G't; 't-'?

4) -It S~
H-W:ti=' I ('-; tj 'U.)
6.1 1. Active 6.6 Old technology:
2. Saturation 10- 3 = 2 x lO' J5 e VsEiVT
3. Active
4. Saturation VSE = 0.025 In ( ,.2 x10-3 )
, 15. = 0.673 V
10-
5. Active
6. Cutoff New technology:
10-3 2 x 1O-18eVllElVT
G h e EB junctions have a 4: 1 area ratio.
10-3 )
Ie = /seVBE/VT VBE = 0.025 In ( 2 x 10- 18 0.846 V

0.5 x 10- 3 lSI x

=> lSI = 4.7 X 10- 17 A

6.7 5 x 10- 3 I s eo,7 6 /Q,025 (I)
16
/s2 = 4Ln 1.87 x 10- A
Ie /se ,70/0,025
Q
(2)
Dividing Eq. (2) by Eq. (1) yields
6.3 Ie IseVsEIVT
Ie 5x
200 x 10- 6 = Ise 30
0,45 rnA
=> Is = 1.87 X 10- 11 A
For Ie = 5 f.LA,
For the transistor that is 32 times larger,
(3)
Is 32 x 1.87 x 10- 17
Dividing Eq. (3) by Eq. (1) yields
= 6 X 10- 16 A 10- 3
At VilE = 30 VT, the larger transistor conducts a
current of
VilE 0.76 + 0.025 In(1O-3 )
= 0.587 V
Ie 32 x 200 ILA = 6.4 rnA

At Ie = 1 rnA. the base-ernitter voltage of the

larger transistor can be found as
6.8 Ia 10 p.A
1 x 10- 3 = 6 x 1O~J6eVBE;VT

Ie = 800 ILA
3
Ix 10- ) Ie
VBE = VT In ( 1
6 x 10- 6
= 0.704 V Ii = - = 80
1/3

= 80 = 0.988

IS! AEI
6.4 - = - - =
200 x 200
250,000
Ii + I 81

IS2 A£2

Ie! ISlevYElIVT

6.9
Ie2 Is2 e VEn/ V,

For In = Iez we have

lSI
250,000
IS2

V ilE2 - V ilEI = 0.025 In(250,000) 6.10

0.31 V

6.5 let = 1O-13 e 700/25 0.145A 145 rnA

Iez = 1O-18e700/25 1.45 f1A

6.11 Ii = (1)
For the first transistor 1 to conduct a current of I-a
1,45 p.A, its VBE must be a -+ a + D.a
6
1.45 x 10- )
VBE1 0.025 In (
-' 10-13­

= 0,412 V
(2)
 Chapter 6-2

Subtracting Eq. (1) from Eq. (2) gives 6.14 For i8 = 10 J..LA,

6fi
a+6a
~.~-- .. -- a
.. ~
Ie iE i8 = 1000 - 10 = 990 J..LA

a 6.a a
f! = Ie := 990 = 99

i8 10

6.fi (3)
- a - 6a)(l a)
iY = ~ =0.99

Dividing Eq. (3) by Eq. (1) gives Ii + 1 100

For in 20 J..LA,

ie = iE - iE = 1000 20 = 980 J..LA

For 6a 1, the second factor on the right-hand Ii = is;. = 980 = 49

side is approximately equal to f!. Thus in 20

49
6fJ : : :. f3 (6iY.) a = - = 0.98
Q.E.D. fi + 1 50

fJ a

For in = 50 J..LA,

613

For /1- := - 10% and Ii 100, ie = iE is 1000 50 950 J..LA

ie 950

-6a -10%
: : :. -~- = -0.1% 13
iE 50

= 19
a 100
19

a 0.95

l'i + 1 20

6.12 Transistor is operating in active region:

13 = 50 -4300 EseeTa~
In = 10 J..LA
6.16 First we determine Is, f3, and a:

Ie = filn 0.5 rnA -4 3 rnA

1 x 10"3 = fse 700/25

Ie = (f3 + l)18 0.51 rnA -4 3.01 rnA

=} Is = 6.91 X 10- 16 A

Maximum power dissipated in transistor is

Ie 1 rnA

fi 100

In x 0.7 V + Ie x Ve 18 10 J..LA

100
0.01 x 0.7 +3 x 10:::::. 30mW a = = --. = 0.99
li+l101

Then we can determine ISE and IsB :

6.13 ie IsevBE/VT
5 xlO-15eO.7/0.025 = 7.2 rnA
ISE Is 6.98 X 10- 16 A

= IiIs = 6.91
.. 7.2 7.2

iB w1ll be m the range 50

rnA to 200 rnA, that is, IS8 X 10-
18
A
144 IlA to 36 J..LA
The figure on next page shows the four
iE will be in the range (7.2 + 0.144) rnA to large-signal models, corresponding to Fig. 6.5(a)
(7.2 + 0.036) rnA, that is, 7.344 rnA to 7.236 rnA. to (d), together with their parameter values,

This table belongs to Problem 6.15.

 Chapter 6-9

G (a)

8) Ie a 0.26
0.255 mA
+1.5 Y

2.7 kn
Ve 1.5 0,255 X 2.7 0.81 V
@In IE
---3>--

..L +
-=- 0.8V
CD r
IE
-O.g - (-1.5)
2,7
0.26mA
tr 2.7kn
VE =-0,8V(D

15 V

1.5 V
(b)
- 1.5 - 0.8
I E- 2
= 0.35 rnA -'-0.8Y G)

Vc -1.5 + 2 x 0.343
f7\ Ie ~ x 0,35
0' f -0.81 Y G)
\V = 0.98 x 035
0= 0.343

-1.5V
(c) +3Y
3 - 1.8 .I.
IE 10 = 0,12mA 'f
10kH

+1 V CD 1.8 Y CD
A O,8Y
..0­
le/50 = 2.4 f.LA

o Ie =
=
0' X 0.12

0,98 x 0,12

=0,118mA

-'-3Y
(d)
t
Ve 3 0,147 x 8,2 1.8 V
L5Y
~

3 f.LA ---3>-­ +
Is
0.8V
V CD 1.5 - 0.8 = 0.7 CD
IE=~
4.7

0.7
4.7 = ° 15mA
., -=­

In all circuits shown in P6,35, we assume indicated on the conesponding circuit diagrams:
active-mode operation and that this is the the order of the steps is shown by the circled
case at the end of the solution, The solutions are numbers,
 Chapter 6-16

Ga) Vs=OV (c)

-1-5 V
+5V

~
1.3 = 2.7 rnA ®
......---.--DVe = 2.7 X I
2.7V®

~0.7V CD
1 kO.

~5V

Figure 1
Figure 3

The analysis is shown on the circuit diagram in

Figure 3 shows the transistor at the edge of
Fig. L The circled numbers indicate the order of
saturation. Here VCE = 0.3 V and
the analysis steps.
Ie ale :::::: Ie. A node eqnation at the emitter
(b) The transistor cuts off at the value of VB that gives
causes the 2-mA current of the current source
h = 2 + VB 0.7 VB + 1.3 rnA
feeding the emitter to flow through the l-krt
resistor connected between the emitter and A node equation at the collector gives
ground. The circuit under these conditions is
shown in Fig. 2. Ie = 4 (VB 0.4) = 4.4 VB rnA

Imposing the condition Ie :::::: Ie gives

+5V
4.4 VB VB + 1.3
=} VB +1.55 V

4mA Correspondingly,

~~F'OVc +4V VE +0.85 V

Ve = +1.15 V
IkD.

VB = -1.5V 6.55

0.1 mA t
lkD.
-5V
VB = +1.2 V
Figure 2

Observe that VE = -2 rnA xl kQ -2 V,

IkD.
h = 0, and VB = VE + 0.5 1.5 V. Since
Ie 0, all the 4 rnA supplied by the current
source feeding the collector flows through the
collector l-kQ resistor, resulting in Ve +4 V Figure 1
 Chapter 6-20 f,5

~
~ +3V (c) +3V

Ie ==.0.5 mAt+ 3.6 kf!

t
V7 = 3 - 0.5 X 3.6

~
V2=3-0'5X3'6
= +L2V
43 kfl -r1.2 V

JO~
-1 ",,0 OV
8
4.7kfl
I ,}IE
0.5 rnA t 4.7
-3V 0.5 rnA

(b) (d)
+3V
+3V

t~'2kD v8-
' - 0..T),O..7

-
V, 3 - 0.5 X 3.6 +0.75 V
1,45 V
= +1.2 V t'" OmA

V9 =, -3+ 0.25 x 10
VE -0.7V -O.5V
lOW
+4.7kfl .
t
! 14 = -0.7 - (-3) = 0.5 rnA
t 4.7 --3 V
-3V

Ce) +3V

1
480
0.0125 rnA
t "--~= = 0.25 rnA

VIl = 0.75 + 0.7 1.45 V

VIO = -3 + 0.0125 X 300

= +0.75 V
300kfl

-3V

For the solutions and answers to parts Ca) through

(e), see the corresponding circuit diagrams.
 Chapter 7-19
f)
Substituting for i from Eq. (2) into Eq. (I) yields @ R e f e r to the circuit of Fig. P7.125.
CI(VBS - VBE )
Ie = ------O~--

Thus RE +
fl+l
Vo gmR~
where
V'ig 1 + flgmR~
R2 15
I/fl VBB Vee - - - = 15 x - - - = 5.357 V
R2 + RI 15 + 27
RB = Rl !i R2 = 151127 = 9.643 kQ
_ 0.99(5.357 -- 0.7) _ 1 8
I+(RdR 1) Ie - 9.643 - . 5 mA
Q.E.D (3)
1 + 1 + RdR 1­ 2.4+ 101

, gmR~

Ie 1.85 mA
74 rnA/V
The input resistance Rin can be obtained as gm = Vr = 0.025 V
follows:
100
1.35 kQ
gm 74
Substituting for i from Eq. (1) yields Replacing the BJT with its hybdd-7T model
results in the equivalent circuit shown at the
bottom of the page:

Rin Rl II R211 rn RR II r;r 9.643 il1.35

and replacing by the inverse of the gain
Va
= 1.18kQ
expression in Eq. (3) gives
VIi Rin 1.18
- - - =0.371 V/V
Rin = RD[ + T+ (;dRl) ] Vsig Rill + R,lg 1.18 + 2

Rin 1 [ 1+ RI ] Q.E.D
gm Rj +R2 -74(39112) = -97.83
(d) Substituting numerical values:
110 = -0.371 x 97.83 -36.3 V IV
'Vsig

1 + 1 x (82.4 111000) 7.126 Refer to the circuit of P7.125.

2 DC design:
-._-:::;-- = L95 VIV

1 + 76.13
VB = 5 V, VBE = 0.7 V

Note that the gain::::: 1 +

Rl
-=R,
= 2, similar to that
VE

For
= 4.3 V

of an op amp connected in the noninverting

configuration! h = 2mA, 2.15 kQ

f[ 05
1 + 1 x (82.41: 1000) -c:-_.--:--: ] 5
25 kQ
0.2
= 39.1 kQ
h 2
IR = - - = - ::::: 0.02 rnA
{3+1101
This figure belongs to Problem 7.125.
R sig

+
V 1T rw g",v"rr Re RL

- - 1- 1-
 Chapter 7-52

This figure belongs to Problem 7.130.

,1

G e f e r to the circuit of Fig. 1'7.130.

Ie 0.1 rnA
gm = V c::: 0.025 V = 4 mA/V
T
(a) DC analysis of each of the two stages:
Note that the emitter has a resistance
Ro 47 Re 250 Q.
V --"-­ ---=4.8V
ee Rl +R2 ­ 100+ 47

Rin 200 kQ I (fJ + 1)(re + R,,)

RB=R11IR2 1001147=32kQ

= 200 11[101 x (0.25 + 0.25)]

200 I 50.5 40.3 kQ
'{)h Rio 40.3
- - - = 0.668 V IV
0.7 V'ig Rio + R,ig 40.3 + 20
--~~ = 0,97 rnA c::: rnA

Vo Total resistance in collector

3.9+­ - =-Ci
101 Vb Total resistance in emitter
Ie = alE c::: 1 rnA 20120
-20 V/V
0.25 + 0.25
Ve Vee - [eRe = 15 1 x 6.8 = 8.2 V
71
(b) See figure above. G" 0 = -0.668 20 -13.4 V IV
VEig

Ie For Vbe to be limited to 5 mV, the signal between

gm = - 40mA/V
VT base and ground will be 10 m V (because of the
5 m V across Re). The limit on Vs;g can be obtained
r" f3 = 2.5 kQ by dividing the 10 mV by

(c) RillI Rl II R211 r" Rs II rJ( = 32112.5 WmV

=15mV
0.668
= 2.32kQ
Correspondingly, at the output we have
VbI Rin 2.32
- - - = 0.32 V IV 13.4 x 15 = 200mV = 0.2 V
Vsig Kn + 2.32 + 5

(d) Rin2 Rl II R211 Y" Rio! = 2.32 kQ

= -40(6.8112.32) = -69.2 V IV

= -40(6.8112) = -61.8 V jV
Vo Va 11h2 Vbl
(f ) - = x-x -61.8 200kD
Vsig Vb2 Vbi Vsig

x 69.2 X 0.32 1368.5 V IV

7.131 Refer to the circuit in Fig. 1'7.131:

h = 0.1 rnA
25mV
= 250Q
0.1 rnA Figure]
 Chapter 7-53

From 1 we see that

Ie 0.495 rnA

Vc = IB X 200 kQ +h x 0.2 kQ + VBE

= 0.005 x 200 + 0.5 x 0.2 + 0.7
= 1.18 V
(b)

v" - v, ~,
200

200kfi

-Vsig
Q 0.1 leQ
At the output node,
Vo = -ai,(5 !I 100)
0.1 (5 i1100)

5 II 100
1J ig = a 0.1 ~ 47.6 V IV
Figure 2 s

From Fig. 2, we have

0.495
0.025 ::::: 20 rnA/V

2.3
-~~=0.78mA
1 +-.
51
VE = !eRE = 0.78 V
0.25 kQ
VB = VE +0.7 = 1.48 V
Node equation at the output:
I') 200:
Vo • ai + v" - v, _ 0
20 200­ 2.3
T
h = ---== = 1.54 rnA
1 •
Vo + 0.99 x 4v +~ - 2 = 0 T 201
20 ' 200 200
1 VE = hRE 1.54 V
VO (2 0+ ~O) = -V; (4 x 0.99 - 2~()) VB = VE + 0.7 2.24 V
v
o = -71.9 V/V
(b) Rill 100 Ii (fJ + 1)[re + (\ ,II)]
Vi

100" (/') + 1)(r, + 0.5)

fJ = 50:
7,133 Refer to the circuit in Fig. P7 .133.
VT 25 mV
The de emitter current is equal to 0.5 rnA, and r, = - = --- 32.1 Q
h 0.78mA
Ie a h ::::: 0.5 rnA; also,
Rill 100" [51 x (0.0321 + 0.5)J
V 25 mV
re = -T = - - = 50 Q 21.3 kQ
h 0.5 mA
Rill re = 50 Q fJ = 200;
VI' 25 mV
re = - = -'-- = 16.2 Q
Iii 1.54 rnA
 Chapter 7-54

~Lp;I~
~Rn = 100 II [201 x (0.0162 + 0.5)] 130.4
= 0.964 x 65.2
Vh 2
50.9 kQ
62.9 A/A
• 100

Rout 3.3!1 (re + fJ + 1

Va (l111) 500
- - - e r e in Q) 100)
Vh (1 111) + re 500 + r" = 3.3 II ( 0.0463 + Wi
{) 50:
= 0.789 kQ 789 Q
Vb
21.3 = 0.68 V V
Vsig 21.3 + 10 /
7.136 Refer to the circuit in Fig. P7.136.
Va 500
0.94 V/V For dc analysis, open-circuit the two coupling
Vb 500 + 32.1
capacitors. Then replace the 9-V source and the
Va
0.68 X 0.94 0.64 V/V two 20-kQ resistors by their Th6venin equivalent,
Vsig namely, a 4.5-V source and a 10-kQ series
fJ = 200: resistance. The latter can be added to the lO-kQ
resistor that is connected to the base. The result is
50.9 the circuit shown in 1, which can be used to
-=-50--.9-+-1-0 = 0.836 V /V
Vsig calculate IE.
Vo 500
= 0.969 V /V
50O + 16 .2 +9V

+4~

Va
- = 0836 X 0,969 = 0.81 V /V
Vsig

7.135 Refer to the circuit in Fig. P7.135.

h=
3 0.7
11£
[nn
3.3 + fJ + 1
2.3

-----,1""'0"0:"0 = 0.54 rnA

Figure 1
3.3 + 101

25mV
4.5 - 0.7
- - - =46.3Q (a) Ie
0.54 rnA 20

2+-­

Rin = (fJ + l)[re + (3.3 12)] fi+l

101 X (0.0463 + 1.245) 3.8
1.73 rnA
20
= BOA kQ 2+
Wi
13004 Ie ale = 0.99 x 1.73 rnA
13004 + 100
= 1.71 rnA
0.566 V/V Ie
gm = V 6804 mA/V
Va 3.3112 T

Vb (3.3 112) + re re VT = 25 mV = 14.5 Q

= 0.964 V/V Ie 1.73 rnA
= 0.0145 kQ
~ = 0.566 0.964 = 0.55 V IV
Vsig r" = (fJ + l)re = 101 x 0.0145
1.4645 kQ
2kQ
. Vi l)b
(b) Replacing the BJT with its T model (without
l[=- ro) and replacing the capacitors with short circuits
Rn BOA kQ