Ch14s

Chapter 14
Problem Solutions

14.1
vo
Ad = = −80
vi
vo (max) = 4.5 ⇒ vi (max) = 56.25 mV
56.25
So vi (max)rms = = 39.77 mV
2

14.2

(a)
4.5
i2 = = 0.028125 mA
160
4.5
iL = = 4.5 mA
1
Output Circuit = 4.528 mA
v −4.5
vi = − o = ⇒ vi = −0.05625 V
A 80
(b)
v 4.5
io ≈ 15 mA = o =
RL RL
⇒ RL (min) = 300 Ω

14.3
(1) vo = 2 V
(2) v2 = 12.5 mV
(3) AOL = 2 × 104
(4) v1 = 8 μ V
(5) AOL = 1000

14.4
From Eq. (14.4)

− R2 / R1
ACL =
1 ⎛ R2 ⎞
1+ ⎜1 + ⎟
AoL ⎝ R1 ⎠
− R2 / R1
−15 =
1 ⎛ R2 ⎞
1+ ⎜1 + ⎟
2 × 103 ⎝ R1 ⎠
⎛R ⎞
15 ⎜ 2 ⎟
⎛ 1 ⎞ ⎝ R1 ⎠ = R2
15 ⎜ 1 + ⎟+
⎝ 2 × 103 ⎠ 2 × 103 R1
R2
15.0075 = (0.9925)
R1
R2
= 15.12
R1

14.5

vI − v1 v1 − v0 v1
= + and v0 = − A0 L v1
R1 R2 Ri
v0
so that v1 = −
A0 L
vI v0 ⎛ 1 1 1⎞
+ = v1 ⎜ + + ⎟
R1 R2 ⎝ R1 R2 Ri ⎠
So
vI ⎡1 1 ⎛ 1 1 1 ⎞⎤
= −v0 ⎢ + ⎜ + + ⎟⎥
R1 ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦
Then
v0 −(1/ R1 )
= = ACL
vI ⎡ 1 1 ⎛ 1 1 1 ⎞⎤
⎢ + ⎜ + + ⎟⎥
⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦
From Equation (14.20) for RL = ∞ and R0 = 0
1 1 1 (1 + A0 L )
= + ⋅
Rif Ri R2 1
a. For Ri = 1 kΩ
−(1/ 20)
ACL =
⎡ 1 1 ⎛ 1 1 1 ⎞⎤
⎢100 + 103 ⎜ 20 + 100 + 1 ⎟ ⎥
⎣ ⎝ ⎠⎦
−0.05
=
[0.01 + 1.06 × 10−3 ]
or

⇒ ACL = −4.52
1 1 1 + 103
= + ⇒ Rif = 90.8 Ω
Rif 1 100
b. For Ri = 10 kΩ
−(1/ 20)
ACL =
⎡ 1 1 ⎛ 1 1 1 ⎞⎤
⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥
⎣ ⎝ ⎠⎦
−0.05
=
[0.01 + 1.6 × 10−4 ]
or
⇒ ACL = −4.92
1 1 1 + 103
= + ⇒ Rif = 98.9 Ω
Rif 10 100
c. For Ri = 100 kΩ
−(1/ 20)
ACL =
⎡ 1 1 ⎛ 1 1 1 ⎞⎤
⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥
⎣ ⎝ ⎠⎦
−0.05
=
[0.01 + 7 × 10−5 ]
or
⇒ ACL = −4.965
1 1 1 + 103
= + ⇒ Rif = 99.8 Ω
Rif 100 100

14.6

⎛ R2 ⎞
⎜1 + ⎟
v
ACL = o = ⎝ R1 ⎠
vi ⎡ 1 ⎛ R2 ⎞ ⎤
⎢1 + ⎜1 + ⎟ ⎥
⎣ AOL ⎝ R1 ⎠ ⎦
For the ideal:
⎛ R2 ⎞ 0.10
⎜1 + ⎟ = = 50
⎝ R1 ⎠ 0.002
vo (actual ) = (0.10)(1 − 0.001) = 0.0999
So

0.0999 50
= = 49.95
0.002 1 + 1
(50)
AOL
which yields
AOL = 1000

14.7
From Equation (14.18)
⎛A 1 ⎞
− ⎜ OL − ⎟
v
Avf 1 = o1 = ⎝ Ro R2 ⎠
v1 ⎛ 1 1 1 ⎞
⎜ + + ⎟
⎝ RL Ro R2 ⎠
Or
⎛ 5 × 103 1 ⎞
−⎜ − ⎟
1 100 ⎠ −(4.99999 × 103 )
vo1 = ⎝ ⋅ v1 = ⋅ v1
⎛1 1 1 ⎞ 1.11
⎜ + + ⎟
⎝ 10 1 100 ⎠
vo1 = −4.504495 × 103 ⋅ v1
Now
i1 vi − v1
= ≡K
v1 R1v1
Then
vi − v1 = KR1v1
which yields
vi
v1 =
KR1 + 1
Now, from Equation (14.20)
⎡ 1 ⎤
1 + 5 × 103 + ⎥
1 1 ⎢ 10
K= + ⎢ ⎥
10 100 ⎢ 1 + 1 + 1 ⎥
⎢
⎣ 10 100 ⎥ ⎦
⎡ 5.0011× 103 ⎤
= (0.1) + (0.01) ⎢ ⎥ = 45.15495
⎣ 1.11 ⎦
Then
vi vi
v1 = =
( 45.15495)(10 ) + 1 452.5495
We find
⎡ vi ⎤
vo1 = −4.504495 × 103 ⎢ ⎥
⎣ 452.5495 ⎦
Or
v
Avf 1 = o1 = −9.9536
vi
For the second stage, RL = ∞

⎛ 5 × 103 1 ⎞
−⎜ − ⎟
1 100 ⎠
vo 2 = ⎝ ⋅ v1′ = −4.950485 × 103 ⋅ v1′
⎛1 1 ⎞
⎜ + ⎟
⎝ 1 100 ⎠
⎡ ⎤
1 1 ⎢1 + 5 × 103 ⎥
K≡ + ⎢ ⎥ = 49.61485
10 100 ⎢ 1 + 1 ⎥
⎢
⎣ 100 ⎥ ⎦
vo1 vo1 vo1
v1′ = = =
KR1 + 1 (49.61485)(10) + 1 497.1485
Then
vo 2 −4.950485 × 103
= = −9.95776
vo1 497.1485
So
v
Avf = o 2 = (−9.9536)(−9.95776) ⇒ Avf = 99.12
vi

14.8
a.

v1 − vI v v −v
+ 1 + 1 0 =0 (1)
R3 + Ri R1 R2
⎡ 1 1 1⎤ v vI
v1 ⎢ + + ⎥= 0 +
⎣ R3 + Ri R1 R2 ⎦ R2 R3 + Ri
v0 v0 − A0 L vd v0 − v1
+ + =0 (2)
RL R0 R2
or
⎡1 1 1⎤ v A v
v0 ⎢ + + ⎥ = 1 + 0L d
⎣ RL R0 R2 ⎦ R2 R0
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ Ri (3)
⎝ R3 + Ri ⎠
So substituting numbers:
⎡ 1 1 1⎤ v vI
v1 ⎢ + + ⎥= 0 + (1)
⎣10 + 20 10 40 ⎦ 40 10 + 20
or
v1[0.15833] = v0 [0.025] + vI [0.03333]
⎡1 1 1 ⎤ v (104 )vd
v0 ⎢ + + ⎥= 1 + (2)
⎣1 0.5 40 ⎦ 40 0.5

or
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd
⎛ v −v ⎞
vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 ) (3)
⎝ 10 + 20 ⎠
So
v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 ) (2)
or
v0 [3.025] = 1.333 ×10 4 vI − 1.333 × 104 v1
From (1):
v1 = v0 ( 0.1579 ) + vI ( 0.2105 )
Then
v0 [3.025] = 1.333 × 104 vI − 1.333 × 104 ⎡ v0 ( 0.1579 ) + vI ( 0.2105 ) ⎤
⎣ ⎦
v0 ⎡ 2.1078 ×103 ⎤ = vI ⎡1.0524 ×104 ⎤
⎣ ⎦ ⎣ ⎦
or
v
ACL = 0 = 4.993
vI
To find Rif : Use Equation (14.27)
⎛ 0.5 0.5 ⎞
iI ⎜ 1 + + ⎟
⎝ 1 40 ⎠
1 ⎞ ⎛ 0.5 0.5 ⎞ 0.5 ⎫ (10 ) vd
3
⎧⎛ 1
= v1 ⎨⎜ + ⎟ ⎜ 1 + + ⎟ − ⎬−
⎩⎝ 10 40 ⎠ ⎝ 1 40 ⎠ (40) 2 ⎭ 40
iI (1.5125) = v1{(0.125)(1.5125) − 0.0003125} − 25vd
or
iI (1.5125) = vI {0.18875} − 25vd
Now
vd = iI Ri = iI (20) and v1 = vI − iI (20)
So
iI (1.5125) = [vI − iI (20)] ⋅ [0.18875] − 25iI (20)
iI [505.3] = vI (0.18875)
or
vI
= 2677 kΩ
iI
Now Rif = 10 + 2677 ⇒ Rif = 2.687 MΩ
To determine R0 f : Using Equation (14.36)
⎡ ⎤ ⎡ ⎤
1 1 ⎢ A0 L ⎥ 1 ⎢ 103 ⎥
= ⋅⎢ ⎥= ⋅⎢ ⎥
′
R0 f R0 ⎢ R2 ⎥ 0.5 ⎢1 + 40 ⎥
⎢1 + R R ⎥ ⎢ 10 20 ⎥
⎣ 1 i ⎦ ⎣ ⎦
′
or R0 f = 3.5 Ω
Then R0 f = 1 kΩ 3.5 Ω
⇒ R0 f = 3.49 Ω
b. Using Equation (14.16)
dACL ⎛ 5 ⎞ dA
= (−10) ⎜ 3 ⎟ ⇒ CL = −(0.05)%
ACL ⎝ 10 ⎠ ACL

14.9

v0 − A0 L vd v0 − vI
+ = 0 and vd = vI − v0
R0 Ri
So
v0 A0 L v v
− ⋅ (vI − v0 ) + 0 − I = 0
R0 R0 Ri Ri
⎡1 A 1⎤ ⎡1 A ⎤
v0 ⎢ + 0 L + ⎥ = vI ⎢ + 0 L ⎥
⎣ R0 R0 Ri ⎦ ⎣ Ri R0 ⎦
⎡ 1 (104 ) 1 ⎤ ⎡ 1 (104 ) ⎤
v0 ⎢ + + ⎥ = vI ⎢ + ⎥
⎣ 0.2 0.2 100 ⎦ ⎣100 0.2 ⎦
v0 [5.000501× 104 ] = vI [5.000001× 10 4 ]
v0
So ACL = = 0.9999
vI
b. Set vI = 0
v0 − A0 L vd v0
i0 = + and vd = −v0
R0 Ri
⎡1 A 1⎤
i0 = v0 ⎢ + 0 L + ⎥
⎣ R0 R0 Ri ⎦
Then
1 1 A0 L 1
= + +
R0 f R0 R0 Ri
or
1 1 (104 ) 1
= + +
R0 f 0.2 0.2 100
which yields
R0 f ≅ 0.02 Ω

14.10

vI 1 − v1 vI 2 − v1 v1 − v0
+ =
20 10 40
vI 1 vI 2 v0 ⎡1 1 1⎤
+ + = v1 ⎢ + + ⎥
20 10 40 ⎣ 20 10 40 ⎦
v
and v0 = − A0 L v1 so that v1 = − 0
A0L
Then
⎧1 1 ⎛ 7 ⎞⎫
vI 1 (0.05) + vI 2 (0.10) = −v0 ⎨ + ⋅
3 ⎜ ⎟⎬
⎩ 40 2 × 10 ⎝ 40 ⎠ ⎭
= −v0 [2.50875 × 10−2 ]
⇒ v0 = −1.993vI 1 − 3.986vI 2
Δv0 2 − 1.993 Δv
= ⇒ 0 = 0.35%
v0 2 v0

14.11

⎛ 40 ⎞ ⎛ 4⎞
vB = ⎜ ⎟ v2 = ⎜ ⎟ v2 = 0.8v2 (1)
⎝ 40 + 10 ⎠ ⎝ 5⎠
v1 − v A v A − v0
=
10 40
v1 v0 ⎛1 1 ⎞
+ = vA ⎜ + ⎟
10 40 ⎝ 10 40 ⎠
v1 (0.1) + v0 (0.025) = vA (0.125) (2)
v0 = A0 L vd = A0 L (vB − v A ) (3)
or
v0 = A0 L [0.8v2 − v A ]
v0
− 0.8v2 = −v A
A0 L
v0
⇒ v A = 0.8v2 −
A0 L
Then

⎡ v ⎤
v1 (0.1) + v0 (0.025) = (0.125) ⎢0.8v2 − 0 ⎥
⎣ A0 L ⎦
⎡ 0.125 ⎤
v1 (0.1) − v2 (0.1) = −v0 ⎢0.025 +
⎣ 103 ⎥⎦
−2
= −v0 [2.5125 × 10 ]
v0
⇒ Ad = = 3.9801
v2 − v1
ΔAd 0.0199
⇒ = ⇒ 0.4975%
Ad 4

14.12
a. Considering the second op-amp and Equation (14.20), we have
⎡ ⎤
1 1 1 ⎢1 + 100 ⎥ 101
= + ⋅⎢ ⎥ = 0.10 +
Rif 2 10 0.1 ⎢ 1 + 1 ⎥ (0.1)(11)
⎢ 0.1 ⎥
⎣ ⎦
So Rif 2 = 0.0109 kΩ
The effective load on the first op-amp is then
RL1 = 0.1 + Rif 2 = 0.1109 kΩ
Again using Equation (14.20), we have
1
1 + 100 +
1 1 1 0.1109 = 0.10 + 110.017
= + ⋅
Rif 10 1 1 + 1 + 1 11.017
0.1109 1
so that
Rif = 99.1 Ω
b. To determine R0 f :
For the first op-amp, we can write, using Equation (14.36)
⎡ ⎤ ⎡ ⎤
1 1 ⎢ A0 L ⎥ 1 ⎢ 100 ⎥
= ⋅⎢ ⎥ = ⋅⎢ ⎥
R0 f 1 R0 ⎢ R2 ⎥ 1 ⎢ 40 ⎥
1+ 1+
⎢ ⎥ ⎢ ⎥
⎣ 1 || 10 ⎦
⎣ R1 || Ri ⎦
which yields R0 f 1 = 0.021 kΩ
For the second op-amp, then
⎡ ⎤
⎢ ⎥
1 1 ⎢ A0 L ⎥
= ⋅
R0 f R0 ⎢ R2 ⎥
⎢1 + ( R + R ) || R ⎥
⎣ 1 0f1 i ⎦
⎡ ⎤
1 ⎢ 100 ⎥
= ⋅⎢ ⎥
1 ⎢1 + 0.10 ⎥
⎢ (0.121) ||10 ⎥
⎣ ⎦
or R0 f = 18.4 Ω
c. To find the gain, consider the second op-amp.

v01 − (−vd 2 ) vd 2 −vd 2 − v02
+ = (1)
0.1 Ri 0.1
v01 ⎛ 1 1 1 ⎞ v02
+ vd 2 ⎜ + + ⎟=−
0.1 ⎝ 0.1 10 0.1 ⎠ 0.1
or
v01 (10) + vd 2 (20.1) = −v02 (10)
v02 − A0 L vd 2 v02 − (−vd 2 )
+ =0 (2)
R0 0.1
v02 ⎛ 100 1 ⎞ v02
− vd 2 ⎜ − ⎟+ =0
1 ⎝ 1 0.1 ⎠ 0.1
v02 (11) − vd 2 (90) = 0
or
vd 2 = v02 (0.1222)
Then Equation (1) becomes
v01 (10) + v02 (0.1222)(20.1) = −v02 (10)
or
v01 = −v02 (1.246)
Now consider the first op-amp.

vI − (−vd 1 ) vd 1 −vd 1 − v01
+ = (1)
1 Ri 1
⎛1 1 1⎞
vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1)
⎝ 1 10 1 ⎠
or
vI (1) + vd 1 (2.1) = −v01 (1)
v01 v − A0 L vd 1 v01 − (−vd 1 )
+ 01 + =0 (2)
0.1109 R0 1
⎛ 1 1 1⎞ ⎛ 100 1 ⎞
v01 ⎜ + + ⎟ − vd 1 ⎜ − ⎟=0
⎝ 0.1109 1 1 ⎠ ⎝ 1 1⎠
v01 (11.017) − vd 1 (99) = 0
or
vd 1 = v01 (0.1113)
Then Equation (1) becomes

vI (1) + v01 (0.1113)(2.1) = −v01
or vI = −v01 (1.234)
We had v01 = −v02 (1.246)
So vI = v02 (1.246)(1.234)
v02
or = 0.650
vI
v02
d. Ideal =1
vI
So ratio of actual to ideal = 0.650.

14.13
(a) For the op-amp. A0 L ⋅ f 3dB = 106
106
f 3dB = = 50 Hz
2 × 104
For the closed-loop amplifier.
106
f 3dB = = 40 kHz
25
(b) Open-loop amplifier.
2 × 104 2 × 104
A= ⇒| A | =
f 2
1+ j ⎛ f ⎞
f3dB 1+ ⎜ ⎟
⎝ f3dB ⎠
2 × 104
f = 0.25 f 3dB ⇒ A = = 1.94 × 104
2
1 + (0.25)
2 × 104
f = 5 f 3− dB ⇒ A = = 3.92 × 103
1 + (5) 2
Closed-loop amplifier
25
f = 0.25 f 3dB ⇒ A = = 24.25
1 + (0.25) 2
25
f = 5 f 3− dB ⇒ A = = 4.90
1 + (5) 2

14.14
The open loop gain can be written as
A0
A0 L ( f ) =
⎛ f ⎞⎛ f ⎞
⎜1 + j ⋅ ⎟⎜ 1 + j ⋅ ⎟
⎝ f PD ⎠ ⎝ 5 × 106 ⎠
where A0 = 2 × 105.
The closed-loop response is
A0 L
ACL =
1 + β A0 L
At low frequency,
2 × 105
100 =
1 + β (2 × 105 )
So that β = 9.995 × 10−3.
Assuming the second pole is the same for both the open-loop and closed-loop, then
⎛ f ⎞ −1 ⎛ f ⎞
φ = − tan −1 ⎜ ⎟ − tan ⎜ 6 ⎟
⎝ f PD ⎠ ⎝ 5 × 10 ⎠

For a phase margin of 80°, φ = −100°.
So
⎛ f ⎞
−100 = −90 − tan −1 ⎜ 6 ⎟
⎝ 5 × 10 ⎠
or
f = 8.816 × 105 Hz
Then
A0 L = 1
2 × 105
=
2 2
⎛ 8.816 × 105 ⎞ ⎛ 8.816 × 105 ⎞
1+ ⎜ ⎟ 1+ ⎜ 6 ⎟
⎝ f PD ⎠ ⎝ 5 × 10 ⎠
or
8.816 × 105
≅ 1.9696 × 105
f PD
or
f PD = 4.48 Hz

14.15
(a) 1st stage
(10) f3− dB = 1 MHz ⇒ f 3− dB = 100 kHz
2nd stage
(50) f3− dB = 1 MHz ⇒ f 3− dB = 20 kHz
Bandwidth of overall system ≅ 20 kHz
(b) If each stage has the same gain, so
K = 500 ⇒ K = 22.36
2

Then bandwidth of each stage
(22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz

14.16

Ao
A=
f
1+ j
f3− dB
Ao 5 × 104
A= ⇒ 200 = ⇒ f3− dB = 40 Hz
2 2
⎛ f ⎞ ⎛ 104 ⎞
1+ ⎜ ⎟ 1+ ⎜ ⎟
⎝ f3− dB ⎠ ⎝ f3− dB ⎠
Then
fT = (5 × 104 )(40) ⇒ fT = 2 MHz

14.17

(5 × 104 ) f PD = 106 ⇒ f PD = 20 Hz
(25) f 3− dB 106 ⇒ f3− dB = 40 kHz
Avo 25
Av = ⇒ Av =
f 2
1+ j ⎛ f ⎞
f3− dB 1+ ⎜ 3 ⎟
⎝ 40 × 10 ⎠
At f = 0.5 f 3− dB = 20 kHz
25
Av = = 22.36
1 + (0.5) 2
At f = 2 f 3− dB = 80 kHz
25
Av = = 11.18
1 + (2) 2

14.18
(20 × 103 ) ⋅ Avf MAX
= 106 ⇒ Avf MAX
= 50

14.19
From Equation (14.55),
SR 10 × 106
F P BW = =
2π VP 0 2π (10)
or
F P BW = f max = 159 kHz

14.20
a. Using Equation (14.55),
8 × 106
VP 0 =
2π (250 × 103 )
or
VP 0 = 5.09 V
b.

1 1
Period T = = = 4 × 10−6 s
f 250 × 103
One-fourth period = 1 μ s
VP 0
Slope = = SR = 8 V/μ s
1μ s
⇒ VP 0 = 8 V

14.21
For input (a), maximum output is 5 V.

S R = 1 V/μs
so

For input (b), maximum output is 2 V.

For input (c), maximum output is 0.5 V so the output is

14.22
For input (a), max v01 = 3 V.

Then v02 max
= 3(3) = 9 V

For input (b), max v01 = 1.5 V.

Then v02 max
= 3 (1.5 ) = 4.5 V

14.23
f MAX = 20 kHz, SR = 0.8 V / μ s
SR 0.8 × 106
V po = = ⇒ V po = 6.37 V
2π f MAX 2π (20 × 103 )

14.24
⎛V ⎞ ⎛V ⎞
I1 = I S 1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟
⎝ VT ⎠ ⎝ VT ⎠
Want I1 = I 2 , so
⎛V ⎞
5 × 10−14 (1 + x) exp ⎜ BE1 ⎟
I1
=1= ⎝ VT ⎠
I2 ⎛V ⎞
5 × 10−14 (1 − x) exp ⎜ BE 2 ⎟
⎝ VT ⎠
(1 + x) ⎛ V −V ⎞
= exp ⎜ BE1 BE 2 ⎟
(1 − x) ⎝ VT ⎠
Or
1+ x ⎛ V − VBE1 ⎞ ⎛ VOS ⎞
= exp ⎜ BE 2 ⎟ = exp ⎜ ⎟
1− x ⎝ VT ⎠ ⎝ VT ⎠
⎛ 0.0025 ⎞
= exp ⎜ ⎟ = 1.10
⎝ 0.026 ⎠
Now
1 + x = (1 − x )(1.10) ⇒
x = 0.0476 ⇒ 4.76%

14.25
⎛ vCE1 ⎞ ⎛ vCE 2 ⎞
⎜1+ V ⎟ I ⎜ 1+ V ⎟
⎜ AN ⎟
= S3 ⋅⎜ AN ⎟
⎜ 1 + vEB ⎟ I S 4 ⎜ 1 + vEC 4 ⎟
⎜ V ⎟ ⎜ ⎟
⎝ AP ⎠ ⎝ VAP ⎠
For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V so

⎛ 5 ⎞ ⎛ 0.6 ⎞
⎜ 1 + 80 ⎟ I S 3 ⎜ 1 + 80 ⎟
⎜ ⎟= ⋅⎜ ⎟
⎜ 1 + 0.6 ⎟ I S 4 ⎜ 1 + 5 ⎟
⎜ ⎟ ⎜ ⎟
⎝ 80 ⎠ ⎝ 80 ⎠
or
I S 3 (1.0625) 2
= = 1.112
I S 4 (1.0075) 2
So
I S 3 = (10−14 ) (1.112 )
or
I S 3 = 1.112 × 10−14 A

14.26

By superposition:
R
vo (vi ) = − 2 ⋅ vi = −50vi
R1
⎛ R ⎞
vo (vos ) = ⎜ 1 + 2 ⎟ ⋅ vos = 51vos
⎝ R1 ⎠
So
vo = vo ( vi ) + vo ( vos ) = −50vi + 51vos
For vi = 20 mV and vos + 2.5 mV
vo = −50(0.02) + 51(0.0025) = −0.8725 V
For vi = 20 mV and vos = −2.5 mV
vo = −50(0.02) + 51(−0.0025) = −1.1275 V
So
−1.1275 ≤ vo ≤ −0.8725 V

14.27
vo = −50vi = −50 ⎡ ±2.5 mV + sin ω t ( mV ) ⎤
⎣ ⎦
vo = [ ±0.125 − 0.25sin ω t ] ( v )

14.28

0.5 × 10−3
I= = 5 × 10−8 A
104
Also
i
dV 1 I
I = C o ⇒ Vo = ∫ Idt = ⋅ t
dt C0 C
Then
5 × 10−8
5= t ⇒ t = 103 s
10 × 10−6

14.29
⎛ 100 ⎞
a. | v01 | = 10 ⎜ 1 + ⎟ or | v01 | = 110 mV
⎝ 10 ⎠
Then
⎛ 50 ⎞
| v02 | = | v01 | (5) + 10 ⎜ 1 + ⎟ = (110)(5) + (10)(6)
⎝ 10 ⎠
or
| v02 | = 610 mV

14.30
v0 due to vI
⎛ 1 ⎞
v0 = (0.5) ⎜1 + ⎟ = 0.9545 V
⎝ 1.1 ⎠
Wiper arm at V + = 10 V, (using superposition)
⎛ R1 || R5 ⎞ ⎛ 0.0909 ⎞
v1 = ⎜ ⎟ (10) = ⎜ ⎟ (10)
⎝ R1 || R5 + R4 ⎠ ⎝ 0.0909 + 10 ⎠
= 0.090
⎛ 1⎞
Then v01 = − ⎜ ⎟ (0.090) = −0.090
⎝ 1⎠
Wiper arm in center, v1 = 0 and v02 = 0
Wiper arm at V − = −10 V, v1 = −0.090
So
v03 = 0.090
Finally, total output v0 : (from superposition)
Wiper arm at V + ,
v0 = 0.8645 V
Wiper arm in center,
v0 = 0.9545 V
Wiper arm at V − ,

v0 = 1.0445 V

14.31
a. R1′ = R2 = 0.5 || 25 = 0.490 kΩ
′
or
R1′ = R2 = 490 Ω
′
b. From Equation (14.75),
⎛ 125 × 10−6 ⎞
(0.026) ln ⎜ −14 ⎟
+ (0.125) R1′
⎝ 2 × 10 ⎠
⎛ 125 × 10−6 ⎞
= (0.026) ln ⎜ −14 ⎟
+ (0.125) R2′
⎝ 2.2 × 10 ⎠
0.586452 + (0.125) R1′ = 0.583974 + (0.125) R2 ′
0.002478 = (0.125)( R2 − R1′)
′
So R2 − R1′ = 0.0198 kΩ ⇒ 19.8 Ω
′
Then
R2 (1 − x) Rx R × Rx
− 1 = 0.0198
R2 + (1 − x) Rx R1 + xRx
(0.5)(1 − x)(50) (0.5)(50) x
− = 0.0198
(0.5) + (1 − x)(50) (0.5) + x(50)
25(1 − x) 25 x
− = 0.0198
50.5 − 50 x 0.5 + 50 x
(0.5 + 50 x)(25 − 25 x) − (25 x)(50.5 − 50 x)
= 0.0198
(50.5 − 50 x )(0.5 + 50 x)
25 {0.5 − 0.5 x + 50 x − 50 x 2 − 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x − 25 x − 2500 x 2 }
25 {0.5 − x} = 0.0198 {25.25 + 2500 x − 2500 x 2 }
0.5 − x = 0.019998 + 1.98 x − 1.98 x 2
1.98 x 2 − 2.98 x + 0.48 = 0
2.98 ± (2.98) 2 − 4(1.98)(0.48)
x=
2(1.98)
So
x = 0.183
and
1 − x = 0.817

14.32
R1′ = R1 ||15 = 0.5 ||15 = 0.4839 kΩ
′
R2 = R2 || 35 = 0.5 || 35 = 0.4930 kΩ
⎛i ⎞ ⎛i ⎞
(0.026) ln ⎜ C1 ⎟ + iC1 R1′ = (0.026) ln ⎜ C 2 ⎟ + iC 2 R2
′
⎝ IS3 ⎠ ⎝ IS 4 ⎠
⎛i ⎞
(0.026) ln ⎜ C1 ⎟ = iC 2 R2 − iC1 R1′
′
⎝ iC 2 ⎠
⎛i ⎞ ⎡ i R′ ⎤
′
(0.026) ln ⎜ C1 ⎟ = iC 2 R2 ⎢1 − C1 ⋅ 1 ⎥
⎝ iC 2 ⎠ ⎣ ′
iC 2 R2 ⎦
⎛i ⎞ ⎡ ⎛ i ⎞⎤
(0.026) ln ⎜ C1 ⎟ = iC 2 (0.4930) ⎢1 − (0.9815) ⎜ C1 ⎟ ⎥
⎝ iC 2 ⎠ ⎣ ⎝ iC 2 ⎠ ⎦
By trial and error:

iC1 = 252 μ A and iC 2 = 248 μ A
or
iC1
= 1.0155
iC 2

14.33
From Eq. (14.79), we have
⎛ R ⎞
vo = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟
⎝ R1 ⎠
I B1 = 1 μ A I B 2 = 2 μ A
Setting vo = 0, we have
⎛ 200 ⎞
0 = (10−6 )( 200 × 103 ) − ( 2 × 10−6 ) R3 ⎜ 1 + ⎟
⎝ 20 ⎠
200 × 10−3
R3 = ⇒ R = 9.09 K
( 2 ×10−6 ) (11) 3
14.34

R2
1+ = 80
R1
R1 = 6.329 K
V f = vI = 5sin ω t ( mV )
I1 + I B = I 2
(a) VI = 0 ⇒ VX = 0 I 2 = I B
VO = (10−6 )(500 × 103 ) ⇒ vo = 0.50 V

VX v − VX
(b) + IB = o
R1 R2
⎛ 1 1 ⎞ v ⎛ 1 1 ⎞
VX ⎜ + ⎟ + I B = 0 ⇒ vo = R2 ⎜ + ⎟ vI + I B R2
⎝ R1 R2 ⎠ R1 ⎝ R1 R2 ⎠
vo = 80 ⎡5sin ω t ( mV ) ⎤ + (10−6 )( 500 × 103 )
⎣ ⎦
vo = [ 0.5 + 0.4sin ω t ] ( v )

14.35
a.

For I B 2 = 1 μ A, then v0 = − (10−6 )(104 )
or v0 = −0.010 V
b. If a 10 kΩ resistor is included in the feedback loop

Now v0 = − I B 2 (10) + I B1 (10) = 0
Circuit is compensated if I B1 = I B 2 .

14.36
From Equation (14.83), we have
v0 = R2 I 0S
where R2 = 40 kΩ and I 0 S = 3 μ A.
Then
v0 = ( 40 × 103 )( 3 × 10−6 )
or
v0 = 0.12 V

14.37
a. Assume all bias currents are in the same direction and into each op-amp.
v01 = I B1 (100 kΩ ) = (10−6 )(105 ) ⇒ v01 = 0.1 V
Then
v02 = v01 ( −5 ) + I B1 ( 50 kΩ )
= ( 0.1)( −5 ) + (10−6 )( 5 × 104 )
= −0.5 + 0.05
or
v02 = −0.45 V
b. Connect R3 = 10 ||100 = 9.09 kΩ resistor to noninverting terminal of first op-amp, and
R3 = 10 || 50 = 8.33 kΩ resistor to noninverting terminal of second op-amp.

14.38
a. For a constant current through a capacitor.
1 t
v0 = ∫ I dt
C 0
0.1× 10−6
or v0 = ⋅ t ⇒ v0 = (0.1)t
10−6
b. At t = 10 s, v0 = 1 V
c. Then
100 × 10−12
v0 = −6
⋅ t ⇒ v0 = (10−4 )t
10
At t = 10 s, v0 = 1 mV

14.39
a. Assume all bias currents are into the op-amp.
v01 = I B1 ( 50 kΩ ) = (10 ×10−6 )( 50 ×103 )
or
v01 = v02 = 0.5 V
v03 = ( −1)( v01 ) + (10 × 10−6 )( 20 × 103 )
or
v03 = −0.3 V
b. RA = 10 || 50 ⇒ RA = 8.33 kΩ
RB = 20 || 20 ⇒ RB = 10 kΩ
c. Assume the worst case offset current, that is, I 0 S = I B1 − I B 2 or I 0 S = I B 2 − I B1 .
v01 = R2 I 0 S = ( 50 × 103 )( 2 × 10−6 )
or
v01 = v02 = 0.1 V
v03 = ( −1) v01 − I 0 S R2
= ( −1)( 0.1) − ( 2 × 10−6 )( 20 × 103 )
or
v03 = −0.14 V

14.40
a. Using Equation (14.79),
Circuit (a),
⎛ 50 ⎞
v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 ×10−6 )( 25 × 103 ) ⎜1 + ⎟
⎝ 50 ⎠
or
v0 = 0
Circuit (b),
⎛ 50 ⎞
v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 × 10 −6 )(103 ) ⎜1 + ⎟
⎝ 50 ⎠
−2
= 4 × 10 − 1.6
or
v0 = −1.56 V
b. Assume I B1 = 0.7 μ A and I B 2 = 0.9 μ A, then using Equation (14.79):
Circuit (a),

⎛ 50 ⎞
v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )( 25 × 103 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.035 − 0.045
or v0 = −0.010 V
Circuit (b),
⎛ 50 ⎞
v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 1.8
or v0 = −1.765 V

14.41
a. If R = 0,
⎛ 100 ⎞
v0,max = ⎜1 + ⎟ V0 S + I B (100 kΩ)
⎝ 10 ⎠
= (11)(10 × 10−3 ) + (2 × 10−6 )(100 × 103 )
v0,max = 0.110 + 0.20
⇒ v0,max = 0.310 V
b.

R = 10.1|| 100 = 9.17 kΩ = R

14.42
⎛ Ri ⎞
a. ⎜ ⎟ (15) = 0.010 V
⎝ Ri + R2 ⎠
15
= 0.0006667
15 + R2
15(1 − 0.0006667) = 0.0006667 R2
Then
R2 = 22.48 MΩ
b. R1 = Ri || RF = 15 || 10 ⇒ R1 = 6 kΩ

14.43
a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias
currents are in the same direction.
Use superposition:

Offset voltages
⎛ 100 ⎞
| v01 | = ⎜1 + ⎟ (10) = 110 mV =| v01 |
⎝ 10 ⎠
⎛ 50 ⎞
| v02 | = (5)(110) + ⎜ 1 + ⎟ (10)
⎝ 10 ⎠
⇒ | v02 | = 610 mV
Bias Currents:
v01 = I B (100 kΩ) = (2 ×10−6 )(100 × 103 ) = 0.2 V
Then
v02 = (−5)(0.2) + (2 × 10−6 )(50 × 103 ) = −0.9 V
Worst case: v01 is positive and v02 is negative, then
v01 = 0.31 V and v02 = −1.51 V
b. Compensation network:

If we want
⎛ RB ⎞ + +
⎜ ⎟ V = 20 mV and V = 10 V
⎝ RB + RC ⎠
⎛ 8.33 ⎞
⎜ ⎟ (10) = 0.020
⎝ 8.33 + RC ⎠
or RC ≅ 4.15 MΩ

14.44
Assume bias currents are in same direction, but assume polarity of offset voltages are such as to
produce the worst case output.
a. Let I B1 = 5.5 μ A, I B 2 = 4.5 μ A
Bias Current Effects:
v01 = I B1 (50 kΩ) = 0.275 V ⇒ v02 = 0.275 V
v03 = I B1 (20 kΩ) − v01 ⇒ v03 = −0.165 V
Offset Voltage Effects:
⎛ 50 ⎞
v01 = (5) ⎜1 + ⎟ = 30 mV ⇒ v02 = 30 mV
⎝ 10 ⎠
⎛ 20 ⎞
v03 = −v01 − 5 ⎜ 1 + ⎟ ⇒ v03 = −40 mV
⎝ 20 ⎠
Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = −0.205 V

14.45
For circuit (a), effect of bias current:
v0 = (50 × 103 )(100 × 10−9 ) ⇒ 5 mV
Effect of offset voltage

⎛ 50 ⎞
v0 = (2) ⎜1 + ⎟ = 4 mV
⎝ 50 ⎠
So net output voltage is v0 = 9 mV
For circuit (b), effect of bias current:
Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79),
⎛ 50 ⎞
v0 = (450 × 10−9 )(50 × 103 ) − (550 × 10−9 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
−2
= 2.25 × 10 − 1.1
or
v0 = −1.0775 V
If the offset voltage is negative, then
v0 = (−2)(2) = −4 mV
So the net output voltage is
v0 = −1.0815 V

14.46
a. At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is
v0 = 4 mV
b. For T = 50°C, the offset voltage for is
V0 S = 2 mV + (0.0067)(25) = 2.1675 mV
so the output voltage for each circuit is
v0 = 4.335 mV

14.47
a. At T = 25°C, V0 S = 1 mV, then
⎛ 50 ⎞
v01 = (1) ⎜ 1 + ⎟ ⇒ v01 = 6 mV
⎝ 10 ⎠
and
⎛ 60 ⎞ ⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + (1) ⎜ 1 + ⎟
⎝ 20 ⎠ ⎝ 20 ⎠
= 6(4) + (1)(4) ⇒ v02 = 28 mV
b. At T = 50°C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then
v01 = (1.0825)(6) ⇒ v01 = 6.495 mV
and
v02 = (6.495)(4) + (1.0825)(4)
or
v02 = 30.31 mV

14.48
25°C; I B = 500 nA, I 0 S = 200 nA
50°C, I B = 500 nA + (8 nA / °C)(25°C) = 700 nA
I 0 S = 200 nA + (2 nA / °C)(25°C) = 250 nA
a. Circuit (a): For I B , bias current cancellation, v0 = 0
Circuit (b): For I B , Equation (14.79),
⎛ 50 ⎞
v0 = (500 × 10−9 )(50 × 103 ) − (500 × 10 −9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.025 − 1.00 ⇒ v0 = −0.975 V
b. Due to offset bias currents.
Circuit (a):

v0 = (200 × 10−9 )(50 × 103 ) ⇒ v0 = 0.010 V
Circuit (b):
Let I B 2 = 600 nA
I B1 = 400 nA
Then
⎛ 50 ⎞
v0 = (400 × 10−9 )(50 × 103 ) − (600 × 10−9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.020 − 1.20 ⇒ v0 = −1.18 V
c. Circuit (a): Due to I B , v = 0
Circuit (b): Due to I B ,
⎛ 50 ⎞
v0 = (700 × 10−9 )(50 × 103 ) − (700 × 10−9 )(106 ) ⎜1 + ⎟
⎝ 50 ⎠
= 0.035 − 1.40 ⇒ v0 = −1.365 V
Circuit (a): Due to I 0 S ,
v0 = (250 × 10−9 )(50 × 103 ) ⇒ v0 = 0.0125 V
Circuit (b): Due to I 0 S ,
Let I B 2 = 825 nA
I B1 = 575 nA
Then
⎛ 50 ⎞
v0 = (575 × 10−9 )(50 × 103 ) − (825 × 10−9 )(106 ) ⎜ 1 + ⎟
⎝ 50 ⎠
= 0.02875 − 1.65 ⇒ v0 = −1.62 V

14.49
25°C; I B = 2 μ A, I 0 S = 0.2 μ A
50°C, I B = 2 μ A + (0.020 μ A / °C)(25°C ) = 2.5 μ A
I 0 S = 0.2 μ A + (0.005 μ A / °C)(25°C) = 0.325 μ A
a. Due to I B : (Assume bias currents into op-amp).
v01 = I B (50 kΩ) = (2 × 10−6 )(50 × 103 )
⇒ v01 = 0.10 V
⎛ 60 ⎞ ⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟
⎝ 20 ⎠ ⎝ 20 ⎠
−6 −6
= (0.1)(4) + (2 × 10 )(60 × 10 ) − (2 × 10 )(60 × 103 )4
3

or v02 = 0.12 V
b. Due to I 0 S :
1st op-amp. Let I B1 = 2.1 μ A
2nd op-amp. Let I B1 = 2.1 μ A
I B 2 = 1.9 μ A
v01 = I B1 (50 kΩ) = (2.1× 10−6 )(50 × 103 )
⇒ v01 = 0.105 V
⎛ 60 ⎞ ⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟
⎝ 20 ⎠ ⎝ 20 ⎠
= (0.105)(4) + (2.1× 10 )(60 × 10 ) − (1.9 × 10 −6 )(50 × 103 )(4)
−6 3

or
v02 = 0.166 V
c. Due to I B :

v01 = (2.5 × 10 −6 )(50 × 103 ) ⇒ v01 = 0.125 V
⎛ 60 ⎞ ⎛ 60 ⎞
v01 = v02 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜1 + ⎟
⎝ 20 ⎠ ⎝ 20 ⎠
−6
= (0.125)(4) + (2.5 × 10 )(60 × 10 ) − (2.5 × 106 )(50 × 103 (4)
3

or v02 = 0.15 V
Due to I 0 S :
Let I B1 = 2.625 μ A
I B 2 = 2.3375 μ A
v01 = I B1 (50 kΩ) = (2.6625 × 10−6 )(50 × 103 )
⇒ v01 = 1.133 V
⎛ 60 ⎞ ⎛ 60 ⎞
v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜ 1 + ⎟
⎝ 20 ⎠ ⎝ 20 ⎠
= (0.133)(4) + (2.6625 × 10−6 )(60 × 103 ) − (2.3375 × 10−6 )(50 × 103 )(4)
or
v02 = 0.224 V

14.50

⎛ R4 ⎞ ⎛ R2 ⎞
vB = ⎜ ⎟ vI 1 and v0 (vI 2 ) = vB ⎜ 1 + ⎟
⎝ R3 + R4 ⎠ ⎝ R1 ⎠
or
⎛ R4 ⎞ ⎛ R2 ⎞
v0 (vI 2 ) = ⎜ ⎟ ⎜ 1 + ⎟ vI 2
⎝ R3 + R4 ⎠ ⎝ R1 ⎠
or vI 1 .
R2
v0 (vI 1 ) = − ⋅ vI
R1
Then
⎛ R4 ⎞⎛ R2 ⎞ R2
v0 = ⎜ ⎟⎜ 1 + ⎟ vI 2 − ⋅ vI 1
⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1
V V
we can write vI 2 = vcm + d and vI 1 = Vcm − d ⋅ Then
2 2
⎛ R4 ⎞⎛ R2 ⎞ ⎛ Vd ⎞ R2 ⎛ Vd ⎞
v0 = ⎜ ⎟⎜ 1 + ⎟ ⎜ Vcm + ⎟ − ⋅ ⎜ Vcm − ⎟
⎝ R3 + R4 ⎠ ⎝ R1 ⎠ ⎝ 2 ⎠ R1 ⎝ 2⎠
Common-mode gain
v ⎛ R4 ⎞ ⎛ R2 ⎞ R2
Acm = 0 = ⎜ ⎟ ⎜1 + ⎟ −
Vcm ⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1

Differential mode gain
v 1 ⎡⎛ R4 ⎞⎛ R2 ⎞ R2 ⎤
Ad = 0 = ⎢⎜ ⎟⎜ 1 + ⎟ + ⎥
Vd 2 ⎣⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 ⎦
Then
A
CM RR = d
Acm
1 ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤
⋅ ⎢⎜ ⎟ ⎜1 + ⎟ + ⎥
2 ⎣⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 ⎦
=
⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤
⎢⎜ ⎟ ⎜1 + ⎟ − ⎥
⎣⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 ⎦
⎡ ⎤
⎢ ⎥
1 ⎢ R4 1 ⎛ R2 ⎞ R2 ⎥
⋅ ⋅ ⋅ ⎜1 + ⎟ +
2 ⎢ R3
⎛ R4 ⎞ ⎝ R1 ⎠ R1 ⎥
⎢ ⎜ 1+ ⎟ ⎥
⎢
⎣ ⎝ R3 ⎠ ⎥
⎦
CM RR =
R4 1 ⎛ R2 ⎞ R2
⋅ ⋅ ⎜1 + ⎟ −
R3 ⎛ R4 ⎞ ⎝ R1 ⎠ R1
⎜1 + ⎟
⎝ R3 ⎠
Minimum CMRR ⇒ Maximum denominator
R R
⇒ maximum 4 and minimum 2 . Then
R3 R1
R4 (1.02)(50)
= = 5.204
R3 (0.98)(10)
R2 (0.98)(50)
= = 4.804
R1 (1.02)(10)
Then
1 ⎡ 5.204 ⎤
⋅ (5.804) + (4.804) ⎥
2 ⎢ 6.204
CMRR = ⎣ ⎦
⎡ 5.204 ⎤
⎢ 6.204 ⋅ (5.804) − (4.804) ⎥
⎣ ⎦
1
⋅ (9.6725)
= 2
(0.06447)
CMRR = 75.0 ⇒ CMRRdB = 20 log10 (75.0)
⇒ CMRRdB = 37.5 dB

14.51
Use the results of Problem 14.50:
R 1 + x ⎛ 50 ⎞
Let 4 = ⋅ ⎜ ⎟ ≈ (1 + 2 x)(5)
R3 1 − x ⎝ 10 ⎠
R 1 − x ⎛ 50 ⎞
Let 2 = ⋅ ⎜ ⎟ ≈ (1 − 2 x)(5)
R1 1 + x ⎝ 10 ⎠
Then

1 ⎡ (1 + 2 x ) 5 ⎤
⎢ ⋅ ( 6 − 10 x ) + (1 − 2 x )( 5 ) ⎥
2 ⎣ 6 + 10 x ⎦
CMRR =
⎡ (1 + 2 x ) 5 ⎤
⎢ ⋅ ( 6 − 10 x ) − (1 − 2 x )( 5 ) ⎥
⎣ 6 + 10 x ⎦
1
⎡30 + 10 x − 100 x + 30 − 10 x − 100 x 2 ⎤
⎣
2
⎦
= 2
⎡30 + 10 x − 100 x − ( 30 − 10 x − 100 x ) ⎤
2 2
⎣ ⎦
1
⋅ [60 − 200 x 2 ]
2 30 − 100 x 2
= =
20 x 20 x
a. For CMRRdB = 90 dB ⇒ CMRR = 31, 623 x will be small, neglect the x 2 term. Then
30
20 x = ⇒ x = 0.0000474 = 0.00474%
31, 623
b. For CMRRdB = 60 dB ⇒ CMRR = 1000. Then
30
20 x = ⇒ x = 0.0015 = 0.15%
1000

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