Microelectronic Circuit Design Third Edition - Part III Solutions to Exercises doc

1
Microelectronic Circuit Design
Third Edition - Part III
Solutions to Exercises

CHAPTER 10

Page 509

€
V
o
= 2P
o
R
L
= 2 20W
( )
16Ω
( )
= 25.3 V A
v
=
V
o
V
i
=
25.3V
0.005V
= 5.06x10

3
I
o
=
V
o
R
L
=
25.3V
16Ω
= 1.58 A I
i
=
V
i
R
S
+ R
in
=
0.005V
10kΩ + 20kΩ
= 0.167
µ
A A
i
=
I
o

I
i
=
1.58 A
0.167
µ
A
= 9.48x10
6
A
P
=
P
o
P
S
=
25.3V 1.58A
( )
0.005V 0.167
µ
A
( )
= 4.79x10
10
Checking : A
p
= 5.06x10
3
( )

9.48x10
6
( )
= 4.80x10
10

Page 510

€
i
( )
A
vdB
= 20 log 5060
( )
= 74.1 dB A
idB
= 20 log 9.48x10
6
( )
= 140 dB A
PdB
= 10 log 4.80x10
10
( )
= 107 dB
ii
( )
A
vdB

= 20 log 4x10
4
( )
= 92.0 dB A
idB
= 20 log 2.75x10
8
( )
= 169 dB A
PdB
= 10 log 1.10x10
13
( )
= 130 dB

Page 511

€
i
( )
The constant slope region spanning a maximum input range is between 0.4 ≤ v
I
≤ 0.65,
and the bias voltage V
I
should be centered in this range : V
I
=
0.4 + 0.65
2

V = 0.525 V .
v
i
≤ 0.65 − 0.525 = 0.125 V and v
i
≤ 0.525 − 0.40 = 0.125 V. For v
I
= 0.8V, the slope is 0. A
v
= 0.
(ii) v
O
= V
O
+ v
o
For v
i
= 0, v
I
= V
I
= 0.6, V
O
= 14V and A
v
= +40. V
o
= A
v

V
i
= 40 0.01V
( )
= 4V
v
O
= 14.0 + 4.00sin1000
π
t
( )
volts V
O
= 14 V

2
Page 519

€
g
11
=
1
20kΩ + 76 50kΩ
( )
= 0.262
µ
S g
21
= 0.262

µ
S 76
( )
50kΩ
( )
= 0.995
g
22
=
1
50kΩ
+
1
20kΩ
+
75
20kΩ
= 3.82 mS g
12
= −
1
g
22
20kΩ
( )
= −
1
3.82mS 20kΩ
( )
= -0.0131

R
in
=
1
g
11
= 3.82 MΩ A = g
21
= 0.995 R
out
=
1
g
22
= 262 Ω

Page 520

€
i
( )
P = A
v
A
i
= A
v
2
R
S

+ R
in
R
L
ii
( )
V
o
= 2 100W
( )
8Ω
( )
= 40 V 40 = 0.001
50kΩ
5kΩ + 50kΩ






A
8Ω
0.5Ω + 8Ω







→ A = 46,800
P =
I
o
2
R
L
2
=
0.5Ω
2
40V
8Ω






2
= 6.25 W A
i
=
40V
8Ω







5kΩ + 50kΩ
0.001V






= 2.75 x 10
8
iii
( )
40 = 0.001
5kΩ
5kΩ + 5kΩ






A
8Ω
8Ω + 8Ω







→ A = 160,000
P =
I
o
2
R
L
2
=
8Ω
2
40V
8Ω






2
= 100 W! A
i
=
40V
8Ω







5kΩ + 5kΩ
0.001V






= 5.00 x 10
7

Page 521

€
A
v
s
( )
=
300s
s + 5000
( )
s +100
( )
Zeros at s = 0 and s = ∞; Poles at s = -5000 and s = −100.

Page 523

€
A
v
s
( )
= −
2
π
x 10
6
s + 5000
π
=
−400
1+
s
5000
π
→ A
mid
= −400 f
H
=
5000
π
2
π
= 2.50 kHz
BW = f
H

− f
L
= 2.50 kHz − 0 = 2.50 kHz GBW = 400
( )
2.50kHz
( )
= 1.00 MHz

3
Page 524

€
i
( )
A
v
j5
( )
= 50
5
2
− 4
5
2
− 2
( )
2
+ 4 5
2
( )

= 41.87 20 log 41.87
( )
= 32.4 dB
∠A
v
j5
( )
= ∠ 5
2
− 4
( )
− tan
−1
−2 5
( )
5
2
− 2








= 0 − −23.5
o
( )
= 23.5

o
A
v
j1
( )
= 50
1
2
− 4
1
2
− 2
( )
2
+ 4 1
2
( )
= 67.08 20 log 41.87
( )
= 36.5 dB
∠A
v
j1
( )
= ∠ 1
2
− 4
( )
− tan
−1

−2 1
( )
1
2
− 2








= 180
o
− −63.43
o
( )
= 243
o
= −117
o

Page 524

€
ii
( )
A
v

j
ω
( )
=
20
1+ j
0.1
ω
1−
ω
2
A
v
j0.95
( )
=
20
1
2
+
0.1
( )
2
0.95
2
( )
1− 0.95
2
( )
2

= 14.3 ∠A
v
j0.95
( )
= ∠20 − tan
−1
0.1 0.95
( )
1− 0.95
2








= 0 − 44.3
o
( )
= −44.3
o
A
v
j1
( )
=
20
1

2
+
0.1
( )
2
1
2
( )
1−1
2
( )
2
= 0 ∠A
v
j1
( )
= ∠20 − tan
−1
0.1 1
( )
1−1
2









= 0 − 90
o
( )
= −90.0
o
A
v
j1.1
( )
=
20
1
2
+
0.1
( )
2
1.1
2
( )
1−1.1
2
( )
2
= 17.7 ∠A
v
j1.1
( )
= ∠20 − tan
−1

0.1 1.1
( )
1−1.1
2








= 0 − −27.6
o
( )
= 27.6
o

Page 526

€
f
H
=
1
2
π
1
1kΩ 100kΩ
( )

200 pF
( )
= 804 kHz

Page 527

€
A
v
s
( )
=
250
1+
250
π
s
A
o
= 250 f
L
=
250
π
2
π
= 125 Hz f
H
= ∞ BW = ∞ −125 = ∞
4

Page 528

€
f
L
=
1
2
π
1
1kΩ 100kΩ
( )
0.1
µ
F
( )
= 15.8 Hz

Page 529

€
i
( )
A
v
s
( )
=
−400
1+

100
s






1+
s
50000






A
o
= 400 or 52 dB
f
L
=
100
2
π
= 15.9 Hz f
H
=
50000

2
π
= 7.96 kHz BW = 7960 −15.9 = 7.94 kHz
ii
( )
∠A
v
j0
( )
= −90 − 0 − 0 = −90
o
∠A
v
j100
( )
= −90
o
− tan
−1
100
100






− tan
−1
100

50000






= −90 − 45 − 0.57 = −136
o
∠A
v
j50000
( )
= −90
o
− tan
−1
50000
100






− tan
−1
50000
50000







= −90 − 89.9 − 45 = −225
o
∠A
v
j∞
( )
= −90 − 90 − 90 = −270
o

Page 531

€
The numerator coefficient should be 6 x10
6
.
A
v
s
( )
= 30
2x10
5
s
s
2

+ 2x10
5
s +10
14
A
o
= 30
f
o
=
1
2
π
10
14
= 1.59 MHz Q =
10
7
2x10
5
= 50 BW =
1.59 MHz
50
= 31.8 kHz

Page 533

€
The transfer fucntion should be A
v

s
( )
=
6.4x10
12
π
2
s
s + 200
π
( )
s + 80000
π
( )
2
.
A
v
s
( )
=
1000
1+
200
π
s







1+
s
80000
π






2
A
o
= 1000 or 60 dB
f
L
=
200
π
2
π
= 100 Hz f
H
= 0.644
80000
π
2
π







= 25.8 kHz BW = 25800 −100 = 25.7 kHz

5
CHAPTER 11

Page 545

€
v
id
=
10V
100
= 0.100V =100 mV v
id
=
10V
10
4
= 0.001 V = 1.00 mV v
id
=
10V
10

6
= 1.00x10
−5
V = 10.0
µ
V

Page 547

€
A
v
= −
360kΩ
68kΩ
= −5.29 v
O
= −5.29 0.5V
( )
= −2.65 V i
S
=
0.5V
68kΩ
= 7.35
µ
A i
O
= −i
2

= −i
S
= −7.35
µ
A

Page 549

€
I
S
=
2V
4.7kΩ
= 426
µ
A I
2
= I
S
= 426
µ
A A
v
= −
24kΩ
4.7kΩ
= −5.11 V
O
= −5.11 2V

( )
= −10.2 V

Page 551

€
A
v
= 1+
36kΩ
2kΩ
= +19.0 v
O
= 19.0 −0.2V
( )
= − 3.80 V i
O
=
−3.80V
36kΩ + 2kΩ
= −100
µ
A

Page 552


€
i
( )

A
v
= 1+
39kΩ
1kΩ
= +40.0 A
vdB
= 20log 40.0
( )
= 32.0 dB R
in
= 100kΩ ∞ =100kΩ
v
O
= 40.0 0.25V
( )
= 10.0 V i
O
=
10.0V
39kΩ + 1kΩ
= 250
µ
A
ii
( )
A
v
= 10
54

20
= 501 1+
R
2
R
1
= 501
R
2
R
1
= 500 i
O
=
v
O
R
2
+ R
1

10
R
2
+ R
1
≤ 0.1 mA
R
1
+ R

2
≥ 100kΩ 501R
1
≥ 100kΩ → R
1
≥ 200 Ω There are many possibilities.
(R
1
= 200 Ω, R
2
= 100 kΩ ), but ( R
1
= 220 Ω, R
2
= 110 kΩ ) is a better solution since
resistor tolerances could cause i
O
to exceed 0.1 mA in the first case.

Page 554

€
Inverting Amplifier : A
v
= −
30kΩ
1.5kΩ
= −20.0 R
in
= R

1
= 1.5 kΩ
v
O
= −20.0 0.15V
( )
= −3.00 V i
O
=
v
O
R
2
=
−3.00V
30kΩ
= −100
µ
A
Non - Inverting Amplifier : A
v
= 1+
30kΩ
1.5kΩ
= +21.0 R
in
=
v
S
i

S
=
0.15V
0 A
= ∞
v
O
= 21.0 0.15V
( )
= 3.15 V i
O
=
v
O
R
2
+ R
1
=
3.15V
30kΩ + 1.5kΩ
= 100
µ
A

6
Page 555

€
V

o1
= 2V −
3kΩ
1kΩ






= −6V V
o2
= 4V −
3kΩ
2kΩ






= −6V v
O
= −6sin1000
π
t − 6sin 2000
π
t
( )
V

The summing junction is a virtual ground : R
in1
=
v
1
i
1
= R
1
= 1 kΩ R
in2
=
v
2
i
2
= R
2
= 2 kΩ
I
o1
=
V
o1
R
3
=
−6V
3kΩ
= −2mA I

o2
=
V
o2
R
3
=
−6V
3kΩ
= −2mA i
O
= −2sin1000
π
t − 2sin2000
π
t
( )
mA

Page 559

€
i
( )
I
2
=
3V
10kΩ + 100kΩ
= 27.3

µ
A
ii
( )
A
v
= −
100kΩ
10kΩ
= −10.0 V
O
= −10 3V − 5V
( )
= 20.0 V I
O
=
V
O
− V
−
100kΩ
=
V
O
− V
+
100kΩ
V
+
= V

2
R
4
R
3
+ R
4
= 5
100kΩ
10kΩ + 100kΩ
= 4.545V I
O
=
20.0 − 4.545
100kΩ
= 155
µ
A I
2
=
5V
10kΩ + 100kΩ
= 45.5
µ
A
iii
( )
A
v
= −

36kΩ
2kΩ
= −18.0 V
O
= −18 8V − 8.25V
( )
= 4.50 V I
O
=
V
O
− V
−
36kΩ
=
V
O
− V
+
36kΩ
V
+
= V
2
R
2
R
1
+ R
2

= 8.25
36kΩ
2kΩ + 36kΩ
= 7.816V I
O
=
4.50 − 7.816
36kΩ
= −92.1
µ
A

Page 560

€
I =
V
A
−V
B
2R
1
=
5.001V − 4.999V
2kΩ
= 1.00
µ
A
V
A

= V
1
+ IR
2
= 5.001V +1.00
µ
A 49kΩ
( )
= 5.05 V
V
B
= V
2
− IR
2
= 4.999V −1.00
µ
A 49kΩ
( )
= 4.95 V
V
O
= −
R
4
R
3







V
A
−V
B
( )
= −
10kΩ
10kΩ






5.05 − 4.95
( )
= −0.100 V

Page 564

€
i
( )
A
v
= −
R

2
R
1
= −10
26
20
= −20.0 R
1
= R
in
= 10kΩ R
2
= 20R
1
= 200kΩ
C =
1
2
π
3kHz
( )
200kΩ
( )
= 265 pF Closest values : R
1
= 10kΩ R
2
= 200kΩ C = 270 pF
7
Page 564

€
ii
( )
R
in
= R
1
= 10 kΩ ΔV = −
I
C
ΔT C =
5V
10kΩ
1
10V






1ms
( )
= 0.05
µ
F
t (msec)
v
O

2
4
6
8
-10V

Page 567

€
v
O
= −RC
dv
S
dt
= − 20kΩ
( )
0.02
µ
F
( )
2.50V
( )
2000
π
( )
cos2000
π
t
( )

= −6.28cos2000
π
t V

Page 569

€
i
( )
A
vA
= A
vB
= A
vC
= −
R
2
R
1
= −
68kΩ
2.7kΩ
= −25.2 R
inA
= R
inB
= R
inC
= R

1
= 2.7 kΩ
The op - amps are ideal : R
outA
= R
outB
= R
outC
= 0
ii
( )
A
v
= A
vA
A
vB
A
vC
= −25.2
( )
3
= −16,000 R
in
= R
inA
= 2.7 kΩ R
out
= R
outC

= 0

Page 570

€
A
v
= −25.2
( )
3
2.7kΩ
R
out
+ 2.7kΩ






2
≥ 0.99 25.2
( )
3

2.7kΩ
R
out
+ 2.7kΩ







2
≥ 0.99
2.7kΩ
R
out
+ 2.7kΩ
≥ 0.9950 R
out
≥13.6 Ω

Page 574

€
i
( )
A
v
0
( )
= +1 A
v
s
( )
=
ω

o
2
s
2
+ s 2
ω
o
+
ω
o
2
A
v
j
ω
( )
=
ω
o
2
j
ω
2
ω
o
+
ω
o
2
−

ω
2
A
v
j
ω
H
( )
=
1
2
→
ω
o
2
ω
o
2
−
ω
H
2
( )
2
+ 2
ω
H
2
ω
o

2
=
1
2
→ 2
ω
o
4
=
ω
o
4
+
ω
H
4
→
ω
o
=
ω
H
ii
( )

1
2
=
C
1

C
2
R
2
2R
→ C
1
= 2C
2
→ C
2
=
1
2 2.26kΩ
( )
20000
π
( )
= 4.98nF
C
2
= 0.005
µ
F C
1
= 0.01
µ
F
8
Page 574

€
iii
( )
To decrease the cutoff frequency from 5kHz to 2 kHz, we must increase the
resistances by a factor of
5kHz
2kHz
= 2.50 → R
1
= R
2
= 2.50 2.26kΩ
( )
= 5.65 kΩ
iv
( )

1
2
=
C
C
R
1
R
2
R
1
+ R

2
→ R
1
2
+ 2R
1
R
2
+ R
2
2
= 2R
1
R
2
→ R
1
2
= −R
2
2
- - can't be done!
Q =
R
1
R
2
R
1
+ R

2

dQ
dR
2
=
1
R
1
+ R
2
( )
2
R
1
R
1
+ R
2
( )
2 R
1
R
2
− R
1
R
2









= 0 → R
2
= R
1
→ Q
max
=
1
2

Page 575

€
S
C
1
Q
=
C
1
Q
dQ
dC
1

=
C
1
Q
1
2 C
1
C
2
R
1
R
2
R
1
+ R
2








=
C
1
Q
Q

2C
1
= 0.5
S
R
2
Q
=
R
2
Q
dQ
dR
2
R
1
= R
2
→ Q =
1
2
C
1
C
2
→ S
R
2
Q
= 0

Page 577

€
i
( )
A
v
j
ω
o
( )
= K
−
ω
o
2
−
ω
o
2
+ j 3 − K
( )
ω
o
2
+
ω
o
2

=
K
3 − K
A
v
j
ω
o
( )
=
K
3 − K
∠90
o
ii
( )
f
o
=
1
2
π
10kΩ 20kΩ
( )
0.0047
µ
F
( )
0.001
µ

F
( )
= 5.19 kHz
Q =
10kΩ
20kΩ
4.7nF +1.0nF
4.7nF 1.0nF
( )
+ 1− 2
( )
20kΩ 1.0nF
( )
10kΩ 4.7nF
( )








−1
= 0.829
iii
( )
S
K
Q

=
K
Q
dQ
dK
Q =
1
3 − K

dQ
dK
=
−1
3 − K
( )
2
−1
( )
= Q
2
S
K
Q
=
K
Q
dQ
dK
= KQ
Q =

1
3 − K
→ KQ = 3Q −1 S
K
Q
= 3Q −1 =1.12

Page 578

€
R
th
= 2kΩ 2kΩ =1kΩ f
o
=
1
2
π
1kΩ 82kΩ
( )
0.02
µ
F
( )
0.02
µ
F
( )
= 879 Hz Q =
1

2
82kΩ
1kΩ
= 4.53

9
Page 582

€
ii
( )
A
BP
j
ω
o
( )
= KQ =
R
2
R
1
10 =
294kΩ
R
1
→ R
1
= 29.4 kΩ
iii

( )
f
o
=
1
2
π
RC
=
1
2
π
40.2kΩ
( )
2nF
( )
= 1.98 kHz BW =
1
2
π
R
2
C
=
1
2
π
402kΩ
( )
2nF

( )
= 198 Hz
A
BP
j
ω
o
( )
= −
R
2
R
1
= −
402kΩ
20.0kΩ
= −20.1
iv
( )
Blindly using the equations at the top of page 580 yields
f
o
min
=
1
2
π
RC
=
1

2
π
1.01
( )
29.4kΩ
( )
1.02
( )
2.7nF
( )
= 1946 Hz
f
o
max
=
1
2
π
RC
=
1
2
π
0.99
( )
29.4kΩ
( )
0.98
( )
2.7nF

( )
= 2067 Hz
BW
min
=
1
2
π
R
2
C
=
1
2
π
1.01
( )
294kΩ
( )
1.02
( )
2.7nF
( )
= 195 Hz
BW
max
=
1
2
π

R
2
C
=
1
2
π
0.99
( )
294kΩ
( )
0.98
( )
2.7nF
( )
= 207 Hz
A
BP
min
= −
R
2
R
1
= −
294kΩ 1.01
( )
14.7kΩ 0.99
( )
= −20.4 A

BP
max
= −
R
2
R
1
= −
294kΩ 0.99
( )
14.7kΩ 1.01
( )
= −19.6
The W/C results are similar if R and C are not the same for example where
ω
o
=
1
R
A
R
B
C
A
C
B
.

Page 583

€
i
( )
- a
( )
R
1
= R
2
= 5 2.26kΩ
( )
= 11.3 kΩ C
1
=
0.02
µ
F
5
= 0.004
µ
F C
2
=
0.01
µ
F
5
= 0.002
µ
F

f
o
=
1
2
π
11.3kΩ
( )
11.3kΩ
( )
0.004
µ
F
( )
0.002
µ
F
( )
= 4980 Hz
Q =
11.3kΩ
11.3kΩ
0.004
µ
F
( )
0.002
µ
F
( )

0.004
µ
F + 0.002
µ
F
= 0.471
b
( )
R
1
= R
2
= 0.885 2.26kΩ
( )
= 2.00 kΩ C
1
=
0.02
µ
F
0.885
= 0.0226
µ
F C
2
=
0.01
µ
F
0.885

= 0.0113
µ
F
f
o
=
1
2
π
2.00kΩ
( )
2.00kΩ
( )
0.0226
µ
F
( )
0.0113
µ
F
( )
= 4980 Hz
Q =
2.00kΩ
2.00kΩ
0.0226
µ
F
( )
0.0113

µ
F
( )
0.0226
µ
F + 0.0113
µ
F
= 0.471
10
Page 583

€
ii
( )
f
o
=
1
2
π
1kΩ
( )
82kΩ
( )
0.02
µ
F
( )
0.02

µ
F
( )
= 879 Hz Q =
82kΩ
1kΩ
0.02
µ
F
( )
0.02
µ
F
( )
0.02
µ
F + 0.02
µ
F
= 4.53
The values of the resistors are unchanged. C
1
= C
2
=
0.02
µ
F
4
= 0.005

µ
F
f
o
=
1
2
π
1kΩ
( )
82kΩ
( )
0.005
µ
F
( )
0.005
µ
F
( )
= 3520 Hz Q =
82kΩ
1kΩ
0.005
µ
F
( )
0.005
µ
F

( )
0.005
µ
F + 0.005
µ
F
= 4.53

Page 585

€
The diode will conduct and pull the output up to v
O
= v
S
= 1.0 V. v
1
= v
O
+ v
D
= 1.0 + 0.6 =1.6 V
For a negative input, there is no path for current through R, so v
O
= 0 V . The op - amp
sees a -1V input so the output will limit at the negative power supply : v
O
= −10 V .
The diode has a 10 - V reverse bias across it, so V
Z

> 10 V .

Page 587

€
i
( )
v
S
= 2 V : Diode D
1
conducts, and D
2
is off. The negative input is a virtual ground.
v
1
= −v
D2
= −0.6 V . The current in R is 0, so v
O
= 0 V .
v
S
= −2 V : Diode D
2
conducts, and D
1
is off. The negative input is a virtual ground.
v
O

= −
R
2
R
1
v
S
= −
68kΩ
22kΩ
−2V
( )
= +6.18 V v
1
= v
O
+ v
D1
= 6.78 V .
v
S
=
15V
−3.09
= −4.85 V v
1
= v
O
+ v
D1

= 6.78 V . When v
O
= 15 V , v
D2
= -15.6 V , so V
Z
= 15.6 V .
ii
( )
v
O
=
20kΩ
20kΩ
10.2kΩ
3.24kΩ






2V
π
= 2.00 V

Page 589

€
V

−
= −
R
1
R
1
+ R
2
V
EE
= −
1kΩ
1kΩ + 9.1kΩ
10V = −0.990 V V
+
=
1kΩ
1kΩ + 9.1kΩ
10V = +0.990 V
V
n
= 0.990V − −0.990V
( )
= 1.98 V

Page 591

€
T = 2 10kΩ
( )

0.001
µ
F
( )
ln
1+ 0.5
1− 0.5






= 21.97
µ
s f =
1
T
= 45.5 kHz

11
Page 594

€
β
= −
22kΩ
22kΩ + 18kΩ
= 0.550 T = 11kΩ
( )

0.002
µ
F
( )
ln
1+
0.7
5
1− 0.550










= 20.4
µ
s
T
r
= 11kΩ
( )
0.002
µ
F
( )

ln
1+ 0.55
5V
5V






1−
0.7
5












= 13.0
µ
s

12

CHAPTER 12

Page 612

€
i
( )
A
ideal
=
1
β
= 100 A
v
=
A
1+ A
β
=
10
5
1+ 10
5
0.01
( )
= 99.90
v
o
= 99.9 0.1V
( )

= 9.99 V v
id
=
v
o
A
=
9.99V
10
5
= 99.9
µ
V
ii
( )
Values taken from OP - 27 specification sheet (www.jaegerblalock.com or www.analog.com)
(iii) Values taken from OP - 27 specification sheet

Page 613

€
A
v
=
−R
2
R
1
A
β

1+ A
β
FGE =
−R
2
R
1






−
−R
2
R
1
A
β
1+ A
β






−R
2

R
1






= 1−
A
β
1+ A
β
=
1
1+ A
β

Page 614

€
FGE =
1
1+ A
β
=
1
1+ 10
4
1kΩ

1kΩ + 39kΩ






= 3.98x10
−3
or 0.398 % FGE ≅
1
A
β
= 0.40 %

Page 615

€
Values taken from OP - 77 specification sheet (www.jaegerblalock.com or www.analog.com)

Page 616

€
A ≅
R
o
β
R
out
=

50Ω
0.025
( )
0.1Ω
( )
= 20,000

Page 617

€
i
( )
A
v
=
A
1+ A
β
=
10
4
1+ 10
4
0.025
( )
= +39.8
ii
( )
A
v

max
=
39kΩ +1kΩ
( )
1.05
( )
1kΩ 0.95
( )
= 44.2 A
v
max
=
39kΩ +1kΩ
( )
0.95
( )
1kΩ 1.05
( )
= 36.2
GE = 44.2 − 40.0 = 4.20 FGE =
4.20
40
= 10.5 % GE = 36.2 − 40.0 = −3.80 FGE =
−3.80
40
= −9.5 %

13
Page 618

€
A ≅
R
o
β
R
out
=
200Ω
0.01
( )
0.1Ω
( )
= 200,000 A
dB
= 20 log 2x10
5
( )
= 106 dB

Page 619

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 620

€
R
in

= R
id
1+ A
β
( )
= 1MΩ 1+ 10
4
10kΩ
10kΩ + 390kΩ












= 251 MΩ
i
−
= −
v
s
R
in
= −

1V
251 MΩ
= −3.98 nA i
1
=
β
v
o
R
1
=
A
β
1+ A
β
v
s
R
1






≅
1V
10kΩ







= 100
µ
A i
1
>> i
−
More exactly, i
1
=
β
v
o
R
1
=
A
β
1+ A
β
v
s
R
1







=
10
4
0.025
( )
1+ 10
4
0.025
( )
1V
10kΩ






= 99.6
µ
A

Page 621

€
R
in
= R

1
+ R
id
R
2
1+ A
= 1kΩ +1MΩ
100kΩ
1+ 10
5
= 1001 Ω R
in
ideal
= R
1
= 1000 Ω 1 Ω or 0.1 %

Page 622

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 626

€
v
o
= A v
id
+

v
ic
CMRR







v
o
min
= A v
id
+
v
ic
CMRR






= 2500 0.002 −
5.000
10
4







= 3.750 V
v
o
max
= A v
id
+
v
ic
CMRR






= 2500 0.002 +
5.000
10
4







= 6.250 V

Page 627

€
A
v
=
A 1+
1
2CMRR






1+ A 1−
1
2CMRR






=
10
4

1+
1
2x10
4






1+ 10
4
1−
1
2x10
4






= 1.000 A
v
=
10
3
1+
1
2x10

3






1+ 10
3
1−
1
2x10
3






= 1.000

14
Page 628

€
GE = FGE A
v
( )
≤ 5x10
−5

1
( )
= 5x10
−5
Worst case occurs for negative CMRR : GE ≅
1
A
+
1
CMRR

If both terms make equal contributions: A = CMRR =
1
2.5x10
−5
= 4x10
4
or 92 dB
For other cases : CMRR = 5x10
−5
−
1
A






−1

or A = 5x10
−5
−
1
CMRR






−1
A = 100dB CMRR = 5x10
−5
−
1
10
5






−1
= 2.5x10
4
or 88 dB
CMRR = 100dB A = 5x10
−5

−
1
10
5






−1
= 2.5x10
4
or 88 dB

Page 630

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 631

€
i
( )
V
O
≤ 50 0.002V
( )
→ −0.100 V ≤ V

O
≤ +0.100 V
ii
( )
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 633

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 634

€
i
( )
R = 39kΩ 1kΩ = 975 Ω
ii
( )
v
O
t
( )
= V
OS
+
V
OS
RC
t +

I
B2
C
t 1.5mV +
1.5mV
10kΩ 100 pF
( )
t +
100nA
100 pF
t =15V → t = 6.00 ms

Page 636

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

Page 637

€
Values taken from op - amp specification sheets (www.jaegerblalock.com or www.analog.com)

15
Page 638

€
R
EQ
= R
L

R
2
+ R
1
( )
≥
20V
5mA
= 4kΩ R
1
+ R
2
≥
1
4kΩ
−
1
5kΩ






−1
= 20kΩ
Including 5% tolerances, R
1
+ R
2

≥ 21kΩ A
v
= 10 → R
2
= 9R
1
A few possibilities : 27 kΩ and 3 kΩ, 270 kΩ and 30 kΩ, 180 kΩ and 20 kΩ, etc.

Page 640

€
i
( )
A
o
= 10
100
20
= 10
5

ω
B
=
ω
T
A
o
=
2

π
5x10
6
( )
10
5
=
10
7
π
10
5
= 100
π
f
B
=
100
π
2
π
= 50 Hz A
v
s
( )
=
10
7
π
s +100

π
ii
( )
A
v
s
( )
=
ω
T
s +
ω
T
A
o
=
2
π
x 10
6
s +
2
π
x 10
6
2x10
5
=
2
π

x 10
6
s +10
π

Page 643

€
i
( )
A
o
= 10
90
20
= 31600 f
B
=
f
T
A
o
=
5x10
6
31600
= 158 Hz A
v
s
( )

=
2
π
5x10
6
( )
s + 2
π
158
( )
=
10
7
π
s + 316
π
f
H
≅
β
f
T
= 0.01 5MHz
( )
= 50 kHz A
v
s
( )
=
2

π
5x10
6
( )
s + 2
π
5x10
4
( )
=
10
7
π
s +10
5
π
ii
( )
For
ω
H
>>
ω
B
: A
β
=
ω
T
s +

ω
B
β
≅
βω
T
j
ω
H
=
1
j
= − j1 since
ω
H
=
βω
T

Page 645

€
i
( )
A
o
= 10
90
20
= 31600 f

B
=
f
T
A
o
=
5x10
6
31600
= 158 Hz A
v
s
( )
=
2
π
5x10
6
( )
s + 2
π
158
( )
=
10
7
π
s + 316
π

f
H
≅
β
f
T
=
5MHz
1+ 10
50
20
= 15.8 kHz A
v
s
( )
=
2
π
5x10
6
( )
s + 2
π
15.8x10
3
( )
=
10
7
π

s + 3.16x10
4
π
ii
( )
f
H
≅
β
f
T
= 1 5MHz
( )
= 5 MHz f
H
≅
β
f
T
=
1
2
5MHz
( )
= 2.5 MHz

16
Page 647

€

A
v
0
( )
= 50 25
( )
= 1250 A
v
ω
H
( )
=
1250
2
= 884
1+
ω
H
2
10000
π
( )
2









1+
ω
H
2
20000
π
( )
2








= 2 →
ω
H
2
( )
2
+ 4.935x10
9
ω
H
2
− 3.896 x10
18

= 0
ω
H
2
= 6.925x10
8
→
ω
H
= 26.3x10
3
→ f
H
=
26.3x10
3
2
π
= 4190 Hz

Page 648

€
i
( )
A
v
0
( )
= −100 66.7

( )
50
( )
= −3.34x10
5
A
v
ω
H
( )
=
−3.34x10
5
2
= −2.36x10
5
1+
ω
H
2
10000
π
( )
2









1+
ω
H
2
15000
π
( )
2








1+
ω
H
2
20000
π
( )
2









= 2
ω
H
6
+ 7.156x10
9
ω
H
4
+ 1.486x10
10
ω
H
2
− 8.562x10
27
= 0
Using MATLAB,
ω
H
= 21.7x10
3
→ f
H
=
21.7x10

3
2
π
= 3450 Hz
ii
( )
A
v
0
( )
= −30
( )
3
= −2.70x10
4
f
H
= 33.3kHz
( )
2
1
3
−1 = 17.0 kHz

Page 653

€
V
M
≤

SR
ω
=
5x10
5
V / s
2
π
20kHz
( )
= 3.98 V f
M
=
SR
2
π
V
FS
=
5x10
5
V / s
2
π
10V
( )
= 7.96 kHz

Page 657

€
A
v1
= 1+
130kΩ
22kΩ
= 6.909 v
O1
= 0.001 6.909
( )
= 6.91 mV v
O2
= 0.001V 6.909
( )
2
= 47.7 mV
v
O3
= 0.001 6.909
( )
3
= 330 mV v
O 4
= 0.001V 6.909
( )
4
= 2.28 V v
O5
= 0.001V 6.909
( )

5
= 15.7 V
v
O5
> 15V → v
O5
= 15 V v
O6
= 15V 6.909
( )
= 104 V v
O6
> 15V → v
O6
= 15 V

17
CHAPTER 13

Page 671

€
i
( )
(a) At the Q - point :
β
F
=
I
C

I
B
=
1.5mA
15
µ
A
= 100 (b) I
S
=
I
C
exp
V
BE
V
T






=
1.5mA
exp
0.700V
0.025V







= 1.04 fA
(c) R
in
=
v
be
i
b
=
8mV
5
µ
A
= 1.6 kΩ (d) With the given applied signal, the smallest value of v
CE
is
v
CE
min
= 5V − 0.5mA 3.3kΩ
( )
= 3.35 V which exceeds v
BE
= 0.708 V.
ii
( )

(a) v
DS
= 10 − 3300i
DS

(b) Using the peak - to - peak voltage swings, A
v
=
v
ds
v
gs
=
2.7 - 6.7
4.0 - 3.0
V
V
= −4.0.
Note that there is some distortion in this amplifier since the negative output voltage excursion is
larger than the positive output change.
(c) v
DS
min
= 2.7V with v
GS
−V
TN
= 4 −1 = 3V , so the transistor has entered the triode region.
(d) Choose two points on the i - v characteristics. For example,
1.56mA =

K
n
2
3.5 −V
TN
( )
2
and 1.0mA =
K
n
2
3.0 − V
TN
( )
2
.
Solving for K
n
and V
TN
yields 500
µ
A
V
2
and 1 V respectively.

Page 673

€

i
( )
V
EQ
=
10kΩ
10kΩ + 30kΩ
12V = 3.00V R
EQ
= 10kΩ 30kΩ = 7.5kΩ
I
C
=
β
F
I
B
=
β
F
V
EQ
−V
BE
R
EQ
+
β
F
+ 1

( )
R
4
= 100
3.0V − 0.7V
7.5kΩ + 101
( )
1.5kΩ
( )
= 1.45 mA
V
CE
= 12 − 4300I
C
−1500I
E
= 12 − 4300 1.45mA
( )
−1500
101
100






1.45mA
( )
= 3.57 V

ii
( )
v
C
t
( )
= V
C
+ v
C
= 5.8 −1.1sin 2000
π
t
( )
V v
E
t
( )
= V
E
+ 0 = 1.45mA 1.5kΩ
( )
= 2.2 V
i
c
=
1.1V
4.3kΩ
= 0.256mA ∠i
c

= 180
o
i
c
t
( )
= −0.26sin 2000
π
t mA
iii
( )
X
C
=
1
ω
C
=
1
2000
π
500
µ
F
( )
= 0.318 Ω

Page 676

€

R
B
= 20kΩ 62kΩ = 15.1 kΩ R
L
= 8.2kΩ 100kΩ = 7.58 kΩ

18
Page 680

€
i
( )
r
d
=
V
T
I
D
+ I
S
r
d
=
0.025V
1 fA
= 25.0 TΩ r
d
=
0.025V

50
µ
A
= 500 Ω
r
d
=
0.025V
2mA
= 12.5 Ω r
d
=
0.025V
3A
= 8.33 mΩ
ii
( )
r
d
=
0.025V
1.5mA
= 16.7 Ω
kT
q
= 8.62x10
−5
V
K







373K
( )
= 0.0322 V r
d
=
0.0322V
1.5mA
= 21.5 Ω

Page 685

€
i
( )
g
m
= 40I
C
= 40 50
µ
A
( )
= 2.00 mS r
π
=

β
o
g
m
=
75
2mS
= 37.5 kΩ
r
o
=
V
A
+ V
CE
I
C
=
60V + 5V
50
µ
A
= 1.30 MΩ
µ
f
= g
m
r
o
= 2mS 1.30 MΩ

( )
= 2600
ii
( )
g
m
= 40I
C
= 40 250
µ
A
( )
= 10.0 mS r
π
=
β
o
g
m
=
50
10mS
= 5.00 kΩ
r
o
=
V
A
+ V
CE

I
C
=
75V +15V
250
µ
A
= 360 kΩ
µ
f
= g
m
r
o
= 10mS 360kΩ
( )
= 3600
iii
( )
The slope of the output characteristics is zero, so V
A
= ∞ and r
o
= ∞.
β
FO
=
β
F
1+

V
CE
V
A
=
β
F
=
I
C
I
B
=
1.5mA
15
µ
A
= 100 g
m
=
Δi
C
Δv
BE
=
0.5mA
8mV
= 62.5 mS
β
o

=
Δi
C
Δi
B
=
500
µ
A
5
µ
A
= 100 r
π
=
β
o
g
m
=
100
62.5mS
= 1.60 kΩ

Page 692

€
A
vt
= −g

m
R
L
= −9.80mS 18kΩ
( )
= −176

19
Page 695

€
Assume the Q - point remains constant.
a
( )
R
L
max
= 19.8kΩ R
L
min
= 16.2kΩ R
E
max
= 3.30kΩ R
L
min
= 2.70kΩ
R
iB
max

= 10.2kΩ + 101 3.3kΩ
( )
= 344kΩ R
iB
min
= 10.2kΩ + 101 2.7kΩ
( )
= 283kΩ
A
v
min
= −
9.80mS 16.2kΩ
( )
1+ 9.80mS 3.30kΩ
( )
104kΩ 344kΩ
1kΩ + 104kΩ 344kΩ








= −4.70
A
v
max

= −
9.80mS 19.8kΩ
( )
1+ 9.80mS 2.70kΩ
( )
104kΩ 283kΩ
1kΩ + 104kΩ 283kΩ








= −6.98
b
( )
r
π
=
125
9.80mS
= 12.8kΩ R
iB
= 12.8kΩ + 126 3.0kΩ
( )
= 391kΩ
A
v

= −
9.80mS 18kΩ
( )
1+ 9.80mS 3.00kΩ
( )
104kΩ 391kΩ
1kΩ + 104kΩ 391kΩ








= −5.73
c
( )
V
CE
= 12V − 22kΩI
C
−13kΩI
E
= 12V − 0.275mA 22kΩ +
101
100
13kΩ







= 2.34 V
g
m
= 40 0.275mA
( )
= 11.0mS r
π
=
100
11.0mS
= 9.09kΩ
R
iB
= 9.09kΩ + 101 3.0kΩ
( )
= 312kΩ A
v
= −
11.0mS 18kΩ
( )
1+ 11.0mS 3.00kΩ
( )
104kΩ 312kΩ
1kΩ + 104kΩ 312kΩ









= −5.75

Page 697

€
i
( )
R
iB
max
= 10.2kΩ + 101 1kΩ
( )
= 111kΩ R
E2
= 13kΩ −1kΩ = 12 kΩ
A
v
= −
9.80mS 18kΩ
( )
1+ 9.80mS 1.00kΩ
( )
104kΩ 111kΩ
1kΩ + 104kΩ 111kΩ









= −16.0
ii
( )
The reference to C
2
should be C
3
. A
v
= −159 as calculated in the Ex. 13.4.
iii
( )
V
T
=
1.38x10
−23
1.602x10
−19
V
K







273K + 27K
( )
= 25.84mV I
S
=
I
C
exp
V
BE
V
T






=
245
µ
A
exp
0.700V
0.02584V







= 0.421fA
iv
( )
From Ex. 13.3, R
iB
= 313 kΩ, and r
π
= 10.2kΩ.

20
Page 700

€
i
( )
R
iC
= 320kΩ 1+
100 2kΩ
( )
1kΩ 104kΩ
( )
+ 10.2kΩ + 2kΩ









= 4.85 MΩ
µ
f
R
E
= 3140 2kΩ
( )
= 6.28 MΩ
R
iC
<
µ
f
R
E
R
out
= 4.85MΩ 22kΩ = 21.9 kΩ
ii
( )
lim R
iC
R

E
→∞
= lim
R
E
→∞
r
o
1+
β
o
R
E
R
th
+ r
π
+ R
E






= r
o
1+
β
o

R
E
R
E






=
β
o
+ 1
( )
r
o

Page 704

€
i
( )
a
( )
g
m
= 2K
n
I

D
1+
λ
V
DS
( )
= 2 1mA/V
2
( )
0.25mA
( )
1+ 0.02 5
( )
[ ]
= 1.73 mS
r
o
=
1
λ
+ V
DS
I
D
=
50V + 5V
250
µ
A
= 220 kΩ

µ
f
= g
m
r
o
= 0.742mS 220kΩ
( )
= 163
g
m
= 2K
n
I
D
1+
λ
V
DS
( )
= 2 1mA/V
2
( )
5mA
( )
1+ 0.02 10
( )
[ ]
= 3.46 mS
r

o
=
1
λ
+ V
DS
I
D
=
50V + 10V
5mA
= 12 kΩ
µ
f
= g
m
r
o
= 3.46mS 12kΩ
( )
= 41.5
b
( )
The slope of the output characteristics is zero, so
λ
= 0 and r
o
= ∞.
For the positive change in v
gs

, g
m
=
Δi
D
Δv
GS
≅
2.1V
3.3kΩ
0.5V
= 1.3 mS
ii
( )
v
gs
≤ 0.2 V
GS
− V
TN
( )
= 0.2
2I
D
K
n
= 0.2
2 25mA
( )
2.0mA/V

2
= 1.00 V v
be
≤ 0.005 V

Page 705

€
η
=
γ
2 V
SB
+ 2
φ
F
=
0.75
2 0 + 0.6
= 0.48
η
=
0.75
2 3+ 0.6
= 0.20

21
Page 713

€

i
( )
V
EQ
=
1.5MΩ
1.5MΩ + 2.2 MΩ
12V = 4.87V R
EQ
= 1.5MΩ 2.2 MΩ = 892kΩ
Neglect
λ
in hand calculations of the Q - point.
4.87 = V
GS
+ 12000I
D
4.87 = V
GS
+ 12000
5x10
−4
2






V

GS
−1
( )
2
3V
GS
2
− 5V
GS
−1.87 = 0 → V
GS
= 1.981V I
D
= 241
µ
A
V
DS
= 12 − 22000I
D
−12000I
D
= 3.81 V Q - point : 241
µ
A, 3.81 V
( )
ii
( )
A
vdB

= 20 log 4.50
( )
= 13.1 dB

Page 714

€
V
DS
≥ V
GS
− V
TH
= 1.981V −1V = 0.981 V

Page 717

€
V
GS
− V
TN
≅
2 241
µ
A
( )
2x10
−3
= 0.491V A

v
CS
≅ −
12V
0.491V
= −24.4 M =
K
n2
K
n1
=
2x10
−3
5x10
−4
= 4

Page 720

€
R
in
CS
= 680kΩ 1.0 MΩ = 405 kΩ V
EQ
=
680kΩ
680kΩ + 1MΩ
12V = 4.86 V

Page 725

€
The gain is proportional to R
L
: R
L
= 10kΩ 220kΩ = 9.57kΩ A
v
CE
= −36.1
9.57kΩ
9.540kΩ






= −36.8

Page 726

€
i
( )

µ
f
= g

m
r
o
= 9.56mS 225kΩ
( )
= 2150
ii
( )
v
be
= v
i
R
in
CE
R
I
+ R
in
CE






1
1+ g
m
R

E






v
i
≤ 5mV
0.33kΩ +14.2kΩ
14.2kΩ






1+ 9.56mS 0.15kΩ
( )
[ ]
= 12.4 mV
v
o
≤12.4mV 36.1
( )
= 0.450 V
iv
( )
R

in
CE
= 100kΩ 6.8kΩ = 6.37 kΩ R
out
CE
= 10kΩ 225kΩ = 9.57 kΩ
A
v
CE
= −g
m
R
L
R
in
CE
R
I
+ R
in
CE






= -9.56mS 9.57kΩ 220kΩ
( )
6.37kΩ

0.33kΩ + 6.37kΩ






= −83.4

22
Page 730

€
i
( )
The gain is proportional to R
L
.
R
L
= 300kΩ R
out
CS
= 300kΩ 28.4kΩ = 25.9kΩ A
v
CS
= −6.85
25.9kΩ
27.3kΩ







= −6.50
The corrected gain agrees more closely with thevalue from SPICE.
ii
( )

µ
f
= g
m
r
o
= 0.515mS 258kΩ
( )
= 133
iii
( )
v
gs
= v
i
R
in
CS
R
I

+ R
in
CS






1
1+ g
m
R
S






v
gs
≤ 0.2 V
GS
−V
TN
( )
v
i
≤ 0.2 1V

( )
10kΩ +1MΩ
1MΩ






1+ 0.515mS 2kΩ
( )
[ ]
= 410 mV v
o
≤ 410mV 6.85
( )
= 2.81 V
iv
( )
A
v
CS
= −
R
in
CS
R
I
+ R
in

CS






g
m
R
L
1+ g
m
R
S






= −
1MΩ
10kΩ + 1MΩ







0.515mS 30kΩ 300kΩ
( )
1+ 0.515mS 32kΩ
( )








= −0.796
The calculation is slightly larger than the SPICE value since it neglects R
out
CS
.

Page 734

€
i
( )
P
D
= I
C
V
CE
+ I

B
V
BE
= 239
µ
A 3.67V
( )
+
239
µ
A
65
0.7V
( )
= 0.880 mW
P
S
= I
C
V
CC
+ I
E
V
EE
= 239
µ
A 5V
( )
+ 239

µ
A 1+
1
65






5V
( )
= 2.41 mW
i
( )
P
D
= I
D
V
DS
= 250
µ
A 4.5V
( )
= 1.13 mW
P
S
= I
D

V
DD
+ I
S
V
SS
= 250
µ
A 10V
( )
+ 250
µ
A 10V
( )
= 5.00 mW

Page 735

€
a
( )
V
M
≤ min I
C
R
C
, V
CE
− V

BE
( )
[ ]
= min 239
µ
A 10kΩ
( )
, 3.67 − 0.7
( )
V
[ ]
= 2.39 V
Limited by the voltage drop across R
C
.
b
( )
V
M
≤ min I
D
R
D
, V
DS
− V
DSSAT
( )
[ ]
= min 250

µ
A 30kΩ
( )
, 4.50 −1
( )
V
[ ]
= 3.50 V
Limited by the value of V
DS
.

23
CHAPTER 14

Page 754

€
R
B
= 160kΩ 300kΩ =104 kΩ R
E
= 3.00 kΩ R
L
= 100kΩ 22kΩ =18.0 kΩ
R
G
= 1.5MΩ 2.2 MΩ = 892 kΩ R
S
= 2.00 kΩ R

L
= 100kΩ 22kΩ = 18.0 kΩ

Page 761

€
i
( )
A
v
CE
= −
0.838V
0.150
= −5.59
ii
( )
R
iB
= 10.2kΩ + 101 1.0kΩ
( )
= 111kΩ
A
v
CE
= −
9.80mS 18kΩ
( )
1+ 9.80mS 1kΩ
( )

104kΩ 111kΩ
2kΩ +104kΩ 111kΩ








= −15.8 R
4
= 13kΩ −1kΩ =12 kΩ
A
v
CS
= −
0.491mS 18kΩ
( )
1+ 0.491mS 1kΩ
( )
892kΩ
2kΩ + 892kΩ






= −5.91 R

4
= 12kΩ −1kΩ =11 kΩ
iii
( )
R
iB
= 10.2kΩ
A
v
CE
= −9.80mS 18kΩ
( )
104kΩ 10.2kΩ
2kΩ +104kΩ 10.2kΩ








= −145
A
v
CS
= −0.491mS 18kΩ
( )
892kΩ
2kΩ + 892kΩ







= −8.82
iv
( )
V
T
=
1.38x10
−23
1.602x10
−19
V
K






273K + 27K
( )
= 25.84mV I
S
=
I

C
exp
V
BE
V
T






=
245
µ
A
exp
0.700V
0.02584V






= 0.421 fA
v
( )
g
m

R
L
= -9.80mS 18kΩ
( )
= −176, A
v
CE
≅ −
18kΩ
3kΩ
= −9.00
g
m
R
L
= -0.491mS 18kΩ
( )
= −8.84, A
v
CE
≅ −
18kΩ
2kΩ
= −9.00
vi
( )
The exercise should refer to Fig. 14.2. R
iB
= 10.2kΩ + 101 3.0kΩ
( )

= 313kΩ
R
in
CE
= R
B
R
iB
= 104kΩ 313kΩ = 78.1 kΩ R
in
CS
= R
G
= 892kΩ R
in
CE
> r
π

24
Page 761

€
i
( )
A
vt
=
2kΩ + 892kΩ
892kΩ

0.956 = 0.958
0.491ms
( )
R
L
1+ 0.491ms
( )
R
L
= 0.958 → R
L
= 46.5kΩ
R
6
100kΩ = 46.5kΩ → R
6
= 86.9kΩ
ii
( )
R
iB
= 10.2kΩ + 101 13kΩ
( )
= 1.32 MΩ R
in
CC
= 104kΩ 1.32 MΩ = 96.4kΩ
A
v
CE

= −
9.80mS 13kΩ
( )
1+ 9.80mS 13kΩ
( )
96.4kΩ
2kΩ + 96.4kΩ






= +0.972
A
v
CS
= −
0.491mS 12kΩ
( )
1+ 0.491mS 12kΩ
( )
892kΩ
2kΩ + 892kΩ







= +0.853
iv
( )
BJT : g
m
R
L
= 9.80mS 11.5kΩ
( )
= 113 FET : g
m
R
L
= 0.491mS 10.7kΩ
( )
= 5.25

Page 767

€
v
i
≤ 0.005V 1+ g
m
R
L
( )
R
I
+ R

B
R
iB
R
B
R
iB
= 0.005V 1+ 9.8mS 11.5kΩ
( )
[ ]
2kΩ + 95.4kΩ
95.4kΩ
= 0.580 V
v
i
≤ 0.2 V
GS
− V
TN
( )
1+ g
m
R
L
( )
R
I
+ R
G
R

G
= 0.2 0.982
( )
1+ 0.491mS 10.7kΩ
( )
[ ]
2kΩ + 892kΩ
892kΩ
= 1.23 V

Page 769

€
R
out
CD
= R
6
1
g
m
12kΩ
1
g
m
= 120Ω → g
m
= 8.25mS 8.25mS = 2 500
µ
A/V

2
( )
I
D
→ I
D
= 68.1 mA
g
m
=
2I
D
V
GS
− V
TN
V
GS
− V
TN
=
2 68.1mA
( )
8.25mS
= 16.5 V
If one neglects R
6
, I
D
= 69.4 mA and V

GS
− V
TN
= 16.8 V

Page 774

€
BJT : v
i
≤ 0.005V 1+ g
m
R
I
( )
R
I
+ R
6
R
6
= 0.005V 1+ 9.8mS 2kΩ
( )
[ ]
2kΩ + 13kΩ
13kΩ







= 119 mV
Neglecting R
6
, v
i
≤ 0.005V 1+ g
m
R
I
( )
= 0.005V 1+ 9.8mS 2kΩ
( )
[ ]
= 103 mV
FET : v
i
≤ 0.2 V
GS
−V
TN
( )
1+ g
m
R
I
( )
R
I

+ R
6
R
6
= 0.2 0.982
( )
1+ 0.491mS 2kΩ
( )
[ ]
2kΩ + 12kΩ
12kΩ
= 454 mV
Neglecting R
6
, v
i
≤ 0.2 V
GS
−V
TN
( )
1+ g
m
R
I
( )
= 0.2 0.982
( )
1+ 0.491mS 2kΩ
( )

[ ]
= 389 mV

25
Page 775

€
R
iC
= r
o
1+
β
o
R
th
R
th
+ r
π






= 219kΩ 1+
100 1.73kΩ
( )
1.73kΩ +10.2kΩ









= 3.40 MΩ
Or more approximately, R
iC
= r
o
1+ g
m
R
th
[ ]
= 219kΩ 1+ 9.8mS 1.73kΩ
( )
[ ]
= 3.93 MΩ
R
iD
= r
o
1+ g
m
R
th

[ ]
= 223kΩ 1+ 0.491 1.71kΩ
( )
[ ]
= 4.10 MΩ

Page 778

€
i
( )
A
v
CB
= g
m
R
L
R
6
1
g
m






R

6
+
1
g
m
R
I
+
R
6
1
g
m






R
6
+
1
g
m
= g
m
R
L
R

6
R
6
+
1
g
m
g
m
R
I
+
R
6
R
6
+
1
g
m
= g
m
R
L
R
6
R
6
1+ g
m

R
I
( )
+ R
I
A
v
CB
= g
m
R
L
R
6
R
6
+ R
I
1
1+
g
m
R
I
R
6
R
6
+ R
I

=
g
m
R
L
1+ g
m
R
th
R
6
R
6
+ R
I






ii
( )
The voltage gains are proportional to the load resistance
A
v
CE
= +8.48
22kΩ
18kΩ







= +10.4 A
v
CG
= +4.12
22kΩ
18kΩ






= +5.02
iii
( )
CB : A
v
CB
≤ g
m
R
L
= 176 A
v

CB
≅
R
L
R
th
=
R
L
R
I
R
6
=
18kΩ
1.73kΩ
= 10.4 8.48 < 10.4 <<176
CG : A
v
CG
≤ g
m
R
L
= 8.84 A
v
CB
≅
R
L

R
th
=
R
L
R
I
R
6
=
18kΩ
1.71kΩ
= 10.5 4.11 < 8.84 < 10.5

Page 786

€
i
( )
Since we need high gain, the emitter should be bypassed, and R
in
CE
= R
B
r
π
= 250kΩ.
If we choose R
B
≅ r

π
, I
C
=
β
o
40r
π
≅
100
40 500kΩ
( )
= 5
µ
A
ii
( )
R
in
CG
≅
1
g
m
I
C
≅
1
40 2kΩ
( )

= 12.5
µ
A