This document contains solutions to exercises from Chapter 14. The solutions involve calculations for amplifier circuits using op amp specifications like gain bandwidth, input offset voltage, input bias current, slew rate, and more. Some key results include: - The input offset voltage for a circuit is calculated as 1.27 mV. - The input resistance of a voltage follower circuit with a large resistor is calculated as 5000 MΩ. - The maximum frequency of a circuit before the gain drops by 3 dB is calculated as 20 kHz.

1. Chapter 14
Exercise Solutions
EX14.1
−50
a. ACL = ⇒ ACL = −49.949
⎡ ⎛ 1 ⎞ ⎤
⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥
⎣ ⎝ ⎠ ⎦
dACL 51 dA
b. = 10 × ⇒ CL = 0.0102%
ACL 5 × 104 ACL
−50
ACL = ⇒ ACL = −49.943
⎡ 51 ⎤
⎢1 + 4.5 × 104 ⎥
⎣ ⎦
EX14.2
a. For R0 = 0
= + (1 + 104 ) = 0.1 + 103
1 1 1
Ri f 10 10
⇒ Ri f = 10−3 kΩ = 1 Ω
b. For R0 = 10 kΩ
1 1 1 ⎡1 + 104 + 1 ⎤ 104
= + ×⎢ ⎥ ≅ 0.1 +
Ri f 10 10 ⎣ 1 + 1 + 1 ⎦ 3 (10 )
Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω
EX14.3
⎛ 40 ⎞
40 (1 + 104 ) + 99 ⎜ 1 + ⎟
Ri f = ⎝ 1 ⎠
99
1+
1
4 × 10 + 4.059 × 103
5
≅
100
Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ
EX14.4
R
1 + 2 = 100
R1
1 1 ⎡ 105 ⎤
a. = ⎢ ⎥ = 10 ⇒ R0 f = 0.1 Ω
R0 f 100 ⎣100 ⎦
1 1 ⎡ 105 ⎤
b. = ⎢ ⎥ = 10
2
R0 f 10 ⎣100 ⎦
R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω
EX14.5
From Equation (14.43)

2. ACL 0
ACL ( f ) =
f
1+ j ⋅
f PD ( A0 /ACL 0 )
25 25
= =
f f
1+ j ⋅ 1+ j ⋅
( 50 ) (104 / 25) 2 × 104
a. f = 2 kHz
v0
= 25 ⇒ v0 ( peak ) = 1.25 mV
vI
b. f = 20 kHz
v0 1
= ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV
vI 2
c. f = 100 kHz
v0 25 25
= = = 4.90
vI 1 + (100 / 20 ) 5.099
2
⇒ v0 = 0.245 mV
EX14.6
Full-scale response = 1× 5 = 5 V
5
t= ⇒ t = 2.5 μ s
2
EX14.7
SR 0.63 × 106
a. FPBW = =
2π V0 ( max ) 2π (1)
FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz
0.63 × 106
b. FPBW = = 1.0 × 104
2π (10 )
⇒ FPBW = 10 kHz
EX14.8
⎛I ⎞ ⎛ 1.85 × 10−14 ⎞
V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜ −14 ⎟
⎝ I S1 ⎠ ⎝ 2 × 10 ⎠
⇒ V0 S = 2.03 mV
EX14.9
We need
iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V
By Equation (14.60(a))
⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
By Equation (14.60(b))

3. ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞
iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞
= I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 + ⎟
⎣ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠
I S1 = I S 2, take the ratio:
⎛v −v ⎞ I
exp ⎜ BE1 BE 2 ⎟ = S 3
⎝ VT ⎠ IS 4
⎛I ⎞
vBE1 − vBE 2 = V0 S = VT ln ⎜ S 3 ⎟
⎝ IS 4 ⎠
= 0.026 ⋅ ln (1.05 )
⇒ V0 S = 1.27 m V
EX14.10
1 I Q ⎛ ΔK n ⎞
VOS = ⋅ ⋅⎜ ⎟
2 2Kn ⎝ Kn ⎠
1 150 ⎛ ΔK n ⎞
0.020 = ⋅ ⋅
2 2 ( 50 ) ⎜ 50 ⎟
⎝ ⎠
⇒ ΔK n = 1.63μ A / V 2
ΔK n 1.63
⇒ = ⇒ 3.26%
Kn 50
EX14.11
⎛ R5 ⎞ +
Want ⎜ ⎟V = 5 m V
⎝ R5 + R4 ⎠
R5
R5 R4 so × V + = 0.005
R4
( 0.005)(100 )
R5 = = 0.05 kΩ
10
⇒ R5 = 50 Ω
EX14.12
R1′ = 25 1 = 0.9615 kΩ
′
R2 = 75 1 = 0.9868 kΩ
For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A
From Equation (14.75)
⎛ 50 × 10−6 ⎞
( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 )
⎝ 10 ⎠
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 )
⎝ IS 4 ⎠
0.58065 + 0.048075
⎛i ⎞
= ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934
⎝ IS 4 ⎠

4. ⎛i ⎞
ln ⎜ C 2 ⎟ = 22.284
⎝ IS 4 ⎠
50 × 10−6
= 4.7625 × 109
IS 4
I S 4 ≅ 1.05 × 10−14 A
EX14.13
From Equation (14.79)
⎛ R ⎞
v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
⎝ R1 ⎠
For v0 = 0
⎛ 100 ⎞
0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 + ⎟
⎝ 10 ⎠
R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ
TYU14.1
v1CM ( max ) = V + − VSD1 ( sat ) − VSG1
v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1
We have:
I REF = 100 μ A, kn = 80 μ A / V 2 , k ′ = 40 μ A / V 2 ,
′ p
⎛W ⎞
⎜ ⎟ = 25
⎝L ⎠
For M1 :
⎛ 40 ⎞
I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP )
2
⎝ 2 ⎠
So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V
2
VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V
Then
vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V
For M 4 :
⎛ 80 ⎞
I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN )
2
⎝ 2⎠
So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V
2
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184
So
V − − 0.184 ≤ vCM ≤ V + − 1.13 V
TYU14.2
vo ( max ) = V + − VSD 8 ( sat ) − VSG10
vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat )
Now

5. 50
VSG 8 = VSG10 = + 0.5 = 0.816 V
( 40/ 2 )( 25)
VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V
So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13
Also
50
VGS 6 = + 0.5 = 0.724 V
(80/ 2 )( 25)
100
VGS 4 = + 0.5 = 0.816 V
(80/ 2 )( 25)
VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V
VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V
So
vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54
Then
V − + 0.54 ≤ vo ≤ V + − 1.13 V
TYU14.3
500
ACL ( ideal ) = −= −25
20
Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975
−25
−24.975 =
⎡ 26 ⎤
⎢1 + A ⎥
⎣ 0L ⎦
26 −25
= − 1 = 0.0010
A0 L −24.975
A0 L = 25.974
TYU14.4
ACL ( ∞ )
a. ACL =
⎡ A (∞) ⎤
1 + ⎢ CL ⎥
⎣ A0 L ⎦
R 495
ACL ( ∞ ) = 1 + 2 = 1 + = 100
R1 5
100
ACL = ⇒ ACL = 99.90
100
1+ 5
10
ACL ( ∞ ) = 100
dACL 100
b. = 10 × 5 = 0.01%
ACL 10
ACL = 99.90 − ( 0.0001)( 99.90 )
⇒ ACL = 99.89
TYU14.5

6. ACL ( ∞ ) − ACL ACL 1
= 1− = 1−
ACL ( ∞ ) ACL ( ∞ ) A (∞)
1 + CL
A0 L
ACL ( ∞ ) ACL ( ∞ )
1+ −1
A0 L A0 L
So 0.001 = =
A (∞) A (∞)
1 + CL 1 + CL
A0 L A0 L
ACL ( ∞ )
0.001 = 0.999 ⋅
A0 L
0.001 0.001
ACL ( ∞ ) = ⋅ A0 L = ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010
0.999 0.999
or
ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0
TYU14.6
ii ⎛ Rif ⎞
=⎜ ⎟
I1 ⎝ Ri ⎠
iI 0.1
a. = = 1× 10−5
i1 104
iI 10
b. = = 1× 10−3
i1 104
TYU14.7
Voltage follower R2 = 0, R1 = ∞
Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 )
≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ
TYU14.8
f3dB =
fT
=
(105 ) (10 ) ⇒ 20 kHz
ACL 0 50
SR
f max = f 3dB =
2π V0 ( max )
SR 0.8 × 106
V0 ( max ) = =
2π f3dB 2π ( 20 × 103 )
⇒ V0 ( max ) = 6.37 V
TYU14.9
a. v0 = I B1 R3 = (10−6 )( 200 × 103 )
⇒ v0 = 0.20 V
b. R4 = R1 R2 R3 = 100 50 200
⇒ R4 = 28.6 kΩ