Handout 2 for EE-203
Bipolar Junction Transistor (BJT)
Sheikh Sharif Iqbal
(Ref: Text book and
KFUPM Online course of EE-203)
(Remember to solve all the related examples,
exercises problems as given in the Syllabus)
Chapter 5: Bipolar Junction Transistor (BJT)
Text book: “Microelectronic Circuits by Sedra and Smith
5.1: Device Structure and Physical Operation
• BJT is a three terminal device that can operate as “Amplifier” or as “Switch”
• Voltage between the two terminals is used to control the current in the third terminal
• BJT consist of three semiconductor regions: NPN or PNP
Table 5.1: BJT Modes of operation
Application Mode EBJ CBJ
Amplifier Active Forward Reverse
Cutoff Reverse Reverse
Switch
Saturation Forward Forward
VBE VCB
C C
Collector
Emitter B B
E E
Base
Figures from text book Symbols: NPN and PNP
Active Mode of Operation of an NPN transistor:
N-Type material: Arsenic, Antimony and Phosphorus (group V materials)
P-Type material: Aluminum, Boron and Gallium (group III materials)
Figure from text book
Heavily Lightly doped
doped
Fig 5.3: Forward current flow in an NPN transistor biased to operate in the active mode
(Very small reverse current, due to drift of thermally generated
minority carriers, are not shown.)
• The EMITTER is heavily doped and have high density of electrons. But the
base is thin, lightly doped and has low density of holes. So, the current flow (iE)
between the forward biased emitter-base junction is mainly due to electronics
flowing from emitter to base. This process emits free electrons into the base.
• Among the emitted electrons in BASE, around 5% recombines with available
holes and escapes into external base lead as iB2, (iB=iB1+iB2 Î iB1 is due to majority holes)
Remaining 95% base electrons acts as a minority carriers and are swept away to
collector region by the electric field of reverse biased collector-base junc. (fig)
• These electrons are then collected by more positive collector terminal that
constitute collector current (ic). THUS: (iE) = (iB) + (iC)
Base Collector
Reversed Biased C. B. J
The Collector Current (iC): Base Current (iB): Emitter Current (iE):
v BE v BE
V T iC IS VT
i C I S⋅e iB ⋅e iE = iC + iB
β β
• Here, n = 1; Saturation current, 10-12> IS>10-14 A and thermal voltage, VT = 25 mV
• Note that ‘iC’ is independent of VCB , for VCB ≥0. So collector behaves as an
ideal constant current source where the current is determined by VBE. (fig 2nd slide)
• Since, iC = αiE ; iC = βiB and α= β/(β+1),
where “β” is common emitter current gain constant for a particular BJT
“α” is common base current gain
• So the Emitter Current is given by:
Exercise BJT-1: For an NPN
⎛ vBE ⎞
⎜ transistor having Is=10-11 A,
β=100 &VT=25mV(at room temperature)
β+1 β+1 ⎜ VT β
iE iC + iB ⋅ iC ⋅ IS⋅ e i α ⋅ I BE α C
β ⎝ ⎠
Calculate V for i =1.5A
C E
β β+1
(Solution: 0.643 V)
The PNP Transistor operation in Active mode:
Figure from text book
Note that the PNP transistors have VEB ; whereas the NPN transistors has VBE
5.2.3: Dependence of current, voltage, temperature and the Early Effect:
0.7v
Exercise BJT-2: If a BJT has
VA=100v and IC=1mA, find r0
5.3.3: Q-point/Biasing point: Location of biasing point affects maximum allowable signal swing
Limited + swing of vce as Q-
point (iC & vCE) is close to vCC
(this case is for Low Rc
Q-point
+ swing
(See figure 5.26)
5.4: How to solve NPN or PNP BJT-DC circuits
Assume
Active
mode
n CBJÎRB
EBJÎFB
p =0.7v
n
Solution:
- Assume Active Mode
of operation Î VBE=0.7
n
p
Solve
- Since CBJ is RB,
examples 5.4 to 5.12 - Initial assumption was correct
5.5: Biasing single stage BJT Amplifiers: - Operating point or Q-point (iC and vCE)
Assume
Active mode
Voltage Divider Biasing
4.75v
Two supply Biasing
VCE=3.57 v
Exercises BJT-3
Exercises BJT-3:
Exercises BJT-4: Solve example 5.13 & related exercises
5.6: BJT Amplifiers, Small-Signal Operation and Models :
DC AC
equivalent equivalent
circuit Î circuit Î
5.6.6 and 5.6.7: Small-Signal Equivalent Circuit Models:
(a) (b)
Two different versions of simplified hybrid-Π model for the small-signal operation of the BJT.
(a) represents the BJT as a voltage-controlled current source ( a transconductance amplifier)
(b) represents the BJT as a current-controlled current source (a current amplifier).
Two slightly different versions of what is
known as the T model of the BJT. The
circuit in (a) is a voltage-controlled
current source representation and that in
(b) is a current-controlled current source
representation. These models explicitly
show the emitter resistance re rather than
the base resistance rΠ featured in the
hybrid-π model.
(a) (b)
Figures from text book Solve example 5.14
5.7.3:Common emitter Amplifier: DC analysis to find IB, IC, IE, VB, VC and VE
IC
IB VC
VB
β
IE
VE
Lets assume the BJT is operating in Active Mode. Thus, IC= α.IE , IC= β.IB and α=β/(β+1)
Since IE = I mA ; IB = IE / (β+1) mA ; VB = 0 - IB.RB ; VE = VB - 0.7 ;
Now IC= α.IE = (β.IE) / (β+1) ; VC = VCC - IC.RC and if CBJ remains RB then assumption is OK
Exercise-5: Find the operating point if VCC=VEE=10V, RC=8 kΩ, RB=100 kΩ, I=1mA & β=100
Solution: Q or operating point is, IC=0.99 mA ; VCE=0.3 v (as VB= -1v, VC = 2v, VE = -1.7v)
5.7.3:Common emitter Amplifier: AC analysis to find Gain, Input & output Impedances
Figure from text book
Remember,
r0=|VA|/IC
gm=IC/VT
rπ=VT/IB
Exercise-6: Find Rin ,Rout ,Av & Gv; if Rsig=RL=5kΩ, RB=100k, RC=8k; IC=1mA, IB=0.01mA, VA=100V, VT=25mV
Solution: Ri=2.43 kΩ; Rout=7.4 kΩ, Av=-119 V/V, Gv=-39 V/V (as rπ=2.5K, gm=40 mA/V, r0=100k)
Review of CE amp:
- S/C DC voltage source
- O/C DC current source
Find model
parameters
(rπ, re, gm)
DC analysis: O/C capacitors AC analysis: S/C capacitors
& find IB, IC, IE, VB, VC, VE & find Î Rin, Rout, AV, Ai
Exercises-7:
Exercises-8:
5.7.4:Common emitter Amplifier with Re (emitter resistor): Controlled voltage amp
DC analysis IC= α.IE ,
IC IC= β.IB
α=β/(β+1)
and
β
IB gm=IC/VT;
IE re=VT/IE ;
Exercise-9: Find the operating point if VCC=VEE=10V, RC=8 kΩ, RB=100 kΩ, I=1mA & β=100
Solution: operating point, IC=0.99 mA ; VCE=3.7 v (as IE=1mA ; VB= -1v, VC = 2v, VE = -1.7v)
AC analysis
Figures from text book
Remember, re=VT/IE & Resistance reflection ruleÎ Rbase≈(β+1)R emitter
; Remember, gm=IC/VT
(as ro of T-model is neglected to ease solution process)
5.7.3:Common Emitter (CE) with Emitter Resistance (Re): AC analysis (cont’d…)
introduce a negative feedback Î see pg 474
Exercise-10: if IE=1mA, IC=0.99mA, Re=225Ω, RB=100kΩ, RC=8kΩ, Rsig=RL=5kΩ, β=100, VT=25mV,
Neglect ro to FIND Rin , Rout , Av,, Gv Î Sol: Ri=20.16 kΩ; Rout=8 kΩ, Av= -12.18V/V, Gv=-9.76v/v
5.7.5:Common Base (CB) Amplifier: Unity-gain-current-amplifier or Current-buffer
DC analysis
Lets assume Active Mode.
VB = 0 ; VE = VB - 0.7 ;
IE = I mA ; IB= IE / (β+1) mA;
IC= α.IE = (β.IE) / (β+1) ;
VC = VCC - IC.RC and
if CBJ remains RB
then assumption is OK
AC analysis: For the AC equivalent circuit given in the figure in the next page,
5.7.5:Common Base (CB) Amplifier: Low Zin makes it not good voltage amplifier
AC analysis
Figures from text book
Exercise-11: Determine the voltage gain of the circuit given in figure (a)
β=100
Figures from text book
Hints: Draw the DC and AC (using T-model) equivalent circuits (as shown in figure)
The DC solutions are also shown in figure (b). Calculated Î re = 27 Ω
The Gain of the circuit, calculated from figure (c) is, Av= vo/vi = 183.3 V/V
5.7.5:Common Collector (CC) Amplifier: Emitter Follower
DC analysis Assume Active Mode.
IE=I mA; IB=IE/(β+1)mA
VB =0–(IB)(RB) ;
VE=VB - 0.7 ;
IC=α.IE = (β.IE)/(β+1) ;
VC = VCC
if CBJ remains RB
then assumption is OK
Figures from text book
AC analysis
Emitter to Base:
AC analysis
Figures from text book
Base to Emitter
Exercises-12
DCÎ use KVL
the overall voltage gain, Gv Exercises-13 VCC, RB, VBE, RE
Gv
Gv
Exercise-14: For the following circuits, find the expressions for Rin , Rout ,Av
Circuit 1 configuration Circuit 2 configuration
Assignment Problems: 5:21, 5.26, 5.72, 5.83(b), 5.130 ,5.134, 5.135,
5.143 and 5.141 Î Due on next week
Design Criteria of a BJT Amplifier (review):
Figures from text book
Simulation Examples using the Spice software:
