 5.143
Two separate plots ofvc and Vs in Fig. 5.139.

The last operation to be introduced in this coverage of graph displays is the use of
the cursor option. The result of the sequence Trace-Cursor-Display is a line at the de
level of the graph of Fig. 5.144 intersecting with a vertical line. The level and time both
appear in the small dialog box in the bottom right comer of the screen. The first number
for Cursor 1 is the time intersection and the second is the voltage level at that instant. A
left-click of the mouse will provide control of the intersecting vertical and horizontal lines
at this level. Clicking on the vertical line and holding down on the clicker will allow you to
move the intersection horizontally along the curve, simultaneously displaying the time and
voltage level in the data box at the bottom right of the screen. If it is moved to the first peak COMPUTER ANALYSIS 357
of the waveform, the time appears as 75.194 µ,s with a voltage level of 13.753 V, as shown
in Fig. 5.144. On right-clicking of the mouse, a second intersection, defined by Cursor 2,
will appear, which can be moved in the same way with its time and voltage appearing in the
same dialog box. Note that if Cursor 2 is placed close to the negative peak, the difference
in time is 49.61 µ,s (as displayed in the same box), which is very close to one-half the period
of the waveform. The difference in magnitude is 591 mV, which is very close to the 600 mV
obtained earlier.

SCHEMATlCl OrCAO 5-1 PSptce AID Demo (OrCAD >-1 (octwe)l

'
'
I
' . .
• •
.
l '
' .
o

--- j-··i··-+--. ' ' .

1~.~u

·r·--·
····•····
I
'
I
' . . . ' . . .
I
'
I
' . . .
I I
' ' . . I

13 . lftU +-~~---'-----1-.........- - - + - - - - + - ~ ~ - - 1 - - - - - - - ,
I I I I t I O t I I I

n, ?ftftu-. .tannu c;; "i fUlu,;;

1:1 U( 01:cl

II OrCI\D5-_
3
-
~

-
Trace Curs.art Cursor-1 Iliff Uax IJin Avg =
X V1luo 75 194u 124 8ll6" -49512u 12< 106" 71 19-lu 100 OOOII
V(Q1 cc) 13 153 1316:2 5~1 OOOm 13.763 13 162 13458
-
C:\ECETll ORCAD\Orcad 5-1-PSpi.eFI X..0.000431 V,13.8 Time• SOO.OE-06 100%

FIC. 5.144
Demonstrating the use of cursors to read specific points on a plot.

Voltage-Divider Configuration-Controlled Source Substitution The results obtained

for any analysis using the transistors provided in the PSpice listing will always be some-
what different from those obtained with an equivalent model that only includes the effect
of beta and re· This was clearly demonstrated for the network of Fig. 5 .139. If a solution is
desired that is limited to the approximate model used in the hand calculations, then the
transistor must be represented by a model such as appearing in Fig. 5.145.

---------------1

r
B I C B C B

ll1b ~r,
I
~r,
+
I

~lb:
I
I
I
I
I
I
- kb + ~lb
- lb

F
_ _ _ _ _ _ _ _J
------

E E E

FIC. 5.145
Using a controlled source to represent the transistor of Fig. 5.139.
358 BJT AC ANALYSIS For Example 5.2, f3 is 90, with f3re = 1.66 kO. The current-controlled current source
(CCCS) is found in the ANALOG library as part F. After selection, an OK, and the graphi-
cal symbol for the CCCS will appear on the screen as shown in Fig. 5.146. Because it does
not appear within the basic structure of the CCCS, it must be added in series with the
controlling current that appears as an arrow in the symbol. Note the added 1.66-kO resis-
tor, labeled beta-re in Fig. 5.146. Double-clicking on the CCCS symbol will result in the
Property Editor dialog box, in which the GAIN can be set to 90. It is the only change to
be made in the listing. Then select Display followed by Name and Value and exit (x) the
dialog box. The result is the GAIN = 90 label appearing in Fig. 5 .146.

OrCAD capture as - Demo Edmon - [/ - (SCHEMATIC1 : ...

etace .Macro f'Spi c.e Accessories Qptions

cadence ~

I ..
.,.

R2

""'

Ill

FIG. 5.146
Substituting the controlled source of Fig. 5.145 for the transistor
of Fig. 5.139.

A simulation and the de levels of Fig. 5.146 will appear. The de levels do not match
the earlier results because the network is a mix of de and ac parameters. The equivalent
model substituted in Fig. 5.146 is a representation of the transistor under ac conditions,
not de biasing conditions. When the software package analyzes the network from an ac
viewpoint it will work with an ac equivalent of Fig. 5.146, which will not include the de
parameters. The Output File will reveal that the output collector voltage is 368.3 mV, or
a gain of 368.3, essentially an exact match with the handwritten solution of 368.76. The
effects of rO could be included by simply placing a resistor in parallel with the controlled
source.

Darlington Configuration Although PSpice does have two Darlington pairs in the
library, individual transistors are employed in Fig. 5.147 to test the solution to Exam-
ple 5.17. The details of setting up the network have been covered in the preceding sec-
tions and chapters. For each transistor ls is set to lO0E-18 and /3 to 89.4. The applied
frequency is 10 kHz. A simulation of the network results in the de levels appearing in
Fig. 5.147a and the Output File in Fig. 5.147b. In particular, note that the voltage drop
between base and emitter for both transistors is 10.52 V - 9.148 V = 1.37 V com-
pared to the 1.6 V assumed in the example. Recall that the drop across Darlington pairs
is typically about 1.6 V and not simply twice that of a single transistor, or
2(0.7 V) = 1.4 V. The output voltage of 99.36 mV is very close to the 99.80 mV
obtained in Section 5 .17.
 Orf.AD [;,pture n~ - n,>mo Fdnion

Iools £lace Macro P!ipice Accessories Qptions BIT MODEL PARAMETERS

cadence
Q2N3904
NPN
LEVEL
IS 100.CKXXXJOE-18
BF 89.4
/ - (SCH EMATIC!: PAGHJ NF
BR
NR
CN 2.42
D .87
vcc
·_ ,av~ SMAIL SIGNAL BIAS SOLlITION TEMPERATURE = 27.000 DEG C

l
r~~.·-
r.1 NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE NODE VOLTAGE
N00218) O.CXXXJ (N00225) 181XXJ0 (N00243) 8.9155 (N00250) 9.6513
(N00291) OlXXXl (N02131) 8.0632

**** AC ANALYSIS IBMPERATURE = 27.CXXl DEG C

l FREQ

l.OOOE+04
VM(N00291)

9.936E-02

(a) (b)

FIC. 5.147
(a) Design Center schematic of Darlington network; (b) output listing for circuit of part (a) (edited).

Multisim
Colledor Feedback Configuration Because the collector feedback configuration gen-
erated the most complex equations for the various parameters of a BJT network, it seems
appropriate that Multisim be used to verify the conclusions of Example 5.9. The net-
work appears as shown in Fig. 5.148 using the "virtual" transistor from the Transistor
family toolbar. Recall from the previous chapter that transistors are obtained by first
selecting the Transistor keypad appearing as the fourth option over on the component

ms S-1 Multtsim (ms S l ')

~ file fdit Yiew £lace MCU Simulate Trc,osfer Iool:; Beports Qplfons '{iindow ]::!elp

D" fol!ll!Q~E?:f[i)"b~ · fflllGD , <J>r:t> • 'e ?

Multimeter-XMMl ~

999 576 uV 2 4

00 00 _lvcc
-=--- 9V
h.~kO XMM2
~~
r
B
@
+
SPI...

Ct
~
H~
100kQ

01
C2

~
10,,~
llJT llPH VIRTUAL"
Al=?DO -

Multimeter XMM2 LJ:LJ

24'.i 100mV

C
GJ 0 ~ ~
Ill "v r=7
+
~m,S-1 *
®
Set. .. I ®,

FIC. 5.148
Network of Example 5.9 redrawn using Multisim.
359
360 BJT AC ANALYSIS toolbar. Once chosen, the Select a Component dialog box will appear; under the Fam-
ily heading, select TRANSISTORS_VIRTUAL followed by BJT_NPN_VIRTUAL.
Following an OK the symbols and labels will appear as shown in Fig. 5.148. We
must now check that the beta value is 200 to match the example under investigation.
This can be accomplished using one of two paths. In Chapter 4 we used the EDIT-
PROPERTIES sequence, but here we will simply double-click on the symbol to obtain
the TRANSISTORS_VIRTUAL dialog box. Under Value, select Edit Model to obtain
the Edit Model dialog box (the dialog box has a different appearance from that obtained
with the other route and requires a different sequence to change its parameters). The
value of BF appears as 100, which must be changed to 200. First select the BF line to
make it blue all the way across. Then place the cursor directly over the 100 value and
select it to isolate it as the quantity to be changed. After deleting the 100, type in the
desired 200 value. Then click the BF line directly under the Name heading and the
entire line will be blue again, but now with the 200 value. Then choose Change Part
Model at the bottom left of the dialog box and the TRANSISTORS-VIRTUAL dialog
box will appear again. Select OK and /3 = 200 will be set for the virtual transistor. Note
the asterisk next to the BJT label to indicate the parameters of the device have been
changed from the default values. The label Bf = 100 was set using Place-Text as
described in the previous chapter.
This will be the first opportunity to set up an ac source. First, it is important to real-
ize that there are two types of ac sources available, one whose value is in rms units, the
other with its peak value displayed. The option under Power Sources uses rms values,
whereas the ac source under Signal Sources uses peak values. Because meters display
rms values, the Power Sources option will be used here. Once Source is selected, the
Select a Component dialog box will appear. Under the Family listing select POWER_
SOURCES and then select AC_POWER under the Component listing. An OK, and
the source will appear on the screen with four pieces of information. The label Vl can
be deleted by first double-clicking on the source symbol to obtain the AC_POWER
dialog box. Select Display and disengage Use Schematic Global Settings. To remove
the label Vl, disengage the Show ReIDes option. An OK, and the Vl will disappear
from the screen. Next the value has to be set at 1 mV, a process initiated by selecting
Value in the AC_POWER dialog box and then changing the Voltage (RMS) to 1 mV.
The units of mV can be set using the scroll keys to the right of the magnitude of the
source. After you change the Voltage to 1 mV, an OK will place this new value on the
screen. The frequency of 1000 Hz can be set in the same way. The 0-degree phase shift
happens to be the default value.
The label Bf= 200 is set in the same way as described in Chapter 4. The two multi-
meters are obtained using the first option at the top of the right vertical toolbar. The meter
faces appearing in Fig. 5.148 were obtained by simply double-clicking on the multimeter
symbols on the schematic. Both were set to read voltages, the magnitudes of which will be
in rms units.
After simulation the results of Fig. 5.148 appear. Note that the meter XMMl is not read-
ing the 1 mV expected. This is due to the small drop in voltage across the input capacitor
at 1 kHz. Certainly, however, it is very close to 1 mV. The output of 245.166 mV quickly
reveals that the gain of the transistor configuration is about 245.2, which is a very close
match with the 240 obtained in Example 5.9.

Darlington Configuration Applying Multisim to the network of Fig. 5.147 with a pack-
aged Darlington amplifier results in the printout of Fig. 5.149. For each transistor the
parameters were changed to Is= lO0E-18 A and Bf= 89.4 using the technique described
earlier. For practice purposes the ac signal source was employed rather than the power
source. The peak value of the applied signal is set at 100 mV, but note that the multimeter
reads the effective or rms value of 99.991 mV. The indicators reveal that the base voltage
of Q1 is 7.736 V, and the emitter voltage of Q2 is 6.193 V. Therms value of the output
voltage is 99.163 mV, resulting in a gain of 0.99 as expected for the emitter follower con-
figuration. The collector current is 16 mA with a base current of 1.952 mA, resulting in a
f3v of about 8200.
 ms 5 2 MulUsim (ms 5 2 •J PROBLEMS

99 163 111V

~ V 0 0
~r=l
Cl
~
0.5pF

100mVnrn,
10kH1
u-

"'
~mss-2•

FICi. 5.149
Network of Example 5.9 redrawn using Multisim.

PROBLEMS
*Note: Asterisks indicate more difficult problems.
5.2 Amplification in the AC Domain
•
1. a. What is the expected amplification of a BJT transistor amplifier if the de supply is set to
zero volts?
b. What will happen to the output ac signal if the de level is insufficient? Sketch the effect on
the waveform.
c. What is the conversion efficiency of an amplifier in which the effective value of the current
through a 2.2-kil load is 5 mA and the drain on the 18-V de supply is 3.8 mA?
2. Can you think of an analogy that would explain the importance of the de level on the resulting
ac gain?
3. If a transistor amplifier has more than one de source, can the superposition theorem be applied
to obtain the response of each de source and algebraically add the results?

5.3 BIT Transistor Modeling

4. What is the reactance of a 10-µ,F capacitor at a frequency of 1 kHz? For networks in which the
resistor levels are typically in the kilohm range, is it a good assumption to use the short-circuit
equivalence for the conditions just described? How about at 100 kHz?
5. Given the common-base configuration of Fig. 5.150, sketch the ac equivalent using the nota-
tion for the transistor model appearing in Fig. 5.7.

R, C1 C2

~
0

+
+
RL Re
V
' '\J Va

FICi. 5.150
Problem 5.
5.4 The re Transistor Model
6. a. Given an Early voltage of VA= 100 V, determine r0 if VCEQ = 8 V and IcQ = 4 mA.
b. Using the results of part (a), find the change in le for a change in VCE of 6 Vat the same
Q-point as part (a).
BJT AC ANALYSIS 7. For the common-base configuration of Fig. 5.18, an ac signal of 10 mV is applied, resulting in
an ac emitter current of 0.5 mA. If a = 0.980, determine:
a. Z;.
b. V0 if RL = l.2 kil.
C. Av = Vo/V;.
d. Zo with To= 00 n.
e. A;= 10 /I;.
f. h-
8. Using the model of Fig. 5.16, determine the following for a common-emitter amplifier if
f3 = 80, h(dc) = 2 mA, and T0 = 40 kil.
a. Z;.
b. h-
e. A; = /0 / I; = h/lb if RL = l.2 kil.
d. Av if RL = l.2 kil.
9. The input impedance to a common-emitter transistor amplifier is 1.2 kil with f3 = 140,
T0 = 50 kil, and RL = 2.1 kil. Determine:

a. Te.
b. hifV; = 30mV.
c. le.
d. A; = 10 /I; = h/h-
e. Av = V0 /V;.
10. For the common-base configuration of Fig. 5.18, the de emitter current is 3.2 mA and a is 0.99.
Determine the following if the applied voltage is 48 m V and the load is 2.2 kil.
a. Te.
b. Z;.
c. le.
d. V0 •
e. Av.
f. h-
s.s Common-Emitter Fixed-Bias Configuration
11. For the network of Fig. 5.151:
a. Determine Z; and Z 0 •
b. FindAv.
c. Repeat parts (a) and (b) with T0 = 20 kil.
12. For the network of Fig. 5.152, determine Vee for a voltage gain of Av = -160.

12V

- - - - - - - o Vcc

2.2 kQ

220 kQ i 10
(----a V 0
lMQ
(----a V 0

V; o-----),_______._ _ fi=90
V; o ) 1------+-------<1
ro = 00 Q
~ P=~
___._ ~=40kQ
Z;

FIC. 5.151 FIC. 5.152

Problem 11. Problem 12.

*13. For the network of Fig. 5.153:

a. Calculate IB, le, and Te-
b. Determine Z; and Z 0 •
c. Calculate Av.
d. Determine the effect of r0 = 30 kil on Av.
14. For the network of Fig. 5.153, what value of Re will cut the voltage gain to half the value
obtained in problem 13?
 12 V

v; ---.-
o--------J - - -
I; j3 = 100
390ill g 0 s=25 µS

8V

FIC. 5.153
Problem 13.
s.& Voltage-Divider Bias
15. For the network of Fig. 5.154:
a. Determine T,.
b. Calculate Z; and Z 0 •
c. FindAv.
d. Repeat parts (b) and (c) with T0 = 25 kil.
Vee= 16V

39ill

lµF
v; a-----}---
---.-
I;

4.7ill

~ !.2k!l r!OµF

-=-
FIC. 5.154
Problem 15.
16. Determine Vee for the network of Fig. 5.155 if Av = -160 and T = 100 kil.
0

17. For the network of Fig. 5.156:

a. Determine T,.
b. Calculate VB and Vc-
Vee =20V
c. Determine Z; and Av = V0 /V;.
~ - - - - - - o Vcc

3.3 ill 4.7ill

82ill
f------<> V 0

Cc
v; o-----)1-------1 J3 = 100 J3 = 180
Cc gos =20 µS g 0 s= 30 µS

5.6ill
1 ill 2.2ill

-=- -=- -=- -=-

FIC. 5.155 FIC. 5.156
Problem 16. Problem 17.
BJT AC ANALYSIS 18. For the network of Fig. 5.157:
a. Determiner,.
b. Find the de voltages V8 , Vc8 , and VCE.
c. Determine Z; and Z 0 •
d. Calculate Av = V 0 /V;.

24V

68k0 ~ Z ;

FIG. 5.157
Problem 18.

5.7 CE Emitter-Bias Configuration

19. For the network of Fig. 5.158:
a. Determiner,.
b. Find Z; and Z 0 •
c. Calculate Av.
d. Repeat parts (b) and (c) with r0 = 20 kil.
20. Repeat Problem 19 with Re bypassed. Compare results.
21. For the network of Fig. 5.159, determine Re and R8 if Av = -10 and r, = 3.8 n. Assume that
zb = f3Re.

20V

---------020 V

8.2 ill
390kQ

Vi o------J l - - - - + - - - . 1 1 /3 = 140 V; o--)1----+------.11 /3 = 120

---.-
I;
r0 = lOOkQ 8os=10µS

FIG. 5.158 FIG. 5.159

Problems 19 and 20. Problem 21.

*22. For the network of Fig. 5.160:

a. Determiner,.
b. Find Z; and Av.
23. For the network of Fig. 5.161:
a. Determiner,.
b. Calculate Vs, Vce, and V CB·
c. Determine Z; and Z 0 •
d. Calculate Av = V 0 /V;.
e. Determine A; = 10 /l;.
 ~-----.----022 V

330 kQ
16V

I;
---.-
Vi o------J l---+----1 /3= 80
Cc r0 =40kQ 4.7 kfl
430kfl

l.2kQ f------o V0

/3 = 200
gos= 20 µS

120 kfl
0.47 kQ 1.2 kfl

-=- -=-
FIC. 5.160 FIC. 5.161
Problem 22. Problem 23.

5.8 Emitter-Follower Configuration 16V

24. For the network of Fig. 5.162:

a. Determiner, and f3r,.
b. Find Z; and Z 0 •
c. Calculate Av. 270kQ

Vi a------} t - - + - - - - - 1 /3 = 110
---.-
I;
r0 = 50 kQ

2.7kQ

-=-
FIC. 5.162
Problem 24.
*25. For the network of Fig. 5.163:
a. Determine Z; and Z 0 •
b. FindAv.
c. Calculate V0 if V; = 1 mV.
*26. For the network of Fig. 5.164:
a. Calculate IB and le. Vcc=20V
b. Determiner,.
c. Determine Z; and Z 0 •
d. FindAv.

12V
56kQ
I;
---.- f3 = 120
Vi a------} l----+------11
r0 =40kQ /3= 200
Vi o------)1------1
---.-
I;
g 0 s = 20 µS

8.2kQ

-=- -8V -=- -=-

FIC. 5.163 FIC. 5.164
Problem 25. Problem 26.
BJT AC ANALYSIS 5.9 Common-Base Configuration
27. For the common-base configuration of Fig. 5.165:
a. Determiner,.
b. Find Z; and Z 0 •
c. Calculate Av.
*28. For the network of Fig. 5.166, determine Av.

8V

3.6kQ

~I 0

+6V -lOV
(----a V0

/3= 75
gos=S µS
6.8 kQ 4.7kQ
I;
___._ ~I 0

Vi a----=-.J ---~ ~------1(--------o Vo

3.9kQ
a =0.998
gos= 10 µS
-SV
-=-
FIC. 5.165 FIC. 5.166
Problem 27. Problem 28.

5.1 o Collector Feedback Configuration

29. For the collector feedback configuration of Fig. 5.167:
a. Determiner,.
b. Find Z; and Z 0 •
c. Calculate Av.
*30. Given r, = 10 n, {3 = 200, Av = -160, and A; = 19 for the network of Fig. 5.168, deter-
mine Re, Rp, and Vcc-
*31. For the network of Fig. 5.49:
a. Derive the approximate equation for Av.
b. Derive the approximate equations for Z; and Z 0 •
c. Given Re = 2.2 kil, Rp = 120 kil, Re = 1.2 kil, {3 = 90, and Vcc = 10 V, calculate
the magnitudes of Av, Z;, and Z 0 using the equations of parts (a) and (b).

12V

~I 0

3.9kQ

r,= lOQ
Vi o-------} Vi o-------)-------11
___._ -------- /3= 120 /3= 200
I; r0 = 80kQ
r0 =40kQ

-=- -=-
FIC. 5.167 FIC. 5.168
Problem 29. Problem 30.

5.11 Collector DC Feedback Configuration

32. For the network of Fig. 5.169:
a. Determine Z; and Z 0 •
b. FindAv.
 9V

,------4~------4'1'-------1(--------o ~
1 µF

I;
---+- /3= 80
V; o---------}--------------11
= 22 µS
g 08
1 µF

FIC. 5.169
Problems 32 and 33.

33. Repeat problem 32 with the addition of an emitter resistor RE = 0.68 kD.

s.12-s.1 s Effect of R, and Rs and Two-Port Systems Approach

*34. For the fixed-bias configuration of Fig. 5.170:
a. Determine AvNL' Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the parameters determined in part (a) in place.
c. Calculate the gain AvL = V0 /V;.
d. Determine the current gain A;L = I0 /I;.

18 V

3.3ill

--
680kQ
1.8 µF /0
- - - - - 1 t - - - - . . - - - o Vo
l.8µF
V; o ___._ ) /3 = 100
I; -- RL 4.7kQ
Zo

-=- -=-
FIC. 5.170
Problems 34 and 35.

35. a. Determine the voltage gainAvJorthe network of Fig. 5.170 for RL = 4.7 kD, 2.2kD, and
0.5 kD. What is the effect of decreasing levels of RL on the voltage gain?
b. How will Z;, Z0 , and AvNL change with decreasing values of RL?
*36. For the network of Fig. 5.171:
a. Determine AVNL, Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the parameters determined in part (a) in place.
c. Determine Av = V0 /V;.
d. Determine Av, = V0 /Vs.
e. Change Rs to 1 kD and determine Av. How does Av change with the level of Rs?
f. Change Rs to 1 kD and determine Av, How does Av, change with the level of Rs?
g. Change Rs to 1 ill and determine AvNL' Z;, and Z0 • How do they change with the change in Rs?
h. For the original network of Fig. 5.171 calculate A; = I 0 /I;.
BJT AC ANALYSIS 12 V

3kQ
lMQ
1 µF
------<••(----a V 0

--
f3 - 180

Zo

FIG. 5.171
Problem 36.

*37. For the network of Fig. 5.172:

a. Determine AvNL' Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the parameters determined in part (a) in
place.
c. Determine AvL and Av,-
d. Calculate A;L-
e. Change RL to 5.6 kD and calculate Av,- What is the effect of increasing levels of RL on the
gain?
f. Change Rs to 0.5 kil (with RL at 2.7 kil) and comment on the effect of reducing Rs on
Av,-
g. Change RL to 5.6 kil and Rs to 0.5 kil and determine the new levels of Z; and Z0 • How are
the impedance parameters affected by changing levels of RL and Rs?

24V

4.3 k.Q

560 k.Q
lo
Vo
v;

--
~ µF /3 = 80
k.Q
+
vs '\, ___._ zo
Z;

-:l- -=- -=-

FIG. 5.172
Problem 37.

38. For the voltage-divider configuration of Fig. 5.173:

a. Determine AVNL, Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the parameters determined in part (a) in
place.
c. Calculate the gain Ave
d. Determine the current gain A;L"
e. Determine AvL, A;L, and Z 0 using the re model and compare solutions.
39. a. Determine the voltage gain AvL for the network of Fig. 5.173 with RL = 4.1 kil, 2.2 kil,
and 0.5 kil. What is the effect of decreasing levels of RL on the voltage gain?
b. How will Z;, Z0 , and AvNL change with decreasing levels of RL?
 16 V

2.2 k.Q
68kil 6.8µF Io
vn
I;
V; 0

6.8µF

__..
Z;
)

16kO
f3 = 100

-- Zo
5.6 k.Q

0.75W
r·µF
'T -=' -=' -='

FICi. 5.173
Problems 38 and 39.

40. For the emitter-stabilized network of Fig. 5.174:

a. Determine A"NL, Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the values determined in part (a).
c. Determine AvL and Av,
d. Change Rs to 1 kD. What is the effect on A"NL, Z;, and Z0 ?
e. Change Rs to 1 kil and determine AvL and Av, What is the effect of increasing levels of Rs
on AvL and Av,?
f. Determine A; = / 0 /I;.

18 V

3 k.Q
680W 1 µF Io
t-----11--
---
- - 1 - - - 0 V0

--
I; R, V.
~ I f3 = 110
+ I
o.6 ill 1 µF
Zo Ri 4.7k0
vs '\, 0.82kO

t T' "="'

FICi. 5.174
Problem 40.

*41. For the network of Fig. 5.175:

a. Determine A"NL, Z;, and Z 0 •
b. Sketch the two-port model of Fig. 5.63 with the values determined in part (a).
c. Determine AvL and Av,
d. Change Rs to 1 kil and determine AvL and Av, What is the effect of increasing levels of Rs
on the voltage gains?
e. Change Rs to 1 kil and determine AvNL, Z;, and Z 0 • What is the effect of increasing levels of
Rs on the parameters?
f. Change RL to 5.6 kil and determine AvL and Av, What is the effect of increasing levels of
RL on the voltage gains? Maintain Rs at its original level of 0.6 kil.
I
g. Determine A; = ~ with Ri = 2.7 kil and Rs = 0.6 kil.
I;
BJT AC ANALYSIS 20V

6.8 kn
91kO

~- 1:056µF! o V"

12kO
l.2kn kn

-=-
FIG. 5.175
Problem 41.

*42. For the common-base network of Fig. 5.176:

a. Determine Z;, Z 0 , and AvNL"
b. Sketch the two-port model of Fig. 5.63 with the parameters of part (a) in place.
c. Determine AvL and Av,
d. Determine AvL and Av, using the r, model and compare with the results of part (c).
e. Change Rs to 0.5 kil and RL to 2.2 kil and calculate AvL and Av, What is the effect of
changing levels of Rs and RL on the voltage gains?
f. Determine Z 0 if Rs changed to 0.5 kil with all other parameters as appearing in Fig. 5.176.
How is Z 0 affected by changing levels of Rs?
g. Determine Z; if RL is reduced to 2.2 kil. What is the effect of changing levels of RL on the
input impedance?
h. For the original network of Fig. 5.176 determine A;= I0 /I;.

6V -22V

2.2kO 4.7kn
a=- 1
~ R, 4.7µF V, 4.7µF Io

~
-
Vo
+
v, '\, Z; -- zo

J -=- -=-
FIG. 5.176
Problem 42.

5.16 cascaded Systems

*43. For the cascaded system of Fig. 5.177 with two identical stages, determine:
a. The loaded voltage gain of each stage.
b. The total gain of the system, Av and Av,
c. The loaded current gain of each stage.
d. The total current gain of the system A;L = I0 / I;.
e. How Z; is affected by the second stage and RL.
f. How Z 0 is affected by the first stage and Rs.
g. The phase relationship between V0 and V;.
 I; R l µF lµF Io
~ s ~ Vo

+ ~ CE amplifier
Z; - 1 kQ
CE amplifier
Z; - lkQ
vs '\J ---+- Z 0 - 3.3 kQ
_.,_ RL 2.7kQ
Z; Z0 - 3.3 kQ
Zo
AVNL =-420 AVNL =-420

"II' "II'

FICi. 5.177
Problem 43.

*44. For the cascaded system of Fig. 5.178, determine:

a. The loaded voltage gain of each stage.
b. The total gain of the system, AvL and Av,-
c. The loaded current gain of each stage.
d. The total current gain of the system.
e. How Z; is affected by the second stage and RL-
f. How Z 0 is affected by the first stage and Rs.
g. The phase relationship between V0 and V;.

I; Rs V

+ ~ · Vo
Emitter - follower CE amplifier
1 kQ lOµF Z; = 50kQ Z; - 1.2kQ
vs '\J ---+- zo - 20 Q Z0 = 4.6 kQ
_.,_
Z; Zo
AVNL "' 1 AVNL - -640

"II' zo) Z;2 "II'

FICi. 5.178
Problem 44.

45. For the BIT cascade amplifier of Fig. 5.179, calculate the de bias voltages and collector current
for each stage.
46. a. Calculate the voltage gain of each stage and the overall ac voltage gain for the BIT cascade
amplifier circuit of Fig. 5.179.
b. FindA;T = 10 /I;.

+15V

5.1 kQ 24kll 5.1 kll

24kll 0.5µF
------11(1------<o Vo

0.SµF
V; o---:--)t-----+---- /3 = 150 /3 = 150
25 µV ~
I

6.2 kll 6.2kll

+
I'°µF l.5kll

"I" ... "I" "I"

FICi. 5.179
Problems 45 and 46.
BJT AC ANALYSIS 47. For the cascade amplifier circuit of Fig. 5.180, calculate the de bias voltages VB!' VB 2 , and Vc 2•
*48. For the cascade amplifier circuit of Fig. 5 .180, calculate the voltage gain Av and output voltage V0 •
49. Calculate the ac voltage across a 10-kil load connected at the output of the circuit in Fig. 5.180.

+20V

1.5 kQ
1 µF
7.5kQ
(----o va

r
50 µF Qz
/3=200

6.2kQ
10 µF Q1
V- ~ /3 = 100
lO~V

3.9kQ

-=-
lkQ

-=-
r lOOµF

FIG. 5.180
Problems 47 and 49.

5.17 Darlington Connection

50. For the Darlington network of Fig. 5.181:

a. Determine the de levels of VB 1, V c 1, VE2 , V cB 1, and VcE2-
b. Find the currents /Bl' IB 2 , and h 2-
c. Calculate Z; and Z 0 •
d. Determine the voltage gain Av = VJV; and current gain A; = / 0 /I;.

+16 V

2.4M!.1

0.1 µF
V; o __._ ) 1-----+-----11
I;

Io i ----11(1------<o V0
+
lOµF
510!.1

FIG. 5.181
Problems 50 through 53.
51. Repeat problem 50 with a load resistor of 1.2 kil.
52. Determine Av = V0 /Vs for the network of Fig. 5 .181 if the source has an internal resistance of
1.2 kil and the applied load is 10 kil.
53. A resistor Re = 470 n is added to the network of Fig. 5.181 along with a bypass capacitor
CE = 5 µ,F across the emitter resistor. If /3v = 4000, VBEr = 1.6 V, and r01 = r02 = 40 kil
for a packaged Darlington amplifier:
a. Find the de levels of VB 1, VE2 , and VcEz·
b. Determine Z; and Z 0 •
c. Determine the voltage gain Av = V0 /V; if the output voltage V0 is taken off the collector
terminal via a coupling capacitor of 10 µ,F.
5.18 Feedback Pair
54. For the feedback pair of Fig. 5.182:
a. Calculate the de voltages VB,, VB2 , Ve,, Ve2, VE,, and VEz·
b. Determine the de currents /Bl' lei' IB2 , le2 , and /e2 •
c. Calculate the impedances Z; and Z 0 •
d. Find the voltage gain Av = V0 /V;.
e. Determine the current gain A; = /0 /I;.

+16 V

68 Q

I; . ~-----+------1{--o V 0

V; o ) t---+-----t1
/31=160
/32 = 200
..,_
l.5MQ zo

FIC. 5.182
Problems 54 and 55.

55. Repeat problem 54 if a 22-il resistor is added between VE2 and ground.
56. Repeat problem 54 if a load resistance of 1.2 kil is introduced.

5.19 The Hybrid Equivalent Model

57. Given fe(dc) = 1.2 rnA, /3 = 120, and r0 = 40 kil, sketch the following:
a. Common-emitter hybrid equivalent model.
b. Common-emitter r, equivalent model.
c. Common-base hybrid equivalent model.
d. Common-base r, equivalent model.
58. Given h;, = 2.4 kil, hfe = 100, h,e = 4 X 10-4 , and h 0 , = 25 µ,S, sketch the following:
a. Common-emitter hybrid equivalent model.
b. Common-emitter r, equivalent model.
c. Common-base hybrid equivalent model.
d. Common-base r, equivalent model.
59. Redraw the cornrnon-ernitter network of Fig. 5.3 for the ac response with the approximate
hybrid equivalent model substituted between the appropriate terminals.
60. Redraw the network of Fig. 5.183 for the ac response with the r, model inserted between the
appropriate terminals. Include r0 •
61. Redraw the network of Fig. 5.184 for the ac response with the r, model inserted between the
appropriate terminals. Include r0 •
62. Given the typical values of h;, = 1 kil, h,, = 2 X 10-4 , and Av = -160 for the input con-
figuration of Fig. 5.185:
a. Determine V0 in terms of V;.
b. Calculate h in terms of V;.
c. Calculate lb if h,, V 0 is ignored.
d. Determine the percentage difference in h using the following equation:
/b(without h,,) - /b(with h,,)
% difference in lb = - - - - - - - - - - - X 100%
/b(without h,,)
e. Is it a valid approach to ignore the effects of h,, V0 for the typical values employed in this
example?
 Vee

Re
VEE -Vee
RB C

r,
Vo RE Re
Cc Cc Cc

r
Vo
B
+
E RL
vs '\, + RL
RE
'\,
l
vs

-=- -=-
.J,. -=- -=-
FICi. 5.183 FICi. 5.184
Problem 60. Problem 61.

+
0
-Ib
h;e
VV\,
lkQ
I +

V; h re V,, '\, 2 X [Q- 4 V,,

FICi. 5.185
Problems 62 and 64.

63. Given the typical values of RL = 2.2 kil and hoe = 20 µ,S, is it a good approximation to
ignore the effects of 1/hoe on the total load impedance? What is the percentage difference in
total loading on the transistor using the following equation?
RL - Rd (l/h 0 e)
% difference in total load = - - - - - - X 100%
RL
64. Repeat Problem 62 using the average values of the parameters of Fig. 5.92 with Av = -180.
65. Repeat Problem 63 for RL = 3.3 kil and the average value of hoe in Fig. 5.92.
5.20 Approximate Hybrid Equivalent Circuit
66. a. Given /3 = 120, re = 4.5 n, and r0 = 40 kil, sketch the approximate hybrid equivalent
circuit.
b. Given h;e = 1 kil, h,e = 2 X 10-4 , hfe = 90, and hoe = 20 µ,S, sketch the re model.
67. For the network of Problem 11:
a. Determine re.
b. Find hfe and h;e•
c. Find Z; and Z 0 using the hybrid parameters.
d. Calculate Av and A; using the hybrid parameters.
e. Determine Z; and Z 0 if h0 e = 50 µ,S.
f. Determine Av and A; if hoe = 50 µ,S.
g. Compare the solutions above with those of Problem 9. (Note: The solutions are available in
Appendix E if Problem 11 was not performed.)
68. For the network of Fig. 5.186:
a. Determine Z; and Z 0 •
b. Calculate Av and A;.
c. Determine re and compare f3re to h;e•
 18 V

2.2 ill
68 ill
i 10
( o vo
l;
5µF
V; 0 )
hfe = 180
~
5µF z0
h;e = 2.75 ill
hoe= 25 µS

---➔- 12 ill
Z; 1.2 k.Q lOµF

FIC. 5.186
Problem 68.

*69. For the common-base network of Fig. 5.187:

a. Determine Z; and Z 0 •
b. Calculate Av and A;.
c. Determine a, {3, r,, and r0 •

hfb= -0.992
h;b = 9.45 Q
hob= l µAJV
l;
---➔-
0 ) ~---11(----- o
+ 10 µF lo 10 µF +
1.2 k.Q

V;
---➔-
Z; -=- 4V

-=-
FIC. 5.187
Problem 69.

5.21 Complete Hybrid Equivalent Model

*70. Repeat parts (a) and (b) of Problem 68 with hr, = 2 X 10-4 and compare results.
*71. For the network of Fig. 5.188, determine:
a. Z;.
b. Av.
c. A;= 10 /I;.
d. Z 0 •
*72. For the common-base amplifier of Fig. 5.189, determine:
a. Z;.
b. A;.
c. Av.
d. Z 0 •
BJT AC ANALYSIS 20V

2.2ill
470ill
i 10
t------1(1--------,0 V0
l
1 ill --+- 5 µF
~
hf,= 140
+ ~µ~F---<+>---------11 zo h;,=0.86 ill
h,, = 1.5 X 10-4
h ,= 25 µS

I
0

V, '\J V;

-1 "II" ""="'
l.2ill

"="'
10 µF

FIG. 5.188
Problem 71.

h;b = 9.45 Q
hfb= -0.997
h0 b=0.5 µAN
h,b = l X 10- 4

1 - - - - 1( 1 - - - - - 0
~ - - - - - - 4>-+--
0

5 µF +

V, '\J ---+-
Z;
V;
-=-4v -;;;;-14V
+
"II"

FIG. 5.189
Problem 72.
5.22 Hybrid rr Model
73. a. Sketch the Giacoletto (hybrid 1r) model for a common-emitter transistor if rb = 4 n,
C7T = 5 pF, Cu = 1.5 pF, hoe = 18 µ,S, /3 = 120, and re = 14.
b. If the applied load is 1.2 kn and the source resistance is 250 n, draw the approximate
hybrid 1T model for the low- and mid-frequency range.

5.23 Variations of Transistor Parameters

For Problems 74 through 80, use Figs. 5.124 through 5.126.
74. a. Using Fig. 5 .124, determine the magnitude of the percentage change in hfe for an le change
from 0.2 mA to 1 mA using the equation
h1,(0.2 mA) - h1eCl mA)
I
% change = - - - - - - - - - X 100%
I
h1,(0.2mA)
b. Repeat part (a) for an le change from 1 mA to 5 mA.
75. Repeat Problem 74 for h;e (same changes in le)-
76. a. If hoe = 20 µ,S at le = 1 mA on Fig. 5.124, what is the approximate value of hoe at
le= 0.2mA?
b. Determine its resistive value at 0.2 mA and compare to a resistive load of 6.8 kn. Is it a
good approximation to ignore the effects of 1/hoe in this case?
77. a. If h0 e = 20 µ,S at le = 1 mA of Fig. 5.124, what is the approximate value of h0 e at
le= lOmA?
b. Determine its resistive value at 10 mA and compare to a resistive load of 6.8 kn. Is it a
good approximation to ignore the effects of I/hoe in this case?
78. a. If h,e = 2 X 10-4 at le = 1 mA on Fig. 5.124, determine the approximate value of h,e at
0.1 mA.
b. For the value of h,e determined in part (a), can h,e be ignored as a good approximation if
Av= 210?
 79. a. Based on a review of the characteristics of Fig. 5.124, which parameter changed the least
for the full range of collector current?
b. Which parameter changed the most?
c. What are the maximum and minimum values of l/h0 ,? Is the approximation l/h 0 , II RL ~ RL
more appropriate at high or low levels of collector current?
d. In which region of current spectrum is the approximation h,, Vee ~ 0 the most appropriate?
80. a. Based on a review of the characteristics of Fig. 5.126, which parameter changed the most
with increase in temperature?
b. Which changed the least?
c. What are the maximum and minimum values of h1,? Is the change in magnitude signifi-
cant? Was it expected?
d. How does r, vary with increase in temperature? Simply calculate its level at three or four
points and compare their magnitudes.
e. In which temperature range do the parameters change the least?

5.24 Troubleshooting
*81. Given the network of Fig. 5.190:
a. Is the network properly biased?
b. What problem in the network construction could cause VB to be 6.22 V and obtain the given
waveform of Fig. 5 .190?

Vee= 14 V

Re 2.2 kQ v, (V)
lSOkQ
10 µF
0 ( ovo 0
V8 =6.22 V C2
{3=70

39kQ
+
V8 E=0.7V
J 't

RE l.SkQ 10 µF

FICi. 5.190
Problem 81.

5.27 Computer Analysis

82. Using PSpice Windows, determine the voltage gain for the network of Fig. 5.25. Display the
input and output waveforms.
83. Using PSpice Windows, determine the voltage gain for the network of Fig. 5.32. Display the
input and output waveforms.
84. Using PSpice Windows, determine the voltage gain for the network of Fig. 5.44. Display the
input and output waveforms.
85. Using Multisim, determine the voltage gain for the network of Fig. 5.28.
86. Using Multisim, determine the voltage gain for the network of Fig. 5.39.
87. Using PSpice Windows, determine the level of V0 for V; = 1 mV for the network of Fig. 5.69.
For the capacitive elements assume a frequency of 1 kHz.
88. Repeat Problem 87 for the network of Fig. 5.71.
89. Repeat Problem 87 for the network of Fig. 5.82.
90. Repeat Problem 87 using Multisim.
91. Repeat Problem 87 using Multisim.