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.