This document contains solutions to problems in Chapter 13 regarding computer simulations and analyses of bipolar junction transistor circuits. The key points summarized are: 1) Problem 13.1 provides a computer simulation of a circuit and calculates the overall gain as -1.59x106. 2) Problem 13.6 calculates various currents in a BJT reference circuit, including the reference current of 0.5 mA and collector current of 21.1 μA. 3) Problem 13.8 calculates currents in another BJT reference circuit, including the reference current of 0.22 mA, collector current of 14.2 μA, and other node currents.

1. Chapter 13
Problem Solutions
13.1 Computer Simulation
13.2 Computer Simulation
13.3
(a) (
Ad = g m1 ro 2 ro 4 Ri 6 )
I C1 20
g m1 = = ⇒ 0.769 mA / V
VT 0.026
VA 2 80
ro 2 = = = 4 MΩ
I C 2 20
VA 4 80
ro 4 = = = 4 MΩ
I C 2 20
Ri 6 = rπ 6 + (1 + β n ) ⎡ R1 rπ 7 ⎤
⎣ ⎦
(120)(0.026)
rπ 7 = = 15.6 k Ω
0.2
V (on) 0.6
I C 6 ≅ BE = = 0.030 mA
R1 20
(120)(0.026)
rπ 6 = = 104 k Ω
0.030
Then
Ri 6 = 104 + (121) ⎡ 20 15.6 ⎤ ⇒ 1.16 M Ω
⎣ ⎦
Then
(
Ad = 769 4 4 1.16 ⇒ Ad = 565 )
Now
⎛ R1 ⎞
Vo = − I c 7 ro 7 = −( β n I b 7 )ro 7 = − β n ro 7 ⎜ ⎟ Ic6
⎝ R1 + rπ 7 ⎠
⎛ R1 ⎞ Vo1
= − β n (1 + β n )ro 7 ⎜ ⎟ I b 6 and I b 6 =
⎝ R1 + rπ 7 ⎠ Ri 6
Then
V − β n (1 + β n )ro 7 ⎛ R1 ⎞
Av 2 = o = ⎜ ⎟
Vo1 Ri 6 ⎝ R1 + rπ 7 ⎠
VA 80
ro 7 = = = 400 k Ω
I C 7 0.2
So
−(120)(121)(400) ⎛ 20 ⎞
Av 2 = ⎜ ⎟ ⇒ Av 2 = −2813
1160 ⎝ 20 + 15.6 ⎠
Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 ×106
(80)(0.026)
(b) Rid = 2rπ 1 and rπ 1 = = 104 k Ω
0.020
Rid = 208 k Ω
1
(c) f PD = and CM = (10)(1 + 2813) = 28,140 pF
2π Req CM
Req = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M Ω
1
f PD = = 7.71 Hz
2π (0.734 × 10 )(28,140 × 10−12 )
6

2. Gain-Bandwidth Product = (7.71)(1.59 × 106 ) ⇒ 12.3 MHz
13.4
a. Q3 acts as the protection device.
b. Same as part (a).
13.5
If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5
So breakdown voltage ≈ 56.4 V.
13.6
15 − 0.6 − 0.6 − (−15)
(a) I REF = = 0.50 ⇒ R5 = 57.6 k Ω
R5
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
0.026 ⎛ 0.50 ⎞
R4 = ln ⎜ ⎟ ⇒ R4 = 2.44 k Ω
0.030 ⎝ 0.030 ⎠
5 − 0.6 − 0.6 − (−5)
(b) I REF = ⇒ I REF = 0.153 mA
57.6
⎛ 0.153 ⎞
I C10 (2.44) = (0.026) ln ⎜ ⎟
⎝ I C10 ⎠
By trial and error, I C10 ≅ 21.1 μ A
13.7
(a) I REF ≅ 0.50 mA
⎛I ⎞ ⎛ 0.50 × 10−3 ⎞
VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜ −14 ⎟ ⇒ VBE11 = 0.641V = VEB12
⎝ IS ⎠ ⎝ 10 ⎠
Then
15 − 0.641 − 0.641 − (−15)
R5 = ⇒ R5 = 57.4 k Ω
0.50
0.026 ⎛ 0.50 ⎞
R4 = ln ⎜ ⎟ ⇒ R4 = 2.44 k Ω
0.030 ⎝ 0.030 ⎠
⎛ 0.030 × 10−3 ⎞
VBE10 = 0.026 ln ⎜ −14 ⎟ ⇒ VBE10 = 0.567 V
⎝ 10 ⎠
(b) From Problem 13.6, I REF ≅ 0.15 mA
⎛ 0.15 × 10−3 ⎞
VBE11 = VEB12 = 0.026 ln ⎜ −14 ⎟ = 0.609 V
⎝ 10 ⎠
5 − 0.609 − 0.609 − (−5)
Then I REF = ⇒ I REF = 0.153 mA
57.4
Then I C10 ≅ 21.1 μ A from Problem 13.6
13.8
5 − 0.6 − 0.6 − (−5)
a. I REF = ⇒ I REF = 0.22 mA
40
⎛I ⎞
I C10 R4 = VT ln ⎜ REF ⎟
⎝ I C10 ⎠
⎛ 0.22 ⎞
I C10 (5) = (0.026) ln ⎜ ⎟
⎝ I C10 ⎠

3. By trial and error;
I C10 ≅ 14.2 μ A
I C10
IC 6 ≅ ⇒ I C 6 = 7.10 μ A
2
I C17 = 0.75 I REF ⇒ I C17 = 0.165 mA
I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA
(b) Using Example 13.4
rπ 17 = 31.5 kΩ
′
RE = 50 [31.5 + (201)(0.1)] = 50 51.6 = 25.4 kΩ
β nVT
rπ 16 = and
I C16
0.165 (0.165)(0.1) + 0.6
I C16 = + = 0.0132 mA
200 50
rπ 16 = 394 kΩ
Then
Ri 2 = 394 + (201)(25.4) ⇒ 5.5 MΩ
rπ 6 = 732 kΩ
0.00710
gm6 = = 0.273 mA / V
0.026
50
r06 = = 7.04 MΩ
0.0071
Then
Ract1 = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ
50
r04 = = 7.04 MΩ
0.0071
Then
⎛ 7.1 ⎞
Ad = − ⎜ ⎟ (7.04 8.96 5.5)
⎝ 0.026 ⎠
or
Ad = −627 Gain of differential amp stage
Using Example 13.5, and neglecting the input resistance to the output stage:
V 50
Ract 2 = A = = 303 kΩ
I C13 B 0.165
−(200)(201)(50)(303) (303)
Av 2 =
(5500)[50 + 31.5 + (201)(0.1)]
or
Av 2 = −545 Gain of second stage
13.9
I C10 = 19 μ A
From Equation (13.6)
⎡ β 2 + 2β P + 2 ⎤ ⎡ (10) 2 + 2(10) + 2 ⎤
I C10 = 2 I ⎢ P ⎥ = 2I ⎢ ⎥
⎣ β P + 3β P + 2 ⎦ ⎣ (10) + 3(10) + 2 ⎦
2 2
⎡122 ⎤
= 2I ⎢ ⎥
⎣132 ⎦
So
⎛ 132 ⎞
2 I = (19) ⎜ ⎟ = 20.56 μ A
⎝ 122 ⎠
I C 2 = I = 10.28 μ A

4. 2I 20.56
IC 9 = = ⇒ I C 9 = 17.13 μ A
⎛ 2 ⎞ ⎛ 2⎞
⎜1 + ⎟ ⎜ 1+ ⎟
⎝ βP ⎠ ⎝ 10 ⎠
I 17.13
I B9 = C9 = ⇒ I B 9 = 1.713 μ A
βP 10
I 10.28
IB4 = = ⇒ I B 4 = 0.9345 μ A
(1 + β P ) 11
⎛ β ⎞ ⎛ 10 ⎞
IC 4 = I ⎜ P ⎟ = (10.28) ⎜ ⎟ ⇒ I C 4 = 9.345 μ A
⎝1+ βP ⎠ ⎝ 11 ⎠
13.10
VB 5 − V − = VBE (on) + I C 5 (1)
= 0.6 + (0.0095)(1) = 0.6095
0.6095
IC 7 = ⇒ I C 7 = 12.2 μ A
50
I C 8 = I C 9 = 19 μ A
I REF = 0.72 mA
I E13 = I REF = 0.72 mA
I C14 = 138 μ A
Power = (V + − V − ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ]
= 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138]
⇒ Power = 48.8 mW
Current supplied by V + and V − = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14
= 1.63 mA
13.11
(a) vcm (min) = −15 + 0.6 + 0.6 + 0.6 + 0.6 = −12.6 V
vcm (max) = +15 − .6 = 14.4 V
So − 12.6 ≤ vcm ≤ 14.4 V
(b) vcm (min) = −5 + 4(0.6) = −2.6 V
vcm (max) = 5 − 0.6 = 4.4 V
So − 2.6 ≤ vcm ≤ 4.4 V
13.12
If v0 = V − = −15 V , the base voltage of Q14 is pulled low, and Q18 and Q19 are effectively cut off. As
a first approximation
0.6
I C14 = = 22.2 mA
0.027
22.2
I B14 = = 0.111 mA
200
Then
I C15 = I C13 A − I B14 = 0.18 − 0.111 = 0.069 mA
Now
⎛I ⎞
VBE15 = VT ln ⎜ C15 ⎟
⎝ 15 ⎠
⎛ 0.069 × 10−3 ⎞
= (0.026) ln ⎜ −14 ⎟
⎝ 10 ⎠
= 0.589 V

5. As a second approximation
0.589
I C14 = ⇒ I C14 = 21.8 mA
0.027
21.8
I B14 = = 0.109 mA
200
and
I C15 = 0.18 − 0.109 ⇒ I C15 = 0.071 mA
13.13
a. Neglecting base currents:
I D = I BIAS
Then
⎛I ⎞
VBB = 2VD = 2VT ln ⎜ D ⎟
⎝ IS ⎠
⎛ 0.25 × 10−3 ⎞
= 2(0.026) ln ⎜ −14 ⎟
⎝ 2 × 10 ⎠
or
VBB = 1.2089 V
⎛V / 2⎞
I CN = I CP = I S exp ⎜ BB ⎟
⎝ VT ⎠
⎛ 1.2089 ⎞
= 5 × 10−14 exp ⎜ ⎟
⎝ 2(0.026) ⎠
So
I CN = I CP = 0.625 mA
b. For vI = 5 V, v0 ≅ 5 V
5
iL ≅ = 1.25 mA
4
As a first approximation
I CN ≈ iL = 1.25 mA
⎛ 1.25 × 10−3 ⎞
VBEN = (0.026) ln ⎜ −14 ⎟
= 0.6225 V
⎝ 5 × 10 ⎠
Neglecting base currents,
VBB = 1.2089 V
Then VEBP = 1.2089 − 0.6225 = 0.5864 V
⎛ 0.5864 ⎞
I CP = 5 × 10−14 exp ⎜ ⎟ ⇒ I CP = 0.312 mA
⎝ 0.026 ⎠
As a second approximation,
I CN = iL + I CP = 1.25 + 0.31 ⇒ I CN ≅ 1.56 mA
13.14
VBB 1.157
R1 + R2 = = = 64.28 kΩ
(0.1) I BIAS 0.018
⎛I ⎞ ⎛ (0.9) I BIAS ⎞
VBE = VT ln ⎜ C ⎟ = (0.026) ln ⎜ ⎟
⎝ IS ⎠ ⎝ IS ⎠
⎛ 0.162 × 10−3 ⎞
= (0.026) ln ⎜ −14 ⎟
⎝ 10 ⎠

6. VBE = 0.6112 V
⎛ R2 ⎞
VBE = ⎜ ⎟ VBB
⎝ R1 + R2 ⎠
⎛ R ⎞
0.6112 = ⎜ 2 ⎟ (1.157)
⎝ 64.28 ⎠
So
R2 = 33.96 kΩ
Then
R1 = 30.32 kΩ
13.15
(a) (
Ad = − g m ro 4 ro 6 Ri 2 )
From example 13.4
9.5
gm = = 365 μ A / V , ro 4 = 5.26 M Ω
0.026
Now
ro 6 = ro 4 = 5.26 M Ω
Assuming R8 = 0, we find
′
Ri 2 = rπ 16 + (1 + β n ) RE
= 329 + (201) ( 50 9.63) ⇒ 1.95 M Ω
Then
( )
Ad = −(365) 5.26 5.26 1.95 ⇒ Ad = −409
(b) From Equation (13.20),
Av 2 =
(
− β n (1 + β n ) R9 Ract 2 Ri 3 R017 )
{
Ri 2 R9 + ⎡ rπ 17 + (1 + β n ) Rg ⎤
⎣ ⎦}
For Rg = 0, Ri 2 = 1.95 M Ω
Using the results of Example 13.5
Av 2 =
(
−200(201)(50) 92.6 4050 92.6 )⇒A = −792
(1950){50 + 9.63}
v2
13.16
Let I C10 = 40 μ A, then I C1 = I C 2 = 20 μ A. Using Example 13.5,
Ri 2 = 4.07 MΩ
(200)(0.026)
rπ 6 = = 260 kΩ
0.020
0.020
gm6 = = 0.769 mA/V
0.026
50
r06 = ⇒ 2.5 MΩ
0.02
Then
Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ
50
r06 = ⇒ 2.5 MΩ
0.02
Then

7. ⎛ I CQ ⎞
Ad = − ⎜ ⎟ (r04 Ract1 Ri 2 )
⎝ VT ⎠
⎛ 20 ⎞
= −⎜ ⎟ (2.5 4.42 4.07)
⎝ 0.026 ⎠
So
Ad = −882
13.17
From Problem 13.8
I1 = I 2 = 7.10 μ A, I C17 = 0.165 mA, I C13 A = 0.055 mA
I E17 R8 + VBE17 0.165 (0.165)(0.1) + 0.6
I C16 ≈ I B17 + = +
R9 200 50
= 0.000825 + 0.01233
I C16 = 0.0132 mA
(200)(0.026)
rπ 17 = = 31.5 K
0.165
RE = R9 [ rπ 17 + (1 + β ) R8 ] = 50 [31.5 + (201)(0.1)]
1
= 50 51.6 = 25.4 K
(200)(0.026)
rπ 16 = = 394 K
0.0132
Then
Ri 2 = rπ 16 + (1 + β ) RE = 394 + (201)(25.4) ⇒ 5.50 MΩ
1
Now
(200)(0.026)
rπ 6 = = 732 K
0.0071
0.0071
gm6 = = 0.273 mA/V
0.026
50
ro 6 = ⇒ 7.04 MΩ
0.0071
Ract1 = ro 6 [1 + g m 6 ( R rπ 6 )]
= 7.04[1 + (0.273)(1 732)] = 8.96 MΩ
50
ro 4 = ⇒ 7.04 MΩ
0.0071
Then
Ad = − g m1 (ro 4 Ract1 Ri 2 )
⎛ 7.10 ⎞
= −⎜ ⎟ (7.04 8.96 5.5)
⎝ 0.026 ⎠
Ad = −627
50 50
Now Ract 2 = ⇒ 303 K Ro17 = = 303 K
0.165 0.165
From Eq. (13.20), assuming Ri 3 → ∞
β (1 + β ) R9 ( Ract 2 R017 )
Av 2 ≅ −
Ri 2 { R9 + [rπ 17 + (1 + β ) R8 ]}
−(200)(201)(50)(303 303) −3.045 × 108
= =
(5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105
Av 2 = −545
Overall gain Av = (−627)(−545) = 341, 715

8. 13.18
Using results from 13.17
⎛ 100 ⎞
Ri 2 = 5.50 MΩ, Ract1 ⎜ ⎟ [1 + (0.273)(1 732)] ⇒ 17.93 MΩ
⎝ 0.0071 ⎠
100
ro 4 = ⇒ 14.08 MΩ
0.0071
⎛ 7.10 ⎞
Ad = − ⎜ ⎟ (14.08 17.93 5.50)
⎝ 0.026 ⎠
Ad = −885
Now
100 100
Ract 2 = = 606 K Ro17 = = 606 K
0.165 0.165
−(200)(201)(50)(606 606) −6.09 × 108
Av 2 = =
(5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105
Av 2 = −1090
Overall gain
Av = (−885)(−1090) = 964, 650
13.19
Now
rπ 14 + R01
Re14 = and R0 = R6 + Re14
1+ βP
Assume series resistance of Q18 and Q19 is small. Then
R01 = r013 A Re 22
rπ 22 + R017 r013 B
where Re 22 =
1+ βP
and R017 = r017 [1 + g m17 ( R8 rπ 17 )]
Using results from Example 13.6,
rπ 17 = 9.63 kΩ rπ 22 = 7.22 kΩ
g m17 = 20.8 mA/V r017 = 92.6 kΩ
Then
R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 kΩ
50
r013 B = = 92.6 kΩ
0.54
Then
7.22 + 283 92.6
Re 22 = = 1.51 kΩ
51
R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ
(50)(0.026)
rπ 14 = = 0.65 kΩ
2
Then
0.65 + 1.50
Re14 = = 0.0422 kΩ
51
or
Re14 = 42.2 Ω
Then
R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω
13.20

9. ⎡ ⎛ r ⎞⎤
Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3 ⎟⎥
⎣ ⎝ 1+ βP ⎠⎦
β n = 200, β P = 10
(a)
I C1 = 9.5 μ A
(200)(0.026)
rπ 1 = = 547 K
0.0095
(10)(0.026)
rπ 3 = = 27.4 K
0.0095
Then
⎡ (201)(27.4) ⎤
Rid = 2 ⎢547 + ⎥
⎣ 11 ⎦
Rid ⇒ 2.095 MΩ
(b)
I C1 = 7.10 μ A
(200)(0.026)
rπ 1 = = 732 K
0.0071
(10)(0.026)
rπ 3 = = 36.6 K
0.0071
⎡ (201)(36.6) ⎤
Rid = 2 ⎢ 732 + ⎥
⎣ 11 ⎦
Rid ⇒ 2.80 MΩ
13.21
We can write
A0
A( f ) =
⎛ f ⎞⎛ f ⎞
⎜1 + j ⎟⎜ 1 + j ⎟
⎝ f PD ⎠⎝ f1 ⎠
181, 260
=
⎛ f ⎞⎛ f ⎞
⎜1 + j ⎟ ⎜1 + j ⎟
⎝ 10.7 ⎠ ⎝ f1 ⎠
Phase:
⎛ f ⎞ −1 ⎛ f ⎞
φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠ ⎝ f1 ⎠
For a Phase margin = 70°, φ = −110°
So
⎛ f ⎞ −1 ⎛ f ⎞
−110° = − tan −1 ⎜ ⎟ − tan ⎜ ⎟
⎝ 10.7 ⎠ ⎝ f1 ⎠
Assuming f 10.7, we have
⎛ f ⎞ f
tan −1 ⎜ ⎟ = 20° ⇒ = 0.364
⎝ f1 ⎠ f1
At this frequency, A( f ) = 1, so

10. 181, 260
1=
2
⎛ f ⎞
1+ ⎜ ⎟ ⋅ 1 + (0.364)
2
⎝ 10.7 ⎠
170,327
=
2
⎛ f ⎞
1+ ⎜ ⎟
⎝ 10.7 ⎠
f
or = 170,327 ⇒ f = 1.82 MHz
10.7
Then, second pole at
f
f1 = ⇒ f1 = 5 MHz
0.364
13.22
a. Original g m1 and g m 2
⎛W ⎞⎛ μ C ⎞
K p1 = K p 2 = ⎜ ⎟⎜ P ox ⎟ = (12.5)(10)
⎝ L ⎠⎝ 2 ⎠
= 125 μ A / V 2
So
⎛ IQ ⎞
g m1 = g m 2 = 2 K p1 ⎜ ⎟ = 2 (0.125)(10)
⎝ 2⎠
= 0.09975 mA/V
⎛W ⎞
If ⎜ ⎟ is increased to 50, then
⎝L⎠
K p1 = K p 2 = (50)(10) = 500 μ A / V 2
So
g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V
b. Gain of first stage
Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025)
or
Ad = 501
Voltage gain of second stage remains the same, or
Av 2 = 251
Then Av = Ad ⋅ Av 2 = (501)(251)
or
Ad = 125, 751
13.24
a. K p = (10)(20) = 200 μ A / V 2 = 0.2 mA / V 2
10 − VSG − (−10)
I REF = I SET =
200
= k P (VSG − 1.5) 2
20 − VSG = (0.2)(200)(VSG − 3VSG + 2.25)
2
40VSG − 119VSG + 70 = 0
2
119 ± (119) 2 − 4(40)(70)
VSG = ⇒ VSG = 2.17 V
2(40)
Then

11. 20 − 2.17
I REF = ⇒ I REF = 89.2 μ A
200
M 5 , M 6 , M 8 matched transistors so that
I Q = I D 7 = I REF = 89.2 μ A
b. Small-signal voltage gain of input stage:
Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 )
1 1
r02 = = = 1.12 MΩ
λP I D ⎛ 89.2 ⎞
(0.02) ⎜ ⎟
⎝ 2 ⎠
1 1
r04 = = = 2.24 MΩ
λn I D ⎛ 89.2 ⎞
(0.01) ⎜ ⎟
⎝ 2 ⎠
Then
Ad = 2(200)(89.2) ⋅ (1.12 2.24)
or
Ad = 141
Small-signal voltage gain of second stage:
Av 2 = g m 7 (r07 r08 )
K n 7 = (20)(20) = 400 μ A / V 2
So
g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.0892) = 0.378 mA/V
1 1
r08 = = = 561 kΩ
λP I D 7 (0.02)(0.0892)
1 1
r07 = = = 1121 kΩ
λn I D 7 (0.01)(0.0892)
So
Av 2 = (0.378)(1121 561) ⇒ Av 2 = 141
Then overall voltage gain
Av = Ad ⋅ Av 2 = (141)(141) ⇒ Av = 19,881
13.25
Small-signal voltage gain of input stage:
Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 )
K p1 = (10)(10) = 100 μ A / V 2
1 1
r02 = = = 1000 kΩ
⎛ IQ ⎞ ⎛ 0.2 ⎞
λP ⎜ ⎟ (0.01) ⎜ ⎟
⎝ 2⎠ ⎝ 2 ⎠
1 1
r04 = = = 2000 kΩ
⎛ IQ ⎞ ⎛ 0.2 ⎞
λn ⎜ ⎟ (0.005) ⎜ ⎟
⎝ 2⎠ ⎝ 2 ⎠
Then
Ad = 2(0.1)(0.2) ⋅ (1000 2000)
or
Ad = 133
Small-signal voltage gain of second stage:
Av 2 = g m 7 ( r07 r08 )
K n 7 = (20)(20) = 400 μ A / V 2
So

12. g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.2) = 0.566 mA/V
1 1
r08 = = = 500 kΩ
λP I D 7 (0.01)(0.2)
1 1
r07 = = = 1000 kΩ
λn I D 7 (0.005)(0.2)
So
Av 2 = (0.566)(1000 500) ⇒ Av 2 = 189
Then overall voltage gain is
Av = Ad ⋅ Av 2 = (133)(189) ⇒ Av = 25,137
13.26
1
f PD =
2π Req Ci
where Req = r04 r02 and Ci = C1 (1 + Av 2 )
We can find that
Av 2 = 251 and r04 = r02 = 5.025 MΩ
Now
Req = 5.025 5.025 = 2.51 MΩ
and
Ci = 12(1 + 251) = 3024 pF
So
1
f PD =
2π (2.51× 106 )(3024 × 10−12 )
or
f PD = 21.0 Hz
13.27
1
f PD =
2π Req Ci
where Req = r04 r02
From Problem 13.22,
r02 = 1.12 MΩ, r04 = 2.24 MΩ and Av 2 = 141
So
1
8=
2π (1.12 2.24) × 106 × Ci
or
Ci = 2.66 × 10−8 = C1 (1 + Av 2 ) = C1 (142)
or
C1 = 188 pF
13.28
R0 = r07 r08
We can find that
r07 = r08 = 2.52 MΩ
Then
R0 = 2.52 2.52
or
R0 = 1.26 MΩ

13. 13.29
a.
V0 = ( g m1Vgs1 )(r01 r02 )
VI = Vgs1 + V0
Then V0 = g m1 (r01 r02 )(VI − V0 )
or
g m1 (r01 r02 )
Av =
1 + g m1 (r01 r02 )
VX VX
b. I X + g m1Vgs1 = + and Vgs1 = −VX
r02 r01
1
R0 = r r
g m1 01 02
13.30
⎛ 80 ⎞
I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ]
2
(a)
⎝ 2⎠
I Q 2 = 180 μ A
⎛ 80 ⎞
I D 6 = ⎜ ⎟ (25) (VGS 6 − 0.7 ) = 25 ⇒ VGS 6 = 0.8581 V
2
(b)
⎝ 2⎠
⎛ 40 ⎞
I D 7 = ⎜ ⎟ (50) (VSG 7 − 0.7 ) = 25 ⇒ VSG 7 = 0.8581 V
2
⎝ 2 ⎠
Set
VSG 8 P = VGS 8 N = 0.8581 V
⎛ 40 ⎞ ⎛ W ⎞ ⎛W ⎞
180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 360
⎝ 2 ⎠ ⎝ L ⎠8 P ⎝ L ⎠8 P
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180
⎝ 2 ⎠ ⎝ L ⎠8 N ⎝ L ⎠8 N
13.31

14. ⎛ 80 ⎞
VGS11 ⇒ 200 = ⎜ ⎟ (20) (VGS 11 − 0.7 )
2
⎝ 2⎠
VGS 11 = 1.20 V
Let M 12 = 2 transistors in series. Than
5 − 1.20
VGS12 = = 1.90 V
2
⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞
200 = ⎜ ⎟⎜ ⎟ (1.90 − 0.7 ) ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.47
2
⎝ 2 ⎠⎝ L ⎠12 ⎝ L ⎠12 A ⎝ L ⎠12 B
13.32
(a)
⎛ 80 ⎞
I Q 2 = 250μ A = ⎜ ⎟ (5) (VGS 8 − 0.7 )
2
⎝ 2⎠
⇒ VGS 8 = 1.818 V
1.818
⇒ VGS 6 = VSG 7 = = 0.909 V
2
⎛ 80 ⎞
I D 6 = I D 7 = ⎜ ⎟ (25)(0.909 − 0.7) 2 = 43.7 μ A
⎝ 2⎠
(b)
⎛ 80 ⎞ ⎛ 250 ⎞
g m1 = 2 ⎜ ⎟ (15) ⎜ ⎟ ⇒ 0.5477 mA/V
⎝ 2⎠ ⎝ 2 ⎠
1
ro 2 = = 800 K
( 0.01)( 0.125)
1
r04 = = 533.3K
( 0.015)( 0.125)
Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800 533.3)
Ad 1 = 175
Second stage:

15. A2 = − g m 5 (ro 5 ro 9 )
⎛ 40 ⎞
g m 5 = 2 ⎜ ⎟ (80)(250) ⇒ 1.265 mA/V
⎝ 2 ⎠
1
r05 = = 266.7 K
(0.015)(0.25)
1
r09 = = 400 K
(0.01)(0.25)
A2 = −(1.265)(266.7 400)
A2 = −202
Assume the gain of the output stage ≈ 1, then
Av = Ad 1 ⋅ A2 = (175)(−202)
Av = −35,350
13.33
(a) Ad = g m1 ( Ro 6 Ro8 )
g m1 = 2 K n I DQ = 2 (0.5)(0.025) ⇒ 224 μ A / V
g m1 = g m8
g m 6 = 2 (0.5)(0.025) ⇒ 224 μ A / V
1 1
ro1 = ro 6 = ro8 = ro10 = = = 2.67 M Ω
λ I DQ (0.015)(25)
1 1
ro 4 = = ⇒ 1.33 M Ω
λ I D 4 ( 0.015 )( 50 )
Now
Ro8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ω
Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) ⇒ Ro 6 = 531 M Ω
Then
Ad = (224)(531 1597) ⇒ Ad = 89, 264
(b) Ro = Ro 6 Ro8 = 531 1597 ⇒ Ro = 398 M Ω
1 1
(c) f PD = = ⇒ f PD = 80 Hz
2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 )
GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz
13.34
(a)
1 1
ro1 = ro8 = ro10 = = = 2 MΩ
λ p I D (0.02)(25)
1 1
ro 6 = = = 2.67 M Ω
λn I D (0.015)(25)
1 1
ro 4 = = = 1.33 M Ω
λn I D 4 (0.015)(50)
⎛ 35 ⎞ ⎛ W ⎞ ⎛W ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ (25) = 41.8 ⎜ ⎟ = g m8
⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1
⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞
g m 6 = 2 ⎜ ⎟⎜ ⎟ (25) = 63.2 ⎜ ⎟
⎝ 2 ⎠⎝ L ⎠6 ⎝ L ⎠6
Ro = Ro 6 Ro8 = [ g m 6 (ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )]

16. ⎛W ⎞ ⎛W ⎞
Define X 1 = ⎜ ⎟ and X 6 = ⎜ ⎟
⎝ L ⎠1 ⎝ L ⎠6
Then
Ro = ⎣ 63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦ ⎡ 41.8 X 1 ( 2 )( 2 ) ⎤
⎡ ⎤ ⎣ ⎦
22,539 X 1 X 6
= 134.8 X 6 167.2 X 1 =
134.8 X 6 + 167.2 X 1
⎛ 22,539 X 1 X 6 ⎞
Ad = g m1 Ro = (41.8 X 1 ) ⎜ ⎟
⎝ 134.8 X 6 + 167.2 X 1 ⎠
= 10, 000
⎛W ⎞ 1 ⎛W ⎞
Now X 6 = ⎜ ⎟ = ⎜ ⎟ = 0.674 X 1
⎝ L ⎠6 2.2 ⎝ L ⎠1
We then find
⎛W ⎞ ⎛W ⎞
X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟
⎝ L ⎠1 ⎝ L ⎠p
and
⎛W ⎞
⎜ ⎟ = 1.85
⎝ L ⎠n
13.35
Let V + = 5V , V − = −5V
P = IT (10) = 3 ⇒ IT = 0.3 mA ⇒ I REF = 0.1 mA = 100 μ A
1
ro1 = ro8 = ro10 = = 1 MΩ
(0.02)(50)
1
ro 6 = = 1.33 MΩ
(0.015)(50)
1
ro 4 = = 0.667 M Ω
(0.015)(100)
⎛ 35 ⎞ ⎛ W ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 59.2 X 1 = g m8
⎝ 2 ⎠ ⎝ L ⎠1
⎛W ⎞
where X 1 = ⎜ ⎟
⎝ L ⎠1
Assume all width-to-length ratios are the same.
⎛ 80 ⎞ ⎛ W ⎞
g m 6 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 89.4 X 1
⎝ 2 ⎠⎝ L ⎠
Now
Ro = Ro 6 Ro8 = ⎡ g m 6 ( ro 6 ) ( ro 4 ro1 ) ⎤ ⎡ g m8 ( ro8 ro10 ) ⎤
⎣ ⎦ ⎣ ⎦
= ⎡89.4 X 1 (1.33) ( 0.667 1) ⎤ ⎡59.2 X 1 (1)(1) ⎤
⎣ ⎦ ⎣ ⎦
( 47.6 X 1 )( 59.2 X 1 )
= [ 47.6 X 1 ] [59.2 X 1 ] =
47.6 X 1 + 59.2 X 1
So Ro = 26.4 X 1
Now
Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000
W
So that X 12 = = 16 for all transistors
L
13.36

17. (a) Ad = Bg m1 (ro 6 ro8 )
1 1
ro 6 = ro8 = = = 0.741 M Ω
λ I DQ (0.015)(90)
⎛ k ′ ⎞⎛ W ⎞
g m1 = 2 ⎜ n ⎟ ⎜ ⎟ I D1 = 2 (500)(30) = 245 μ A / V
⎝ 2 ⎠⎝ L ⎠
Ad = (3)(245)(0.741 0.741) ⇒ Ad = 272
(b) Ro = ro 6 ro8 = 0.741 0.741 ⇒ Ro = 371 k Ω
1 1
(c) f PD = = ⇒ f PD = 85.8 kHz
2π Ro C 2π (371× 103 )(5 × 10−12 )
GBW = (272)(85.8 × 103 ) ⇒ GBW = 23.3 MHz
13.37
1
(a) ro 6 = = 0.5 M Ω
(0.02)(2.5)(40)
1
ro8 = = 0.667 M Ω
(0.015)(2.5)(40)
Ad = Bg m1 ( ro 6 ro8 )
400 = (2.5) g m1 ( 0.5 0.667 ) ⇒ g m1 = 560 μ A / V
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
g m1 = 560 = 2 ⎜ ⎟ ⎜ ⎟ (40) ⇒ ⎜ ⎟ = 49
⎝ 2 ⎠⎝ L ⎠ ⎝L⎠
Assume all (W/L) ratios are the same except for
⎛W ⎞ ⎛W ⎞
M 5 and M 6 . ⎜ ⎟ = ⎜ ⎟ = 122.5
⎝ L ⎠5 ⎝ L ⎠ 6
(b) Assume the bias voltages are
V + = 5V , V − = −5V .
⎛W ⎞ ⎛W ⎞
Assume ⎜ ⎟ = ⎜ ⎟ = 49
⎝ L ⎠ A ⎝ L ⎠B
⎛ 80 ⎞
I Q = ⎜ ⎟ (49)(VGSA − 0.5) 2 = 80 ⇒ VGSA = 0.702 V
⎝ 2⎠
Then
⎛ 80 ⎞ ⎛ W ⎞
I REF = 80 = ⎜ ⎟ ⎜ ⎟ (VGSC − 0.5) 2
⎝ 2 ⎠ ⎝ L ⎠C
For four transistors

18. 10 − 0.702
VGSC = = 2.325 V
4
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
80 = ⎜ ⎟ ⎜ ⎟ (2.325 − 0.5) 2 ⇒ ⎜ ⎟ = 0.60
⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C
1
(c) f 3− dB = Ro = 0.5 0.667 = 0.286 M Ω
2π Ro C
1
f 3− dB = = 185 kHz
2π (286 × 103 )(3 × 10−12 )
GBW = (400)(185 × 103 ) ⇒ 74 MHz
13.38
(a) From previous results, we can write
Ro10 = g m10 (ro10 ro 6 )
Ro12 = g m12 (ro12 ro8 )
Ad = Bg m1 ( Ro10 Ro12 )
Now
1 1
ro10 = ro 6 = = = 0.5 M Ω
λP B ( I Q / 2 ) (0.02)(2.5)(40)
1 1
ro12 = ro8 = = = 0.667 M Ω
λn B ( I Q / 2 ) (0.015)(2.5)(40)
Assume all transistors have the same width-to-length ratios except for M 5 and M 6 .
⎛W ⎞
⎟= X
2
Let ⎜
⎝L ⎠
Then
⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞
g m10 = 2 ⎜ ⎟ ⎜ ⎟ ( I DQ10 ) = 2 ⎜ ⎟ X 2 (2.5)(40)
p
⎝ 2 ⎠ ⎝ L ⎠10 ⎝ 2⎠
= 83.67 X
⎛ k′ ⎞⎛ W ⎞ ⎛ 80 ⎞
g m12 = 2 ⎜ n ⎟ ⎜ ⎟ ( I DQ12 ) = 2 ⎜ ⎟ X 2 (2.5)(40)
⎝ 2 ⎠ ⎝ L ⎠12 ⎝ 2⎠
= 126.5 X
⎛ 80 ⎞
g m1 = 2 ⎜ ⎟ X 2 (40) = 80 X
⎝ 2⎠
Then
Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ω
Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M Ω
We want
20, 000 = (2.5)(80 X )[20.9 X 56.3 X ]
⎡ (20.9 X )(56.3 X ) ⎤
= 200 X ⎢ ⎥ = 3048 X
2
⎣ 20.9 X + 56.3 X ⎦
Then
⎛W ⎞
X 2 = 6.56 = ⎜ ⎟
⎝L⎠
Then
⎛W ⎞ ⎛W ⎞
⎜ ⎟ = ⎜ ⎟ = (2.5)(6.56) = 16.4
⎝ L ⎠ 6 ⎝ L ⎠5
(b) Assume bias voltages are V + = 5V , V − = −5V

19. ⎛W ⎞ ⎛W ⎞
Assume ⎜ ⎟ = ⎜ ⎟ = 6.56
⎝ L ⎠ A ⎝ L ⎠B
⎛ 80 ⎞
I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V
⎝ 2⎠
Need 5 transistors in series
10 − 1.052
VGSC = = 1.79 V
5
Then
⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞
I REF = 80 = ⎜ ⎟ ⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20
⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C
1
(c) f 3− dB = where Ro = Ro10 Ro12
2π Ro C
Now
Ro10 = 20.9 6.56 = 53.5 M Ω
Ro12 = 56.3 6.56 = 144 M Ω
Then
Ro = 53.5 144 = 39 M Ω
1
f 3− dB = = 1.36 kHz
2π (39 × 106 )(3 × 10−12 )
GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz
13.39
Ad = g m ( M 2 ) ⋅ ⎡ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤
⎣ ⎦
⎛ 40 ⎞
g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 447 μ A / V
⎝ 2 ⎠
1 1
ro 2 ( M 2 ) = = = 500 k Ω
λ I DQ (0.02)(0.1)
VA 120
ro 2 (Q2 ) = = = 1200 k Ω
I CQ 0.1
Then
Ad = 447(0.5 1.2) ⇒ Ad = 158
13.40

20. Ad = g m ( M 2 ) ⋅ ⎡ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤
⎣ ⎦
⎛ 80 ⎞
g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 632 μ A / V
⎝ 2⎠
1 1
ro 2 ( M 2 ) = = = 667 k Ω
λ I DQ (0.015)(0.1)
VA 80
ro 2 (Q2 ) = = = 800 k Ω
I CQ 0.1
Ad = (632) ( 0.667 0.80 ) ⇒ Ad = 230
13.41
(a) I REF = 200 μ A K n = K p = 0.5 mA / V 2
λn = λ p = 0.015 V −1
Ad = g m1 ( Ro 6 Ro8 )
where
Ro8 = g m8 (ro8 ro10 )
Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 )
Now
g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V
1 1
ro8 = = = 667 k Ω
λP I D 8 (0.015)(0.1)
1
ro10 = = 667 k Ω
λP I D 8
IC 6 0.1
gm6 = = = 3.846 mA/V
VT 0.026
VA 80
ro 6 = = = 800 k Ω
I C 6 0.1
1 1
ro 4 = = = 333 k Ω
λn I D 4 (0.015)(0.2)
1 1
ro1 = = = 667 k Ω
λ p I D1 (0.015)(0.1)
g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V
So
Ro8 = (0.447)(667)(667) ⇒ 198.9 M Ω
Ro 6 = (3.846)(800)(333 667) ⇒ 683.4 M Ω
Then
Ad = 447(198.9 683.4) ⇒ Ad = 68,865
13.42
Assume biased at V + = 10V , V − = −10V .
P = 3I REF (20) = 10 ⇒ I REF = 167 μ A
Ad = g m1 ( Ro 6 Ro8 ) = 25, 000
kn = 80 μ A / V 2 , k ′ = 35 μ A / V 2
′ p
λn = 0.015V −1 , λ p = 0.02 V −1
⎛W ⎞ ⎛W ⎞
Assume ⎜ ⎟ = 2.2 ⎜ ⎟
⎝ L ⎠p ⎝ L ⎠n

21. Ro8 = g m8 ( ro8 ro10 )
Ro 6 = g m 6 (ro 6 )(ro 4 ro1 )
1 1
ro8 = = = 0.60 M Ω
λP I D 8 (0.02)(83.3)
1
ro10 = = 0.60 M Ω
λP I D 8
⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞
g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3)
p
⎝ 2 ⎠ ⎝ L ⎠8 ⎝ 2⎠
= 113.3 X
⎛W ⎞
where X 2 = ⎜ ⎟
⎝ L ⎠n
VA 80
ro 6 = = = 0.960 M Ω
I C 6 83.3
1 1
ro 4 = = = 0.40 M Ω
λn I D 4 (0.015)(167)
1 1
ro1 = = = 0.60 M Ω
λ p I D1 (0.02)(83.3)
IC 6 83.3
gm6 = = = 3204 μ A / V
VT 0.026
′
⎛ kp ⎞⎛ W ⎞ ⎛ 35 ⎞
g m1 = 2 ⎜ ⎟ ⎜ ⎟ I D1 = 2 ⎜ ⎟ (2.2) X 2 (83.3)
⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 2⎠
= 113.3 X
Now
Ro 6 = (3204)(0.960) ⎡0.40 0.60 ⎤ = 738 M Ω
⎣ ⎦
Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Ω
Then
Ad = 25, 000 = (113.3 X ) ⎡ 738 40.8 X ⎤
⎣ ⎦
⎡ 30,110 X ⎤
= (113.3 X ) ⎢ ⎥
⎣ 738 + 40.8 X ⎦
which yields X = 2.48
or
⎛W ⎞
X 2 = 6.16 = ⎜ ⎟
⎝ L ⎠n
and
⎛W ⎞
⎜ ⎟ + (2.2)(6.16) = 12.3
⎝ L ⎠P
13.43
For vcm (max), assume VCB (Q5 ) = 0. Then
VS = 15 − 0.6 − 0.6 = 13.8 V
0.236
I D 9 = I D10 = = 0.118 mA
2
Using parameters given in Example 13.11
I 0.118
VSG = D 9 − VTP = + 1.4 = 2.17 V
KP 0.20
Then
vcm (max) = 13.8 − 2.17 ⇒ vcm (max) = 11.6 V

22. For
vcm (min) , assume
VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 − 1.4 = 0.77 V
Now
VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) − 15
= 0.118 + 0.6 − 15 ⇒ VD10 = −14.28 V
Then
vcm (min) = −14.28 + VSD (sat) − VSG
= −14.28 + 0.77 − 2.17 = −15.68 V
Then, common-mode voltage range
−15.68 ≤ vcm ≤ 11.6
Or, assuming the input is limited to ±15 V, then
−15 ≤ vcm ≤ 11.6 V
13.44
For I1 = I 2 = 300 μ A,
VSG = VBE + (0.3)(8) = 0.6 + 2.4 = 3.0 V
Then
I 2 = K P (VSG + VTP ) 2
0.3 = K P (3 − 1.4)2 ⇒ K P = 0.117 mA / V 2
13.45
For VCB = 0 for both Q6 and Q7 , then
VS = 0.6 + 0.6 + VSG + (−VS )
So 2VS = 1.2 + VSG
Now
I1
0.6 + I 2 R1 = VSG = + VTP and I1 = I 2
KP
Also I1 = I 2 = K P (VSG + VTP ) 2 so
0.6 + (0.25)(8)(VSG − 1.4) 2 = VSG
0.6 + 2(VSG − 2.8VSG + 1.96) = VSG
2
2VSG − 6.6VSG + 4.52 = 0
2
6.6 ± (6.6) 2 − 4(2)(4.52)
VSG = = 2.33 V
2(2)
Then 2VS = 1.2 + 2.33 = 3.53 and
VS = 1.765 V
13.46
I C 5 = I C 4 = 300 μ A
Using the parameters from Examples 13.12 and 13.13, we have
βV (200)(0.026)
Ri 2 = rπ 13 = n T = = 17.3 kΩ
I C13 0.3
Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3)
or
Ad = 10.38
Now

23. I C13 0.3
g m13 = = = 11.5 mA/V
VT 0.026
VA 50
r013 = = = 167 kΩ
I C13 0.3
Then
| Av 2 | = g m13 ⋅ r013 = (11.5)(167)
or
Av 2 = 1917
Overall gain:
Av = (10.38)(1917) = 19,895
13.47 Assuming the resistances looking into Q4 and into the output stage are very large, we have
β R013
| Av 2 | =
rπ 13 + (1 + β ) RE13
where R013 = r013 ⎡1 + g m13 ( RE13 rπ 13 ) ⎤
⎣ ⎦
50
I C13 = 300 μ A, r013 = = 167 kΩ
0.3
0.3
g m13 = = 11.5 mA / V
0.026
(200)(0.026)
rπ 13 = = 17.3 kΩ
0.3
So
R013 = (167) ⎡1 + (11.5) (1 17.3) ⎤ ⇒ 1.98 MΩ
⎣ ⎦
Then
(200)(1980)
| Av 2 | = = 1814
17.3 + (201)(1)
Now
Ci = C1 (1 + Av 2 ) = 12 [1 + 1814]
⇒ Ci = 21, 780 pF
1
f PD =
2π Req Ci
Req = Ri 2 r012 r010
Neglecting R3 ,
1 1
r010 = = = 333 kΩ
λ I D10 (0.02)(0.15)
Neglecting R5 ,
50
r012 = = 333 kΩ
0.15
Ri 2 = rπ 13 + (1 + β ) RE13 = 17.3 + (201)(1)
= 218 kΩ
Then
1
f PD =
2π ⎡ 218 333 333⎤ × 103 × ( 21, 780 ) × 10−12
⎣ ⎦
or
f PD = 77.4 Hz
Unity-Gain Bandwidth
Gain of first stage:

24. Ad = 2 K n I Qs ⋅ ( R12 ro12 ro10 )
= 2(0.6)(0.3) ⋅ (218 333 333)
= (0.6)(218 333 333)
or Ad = 56.6
Overall gain:
Av = (56.6)(1814) = 102, 672
Then unity-gain bandwidth = (77.4)(102, 672)
⇒ 7.95 MHz
13.48
Since VGS = 0 in J 6 , I REF = I DSS
⇒ I DSS = 0.8 mA
13.49
a. Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ]
(100)(0.026)
rπ 6 = = 13 kΩ
0.2
I 200 μ A
IC 5 ≅ C6 = = 2 μA
β 100
So
(100)(0.026)
rπ 5 = = 1300 kΩ
0.002
Then
Ri 2 = 1300 + (101) [13 + (101)(0.3) ]
or
Ri 2 = 5.67 MΩ
b. Av = g m 2 ( r02 r04 Ri 2 )
2 2
gm2 = ⋅ I D ⋅ I DSS = ⋅ (0.1)(0.2)
VP 3
= 0.0943 mA / V
1 1
r02 = = = 500 kΩ
λ I D (0.02)(0.1)
VA 5.0
r04 = = = 500 kΩ
I C 4 0.1
Then
Av = (0.0943)[500 || 500 || 5670]
or
Av = 22.6
13.50
a. Need VSD (QE ) ≥ VSD ( sat ) = VP For minimum bias ±3 V
Set VP = 3 V and VZK = 3 V
VZK − VD1
I REF 2 =
R3
3 − 0.6
so that R3 = ⇒ R3 = 24 kΩ
0.1
Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA
Therefore,

25. I DSS = 0.2 mA
b. Neglecting base currents
12 − 0.6
I 01 = I REF 1 = 0.5 mA =
R4
so that
R4 = 22.8 kΩ
13.51
a. We have
2 2
gm2 = ⋅ I D ⋅ I DSS = ⋅ (0.5)(1)
| VP | 4
= 0.354 mA/V
1 1
r02 = = = 100 kΩ
λ ID (0.02)(0.5)
VA 100
r04 = = = 200 kΩ
I D 0.5
0.5
gm4 = = 19.23 mA/V
0.026
(200)(0.026)
rπ 4 = = 10.4 kΩ
0.5
So
R04 = r04 ⎡1 + g m 4 ( rπ 4 R2 ) ⎤
⎣ ⎦
= 200 ⎡1 + (19.23) (10.4 0.5 ) ⎤
⎣ ⎦
= 2035 kΩ
Ad = g m 2 ( r02 R04 RL )
For RL → ∞
Ad = 0.354 (100 || 2035 ) = 33.7
With these parameter values, gain can never reach 500.
b. Similarly for this part, gain can never reach 700.