The document contains solutions to example problems from Chapter 9 of an exercise book. The examples calculate resistor values for circuits involving operational amplifiers to achieve specified voltage gains, current values, and time constants. Computer analysis examples also calculate output voltages and currents for various input conditions in operational amplifier circuits.

1. Chapter 9
Exercise Solutions
EX9.1
R2
Av = −15 = −
R1
R1 = Ri = 20 K ⇒ R2 = 15 R1 = 15 ( 20 ) = 300 K
EX9.2
R2 ⎛ R3 ⎞ R3
We can write ACL = − ⎜1 + ⎟ −
R1 ⎝ R4 ⎠ R1
R1 = R1 = 10 kΩ
Want ACL = −50 Set R2 = R3 = 50 kΩ
⎛ R ⎞
ACL = −50 = −5 ⎜ 1 + 3 ⎟ − 5
⎝ R4 ⎠
R R 50
1+ 3 = 9 ⇒ 3 = 8 =
R4 R4 R4
R4 = 6.25 kΩ
EX9.3
R2 1
We have ACL = − ⋅
R1 ⎡ 1 ⎛ R2 ⎞ ⎤
⎢1 + ⎜1 + ⎟ ⎥
⎣ Ad ⎝ R1 ⎠ ⎦
R
Ri = R1 = 25 kΩ Let 2 = x
R1
x
−12 = −
1
1+ (1 + x )
5 × 103
−x
=
x
1.0002 +
5 × 103
⎛ x ⎞
12 ⎜ 1.0002 + ⎟=x
⎝ 5 × 103 ⎠
12.0024 = x − ( 2.4 × 10 −3 ) x
12.0024 R2
x= = 12.0313 =
0.9976 25 kΩ
R2 = 300.78 kΩ
EX9.4
⎛R R R R ⎞
v0 = − ⎜ F vI 1 + F vI 2 + F vI 3 + F VI 4 ⎟
⎝ R1 R2 R3 R4 ⎠

2. RF R R R
We need = 7, F = 14, F = 3.5, F = 10
R1 R2 R3 R4
Set RF = 280 kΩ
280
Then R1 = = 40 kΩ
7
280
R2 = = 20 kΩ
14
280
R3 = = 80 kΩ
3.5
280
R4 = = 28 kΩ
10
EX9.5
R3 3 R 20 R R
We may note that = = 2 and F = = 2 so that 3 = F
R2 1.5 R1 10 R2 R1
Then
− vI − ( −3)
iL = = ⇒ iL = 2 mA
R2 1.5 kΩ
vL = iL Z L = ( 2 × 10−3 ) ( 200 ) = 0.4 V
vL 0.4
i4 = = = 0.267 mA
R2 1.5 kΩ
i3 = i4 + iL = 0.267 + 2 = 2.267 mA
v0 = i3 R3 + vL = ( 2.267 × 10−3 )( 3 × 103 ) − 0.4 ⇒ v0 = 7.2 V
EX9.6
Refer to Fig. 9.24
R1 = 2 R1 = 5 kΩ
Let R1 = R3 = 2.5 kΩ
Set R2 = R4
v0 R2 R2
Differential Gain = = = 100 = ⇒ R2 = R4 = 250 kΩ
v1 R1 2.5 kΩ
EX9.7

3. We have the general relation that
⎛ R ⎞ ⎛ [ R4 / R3 ] ⎞ R
v0 = ⎜ 1 + 2 ⎟ ⎜ v − 2v
⎜ 1 + [ R / R ] ⎟ I 2 R I1
⎟
⎝ R1 ⎠ ⎝ 4 3 ⎠ 1
R1 = R3 = 10 kΩ, R2 = 20 kΩ, R4 = 21 kΩ
⎛ 20 ⎞ ⎛ [ 21/10] ⎞ ⎛ 20 ⎞
v0 = ⎜ 1 + ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1
⎝ 10 ⎠ ⎜ 1 + [ 21/10] ⎟
⎝ ⎠ ⎝ 10 ⎠
v0 = 2.0323vI 2 − 2.0vI
a. vI 1 = 1, vI 2 = −1
v0 = −2.0323 − 2.0 ⇒ v0 = −4.032 V
b. vI 1 = vI 2 = 1 V
v0 = 2.0323 − 2.0 ⇒ v0 = 0.0323 V
c. vcm = vI 1 = vI 2 so common-mode gain
v0
Acm = = 0.0323
vcm
d.
⎛ A ⎞
C M R RdB = 20 log10 ⎜ d ⎟
⎝ Acm ⎠
2.0323 ⎛ 1⎞
Ad = − ( 2.0 ) ⎜ − ⎟ = 2.016
2 ⎝ 2⎠
⎛ 2.016 ⎞
C M R RdB = 20 log10 ⎜ ⎟ = 35.9 d B
⎝ 0.0323 ⎠
EX9.8
R4 ⎛ 2 R2 ⎞
v0 = − ⎜1 + ⎟ ( vI 1 − vI 2 )
R3 ⎝ R1 ⎠
R4 ⎛ 2 R2 ⎞
Differential gain (magnitude) = ⎜1 + ⎟
R3 ⎝ R1 ⎠
Minimum Gain ⇒ Maximum R1 = 1 + 50 = 51 kΩ
20 ⎛ 2 (100 ) ⎞
So Ad = ⎜1 + ⎟ ⇒ Ad = 4.92
20 ⎝ 51 ⎠
Maximum Gain ⇒ Minimum R1 = 1 kΩ
20 ⎛ 2 (100 ) ⎞
Ad = ⎜1 + ⎟ ⇒ Ad = 201
20 ⎝ 1 ⎠
Range of Differential Gain = 4.92 − 201
EX9.9
Time constant = r = R1C2 = (104 )( 0.1×10−6 )
= 1 m sec
−1
0 ≤ t ≤ 1 ⇒ v0 = ×t
R1 C2
At t = 1 m sec ⇒ v0 = −1 V
1
0 ≤ t ≤ 2 ⇒ v0 = −1 + × ( t − 1)
R1 C2
( 2 − 1)
At t = 2 m sec ⇒ v0 = −1 + =0
1

4. EX9.10
v0 = vI 1 + 10vI 2 − 25vI 3 − 80vI 4
From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB.
From Equation (9.94)
RF R
= 25 and F = 80
R1 R2
Set RF = 500 kΩ, then R1 = 20 kΩ, and R2 = 6.25 kΩ.
⎛ R ⎞⎛ R ⎞ ⎛ R ⎞⎛ R ⎞
Also ⎜1 + F ⎟⎜ P ⎟ = 1 and ⎜1 + F ⎟ ⎜ P ⎟ = 10
⎝ RN ⎠ ⎝ RA ⎠ ⎝ RN ⎠ ⎝ RB ⎠
where RN = R1 R2 = 20 6.25 = 4.76 kΩ and RP = RA RB RC
RA
We find that = 10
RB
Let RA = 200 kΩ, RB = 20 kΩ
⎛ 500 ⎞ ⎛ RP ⎞ ⎛ RP ⎞
Now ⎜1 + ⎟⎜ ⎟ = 1 = (106 ) ⎜ ⎟
⎝ 4.76 ⎠ ⎝ RA ⎠ ⎝ 200 ⎠
Then RP = 1.89 kΩ
RA RB = 200 20 = 18.2 kΩ
18.2 RC
So RP = 1.89 = ⇒ RC = 2.11 kΩ
18.2 + RC
EX9.11 Computer Analysis
TYU9.1
R2 −100 kΩ
ACL = − = ⇒ ACL = −10
R1 10 kΩ
vI = 0.25 V ⇒ vo = −2.5 V
vI 0.25
i1 = = = 0.025 mA ⇒ i1 = 25 μ A
R1 10 kΩ
i2 = i1 = 25 μ A
Ri = R1 = 10 kΩ
TYU9.2
(a)
− R2
Av =
R1 + RS
−100
Av ( min ) = = −4.926
19 + 1.3
−100
Av ( max ) = = −5.076
19 + 0.7
so 4.926 ≤ Av ≤ 5.076
(b)

5. 0.1
i1 ( max ) = = 5.076 μ A
19 + 0.7
0.1
i1 ( min ) = = 4.926 μ A
19 + 1.3
so 4.926 ≤ i1 ≤ 5.076 μ A
(c) Maximum current specification is violated.
TYU9.3
v0 = Ad ( v2 − v1 ) Ad = 103
a.
v2 = 0, v0 = 5
v 5
v1 = − 0 = − 3 ⇒ v1 = −5 mV
Ad 10
b.
v1 = 5, v0 = −10
v0
= v2 − v1
Ad
−10
= v2 − 5 ⇒ v2 = 4.99 V
103
c.
v1 = 0.001, v2 = −0.001
v0 = 103 ( −0.001 − 0.001)
v0 = −2 V
d.
v2 = 3, v0 = 3
v0 = Ad ( v2 − v1 )
v0
= v2 − v1
Ad
3
= 3 − v1 ⇒ v1 = 2.997 V
103
TYU9.4
⎛R R R ⎞
v0 = − ⎜ 4 vI 1 + 4 vI 2 + 4 vI 3 ⎟
⎝ R1 R2 R3 ⎠
⎡⎛ 40 ⎞ ⎛ 40 ⎞ ⎛ 40 ⎞ ⎤
v0 = − ⎢⎜ ⎟ ( 250 ) + ⎜ ⎟ ( 200 ) + ⎜ ⎟ ( 75 ) ⎥
⎣⎝ 10 ⎠ ⎝ 20 ⎠ ⎝ 30 ⎠ ⎦
v0 = − [1000 + 400 + 100]
v0 = −1500 μ V = −1.5 mV
TYU9.5

6. vI 1 + vI 2 + vI 3 RF
vO = = ( vI 1 + vI 2 + vI 3 )
3 R
RF 1
= ⇒ R1 = R2 = R3 ≡ R = 1 M Ω
R 3
1
Then RF = M Ω = 333 k Ω
3
TYU9.6
v ⎛ R ⎞ R
Av = 0 = ⎜ 1 + 2 ⎟ = 5 so that 2 = 4
vI ⎝ R1 ⎠ R1
For v0 = 10 V, vI = 2 V
2
Then i1 = = 50 μ A ⇒ R1 = 40 kΩ
R1
v0 − vI 10 − 2
Then R2 = 160 kΩ we find i2 = = = 50 μ A
R2 160
TYU9.7
v0 = Aod ( v2 − v1 ) = Aod ( vI − v1 )
v0 v
− vI = −v1 or v1 = vI − 0
Aod Aod
v1 v −v
i1 = = i2 and i2 = 0 1
R1 R2
⎛ 1 ⎞ v −v
Then v1 ⎜ ⎟ = 0 1
⎝ R1 ⎠ R2
⎛ 1 1 ⎞ v
v1 ⎜ + ⎟ = 0
⎝ R1 R2 ⎠ R2
⎛ R ⎞ ⎛ R ⎞⎛ v ⎞
v0 ⎜ 1 + 2 ⎟ v1 = ⎜ 1 + 2 ⎟ ⎜ vI − 0 ⎟
⎝ R1 ⎠ ⎝ R1 ⎠⎝ Aod ⎠
⎛ R2 ⎞
⎜1 + ⎟
So Av = =
v0 ⎝ R1 ⎠
vI 1 ⎛ R2 ⎞
1+ ⎜1 + ⎟
Aod ⎝ R1 ⎠
TYU9.8

7. ⎛ Rb ⎞
For vI 2 = 0, v2 = ⎜ ⎟ vI 1 and
⎝ Rb + Ra ⎠
⎛ R ⎞ ⎛ Rb ⎞
v0 ( vI 1 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 1
⎝ R1 ⎠⎝ Rb + Ra ⎠
⎛ 70 ⎞⎛ 50 ⎞
= ⎜ 1 + ⎟⎜ ⎟ vI 1
⎝ 5 ⎠⎝ 50 + 25 ⎠
= 10vI 1
⎛ Ra ⎞
For vI 1 = 0, v2 = ⎜ ⎟ vI 2
⎝ Rb + Ra ⎠
⎛ R ⎞ ⎛ Ra ⎞
v0 ( vI 2 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2
⎝ R1 ⎠⎝ Rb + Ra ⎠
⎛ 70 ⎞⎛ 25 ⎞
= ⎜ 1 + ⎟⎜ ⎟ vI 2
⎝ 5 ⎠⎝ 25 + 50 ⎠
= 5vI 2
Then
v0 = v0 ( vI 1 ) + v0 ( vI 2 )
v0 = 10vI 1 + 5vI 2
TYU9.9
RS Ri so i1 = i2 = iS = 100 μ A
v0 = −iS RF
We want −10 = − (100 × 10−6 ) RF ⇒ RF = 100 kΩ
TYU9.10
We want iL = 1 mA when vI = −5 V
−VI −v − ( −5 )
iL = ⇒ R2 = I = ⇒ R2 = 5 kΩ
R2 i2 10−3
vL = iL Z L = (10−3 ) ( 500 ) ⇒ vL = 0.5 V
vL 0.5
i4 = = ⇒ i4 = 0.1 mA
R2 5 kΩ
i3 = i4 + iL = 0.1 + 1 = 1.1 mA
If op-amp is biased at ±10 V, output must be limited to ≈ 8 V.
So v0 = i3 R3 + vL
8 = (1.1× 10−3 ) R3 + 0.5 ⇒ R3 = 6.82 kΩ
Let R3 = 7.0 kΩ

8. R3 RF 7
Then we want = = = 1.4
R2 R1 5
Can choose R1 = 10 kΩ and RF = 14 kΩ
TYU9.11
a.
v −v
i1 = I 1 I 2
R1
R4
′
v01 = vI 1 + i1 R2 , v02 = vI 2 − i1 R2 and v0 = ( v02 − v01 )
R3
R4
v0 = [vI 2 − i1 R2 − vI 1 − i1 R2′ ]
R3
R4
v0 = ⎡( vI 2 − vI 1 ) − i1 ( R2 + R2 ) ⎤
′ ⎦
R3 ⎣
R4 ⎡ ⎛ vI 2 − vI 1 ⎞ ⎤
v0 = ⎢ ( vI 2 − vI 1 ) − ⎜ ⎟ ( R2 + R2 ) ⎥
′
R3 ⎣ ⎝ R1 ⎠ ⎦
For common-mode input vI 2 = vI 1 ⇒ v0 = 0 ⇒ Common Gain = 0, C M R R = ∞
b. Ad ( min ) ⇒ R2 min, R1 max
′
⎛ 20 ⎞ ⎡ 100 + 95 ⎤
Ad = ⎜ ⎟ ⎢1 + ⎥ = 4.82
⎝ 20 ⎠ ⎣ 51 ⎦
⎛ 20 ⎞ ⎡ 100 + 105 ⎤
Ad ( max ) = ⎜ ⎟ ⎢1 + ⎥ = 206
⎝ 20 ⎠ ⎣ 1 ⎦
c.
A
CM RR = d
Acm
Acm = 0 ⇒ C M R R = ∞
TYU9.12
R4 ⎛ 2 R2 ⎞
Differential Gain = ⎜1 + ⎟
R3 ⎝ R1 ⎠
Let R3 = R4 so the difference amplifier gain is unity.
Minimum Gain ⇒ Maximum R1
⎛ 2 R2 ⎞
So ⎜1 +
⎜ R ( max ) ⎟ = 2
⎟
⎝ 1 ⎠
We want 2 R2 = R1 ( max )
Maximum Gain ⇒ Minimum R1
⎛ 2 R2 ⎞
⎜ R ( min ) ⎟ = 1000 or 2 R2 = 999 R1 ( min )
So ⎜1 + ⎟
⎝ 1 ⎠
If R2 = 50 kΩ, let R1 ( min ) = 100 Ω fixed resistor and let R1 ( max ) = 100 kΩ + 100 Ω = 100.1
pot
Then actual differential gain is in the range of 1.999 − 1001
TYU9.13
−1 10 μ sec −10 × 10−6
End of 1st pulse: v0 = ×t 0 =
r r
− (10 ) (10 × 10−6 )
After 10 pulses: v0 = −5 =
r

9. 100 × 10−6
So r = = 20 μ sec = r
5
r = R1C2 = 20 μ sec = 20 × 10−6
For example, C2 = 0.01× 10−6 = 0.01 μ F ⇒ R1 = 2 kΩ
TYU9.14
⎡ R − ΔR R + ΔR ⎤ +
v01 = ⎢ − ⎥V
⎢ ( R − ΔR ) + ( R + ΔR ) ( R + ΔR ) + ( R − ΔR ) ⎥
⎣ ⎦
⎡ R − ΔR R + ΔR ⎤ +
=⎢ − V
⎣ 2R 2R ⎥ ⎦
⎛ R − ΔR − R − ΔR ⎞ +
=⎜ ⎟V
⎝ 2R ⎠
⎛ ΔR ⎞ +
v01 = − ⎜ ⎟V
⎝ R ⎠
For V + = 3.5 V, ΔR = 50, R = 10 × 103
⎛ 50 ⎞
v01 = − ⎜ 4 ⎟ ( 3.5 ) = −1.75 × 10−2
⎝ 10 ⎠
v0 5
Need an amplifier with a gain of Ad = = ⇒ Ad = −285.7
vi −1.75 × 10−2
Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( −v01 ) to vI2.
⎛ R ⎞ ⎛ 2R ⎞
Ad = ⎜ 4 ⎟ ⎜ 1 + 2 ⎟ = 285.7
⎝ R3 ⎠ ⎝ R1 ⎠
R2
Let R4 = 150 kΩ, R3 = 10 kΩ Then = 9.02
R1
Let R2 = 100 kΩ, R1 = 11.1 kΩ
TYU9.15
⎡1 R ⎤ +
v01 = ⎢ − ⎥V
⎢ 2 R + R (1 + δ ) ⎥
⎣ ⎦
⎡ R + R (1 + δ ) − 2 R ⎤ +
=⎢ ⎥V
⎢ 2 ( R + R (1 + δ ) ) ⎥
⎣ ⎦
Rδ
= ×V +
2R ( 2 + δ )
⎛δ ⎞
v01 ≈ ⎜ ⎟ V +
⎝4⎠
V = 5 For δ = 0.01
+
⎛ 0.01 ⎞
v01 = ⎜ ⎟ ( 5 ) = 0.0125
⎝ 4 ⎠
v 5
Need a gain Ad = 0 = = 400
v01 0.0125
⎛ R ⎞ ⎛ 2R ⎞
Use an instrumentation amplifier Ad = 400 = ⎜ 4 ⎟ ⎜1 + 2 ⎟
⎝ R3 ⎠ ⎝ R1 ⎠
R
Let R4 = 150 kΩ, R3 = 10 kΩ then 2 = 12.8
R1
Let R2 = 150 kΩ, R1 = 11.7 kΩ