CHAPTER 14
14.1 (a) Common-collector Amplifier (Emitter-follower)
14.1 (b) Common-gate Amplifier
14.1 (c) Common-collector Amplifier (Emitter-follower)
14.1 (d) Not a useful circuit because the signal is injected into the drain of the transistor.
14.1 (e) Common-emitter Amplifier
14.1 (f) Common-source Amplifier
14.1 (g) Common-gate Amplifier
14.1 (h) Not a useful circuit since the signal is being taken out of the base terminal.
1
14.1 (i) Common-source Amplifier
14.1 (j) Common-source Amplifier
14.1 (k) Common-base Amplifier
14.1 (l) Not a useful circuit because the signal is injected into the drain of the transistor.
Note that the resistor labels on RD and R1 were reversed in the first printing.
2
14.1 (m) Common-source Amplifier
14.1 (n) Common-emitter Amplifier
14.1 (o) Common-drain Amplifier (Source-follower)
14.2
3
14.3
14.4
14.5
g R
a Neglecting R OUT : A Vth   m L
1  gm R 5

0.005S2000  5.00 | R  
IN
1  0.005S200
R OUT  r o 1  g m R 5   10k1  0.005S200  20.0 k  2k | A Ith  
b  A Vth  g m R L  0.005S2k 10k  8.33 | R IN   | R OUT  r o  10.0 k | A Ith  
14.6
a g m  0.02S | r  
Neglecting R OUT : A Vth
75
 3750 | R IN  r    o  1R 5  3750  76 200  19.0 k
.02
o R L
75 12k


 46. 3
R th  r    o  1R 5
500  3750  76200



75200

o R 5
R OUT  r o 1 
  100k 1 
  437 k  12k
 R th  r   R 5 
 500  3750  200 
A Ith   o  75
7512k
b  A Vth  
 19.2
500  3750  76560
R IN  r    o  1R 5  3750  76 560  46.3 k | A Ith   o  75



75560

o R 5
R OUT  r o 1 
  100k 1 
  973 k
 R th  r   R 5 
 500  3750  560 
14.7
a For large o : A V  
RL
8.2k 47k

 6.91 | b Place a bypass capacitor in
R5
330  680
parallel with the 330 resistor. Then AV  
8. 2k 47k
RL

 10.3 | c Place a
R5
680
8.2k 47k
 21. 2
330
 10 V CC  120.
bypass capacitor in parallel with the 680 resistor. Then AV  
d Place a bypass capacitor from the emitter to ground. e A V
14.8
R IN  r    o  1R 5  500 k
A Vth  
R th
 oR L
75R L

 10  R L  66.7 k
 r    o  1R 5
100  500k
Assuming  o 1R 5  r  , R 5 
500 k 500 k

 6.58 k
o  1
76
Note: The nearest5% values would be 68 k and 6. 8 k.
14.9
4
R IN  r   500 k
 oR L
75R L
A Vth  

 10 | R L  66.7 k
R th  r 
100  500k
r 
o
 V
 V
75 0. 025V 
 o T | IC  o T 
 3. 75 A
gm
IC
r
500k
5
14.10
i x  g m v gs   g o
 g v
  
m gs  g o

i x   g o
0   g
   o
g o v d 
  | v gs   v s
g o  G5 
v s 
 g m  g o  v d 
 v 
g m  g o  G5 
 s 
 = g oG5 | v d  g m  g o  G5 
R OUT 
ix
g  g o  G5 
 m
ix

g oG5
 g

vd
G 
r 
 R 5 1  m  5   R 5 1   f  o 
ix
go
g o 
R 5 


R OUT  r o  1   f R 5  r o  f R 5  r o 1  g m R 5 
14.11
i x  g 'm v e   g o

  
 g 'm v e  g o
g o
v x 
i x   g
|     o



g o  G5  G1 v e 
0  
g o
 = g o G 5 + G1  | v x 
R OUT 

g 'm
 g o  G5  G1


g 'm

v x 
 
 g o  G5  G1 
v e 
 g 'm  g o

g 'm  g o  G 5 + G1 
ix

ix

g o G5 + G1 



 g '
vx
G  G1 
ro
'
 R 5 R 1 1  m  5
 R 5 R1 
1
g
r



m
o

ix
go
go


R 5 R1 

R OUT  R 5 R1   r o  r o


g m r  R 5 R th  r  
o R 5
 R 5 R1   r o 1 

R th  r  R th  r   R 5
 R th  r   R 5 
14.12
62k
 9. 07V | R EQ  20k 62k  15.1k
20k  62k
12  0.7  9.07 V  7.16A | I  537 A | V  12  3900I  8200I  5. 47 V
IB 
C
EC
E
C
15.1k  75  13.9k
V EQ  12
Forward  active region is correct. | r  
750.025 V
 3.49k
537A
15.1k
 0.938v s | R th  1k 15.1k  0.938k
1k  15.1k
75 7.58k 
R L  r o 8. 2k 100k 8.2k 100k  7.58k | A Vth  
 128
0.938k  3.49k
15.1k
A V  0.938 A Vth  120 | A I  75
 60. 9
15.1k  3.49k
R IN  15.1k 3.49k  2.83 k | R OUT  r o 8.2k  8.2k
vth  v s
v be  0. 938v s
A V  10V CC
6
3.49k
5.00mV
 0.739 vs | v s 
 6.76 mV
0. 938k  3.49k
0.739
 1012  120. | Gain is identical to the rule - of - thumb estimate.
V EQ
14.13
500k
 18
 4.74 V | R EQ  500k 1. 4M  368k
1.4M  500k
2I DS
 27000I DS  I DS  104A
250x10 6
 18  IDS 75k  27k   7.39V | Saturation region is correct.
4.74  V GS  27000I DS  1 
V DS



g m = 2 250x10 6 104 x10 6  0. 228mS
368k
 0.997 vs | R th  1k 368k  0.997k
1k  368k
R L  r o 75k 470k 75k 470k  64.7k | A Vth   0.228mS 64.7k   14.8
vth  v s
A V  0.997 A Vth  14.7 | A I  368k g m 
75k
 11.6
75k  470k
R IN  368 k | R OUT  r o 75k  75k
vgs  0.997v s | VGS  V TN =
AV  
2 104A 
250A / V 2
= 0.912 V | vs  0.2
0.912V
 0.183 V
0.997
V DD
18
V

 19.7 | The rule- of - thumb estimate assumes VR L  DD .
VGS  V TN
0. 912
2
We have VR L  104A75k   7.80 V  0.433V DD
The estimate also doesn't account for the presence of R3 .
14.14
V EQ  18
2.2M
 9.00 V | R EQ  2.2M 2.2M  1.10M
2.2M  2.2M
18  22000I SD  V SG  9 | 9 = 22000I SD  1 
2I SD
 I SD  307A
400x10 6
V SD  18  I SD 22k  18k  5. 72V | Saturation region is correct.
gm =



2 400x10 6 307x10 6  0.496mS
vth  v s
1.1M
 0.999v s | R th  1k 1.1M  0.999k
1k  1.1M
R L  r o 18k 470k 18k 470k  17.3k | A Vth   0.496mS 17. 3k   8.60
A V  0.999 A Vth  8.59 | R IN  1.10 M | R OUT  r o 18k  18k
AI  A V
R IN
1.1M
 8.59
 20.1
R3
470k
vgs  0.999v s | V GS  V TN =
2307A 
400A / V
2
= 1.24 V | vs  0.2
1.24V
 0. 248 V
0.999
14.15
V 2
V
V GS   11kI DS   11k20mA 
1  GS   V GS  3.50V, I DS   GS  318 A

4 
11k
V DS  20  I DS 11k  39k  4.10V | Saturation region is correct.
2
1M
gm =
20mA 318A   1.26mS | v th  v s
 1. 00v s
4
0. 5k 1M
7
R th  0.5k 1M  0.500 k | R L  39k 500k  36. 2k
A V  A Vth  
A I  R G
1.26mS 36.2k
 3.07 | R IN  1.00 M | R OUT  39k
1 1. 26mS 11k
 
gm
1. 26mS
  106
 84. 8
1  g mR 1
1  1.26mS 11k
vth  1.00v s | V GS  V P = 3.5  4  = 0.500 V | vs  0.20.51 1. 26mS 11k  1.49 V
14.16
V SG  0  I SD  I DSS  5.00mA | V DS  16  1800I DS  7. 00V
Saturation region is correct. | g m =
2
5
5mA 5mA   2.00mS
10M
10M
 1. 00v s | A V =
2.00mS 1. 8k 36k = 3. 43
5k 10M
10M + 5k
1.8k
R IN  10.0 M | R OUT  1.80 k | A I  10M 2.00mS 
 952
1.8k  36k
10M
vgs  v s
 0.2 V SG  V P  v s  1 V
10M + 5k
vth  v s
14.17
10  0.7 V
 12. 3A | I C  983 A | V CE = 20  9100I E = 11. 0 V | Forward 20k  80  19.1k
800.025 V 
100  11. 0V  113k
active region is correct. | r 
 2.04k | r 
IB 

983A
o
983A
20k
 0.988 vs | R th  20k 250  247
20k  250
80 102k
R L  r o 1M 113k 1M  102k | A Vth  
 3570
247  2.04k
20k
113k
A V  0.988 A Vth  3530 | A I 
 7.37
80
20k  2. 04k
113k  1M
R IN  20k 2.04k  1. 85 k | R OUT  113 k
vth  v s
v be  0. 988v s
2.04k
5.00mV
 0.881vs | v s 
 5.67 mV
247  2. 04k
0.881
14.18
A Ith  81 | r  
80
811k
 200 | A Vth  
 0.632
0. 4S
47k  200  811k
R IN  200  811k  81.2 k | R OUT 
14.19
A Vth  
8
0.988 47k

 583 
0. 4S
81
0.011k
1
 0. 909 | A Ith   | R IN   | R OUT 
 100 
1  0. 011k
10mS
14.20
Defining v1 as the source node:
a 2k 100k  1.96k
v s  v1  3.54x10 3 v
v1
s  v1  
1960
10 6
3.541x10 3 vs  4.051x10 3 v1
v1  0. 874v s | A V  0. 874
v
vs
R IN  s 
 7.94 M
i s 10 6 vs  v1 
Driving the output with current source ix :
v
v1
R OUT : i x  16 
 3.54x10 3 v1
2000
10
v1
R OUT 
 247  | b R IN  
ix
14.21
51k
 6. 08V | R EQ  51k 100k  33.8k
51k  100k
6.08  0.7  18V  37. 3A | I  4.67 mA | V = 36  2000I  4700I = 4.54 V
IB 
C
CE
C
E
33.8k  126 4.7k 
V EQ  18
Forward - active region is correct. | r  
125 0. 025V 
50  4.54 V
 669 | r o 
 11.7k
4.67mA
4.67mA
33. 8
 0.985v s | R th  33.8k 500  493
500  33.8k
126 2.94k 
R L  24k 4.7k 11.7k  2.94k | A Vth  
 0.997
0. 493k  0.669k 126 2.94k
vth  v s
A V  0.985 A Vth  0.982 | R IN  33. 8k 0. 669k  126 2.94k   31.0 k
AI  A V
R S  R IN
0.5k  31. 0k
493  669
 0. 982
 1.29 | R OUT 
2.94k = 9.19 
R3
24. 0k
126
v be  0. 982v s
0.669k
5. 00mV
 0.00177 vs | v s 
 2. 83 V
0. 493k  0.669k 126 2.94k
0.00177
14.22
5  0.7V
 96.8nA | IC  9.68A | V CE = 10  430000I E = 5. 80V | Forward 1M  100  1430k
1000.025 V 
60  5.80 V  6.80M - neglected
active region is correct. r 
 258k | r 
IB 

9.68A
o
9.68A
In the ac model, R 1 appears in parallel with r . The circuit appears to have a transistor
with r'  500k r   170k and 'o  g m r '  40 9.68A 170k  65.8
vth  v s | R th  500 | R L  500k 430k 500k  158k
AV  
66.8158k
 0. 984 | R IN  170k  66.8 158k   10.7 M
0.500k  170k  66.8158k
9
R S  R IN
500  10.7M
 0. 984
 21.1
R3
500 k
500  170k
R OUT 
500k 430 k = 2.53 k
66.8
170k
5. 00mV
v be  v s
 0.0159v s | v s 
 0.315 V
0.500k 170 k  66.8 158k
0.0159
AI  A V
14.23
V GS  5 V | I DS 
4x10 4
2
5  1  3.2mA | V DS  5  5  10 V - Saturation region
2


operation is correct. | g m  2 4x10 4 3.2mA 1  0.02 10   1.75mS
1
 10
0.02
ro 
 18.8k  Cannot neglect! | R L =18.8k 100k =15.8k
3. 2mA
106
1.75mS 15. 8k
1.75mS 15.8k 1
AV 
 0. 956 | A I  10 6
 9.56
6
1 1.75mS 15.8k 105
10 10 4 1  1.75mS 15.8k
1
R IN  R G  1 M | R OUT 
r  555 
gm o
6
0.25  1
10
1
vgs  v s
 0.0346 vs | v s 
 23. 2 V But,
6
4
10  10 1 1.75mS 15.8k
0. 0346
v DS must exceed vGS  V TN  V GS  V TN = 4 V for saturation.
V DS  10  v o  10  0. 956v s  4  v s  6.28 V  Limited by the Q - point voltages
14.24
 o  g m r   3.54mS 1M  3540 | R L = 2k 100k = 1.96k
 o  1R L
r   o  1R L
R IN  r    o  1R L
AV 
R OUT 
14.25
3540  11.96k
 0.874
1M  3540  11.96k
 1M  3540  11. 96k  7. 94 M

r
106
2k 
2k  247 
 o 1
3541
vs  0. 0051  g m R L  | R L  R 4 R 7  R 4
 I R 
vs  0. 0051  g m R L   0. 0051  g m R 4  = 0.005 1  C 4 
V T 


 I R 
I R 
vs  0. 0051   F E 4   0.005 1 E 4 
V
V T 


T 
V R 
V R 
VR


4
4
4
vs  0. 0051 
 0.005 1 
 0.005 


V T 
5

 0. 025 
10
14.26
a v be  v s  v o | 0. 005  5  v o  A V 
b  A V 
o  1R E 
r    o 1R E 1
v o 4.995

 0. 999
vs
5
1
1
1
1



r
r
o
VT
o
1
1
1


1
R
g
R
I
o  E
 o 1  oR E
m E
ERE
1
VT
0. 025V
 0. 999 
 0.001  I E R E 
 25. 0 V
VT
I
R
0.001
E E
1
I ER E
14.27
vo 7.5  0. 005

 0.999333
vs
7.5
500R E
1
500 
| AV 
V
500  R E
T 500  R E 
1


IE R E  500 
v be  v s  v o = 1  A V vs | 0. 005  1  A V 7.5  A V 
From Prob. 14.26, A V 
1
| RL  RE
VT
1
IE R L
1
V T 500  R E 
 0.999333 

  6.67x104
V T 500  R E 
I E R E  500 
1


I ER E  500 
500I E R E
0.025V

 37. 5V | V CC  I ER E + 0.7 + 7.5
500  R E 6.67x104
Some design possibilities are listed in the table below.
RE
IE
VCC
VCC IE
100 
450 mA
53 V
24 W
250 
225 mA
64 V
16 W
360 
179mA
73V
13 W
500 
150 mA
83 V
12 W
750 
125 mA
102 V
13 W
1000 
113mA
120 V
14 W
2000 
93.8 mA
196 V
18 W
Using a result near the minimum power case in the table:
RE = 510 E= 149mA and VCC = 85V.
149mA
 2.92 mA | Set IR 1  5I B  14.6mA  15mA
51
149mA 510  0.7

 5. 07k  5.1 k | I R 2  I R 1  I B  18mA
15mA
Assumining  F  50: I B 
R1 
V E  V BE
I R1
R2 
85  V BE  V BE
8. 3V

 462  470 
IR 2
18mA
It is very difficult to achieve the required level of linearity!
11
14.28
a A Vth 
0.5mS 100k
g mR L
1
1

 48.8 | R IN 

 2.00 k
1  g m R th 1  0.5mS 50
g m 0. 5mS
R OUT   (since  = 0) | A Ith  1
b  A Vth 
14.29
a r  
0.5mS 100k
 14. 3 | R IN  2.00 k | R OUT   | A Ith  1
1  0.5mS 5k 
100 0. 025V 
 200k | g m  40 12.5A  0.5mS
12.5A
0. 5mS 100k 
r
gm R L

 48.8 | R IN 
 1.98 k
1  g m R th 1  0.5mS 50 
 o 1
60V
 r o 1+ g m R th  =
1  0.5mS50= 4.92 M | A Ith   o  0. 990
12.5A
A Vth 
R OUT
0.5mS 100k 
 23. 8 | R IN  1.98 k
1  0.5mS 2.2k 
60V
=
1  0.5mS2. 2k  10.1 M
12.5A
b  A Vth 
R OUT
14.30 AV ≈ 0 ; The signal is injected into the collector and taken out of the emitter. This is
not a useful amplifier circuit.
14.31
3.3x10 2x10 V
4
V SG  12  33kI SD | V SG  12 
2x10 V

4
2
1
2
SG
SG  1  282A
2
 24  I SD 33k  24k   7. 93 V - Saturation region operation is correct.
V SG  2.68V & I SD
V SD
4


2

g m  2 2x10 4 2. 82x10 4  3.36x10 4 S | v th 
33k
v s  0. 985v s
0.5k  33k
R th  0.5k 33k  493
R L = 24k 100k  19.4k | A V  0.985
AI 
4
3. 36x10 S 19.4k
 5. 51
1 3. 36x10 4 S 493 
24k
1
 0.178 | R IN  33k
 2.73k | R OUT  R D  24k
1 24k  100k
g
m
33k 
gm
vsg  v s
33k
R IN
2.73k
 0.2 V SG 1 | v s
 0.2 2.68  1  vs  0. 398 V
R S  R IN
0.5k  2.73k
14.32 Note: This problem should refer to Fig. P14.1(g).
12
5x10 V
 3900
4
V GS  3900 IDS
5x10 V
2
4
I SD 
gm
GS
 2 | V GS  0.975 V GS  2   V GS  0. 9915V
2
2
 2
2
 254A | V DS  15  23. 9kI DS  8.92V - Saturated.
2
2254A
1

 0.504mS | R IN  3.9k
 1.32k | R OUT  R D  20k
2  0.992
gm
GS
R L = 20k 51k  14.4k | A V 
3.9k
AI 
1
3. 9k 
gm
vsg  v s
1
R IN
1.32k
g R 
0. 504mS14. 4k  4.11
R S  R IN m L 1k  1. 32k
20k
 0.187
20k  51k
1.32k
1. 32k
 0.2V GS  2  | vs
 0.2 0.992  2   vs  0. 354 V
1k 1.32k
1k  1.32k
14.33
IB 
9  0.7 V
 1.94A | I C  96. 9 A
100k  50  182k
V CE =18  82000I E  39000I C = 6.12 V | Forward - active region is correct.
g m = 40IC = 3.88mS | r  
vth  v s
o
50  6.12 V  579k - neglected
 12.9k | r o 
gm
96.9A
82k
= 0.994v s | R th  0.5k 82k = 497 | R L  39k 100k  28.1k
0.5k  82k
A V  0.994
5028.1k 
r
 36.5 | R IN  82k
 253
12.9k  510. 497k 
o  1
AI  A V
R S  R IN
500  253
 36.5
 0.275 | R OUT  R C = 39.0 k
R3
100k
veb  vs
R IN
 5.00mV | 0.336v s  5.00mV | vs  14. 9 mV
R S  R IN
14.34
9  0. 7V
IB 
 194nA | IC  9.69 A
1000k  50  1820k
V CE =18  820000 IE  390000 IC = 6.12 V | Forward - active region is correct.
g m = 40IC = 0.388mS | r  
vth  v s
o
50  6.12V  5.79M - neglected
 129k | r o 
gm
9.69A
820k
= 0.994v s | R th  5k 820k = 4.97k | R L  390k 1M  281k
5k  820k
A V  0.994
50281k 
r
 36.5 | R IN  820k
 2.52k
129k  514. 97k 
 o 1
AI  A V
R S  R IN
5k  2.52k
 36.5
 0.275 | R OUT  R C = 390 k
R3
1M
veb  vs
R IN
 5.00mV | 0.335v s  5.00mV | v s  14. 9 mV
R S  R IN
13
14.35 Note: VTP = -1 V.
2x10 V

4
I SD
2
15  V SG
I SD 
 1
2
SG
|
15  V SG
68k
 10
4
V SG  12  V SG
 2.363 V
 186A | V SD  30  68k  43k I SD  9.35V | Saturation region
68k
is correct . | g m 
2186A 
1
 0.274mS | R IN  68k
 3. 46k
2. 36 1
gm
R OUT  R D  43k | R L = 43k 200k  35. 4k
AV 
AI 
vs
R IN
3.46k
gm R L 
0.274 mS 35. 4k  9.05
R S  R IN
0.250 k  3.46k
68k
1
68k 
gm
1
43k
3.46k
 0.168 | v sg  vs
 0.2 V SG 1
43k  200k
0.250k  3. 46k
3.46k
 0.22.36  1  v s  0.292 V
0.250k  3. 46k
14.36
For R th 
1
, A Vth  g m R L All of the input voltage appeas across the gate- source
gm
terminals of the transistor.
For R th 
current,
14.37
R IN 
1
R
, A Vth  L
gm
R th
For large R th , all of the Thevenin equivalent source
v th
, goes into the transistor source terminal.
R th
r   1.5k
75 0.025V 
1. 88k  1.5k
| r 
 1.88k | R IN 
 44.5 
o  1
1mA
76
14.38
gm 
2
1
1mA 5mA   2.24mS | R IN 
 447 
2
gm
14.39
g m  21.25mA 1mA   1.58mS | R IN 
14.40

a R OUT  r o 1 

r 
 oR E 
15V  0. 7V
 100A | For F  100, I C  99.0A
 | I E 
r   R E 
143k
100 0. 025V 
50V
 25.3k | r o 
 505k
99. 0A
99. 0A

100 143k  
R OUT  505k1 
  43. 4 M
 25.3k  143k 
14
1
 633 
gm
b  0 V
15V  0.7V
1000.025 V 
 953A | For  F  100, IC  944A | r  
 2.65k
c I E 
15k
ro 
944A
50V

10015k 
 53.0k | R OUT  53.0k 1 
 4.56 M | V CB  0 V
944A
15k
 2.65k 


14.41
R OUT   o 1r o   o  1
14.42
V A  V CE
50  10.7
 126
 154 M
IC
49. 6A
30
 10 20
R IN  5M | A V
 31.6 | Large R IN , moderate gain
These requirements are easily met by a common- source amplifier.
V DD
15 V
For example, A V 

 30.
V GS V TN 0.5V
A common - emitter stage operating at a low collector current is a second possibility.
14.43
52
R IN  1M | A V  10 20  398 | Large R IN , large gain
A common - emitter amplifier operating at a low current can achieve both a
large gain and input resistance. A V  20V CC  V CC  20V
Achieving this gain with an FET is much more difficult:
V DD
V DD
AV 

 V DD  100V which is unreasonably large.
V GS V TN 0.25V
14.44
An inverting amplifier with a gain of 40 dB is most easily achieved with a common - emitter
stage: A V  10V CC  V CC =10 V. The input resistance can be achieved by shunting the
100 0. 025V 
= 0.5 A and would
5
waste a large amount of power to achieve the required input resistance
.
input with a 5 resistor. Setting r  = 5 would require IC 
14.45
0 dB corresponds to a follower.
For an emitter - follower, R IN  o  1R L  10110k  1. 01M. So an BJT does
not meet the input resistance requirement
. A source follower provides a gain of
approximately 1 and can easily achieve the required input resistance
.
14.46
A gain of 0. 98 and an input resistance of 250k should be achievable with
either a source- follower or emitter- follower. For the FET , A V 
requires g m R L  49:
g mR S
 0. 98
1  g mR S
2I DSR S
 49  IDS R S  12.3V for a design with V GS  V TN  0.5V.
VGS  V TN
15
The BJT can achieve the required gain with a much lower power supply and
can still meet the RIN requirement: R IN  o R L  100 5k = 500k.
g mR E
 o  1R E

o
g R
AV =

 0. 98 | m E  49  IE R E  49 0. 025   1.23 V.
g
R
r    o  1R E 1  m E
o
o
14.47
A non - inverting amplifier with a gain of 10 and an input resistance of 2k should
be readily achievable with either a common- base or common - gate amplifier with
proper choice of operating point. The gain of 10 is easily achieved with either the
V DD
1
FET or BJT design estimate: A V 
or A V  10 V CC . R IN 
 2k is within
V GS  V TN
gm
easy reach of either device.
14.48
80
A V  10 20  10, 000. This value of voltage gain excceds the amplification factor
of even the best BJTs: A V   f = 40V A = 40 150 V   6000. Such a large gain
requirement cannot be met with a single- transistor amplifier. An FET
typically has a much lower  f and is at an even worse disadvantage.
14.49
R OUT 
R th  r 
R th
250


 1. 66
 o 1
 o 1 151
14.50
R IN  r    o  1R E  r    oR E  r  1  g m R E  | r '  r  1  g m R E 
g 'm 
r 'o 
ic
o
o
o
gm




vs
r    o  1R E r   o R E r  1  g m R E  1  g m R E
ic
vc
vs  0

  oR E 
 oR E 
 r o 1 
  r o 1 
  r o 1  g m R E  for r   R E
r  
 r   R E 

 g m

 g m

'
' '
 'o  g 'm r '  
r  1 g m R E   o |  f  g m r o  
r o 1  g m R E   f
1  g m R E 
1  g m R E 
14.51
*Problem 14.51 - Common-Emitter Amplifier 5mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.005 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
16
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
*Problem 14.51 - Common-Emitter Amplifier 10mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.01 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
*Problem 14.51 - Common-Emitter Amplifier 15mV
VCC 6 0 DC 9
VEE 4 0 DC -9
VS 1 0 SIN(0 0.015 1K)
C1 1 2 1U
RB 2 0 10K
RC 6 5 3.6K
RE 3 4 2K
C2 3 0 50U
C3 5 7 1U
R3 7 0 10K
Q1 5 2 3 NBJT
.OP
.TRAN 1U 5M
.FOUR 1KHZ V(7)
.MODEL NBJT NPN IS=1E-16 BF=100 VA=70
.PROBE V(7)
.END
Results:
5 mV
10 mV
15 mV
1 kHz
5.8 mV
12.4 mV
20.6 mV
2 kHz
0.335 mV (5.7%)
1.54 mV (12.5%)
4.32 mV (21%)
3 kHz
0.043 mV (0.74%
0.258 mV (2.1%)
1.18 mV (5.4%)
THD
5.9%
12.8%
22%
14.52
*Problem 14.52 - Output Resistance
VCC 2 0 DC 10
IB1 0 1 DC 10U
Q1 2 1 0 NBJT
IB2 0 3 DC 10U
RE 4 0 10K
Q2 2 3 4 NBJT
.OP
.DC VCC 10 20 .025
17
.MODEL NBJT NPN IS=1E-16 BF=60 VA=20
.PRINT DC IC(Q1) IC(Q2)
.PROBE IC(Q1) IC(Q2)
.END
Results: A small value of Early voltage has been used deliberately to accentuate the results.
Note that the transistors have significantly different values of F because of the collectoremitter voltage differences and low value of VA.
NAME
MODEL
IB
IC
VBE
VBC
VCE
BETADC
GM
RPI
RO
From SPICE : R OUT1 
Q1
Q2
NBJT
NBJT
1.00E-05 1.00E-05
8.77E-04 6.72E-04
7.61E-01 7.61E-01
-9.24E+00 -2.41E+00
1.00E+01 3.18E+00
8.77E+01 6.72E+01
3.39E-02 2.60E-02
2.59E+03 2.59E+03
3.33E+04 3.33E+04
20 10 
V
20  10 V  43.5 k
 34.1 k | R OUT2 
1.17  0.877  mA
903  673  A
For circuit 1: R OUT1  r o1  33. 3 k


o R E
For circuit 2: R OUT2  r o 2 1 
 + R th  r   R E (See Eq. 14.26)
R

r

R

th

E 
But R th    R OUT2  r o2  R E  33.3k  10k  43.3 k
14.53
vth  G mR th  
14.54
vth  v s
14.55
vth  v s
R th 
gm
r 1  g mR 5  v s  f v s | R th  r o 1  g mR 5   r o   f R 5
1  g mR 5 o




r
1  g mr o   R r r 1  f vs | R th  r o 1  RoR Sr 
RS  r

S

S
 
 o  1r o
R s  r    o  1r o
1
1
 vs
 vs
Rs
r
g mR s
o

1

1
 o 1r o  o  1r o
 o  1 f  o  1 f
 vs
R s  r
R  r
ro  s
o  1
o  1
14.56
a y 21 
y12 
18
i1
v2
i2
v1
| i 2  g m v1 | y 21  g m
v2  0
| i1  0 | y12  0 |
v1 0
y12
0
y 21
b  y 21 
y12 
i1
v2
i2
v1
| i 2    o  1i b    o  1
v2  0
| i 1  i b  
v1 0
i2
v1
c y 21 
y12 
i1
v2
| i 2   g m  g o v1 | y 21   g m  g o 
| i1  g ov 2 | y 21   g o |
v1 0
y12 
y12 
i1
v2
y12  
i2
v1
v2
y 12


1
| y 12  g  |
 o g  o 
 1
r
y 21 g m
o
o  1
v2  0
d y 21 
e y 21 
v1
 1  o   g m  g
| y 21   o
m
r
o
r
o
i1
v2
i2
v1
y 12
 go
1


 1
y 21  g m  g o   f 1
| i 2  g m v1 | y 21  g m
v2  0
| i1  g ov 2 | y 21   g o |
v1 0
| i2 
v2  0
| i 1  i 2
v1 0
y 12
g o
1


 1
y 21  g m  f
o
o
v | y 21 
r    o  1R 5 1
r    o  1R 5
R5
v2
R5


R5  r 
R OUT R 5  r 
v2
R5

 oR 5  R 5  r 
r o 1 

 R 5  r  
1
for  o  1R 5 >> r 


 o 1r o
r   o  1R 5
y12
1
R5
R5
g R


for  o  1R 5 >> r  |
 m 5  1
y 21
o
 or o
 or o
 o f
 o  1r o
f 
y 21 
y12 
i1
v2
i2
v1
| i2 
v2 0
gm
gm
v | y 21 
1  gm R 5 1
1  gm R 5
| i1  0 | y12  0 |
v1 0
y12
0
y 21
14.57
At the output node vo : g m v x  g o  G L vo  g ov x | v o 
i x  g m v x  g o v x  v o  |
g m  go
vx
go  G L
 g  g o 
G  g m 
ix
 g m  g o 1  m
 g m  g o  L


vx
 g o  G L 
 g o  G L 
R
1 L
GL  g m 
ix
gm  go
vx
ro
1
1  R L 
 g m  g o 
| R IN 


  G L
1 

1
vx
go  G L
ix
g m 1
g m 
r o 
 g o  G L 
f
19
14.58
I C  100
5  0.7
 
10 4 101 103
 3.87mA | g m  40IC  0.155S | r  
100
 645
gm
R L  1k 20k = 952 | R E  1k 20k = 952
A V1  
AV2 
 oR L
100 952 

 0.984
r    o  1R E
645  101952
 o  1R E
r    o  1R E

101952 
645  101952
 0.993
The small - signal requirement limits the output signal to:
0.005
v be = v s  v o2 = v s 1  0.993  = 0. 007v s | v s 
= 0.714 V
0.007
v o1  0. 984 0.714 V   0. 703V
We also need to check VCB : V C  5  3.87mA 1k   1.13V and V B = 10 4 IB  0.387 V.
The total collector- base voltage of the transistor is therefore: V CB  1.52V  0.984 v s  v s .
We require VCB  0 for forward - active region operation. Therefore: v s  0.766 V.
The small - signal limit is the most restrictive.
14.59
*Problem 14.59 - Common-Collector Amplifier 14.1(a)
VCC 6 0 DC 18
VEE 7 0 DC -18
VS 1 0 AC 1
*For Output Resistance
*VS 1 0 AC 0
*VO 8 0 AC 1
RS 1 2 500
C1 2 3 10UF
R1 3 0 51K
R2 6 3 100K
RE 4 7 4.7K
RC 6 5 2K
C3 5 0 10UF
C2 4 8 47UF
R3 8 0 24K
Q1 5 3 4 NBJT
.OP
.AC LIN 1 10KHZ 10KHZ
.MODEL NBJT NPN IS=1E-16 BF=125 VA=50
.PRINT AC VM(8) VP(8) VM(3) VP(3) IM(VS) IP(VS) IM(C2) IP(C2)
.END
Results: Q-point: (4.67 mA, 4.57 V), AV = 0.982, RIN = 31.1 kROUT= 9.12 
For hand calculations see Problem 14.21: AV = 0.982, RIN = 31.0 kROUT= 9.19 
14.60
*Problem 14.60 - Common-Gate Amplifier - 14.1(b)
VSS 5 0 DC 12
VDD 6 0 DC -12
VS 1 0 AC 1
*For Output Resistance
*VS 1 0 AC 0
20
*VO 7 0 AC 1
RS 1 2 500
C1 2 3 10U
R1 5 3 33K
RD 6 4 24K
C2 4 7 47U
R3 7 0 100K
M1 4 0 3 3 PFET
.OP
.AC LIN 1 50KHZ 50KHZ
.MODEL PFET PMOS KP=200U VTO=-1 LAMBDA=0.02
.PRINT AC VM(7) VP(7) VM(3) VP(3) IM(VS) IP(VS) IM(C2) IP(C2)
.END
Results: Q-point: (286 A, 7.72V), AV = +5.50, RIN = 2.73 kROUT= 21.8 k
For hand calculations see Problem 14.31: AV = +5.51, RIN = 2.73 k
50  7. 93
R OUT  R CG
 205k | R CG
O RD | r o 
OUT  r o 1  g m R th 
282x10 6
 


4
R CG
493  239k | R OUT  239k 24k  21. 8 k
OUT  205k 1  3. 36x10
14.61
*Problem 14.61 - Common-Collector Amplifier - 14.1(c)
VCC 7 0 DC 5
VEE 8 0 DC -5
VS 1 0 AC 1
*For Output Resistance
*VS 1 0 AC 0
*VO 6 0 AC 1
RS 1 2 500
C1 2 3 10UF
R1 3 5 500K
R2 5 0 500K
RE 4 8 430K
C2 4 5 47UF
C3 4 6 10UF
R3 6 0 500K
Q1 7 3 4 NBJT
.OP
.AC LIN 1 10KHZ 10KHZ
.MODEL NBJT NPN IS=1E-16 BF=100 VA=60
.PRINT AC VM(6) VP(6) VM(3) VP(3) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (9.81 A, 5.74V), AV = 0.983, RIN = 11.0 MROUT= 2.58 k
For hand calculations see Problem 14.22: AV = 0.984, RIN = 10.7 MROUT= 2.53 k
14.62
*Problem 14.62 - Common-Emitter Amplifier - 14.1(e)
VCC 6 0 DC 12
VS 1 0 AC 1
*VS 1 0 AC 0
*VO 7 0 AC 1
RS 1 2 1K
C1 2 3 2.2UF
R1 6 3 20K
R2 3 0 62K
21
RE 6 4 3.9K
C2 6 4 47UF
RC 5 0 8.2K
C3 5 7 10UF
R3 7 0 100K
Q1 5 3 4 PBJT
.OP
.AC LIN 1 5KHZ 5KHZ
.MODEL PBJT PNP IS=1E-16 BF=75 VA=60
.PRINT AC VM(7) VP(7) VM(3) VP(3) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (525 A, 5.62V), AV = -110, RIN = 3.16kROUT= 7.69 k
For hand calculations see Problem 14.12: Q-point: (537 A, 5.47V), RIN = 2.84 k
60  5. 47
 122k | R L  r o 8.2k 100k  122k 8.2k 100k  7.14k
537x10 6
75 7.14k
A Vth  
 121 | A V  0. 938A Vth  113 | R OUT  r o 8.2k  7.68k
0.938k  3. 49k
ro 
14.63
*Problem 14.63 - Common-Source Amplifier - 14.1(f)
VDD 4 0 DC 18
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
RS 1 2 1K
C1 2 3 2.2U
R1 3 0 500K
R2 4 3 1.4MEG
R4 6 0 27K
C2 6 0 47U
RD 4 5 75K
C3 5 7 10U
R3 7 0 470K
M1 5 3 6 6 NMOSFET
.OP
.AC LIN 1 5KHZ 5KHZ
.MODEL NMOSFET NMOS VTO=1 KP=250U LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (106A, 7.14V), AV = -14.2, RIN = 369kROUT= 65.8 k
For hand calculations see Problem 14.13: Q-point: (104 A, 7.39V), RIN = 368 k
50  7.39
 552k | R L  r o R D 470k  552k 75k 470k  57. 9k
104x10 6
368k
R OUT  r o 75k  66.0 k | v th  v s
 0.997 v s
1k  368k
A V  0.997g m R L  0.997 0.228mS 57.9k  13.2
ro 
14.64
22
*Problem 14.64 - Common-Gate Amplifier - 14.1(g)
VDD 5 0 DC 15
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 6 0 AC 1
RS 1 2 1K
C1 2 3 2.2U
R1 3 0 3.9K
RD 4 5 20K
C2 4 6 47U
R3 6 0 51K
M1 4 0 3 3 NMOSFET
.OP
.AC LIN 1 20KHZ 20KHZ
.MODEL NMOSFET NMOS VTO=-2 KP=500U LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VS) IP(VS) IM(C2) IP(C2)
.END
Results: Q-point: (268A, 8.60V), AV = 4.26, RIN = 1.27 kROUT= 18.8 k
For hand calculations see Problem 14.32: Q-point: (254 A, 8.92V), RIN = 1.32 k
50  8.92


CG
6  232k | R OUT = r o 1  g m R th   232k 1  0.504mS 3.9k 1k   325k
254 x10
R L  R D R 3  51k 20k  14.4k | R OUT  R CG
OUT 20k  18. 8 k
ro 
3.9k
 0.796 v s
1k  3.9k
g mR L
0.504mS 14. 4k
 0.796
 0.796
 4.12
1  g m R th
1  0.504mS 3.9k 1k
v th  v s
AV
14.65
*Problem 14.65 - Common-Source Amplifier 14.1(i)
VDD 6 0 DC 18
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
RS 1 2 1K
C1 2 3 2.2UF
R2 6 3 2.2MEG
R1 3 0 2.2MEG
R4 6 4 22K
C2 6 4 47UF
RD 5 0 18K
C3 5 7 10UF
R3 7 0 470K
M1 5 3 4 4 PFET
.OP
.AC LIN 1 7.5KHZ 7.5KHZ
.MODEL PFET PMOS KP=400U VTO=-1 LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (268A, 8.60V), AV = -8.29, RIN = 1.10kROUT= 16.4 k
For hand calculations see Problem 14.32: Q-point: (307 A, 5.72V), RIN = 1.10 k
23
50  5. 72
 182k | R L  r o R D 470k  182k 18k 470k  15. 8k
307x10 6
1.1M
R OUT  r o 18k  16.4 k | v th  v s
 0.999v s
1k 1.1k
A V  0.999g m R L  0.999 0. 496mS 15.8k  7.82

ro 
14.66
*Problem 14.66 - Common-Source Amplifier - 14.1(j)
VDD 6 0 DC 20
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
RS 1 2 500
C1 2 3 2.2UF
RG 3 0 1MEG
R4 4 0 11K
C2 5 7 47UF
RD 6 5 39K
R3 7 0 500K
J1 5 3 4 NJFET
.OP
.AC LIN 1 4KHZ 4KHZ
.MODEL NJFET NJF BETA=1.25M VTO=-4 LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VS) IP(VS) IM(C2) IP(C2)
.END
Results: Q-point: (319 A, 4.03V), AV = -3.02, RIN = 1.00 MROUT= 38.4 k
For hand calculations see Problem 14.15: Q-point: (318 A, 4.10V), RIN = 1.00 M
ro 
50  4.10
 170k | R CS
OUT = r o 1 g m R 1  =170k 1  1.26mS 11k   253k
318x10 6
CS
R L  R CS
OUT R D R 3  253k 39k 500k  31.7k | R OUT  R OUT 39k  33.8 k
v th  v s
14.67
24
1M
 v s | A V  g m R L   1.26mS 31. 7k  3.99
0.5k  1M
*Problem 14.67 - Common-Base Amplifier - 14.1(k
VCC 7 0 DC -9
VEE 4 0 DC 9
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 8 0 AC 1
RS 1 2 500
C2 2 3 47UF
RB 5 0 100K
RE 3 4 82K
C1 5 0 4.7UF
RC 7 6 39K
C3 6 8 10UF
R3 8 0 100K
Q1 6 5 3 PBJT
.OP
.AC LIN 1 12KHZ 12KHZ

.MODEL PBJT PNP IS=1E-16 BF=50 VA=50
.PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (97.2 A, 6.10V), AV = 35.5, RIN = 273 ROUT= 38.1 k
For hand calculations see Problem 14.33: Q-point: (96.9 A, 6.12V), RIN = 253 
ro 

50  6.12
 o R th 

50 497 
CB
  579k1  12.9k  497   1. 65M
6  579k | R OUT = r o 1 
r

R
96. 9x10




th 
CB
R L  R CB
OUT R D R 3  1.65M 39k 100k  26. 6k | R OUT  R OUT 39k  38.1 k
82k
 0. 994v s
0.5k  82k
3.88mS 26.6k
g mR L
 0.994
 0. 994
 34. 9
1  g m R th
1  3.88mS 0.497k
v th  v s
AV
14.68
*Problem 14.68 - JFET Common-Source Amplifier - 14.1(m)
VDD 5 0 DC -16
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 6 0 AC 1
RS 1 2 5K
C1 2 3 2.2U
R1 3 0 10MEG
RD 4 5 1.8K
C2 4 6 10U
R3 6 0 36K
J1 4 3 0 PJFET
.OP
.AC LIN 1 3KHZ 3KHZ
.MODEL PJFET PJF BETA=200U VTO=-5 LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(6) VP(6) IM(VS) IP(VS) IM(C2) IP(C2)
.END
Results: Q-point: (5.59 A, 5.93 V), AV = -3.27, RIN = 10.0 MROUT= 1.53 k
For hand calculations see Problem 14.16: Q-point: (5.00 mA, 7.00V), RIN = 10.0 M
50  7.00
 11.4k | R L  r o R D R 3  11.4k 1.8k 36k  1. 48k
5x103
R OUT  r o R D  11. 4k 1.8k  1.56k
ro 
v th  v s
14.69
10M
 v s | A V  g m R L   2. 00mS 1.48k  2.96
5k  10M

*Problem 14.69 - Common-Emitter Amplifier - 14.1(n)
VCC 7 0 DC 10
VEE 6 0 DC -10
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 8 0 AC 1
RS 1 2 250
25
C1 2 3 4.7UF
RB 3 0 20K
R4 4 6 9.1K
C2 4 0 100UF
L 7 5 1H
C3 5 8 1UF
R3 8 0 1MEG
Q1 5 3 4 NBJT
.OP
.AC LIN 1 500KHZ 500KHZ
.MODEL NBJT NPN IS=1E-16 BF=80 VA=100
.PRINT AC VM(3) VP(3) VM(8) VP(8) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Note: L needs to be 1H to reach midband performance at 500kHz.
Results: Q-point: (979 A, 11.0 V), AV = -3420, RIN = 2.09 kROUT= 113 k
For hand calculations see Problem 14.17: Q-point: (983 A, 11.0V), RIN = 1.85 k
100  11.0
 113k | R L  r o R 3  113k 1M  102k | R OUT  r o  113k
983x10 6
 oR L
80 102k 
20k
 vs
 0. 988v s | A V  0.988
 0.988
 3530
250  20k
R th  r 
247  2. 04k

ro 
v th
14.70
*Problem 14.70 - Source Follower - 14.1(o)
VDD 5 0 DC 5
VSS 6 0 DC -5
VS 1 0 AC 1
*For output resistance
*VS 1 0 AC 0
*VO 7 0 AC 1
RS 1 2 10K
C1 2 3 2.2U
R1 3 0 1MEG
L 4 6 100mH
C3 4 7 4.7U
R3 7 0 100K
M1 5 3 4 4 NMOSFET
.OP
.AC LIN 1 100KHZ 100KHZ
.MODEL NMOSFET NMOS VTO=1 KP=400U LAMBDA=0.02
.PRINT AC VM(3) VP(3) VM(7) VP(7) IM(VS) IP(VS) IM(C3) IP(C3)
.END
Results: Q-point: (3.84 mA, 10.0 V), AV = 0.953, RIN = 1.00 MROUT= 504 
For hand calculations see Problem 14.23: Q-point: (3.20 mA, 10.0V),
AV = 0.956, RIN = 1.00 MROUT= 555 
14.71
AV 
g mR L
 0.95  g m R L  19 | g m  2K n I DS | V GS  V TN 
1  gm RL
I DS 
0.5  0.03   3.75mA | R 
L
2
26
2
19
2 0. 030.00375 
 1.27k
2I DS
 0. 5V
Kn
R L  R S 3k  R S  2.19k | V SS  V GS  3.75mA R S | Possible designs:
2.4k, 11. 5V ; 2.7k, 12.6V ; 3.0k, 13.75V - Making a choice which uses
a nearly minimum value of supply voltage gives: V SS  12 V, R S  2. 4 k.
14.72
R IN  r  
o V T
1000.025 V 
| IC 
 33. 3 mA
IC
75
14.73 The base voltage should remain half way between the positive and negative power
supply voltages. If VEE = +10V and VCC = 0V, then VB should = 5 V which can be
obtained using a resistive voltage divider from the +10V supply. We now have the
standard four-resistor bias circuit. The base curent is 327 A/80 = 4 A.
Setting the current in R1 to 10IB  40A, R1 
current in R 2  11I B  44A, and R 2 
5V
 125k 120k. The
40A
5V
 114k 110 k.
44A
Note that the base terminal must now be bypassed with a capacitor.
14.74
1
0. 01
0.01
 g m  2K n IDS  0.1S | I DS 
| a I DS 
1 A
gm
2K n
20.005 
0.01
b  I DS 
 10.0 mA | The second FET achieves the desired input resistance
20.5 
R IN 
at much lower current and hence much lower power for a given supply voltage.
14.75
This analysis assumes that the source and load resistors are fixed
, and that only
the amplifier parameters are changing. A Vth 
RL
1
R
 th
gm
o
| o 
80
 0. 988 | R th  R E 75
81
max
Since RE  75, R th  75 is essentially constant. To achieve Amax
,g m  g max
Vth , R L  R L
m
which requires IC  Imax
 0.988
C
5.25  0.7 V
 364A | R max
 8.2k1.05  100k  7. 93k
L
13k0. 95
27
A max
Vth 
7. 93k
min
min
 54. 8 | To achieve Amin
Vth , R L  R L , g m  g m which requires
1
75

40364A  0.988
IC  I min
 0.988
C
A min
Vth 
4. 75  0.7 V
 293A | R min
 8.2k0. 95 100k  7.23k
L
13k 1.05 
7.23k
 44.8 | 44.8  A V  54.8 | The range is only slightly larger
1
75

40 293A  0. 988
than that observed in the Monte Carlo analysis in Table14.16.
14.76
*Problem 14.76 - Common-Base Amplifier - Monte Carlo Analysis
*Generate Voltage Sources with 5% Tolerances
IEE 0 8 DC 5
REE 8 0 RTOL 1
EEE 6 0 8 0 1
*
ICC 0 9 DC 5
RCC 9 0 RTOL 1
ECC 7 0 9 0 -1
*
VS 1 0 AC 1
RS 1 2 75
C1 2 3 47U
RE 3 6 RTOL 13K
Q1 4 0 3 PBJT
RC 4 7 RTOL 8.2K
C2 4 5 4.7U
R3 5 0 100K
.OP
.AC LIN 1 10KHZ 10KHZ
.PRINT AC VM(5) VP(5)
.MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%)
.MODEL RTOL RES (R=1 DEV 5%)
.MC 1000 AC VM(5) YMAX
.END
Results: Mean value AV = 47.5; 3 limits: 42.5 ≤ AV ≤ 52.5. However, the worst-case values
min
observed in the analysis are A V
 43.2 and A max
 51.9. The mean is 5% lower than the
V
design value. The width of the distribution is approximately the same as that in Table 14.16. 
14.77 (a)
This analysis assumes that the source and load resistors are fixed
, and that only
the amplifier parameters are changing. A Vth 
RL
1
R
 th
gm
o
| o 
80
 0. 988 | R th  R E 75
81
max
Since RE  75, R th  75 is essentially constant. To achieve Amax
,g m  g max
Vth , R L  R L
m
which requires IC  Imax
 0.988
C
28
5.10  0.7V
 330A | R max
 8. 25k1.01 100k  7.69k
L
13. 3k0.99 
A max
Vth 
7. 69k
min
min
 50.7 | To achieve Amin
Vth , R L  R L ,g m  g m which requires
1
75

40330A  0.988
IC  I min
 0.988
C
A min
Vth 
4. 90  0.7V
 309A | R min
 8.25k0.99  100k  7.55k
L
13.3k1. 01
7.55k
 48.2 | 48.2  A V  50.7
1
75

40 309A  0. 988
Using a Spreadsheet similar to Table 14.16: Mean value AV = 49.6; 3 limits: 48.2 ≤ AV ≤
min
50.9. The worst-case values observed in the analysis are A V
14.78
 48. 4 and A max
 50.8.
V
*Problem 14.76 - Common-Base Amplifier - Monte Carlo Analysis
*Generate Voltage Sources with 2% Tolerances
IEE 0 8 DC 5
REE 8 0 RTOL 1
EEE 6 0 8 0 1
*
ICC 0 9 DC 5
RCC 9 0 RTOL 1
ECC 7 0 9 0 -1
*
VS 1 0 AC 1
RS 1 2 75
C1 2 3 100U
RE 3 6 RR 13.3K
Q1 4 0 3 PBJT
RC 4 7 RR 8.25K
C2 4 5 1U
R3 5 0 100K
.OP
.AC LIN 1 10KHZ 10KHZ
.PRINT AC VM(5) VP(5) IM(VS) IP(VS)
.MODEL PBJT PNP (BF=80 DEV 25%) (VA = 60 DEV 33.33%)
.MODEL RTOL RES (R=1 DEV 2%)
.MODEL RR RES (R=1 DEV 1%)
.MC 1000 AC VM(5) YMAX
*.MC 1000 AC IM(VS) YMAX
.END
Results:
Mean value: AV = 47.2; 3 limits: 45.7 ≤ AV ≤ 48.5
Mean value: RIN = 83.4 ; 3 limits: 79.5  ≤ AV ≤ 87.6 
29
14.79
1
RE
RE
RE
RE




80
g m 1 g m R E 1  40I CR E 1  40
1  39.5I E R E
I R
81 E E
RE
I E R E  2.5  0.7  1.8V | 75 
 R E  5.41k
1  39. 51. 8
80 1.8V
IC =
= 329A | R th  75 5.41k  74.0 | g m  40 329A   13.2mS
81 5. 41k
5410
R
v th  v s
 0.986 v s | A V  0. 986 1 LR  R L  7.64k
75  5410
 th
gm
o
R IN  R E
7.64k  R C 100k  R C  8. 27k | V C  2. 5V  I CR C  0.221V | Oops! We are violating
our definition of the forward- active region. If we use the nearest 5% values, R E  5.6k and
R C  8.2k, IC = 318A and V C = +0.108V. The transistor is entering saturation.
14.80
Using a Spreadsheet similar to Table 14.16:
Mean value: AV = 0.960; 3 limits: 0.942 ≤ AV ≤ 0.979
Mean value: IDS = 4.91 mA; 3 limits: 4.27 mA ≤ IDS ≤ 5.55 mA
Mean value: VDS = 7.03 V; 3 limits: 4.52 V ≤ VDS ≤ 9.54 V
*Problem 14.80 - Common-Drain Amplifier - fig. 14.40
*Generate Voltage Sources with 5% Tolerances
IDD 0 7 DC 5
RDD 7 0 RTOL 1
EDD 5 0 7 0 1
*
ISS 0 8 DC 20
RSS 8 0 RTOL 1
ESS 6 0 8 0 -1
*
VGG 1 0 DC 0 AC 1
C1 1 2 4.7U
RG 2 0 RTOL 22MEG
RS 3 6 RTOL 3.6K
C2 3 4 68U
R3 4 0 3K
M1 5 2 3 3 NMOSFET
.OP
.AC LIN 1 10KHZ 10KHZ
.DC VGG 0 0 1
.MODEL NMOSFET NMOS (VTO=1.5 DEV 33.33%) (KP=20M DEV 50%)
LAMBDA=0.02
.MODEL RTOL RES (R=1 DEV 5%)
.PRINT AC VM(4) VP(4) IM(VGG) IP(VGG)
.MC 1000 DC ID(M1) YMAX
*.MC 1000 DC VDS(M1) YMAX
*.MC 1000 AC IM(VGG) YMAX
*.MC 1000 AC VM(4) YMAX
.END
Results:
30
Mean value: IDS = 4.97 mA; 3 limits: 4.32 mA ≤ IDS ≤ 5.62 mA
Mean value: VDS = 7.19 V; 3 limits: 6.18 V ≤ VDS ≤ 8.20 V
Mean value: RIN = 22.0 M; 3 limits: 20.3  ≤ RIN ≤ 24.0 
Mean value: AV = 0.956; 3 limits: 0.936 ≤ AV ≤ 0.976
14.81
a A Vth 
R OUT 
g mR L
0.01S 1k

 0.625 | R IN  
1  g m 1  R L 1  0.01S 1  0.5 1k
1
1

 66.7 | A Ith  
g m 1   0.01S 1  0.5 
b  A Vth is worse; R OUT is improved.
14.82
a A Vth  
0.01S 2k
g mR L

 4. 44 | R IN  
1  g m 1  R 5
1  0.01S1  0.75 200
R OUT  r o 1 g m 1  R 5   10k1  0.01S1. 75200  45.0 k | A Ith  
b  A Vth is reduced; R OUT is increased.
14.83
a A Vth 
R IN 
g m 1  R L
0.5mS 1  1100k

 95.2
1  g m 1  R th 1  0.5mS 1  150
1
1

 1000 
g m 1   0.5mS 1  1
R OUT  r o 1 g m 1  R th   r o 1  0.5mS 1  150  1.05r o (  for  = 0)
b  A V is larger; R IN is smaller.
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