2. ⎛ Ri ⎞
vid = ⎜ ⎟ vI
⎝ Ri + 25 ⎠
⎛ Ri ⎞
0.790 = ⎜ ⎟ ( 0.80 )
⎝ Ri + 25 ⎠
0.9875 ( Ri + 25 ) = Ri
24.6875 = 0.0125 Ri
Ri = 1975 K
9.5
200 ⎫
Av = − = −10 ⎪
20
⎪
and ⎬ for each case
Ri = 20 kΩ ⎪
⎪
⎭
9.6
a.
100
Av = − = −10
10
Ri = R1 = 10 kΩ
b.
100 100
Av = − = −5
10
Ri = R1 = 10 kΩ
c.
100
Av = − = −5
10 + 10
Ri = 10 + 10 = 20 K
9.7
3. vI 0.5
I1 = ⇒ R1 = ⇒ R1 = 5 K
R1 0.1
R2
= 15 ⇒ R2 = 75 K
R1
9.8
R2
Av = −
R1
(a) Av = −10
(b) Av = −1
(c) Av = −0.20
(d) Av = −10
(e) Av = −2
(f) Av = −1
9.9
R2
Av = −
R1
(a) R1 = 20 K, R2 = 40 K
(b) R1 = 20 K, R2 = 200 K
(c) R1 = 20 K, R2 = 1000 K
(d) R1 = 80 K, R2 = 20 K
9.10
R2
Av = − = −8 ⇒ R2 = 8 R1
R1
1
For vI = −1, i1 = = 15 μ A ⇒ R1 = 66.7 kΩ ⇒ R2 = 533.3 kΩ
R1
9.11
R2
Av = − = −30 ⇒ R2 = 30 R1
R1
Set R2 = 1 MΩ ⇒ R1 = 33.3 kΩ
9.12
a.
R2 1.05R2 ⎛R ⎞
Av = ⇒ = 1.105 ⎜ 2 ⎟
R1 0.95 R1 ⎝ R1 ⎠
0.95R2 ⎛R ⎞
= 0.905 ⎜ 2 ⎟
1.05R1 ⎝ R1 ⎠
Deviation in gain is +10.5% and − 9.5%
b.
1.01R2 ⎛R ⎞ 0.99 R2 ⎛R ⎞
Av ⇒ = 1.02 ⎜ 2 ⎟ ⇒ = 0.98 ⎜ 2 ⎟
0.99 R1 ⎝ R1 ⎠ 1.01R1 ⎝ R1 ⎠
Deviation in gain = ±2%
9.13
(a)
4. vO −15
Av = = = −15
vl 1
vO = −15vl ⇒ vO = −150sin ω t ( mV )
(b)
vI
i2 = i1 = = 10sin ω t ( μ A )
R1
vO
iL = ⇒ iL = −37.5sin ω t ( μ A )
RL
iO = iL − i2
iO = −47.5sin ω t ( μ A )
9.14
R2
Av = −
R1 + R5
Av = −30 ± 2.5% ⇒ 29.25 ≤ Av ≤ 30.75
R2 R2
So = 29.25 and = 30.75
R1 + 2 R1 + 1
We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1)
Which yields R1 = 18.5 k Ω and R2 = 599.6 k Ω
For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V
9.15
R2 120
vO1 = − , vI = − ( 0.2 ) ⇒ vO1 = −1.2 V
R1 20
R4 ⎛ −75 ⎞
vO = − , vO1 = ⎜ ⎟ ( −1.2 ) ⇒ vO = +6 V
R3 ⎝ 15 ⎠
0.2
i1 = i2 = ⇒ i1 = i2 = 10 μ A
20
v −1.2
i3 = i4 = O1 = ⇒ i3 = i4 = −80 μ A
R3 15
1st op-amp: 90 μ A into output terminal
2nd op-amp: 80 μ A out of output terminal.
9.16
(a)
R2 22
Av = − =− ⇒ Av = −22
R1 1
(b) From Eq. (9.23)
R2 1 1
Av = − ⋅ = −22 ⋅
R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎡ 1 ⎤
⎢1 + ⎜1 + ⎟ ⎥ ⎢1 + 104 ( 23) ⎥
⎣ ⎦
⎣ Aod ⎝ R1 ⎠ ⎦
Av = −21.95
(c)
6. R2 ⎛ R3 R3 ⎞
Av = − ⎜1 + + ⎟
R1 ⎝ R4 R2 ⎠
R1 = 500 kΩ
R2 ⎛ R3 R3 ⎞
80 = ⎜1 + + ⎟
500 ⎝ R4 R2 ⎠
Set R2 = R3 = 500 kΩ
⎛ 500 ⎞ 500
80 = 1⎜ 1 + + 1⎟ = 2 + ⇒ R4 = 6.41 kΩ
⎝ R4 ⎠ R4
b.
For vI = −0.05 V
−0.05
i1 = i2 = ⇒ i1 = i2 = −0.1 μ A
500 kΩ
v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05
vX 0.05
i4 = − =− ⇒ i4 = −7.80 μ A
R4 6.41
i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 μ A
9.21
(a)
− R2 −500
Av = −1000 = =
R1 R1
R1 = 0.5 K
(b)
− R2 ⎛ R3 R3 ⎞
Av = ⎜1 + + ⎟
R1 ⎝ R4 R2 ⎠
−250 ⎛ 500 500 ⎞ −1250
−1000 = ⎜1 + + ⎟=
R1 ⎝ 250 250 ⎠ R1
R1 = 1.25 K
9.22
vI
i1 = = i2
R
⎛v ⎞
v A = −i2 R = − ⎜ I ⎟ R = −vI
⎝R⎠
v v
i3 = − A = I
R R
7. vA vA 2v 2v
i4 = i2 + i3 = − − =− A = I
R R R R
⎛ 2vI ⎞
vB = v A − i4 R = −vI − ⎜ ⎟ ( R ) = −3vI
⎝ R ⎠
vB ( −3vI ) 3vI
i5 = − =− =
R R R
2vI 3vI 5vI
i6 = i4 + i5 = + =
R R R
⎛ 5vI ⎞ v0
v0 = vB − i6 R = −3vI − ⎜ ⎟ R ⇒ v = −8
⎝ R ⎠ I
From Figure 9.12 ⇒ Av = −3
9.23
(a)
R2 1
Av = − ⋅
R1 ⎡ 1 ⎛ R2 ⎞ ⎤
⎢1 + ⎜1 + ⎟ ⎥
⎣ Aod ⎝ R1 ⎠ ⎦
50 1
=− ⋅ ⇒ Av = −4.99985
10 ⎡ 1 ⎛ 50 ⎞ ⎤
1+ 1 + ⎟⎥
⎢ 2 × 105 ⎜ 10
⎣ ⎝ ⎠⎦
(b) vO = − ( 4.99985 ) (100 × 10−3 ) ⇒ vO = −499.985 mV
0.5 − 0.499985
(c) Error = × 100% ⇒ 0.003%
0.5
9.24
a. From Equation (9.23)
R2 1
Av = − ⋅
R1 ⎡ 1 ⎛ R2 ⎞ ⎤
⎢1 + ⎜1 + ⎟ ⎥
⎣ Aod ⎝ R1 ⎠ ⎦
100 1
=− ⋅ = −0.9980
100 ⎡ 1 ⎛ 100 ⎞ ⎤
1 + 3 ⎜1 +
⎢ 10 ⎟⎥
⎣ ⎝ 100 ⎠ ⎦
Then v0 = Av ⋅ vI = ( −0.9980 )( 2 ) ⇒ v0 = −1.9960 V
b.
v0 = Aod ( v A − vB )
vB v0 − vB ⎛ 1 1 ⎞ v
= ⇒ vB ⎜ + ⎟ = 0
R1 R2 ⎝ R1 R2 ⎠ R2
v0
vB =
⎛ R2 ⎞
⎜1 + ⎟
⎝ R1 ⎠
8. Aod v0
Then v0 = Aod v A −
⎛ R2 ⎞
⎜1 + ⎟
⎝ R1 ⎠
⎡ ⎤
⎢ ⎥
v0 ⎢1 +
Aod ⎥
= Aod v A
⎢ ⎛ R ⎞⎥
⎢ ⎜1 + ⎟ ⎥
2
⎢ ⎝
⎣ R1 ⎠ ⎥
⎦
⎡ ⎛ R2 ⎞ ⎤
⎢ ⎜ 1 + ⎟ + Aod ⎥
v0 ⎢ ⎝
R1 ⎠ ⎥=A v
⎢ ⎛ R ⎞ ⎥ od A
⎢ ⎜1 + 2 ⎟ ⎥
⎢ ⎝
⎣ R1 ⎠ ⎥ ⎦
⎛ R2 ⎞
Aod ⎜ 1 + ⎟ v A
v0 = ⎝ R1 ⎠
⎛ R ⎞
Aod + ⎜ 1 + 2 ⎟
⎝ R1 ⎠
⎛ R2 ⎞
⎜1 + ⎟ vA
v0 = ⎝
R1 ⎠
1 ⎛ R2 ⎞
1+ ⎜1 + ⎟
Aod ⎝ R1 ⎠
⎛ 10 ⎞ ⎛ vI ⎞
⎜1 + ⎟ ⎜ ⎟
So v0 = ⎝
10 ⎠ ⎝ 2 ⎠
= 0.9980vI
1 ⎛ 10 ⎞
1 + 3 ⎜1 + ⎟
10 ⎝ 10 ⎠
For vI = 2 V
v0 = 1.9960 V
9.25
vl v v R
(a) ii = = i2 = − O ⇒ O = − 2
R1 R2 vl R1
(b)
vl v 1 ⎛ R2 ⎞
i2 = i1 = = i3 + O = i3 + ⎜ − ⋅ vl ⎟
R1 RL RL ⎝ R1 ⎠
v ⎛ R ⎞
Then i3 = l ⎜ 1 + 2 ⎟
R1 ⎝ RL ⎠
9.26
⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞
⎜ 0.1 1 + 10 ⎟ ( )
VX .max = ⎜ ⋅V = ⎜ 10 ⇒ VX .max = 0.09008 V
⎜R R +R ⎟ ⎟ ⎟
⎝ 3 1 4 ⎠ ⎝ ⎠
R
vO = 2 ⋅ VX .max
R1
R2 R
10 = ( 0.09008 ) ⇒ 2 = 111
R1 R1
So R2 = 111 k Ω
9.27
(a)
9. ⎛R R R ⎞
vO = − ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟
⎝ R1 R2 R3 ⎠
⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤
= − ⎢⎜ ⎟ ( 0.5 ) + ⎜ ⎟ ( 0.75 ) + ⎜ ⎟ ( 2.5 ) ⎥
⎣ ⎝ 50 ⎠ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦
= − [1 + 3.75 + 2.5]
vO = −7.25 V
(b)
⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤
−2 = − ⎢⎜ ⎟ (1) + ⎜ ⎟ ( 0.8 ) + ⎜ ⎟ vI 3 ⎥
⎣⎝ 50 ⎠ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦
2 = 2 + 4 + vI 3
vI 3 = −4 V
9.28
− RF R R
vo = ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1 R2 R3
= −4vI 1 − 8vI 2 − 2vI 3
RF RF RF
=4 =8 =2
R1 R2 R3
Largest resistance = RF = 250 K ⇒ R1 = 62.5 K R2 = 31.25 K R3 = 125 K
9.29
RF R
v0 = −4vI 1 − 0.5vI 2 = − vI 1 − F vI 2
R1 R2
RF RF
=4 = 0.5 ⇒ R1 is the smallest resistor
R1 R2
vI 2
i = 100 μ A = = ⇒ R1 = 20 kΩ
R1 R1
⇒ RF = 80 kΩ
⇒ R2 = 160 kΩ
9.30
vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft )
1 1
f = 1 kHz ⇒ T = 3 ⇒ 1 ms vI 2 ⇒ T2 = ⇒ 10 ms
10 100
R R 10 10
vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2
R1 R2 1 5
vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V )
vO = −0.707 sin ( 2π ft ) − ( ±2 V )
10. 9.31
RF R R
vO = − ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1 R2 R3
20 20 20
vO = − ⋅ vI 1 − ⋅ vI 2 − ⋅ vI 3
10 5 2
K sin ω t = −2vI 1 − 4 [ 2 + 100sin ω t ] − 0
Set vI 1 = −4 mV
9.32
Only two inputs.
⎡R R ⎤
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
⎣ R1 R2 ⎦
⎡ 1 ⎤
= − ⎢3vI 1 + ⋅ vI 2 ⎥
⎣ 4 ⎦
RF RF 1
=3 =
R1 R2 4
Smallest resistor = 10 K = R1
RF = 30 K R2 = 120 K
9.33
⎡R R ⎤
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
⎣ R1 R2 ⎦
− RF −R R RF
−5 − 5sin ω t = ( 2.5sin ω t ) F ⋅ ( 2 ) ⇒ F = 2 = 2.5
R1 R2 R1 R2
RF = largest resistor ⇒ RF = 200 K
R1 = 100 K R2 = 80 K
9.34
a.
RF R R R
v0 = − ⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 )
R3 R2 R1 R0
RF ⎡ a3 a2 a1 a0 ⎤
So v0 = + + + ( 5)
10 ⎢ 2 4 8 16 ⎥
⎣ ⎦
R 1
b. v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ
10 2
c.
10 1
i. v0 = ⋅ ⋅ 5 ⇒ v0 = 0.3125 V
10 16
11. 10 ⎡ 1 1 1 1 ⎤
ii. v0 = + + + ( 5 ) ⇒ v0 = 4.6875 V
10 ⎢ 2 4 8 16 ⎥
⎣ ⎦
9.35
(a)
10
vO1 = − ⋅ vI 1
1
20 20
vO = − ⋅ vO1 − ⋅ vI 2 = − ( 20 )( −10 ) vI 1 − ( 20 ) vI 2
1 1
vO = 200vI 1 − 20vI 2
(b)
vI 1 = 1 + 2sin ω t ( mV )
vI 2 = −10 mV
Then vO = 200 (1 + 2sin ω t ) − 20 ( −10 )
So vO = 0.4 + 0.4sin ω t (V )
9.36
For one-input
v0
v1 = −
Aod
vI 1 − v1 v1 v −v
= + 1 0
R1 R2 R3 RF
VI 1 ⎡1 1 1 ⎤ v0
= v1 ⎢ + + ⎥−
R1 ⎣ R1 R2 R3 RF ⎦ RF
v0 ⎡1 1 1 ⎤ v0
=− ⎢ + + ⎥−
Aod ⎣ R1 R2 R3 RF ⎦ RF
⎧ 1
⎪ 1 1 ⎛ 1 1 ⎞⎫ ⎪
= −v0 ⎨ + + ⎜ + ⎟⎬
⎪ Aod RF RF Aod ⎝ R1 R2 R3 ⎠ ⎪
⎩ ⎭
v0 ⎧ 1 1 RF ⎫
=− ⎨ +1+ ⋅ ⎬
RF ⎩ Aod Aod R1 R2 R3 ⎭
⎧ ⎫
⎪ ⎪
R ⎪ 1 ⎪
v0 = − F ⋅ vI 1 ⋅ ⎨ ⎬ where RP = R1 R2 R3
R1 ⎪1 + 1 ⎛ RF ⎞ ⎪
1+
⎪ Aod ⎜ RP ⎟ ⎪
⎝ ⎠⎭
⎩
−1 ⎛R R R ⎞
Therefore, for three-inputs v0 = × ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟
1 ⎛ RF ⎞ ⎝ R1 R2 R3 ⎠
1+ ⎜1 + ⎟
Aod ⎝ RP ⎠
9.37
12. ⎛ R ⎞ R
Av = 12 = ⎜ 1 + 2 ⎟ ⇒ 2 = 11
⎝ R1 ⎠ R1
v v 0.5
i1 = I ⇒ R1 = I =
R1 i1 0.15
R1 = 3.33 K
R2 = 36.7 K
9.38
⎛ 1 ⎞ ⎛ 1 ⎞
vB = ⎜ ⎟ vI v0 = Aod ⎜ ⎟ vi
⎝ 1 + 500 ⎠ ⎝ 501 ⎠
⎛ 1 ⎞
a. 2.5 = Aod ⎜ ⎟ ( 5 ) ⇒ Aod = 250.5
⎝ 501 ⎠
⎛ 1 ⎞
b. v0 = 5000 ⎜ ⎟ ( 5 ) ⇒ v0 = 49.9 V
⎝ 501 ⎠
9.39
⎛ R ⎞
Av = ⎜ 1 + 2 ⎟
⎝ R1 ⎠
(a) Av = 11
(b) Av = 2
(c) Av = 1.2
(d) Av = 11
(e) Av = 3
(f) Av = 2
9.40
R2
(a) = 1 ⇒ R1 = R2 = 20 K
R1
R2
(b) = 9 ⇒ R1 = 20 K, R2 = 180 K
R1
R2
(c) = 49 ⇒ R1 = 20 K, R2 = 980 K
R1
R2
(d) = 0 can set R2 = 20 K, R1 = ∞ (open circuit)
R1
9.41
⎛ 50 ⎞ ⎡⎛ 20 ⎞ ⎛ 40 ⎞ ⎤
v0 = ⎜ 1 + ⎟ ⎢⎜ ⎟ vI 2 + ⎜ ⎟ vI 1 ⎥
⎝ 50 ⎠ ⎣⎝ 20 + 40 ⎠ ⎝ 20 + 40 ⎠ ⎦
v0 = 1.33vI 1 + 0.667vI 2
9.42
(a)
13. vI 1 − v2 vI 2 − v2 v2
+ =
20 40 10
⎛ 100 ⎞
vO = ⎜ 1 + ⎟ v2 = 3v2
⎝ 50 ⎠
Now 2vI 1 − 2v2 + vI 2 − v2 = 4v2
⎛v ⎞
2vI 1 + vI 2 = 7v2 = 7 ⎜ o ⎟
⎝3⎠
6 3
So vO = ⋅ vI 1 + ⋅ vI 2
7 7
( 0.2 ) + ⎛ ⎞ ( 0.3) ⇒ vO = 0.3 V
6 3
(b) vO = ⎜ ⎟
7 ⎝ 7⎠
⎛6⎞ ⎛ 3⎞
(c) vO = ⎜ ⎟ ( 0.25 ) + ⎜ ⎟ ( −0.4 ) ⇒ vO = 42.86 mV
⎝7⎠ ⎝7⎠
9.43
⎛ R4 ⎞
v2 = ⎜ ⎟ vI
⎝ R3 + R4 ⎠
⎛ R ⎞ ⎛ R ⎞ ⎛ R4 ⎞
vO = ⎜1 + 2 ⎟ v2 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI
⎝ R1 ⎠ ⎝ R1 ⎠⎝ R3 + R4 ⎠
vO ⎛ R2 ⎞ ⎛ R4 ⎞
Av = = ⎜1 + ⎟ ⎜ ⎟
vI ⎝ R1 ⎠ ⎝ R3 + R4 ⎠
9.44
(a)
vO ⎛ 50 x ⎞
= ⎜1 +
⎜ (1 − x ) 50 ⎟
⎟
vI ⎝ ⎠
vO ⎛ x ⎞ 1− x + x
= ⎜1 + ⎟=
vI ⎝ 1 − x ⎠ 1− x
v 1
Av = O =
vI 1 − x
(b) 1 ≤ Av ≤ ∞
(c) If x = 1, gain goes to infinity.
9.45
Change resister values as shown.
14. vI
i1 = = i2
R
⎛v ⎞
vx = i2 2 R + vI = ⎜ I ⎟ 2 R + vI = 3vI
⎝R⎠
v x 3I
i3 = =
R R
v 3v 4v
i4 = i2 + i3 = I + I = I
R R R
⎛ 4vI ⎞
v0 = i4 2 R + vx = ⎜ ⎟ 2 R + 3vI
⎝ R ⎠
v0
= 11
vI
9.46
vO
(a) =1
vI
(b) From Exercise TYU9.7
⎛ R2 ⎞
⎜1 + ⎟
vO
= ⎝ R1 ⎠
vI ⎡ 1 ⎛ R2 ⎞ ⎤
⎢1 + ⎜1 + ⎟ ⎥
⎣ Aod ⎝ R1 ⎠ ⎦
But R2 = 0, R1 = ∞
vO 1 1 v
= = ⇒ O = 0.999993
vI 1 + 1 1 vI
1+
Aod 1.5 × 105
vO 1
(b) Want = 0.990 = ⇒ Aod = 99
vI 1
1+
Aod
9.47
15. v0 = Aod ( vI − v0 )
⎛ 1 ⎞
⎜ + 1⎟ v0 = vI
⎝ Aod ⎠
v0 1
=
vI ⎛ 1 ⎞
⎜1 + ⎟
⎝ Aod ⎠
v
Aod = 104 ; 0 = 0.99990
vI
v0
Aod = 103 ; = 0.9990
vI
v0
Aod = 102 ; = 0.990
vI
v0
Aod = 10; = 0.909
vI
9.48
⎛ R ⎞
v0 A = ⎜ 1 + 2 ⎟ vI
⎝ R1 ⎠
⎛ R ⎞ ⎛ R ⎞
v01 = ⎜1 + 2 ⎟ vI , v02 = − ⎜ 1 + 2 ⎟ vI
⎝ R1 ⎠ ⎝ R1 ⎠
So v01 = −v02
9.49
vI
(a) iL =
R1
(b)
vO1 = iL RL + vI = iL RL + iL R1
vOI ( max ) ≅ 10 V = iL (1 + 9 ) = 10iL
So iL ( max ) ≅ 1 mA
Then vI ( max ) ≅ iL R1 = (1)( 9 ) ⇒ vI ( max ) ≅ 9 V
9.50
(a)
⎛ 20 ⎞ ⎛ 20 ⎞
vX = ⎜ ⎟ ⋅ vI = ⎜ ⎟ ( 6 ) = 2
⎝ 20 + 40 ⎠ ⎝ 60 ⎠
vO = 2 V
(b) Same as (a)
(c)
⎛ 6 ⎞
vX = ⎜ ⎟ ( 6 ) = 0.666 V
⎝ 6 + 48 ⎠
⎛ 10 ⎞
vO = ⎜ 1 + ⎟ ⋅ v X ⇒ vO = 1.33 V
⎝ 10 ⎠
9.51
a.
16. v1 v −v
Rin = and 1 0 = i1 and v0 = − Aod v1
i1 RF
v1 − ( − Aod v1 ) v1 (1 + Aod )
So i1 = =
RF RF
v1 RF
Then Rin = =
i1 1 + Aod
b.
⎛ RS ⎞ RF
i1 = ⎜ ⎟ iS and v0 = − Aod ⋅ ⋅ i1
⎝ RS + Rin ⎠ 1 + Aod
⎛ A ⎞⎛ RS ⎞
So v0 = − RF ⎜ od ⎟⎜ ⎟ iS
⎝ 1 + Aod ⎠⎝ RS + Rin ⎠
RF 10
Rin = = = 0.009990
1 + Aod 1001
⎛ 1000 ⎞ ⎛ RS ⎞
v0 = − RF ⎜ ⎟⎜ ⎟ iS
⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠
⎛ 1000 ⎞ ⎛ RS ⎞
Want ⎜ ⎟⎜ ⎟ ≤ 0.990
⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠
which yields RS ≥ 1.099 kΩ
9.52
vO = iC RF , 0 ≤ iC ≤ 8 mA
For vO ( max ) = 8 V, Then RF = 1 k Ω
9.53
v 10
i = I so 1 = ⇒ R = 10 kΩ
R R
In the ideal op-amp, R1 has no influence.
⎛ R ⎞
Output voltage: v0 = ⎜1 + 2 ⎟ vI
⎝ R⎠
v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input
voltage vI in which the output is valid.
9.54
(a)
17. − vI
iL =
R2
− ( −10V )
10mA =
R2
R2 = 1 K
1 R
Also = F
R2 R1 R3
vL = (10mA )( 0.05k ) = 0.5 V
0.5
i2 = = 0.5 mA
1
iR 3 = 10 + 0.5 = 10.5 mA
v −v 13 − 0.5
Limit vo to 13V ⇒ R3 = O L = R3 = 1.19 K
iR 3 10.5
RF R3 1.19 R
Then = = = 1.19 = F
R1 R2 1 R1
For example, RF = 119 K, R1 = 100 K
(b) From part (a), vO = 13 V when vI = −10 V
9.55
(a)
vx
i1 = i2 and i2 = + iD , vx = −i2 RF
R2
⎛R ⎞
Then i1 = −i1 ⎜ F ⎟ + iD
⎝ R2 ⎠
⎛ R ⎞
Or iD = i1 ⎜ 1 + F ⎟
⎝ R2 ⎠
(b)
vI 5
R1 = = ⇒ R1 = 5 k Ω
i1 1
⎛ R ⎞ R
12 = (1) ⎜ 1 + F ⎟ ⇒ F = 11
⎝ R2 ⎠ R2
For example, R2 = 5 k Ω, RF = 55 k Ω
9.56
VX VX − vO
(1) IX = +
R2 R3
VX VX − vO
(2) + =0
R1 RF
18. ⎛ R ⎞
From (2) vO = VX ⎜ 1 + F ⎟
⎝ R1 ⎠
⎛ 1 1 ⎞ 1 ⎛ R ⎞
Then (1) I X = VX ⎜ + ⎟ − ⋅ VX ⎜1 + F ⎟
⎝ R2 R3 ⎠ R3 ⎝ R1 ⎠
IX 1 1 1 1 R 1 R
= = + − − F = − F
VX R0 R2 R3 R3 R1 R3 R2 R1 R3
R1 R3 − R2 RF
=
R1 R2 R3
R1 R2 R3
or Ro =
R1 R3 − R2 RF
RF 1
Note: If = ⇒ R2 RF = R1 R3 then Ro = ∞, which corresponds to an ideal current source.
R1 R3 R2
9.57
R2 R4
Ad = = =5
R1 R3
Minimum resistance seen by vI1 is R1.
Set R1 = R3 = 25 kΩ Then R2 = R4 = 125 kΩ
v0
iL = ⇒ v0 = iL RL = ( 0.5 )( 5 ) = 2.5 V
RL
v0 = 5 ( vI 2 − vI 1 )
2.5 = 5 ( vI 2 − 2 ) ⇒ vI 2 = 2.5 V
9.58
R2
vO = ( vI 2 − vI 1 )
R1
R2 R R
Ad = and 2 = 4 with R2 = R4 and R1 = R3
R1 R1 R3
Differential input resistance
R 20
Ri = 2 R1 ⇒ R1 = i = = 10 K
2 2
R2
(a) = 50 ⇒ R2 = R4 = 500 K
R1
R1 = R3 = 10 K
R2
(b) = 20 ⇒ R2 = R4 = 200 K
R1
R1 = R3 = 10 K
R2
(c) = 2 ⇒ R2 = R4 = 20 K
R1
R1 = R3 = 10 K
R2
(d) = 0.5 ⇒ R2 = R4 = 5 K
R1
R1 = R3 = 10 K
9.59
19. We have
⎛ R ⎞⎛ R / R ⎞ ⎛R ⎞ ⎛ R ⎞⎛ 1 ⎞ ⎛ R2 ⎞
vO = ⎜ 1 + 2 ⎟ ⎜ 4 3 ⎟ vI 2 − ⎜ 2 ⎟ vI 1 or vO = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1
⎝ R1 ⎠ ⎝ 1 + R4 / R3 ⎠ ⎝ R1 ⎠ ⎝ R1 ⎠ ⎝ 1 + R3 / R4 ⎠ ⎝ R1 ⎠
Set R2 = 50 (1 + x ) , R1 = 50 (1 − x )
R3 = 50 (1 − x ) , R4 = 50 (1 + x )
⎡ ⎤
⎡ ⎛ 1 + x ⎞⎤ ⎢ ⎥ 1+ x ⎞
⎥ vI 2 − ⎛
1
vO = ⎢1 + ⎜ ⎟⎥ ⎢ ⎜ ⎟ vI 1
⎣ ⎝ 1 − x ⎠ ⎦ ⎢1 + ⎛ 1 − x ⎞ ⎥ ⎝ 1− x ⎠
⎢ ⎜ 1+ x ⎟ ⎥
⎣ ⎝ ⎠⎦
⎡1 − x + (1 + x ) ⎤ ⎡ 1+ x ⎤ ⎛ 1+ x ⎞
vO = ⎢ ⎥⋅⎢ ⎥ vI 2 − ⎜ ⎟ vI 1
⎣ 1− x ⎦ ⎢1 + x + (1 − x ) ⎥
⎣ ⎦ ⎝ 1− x ⎠
⎛ 1+ x ⎞ ⎛1+ x ⎞
=⎜ ⎟ vI 2 − ⎜ ⎟ vI 1
⎝1− x ⎠ ⎝ 1− x ⎠
For vI 1 = vI 2 ⇒ vO = 0
Set R2 = 50 (1 + x ) R1 = 50 (1 − x )
R3 = 50 (1 + x ) R4 = 50 (1 − x )
⎛ ⎞
⎛ 1+ x ⎞⎜ 1 ⎟ ⎛ 1+ x ⎞
vO = ⎜1 + ⎟ ⎜ 1 + x ⎟ vI 2 − ⎜ ⎟ vI 1
⎝ 1− x ⎠⎜1+ ⎟ ⎝ 1− x ⎠
⎜ ⎟
⎝ 1− x ⎠
⎛ 1+ x ⎞
= vI 2 − ⎜ ⎟ vI 1
⎝ 1− x ⎠
vI 1 = vI 2 = vcm
vO 1 + x 1 − x − (1 + x ) −2 x
= 1− = =
vcm 1− x 1− x 1− x
Set R2 = 50 (1 − x ) R1 = 50 (1 + x )
R3 = 50 (1 − x ) R4 = 50 (1 + x )
⎛ ⎞
⎛ 1− x ⎞⎜ 1 ⎟ ⎛ 1− x ⎞
vO = ⎜ 1 + ⎟ ⎜ 1 − x ⎟ vI 2 − ⎜ ⎟ vI 1
⎝ 1+ x ⎠⎜ 1+ ⎟ ⎝ 1+ x ⎠
⎜ ⎟
⎝ 1+ x ⎠
⎛ 1− x ⎞
= ⎜1 − ⎟ vcm
⎝ 1+ x ⎠
1 + x − (1 − x )
2x
Acm = =
1+ x 1+ x
Worst common-mode gain
−2 x
Acm =
1− x
(b)
20. −2 x −2 ( 0.01)
For x = 0.01, Acm = = = −0.0202
1 − x 1 − 0.01
−2 ( 0.02 )
For x = 0.02, Acm = = −0.04082
1 − 0.02
−2 ( 0.05 )
For x = 0.05, Acm = = −0.1053
1 − 0.05
1 1
For this condition, set vI 2 = + , vI 1 = − ⇒ vd = 1 V
2 2
1 ⎡ ⎛ 1 + x ⎞ ⎤ 1 ⎡1 − x + (1 + x ) ⎤ 1 2 1
Ad = ⎢1 + ⎜ ⎟⎥ = ⎢ ⎥= ⋅ =
2 ⎣ ⎝ 1 − x ⎠⎦ 2 ⎣ 1− x ⎦ 2 1− x 1− x
1.010
For x = 0.01 Ad = 1.010 C M R RdB = 20 log10 = 33.98 dB
0.0202
1 1.020
For x = 0.02, Ad = = 1.020 C M R RdB = 20 log10 = 27.96 dB
0.98 0.04082
1 1.0526
For x = 0.05 Ad = = 1.0526 C M R RdB = 20 log10 ≅ 20 dB
0.95 0.1053
9.60
⎛ 10R ⎞ ⎛ 10 ⎞
vy = ⎜ ⎟ v2 = ⎜ ⎟ ( 2.65 ) ⇒ v y = vx = 2.40909 V
⎝ 10R+R ⎠ ⎝ 11 ⎠
v2 − v y 2.65 − 2.40909
i3 = i4 = = = 0.0120 mA
R 20
v −v 2.50 − 2.40909
i1 = i2 = 1 x = = 0.0045455 mA
R 20
vO = vx − i2 (10R ) = ( 2.40909 ) − ( 0.0045455 )( 200 )
vO = 1.50 V
9.61
10
iE = (1 + β )( iB ) = ( 81)( 2 ) = 162 mA =
R
R = 61.73 Ω
9.62
a. From superposition:
R2
v01 = − ⋅ vI 1
R1
⎛ R ⎞ ⎛ R1 ⎞
v02 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2
⎝ R1 ⎠ ⎝ R3 + R4 ⎠
Setting vI 1 = vI 2 = vcm
⎡ ⎛ ⎞ ⎤
⎢⎛ R ⎞ ⎜ 1 ⎟ R ⎥
v0 = v01 + v02 = ⎢⎜ 1 + 2 ⎟ ⎜ ⎟ − 2 ⎥ vcm
⎢⎝ R1 ⎠ ⎜ R3 ⎟ R1 ⎥
⎢ ⎜ 1+ R ⎟ ⎥
⎣ ⎝ 4 ⎠ ⎦
21. ⎛ ⎞
v R ⎛ R ⎞⎜ 1 ⎟ R
Acm = 0 = 4 ⋅ ⎜ 1 + 2 ⎟ ⎜ ⎟− 2
vcm R3 ⎝ R1 ⎠ ⎜ R4 ⎟ R1
⎜ 1+ R ⎟
⎝ 3 ⎠
R4 ⎛ R2 ⎞ R2 ⎛ R4 ⎞
⎜1 + ⎟ − ⎜1 + ⎟
= 3⎝
R R1 ⎠ R1 ⎝ R3 ⎠
⎛ R4 ⎞
⎜1 + ⎟
⎝ R3 ⎠
R4 R2
−
R R1
Acm = 3
⎛ R4 ⎞
⎜1 + ⎟
⎝ R3 ⎠
R4 R
b. Max. Acm ⇒ Min. and Max. 2
R3 R1
47.5 52.5
−
Max. Acm = 10.5 9.5 = 4.5238 − 5.5263 ⇒ A = 0.1815
1 + 4.5238
cm
47.5 max
1+
10.5
9.63
vI 1 − v A v A − vB vA − v0
= + (1)
R1 + R2 Rv R2
vI 2 − vB vB − v A vB
= + (2)
R1 + R2 Rv R2
⎛ R1 ⎞ ⎛ R2 ⎞
v− = ⎜ ⎟ vA + ⎜ ⎟ vI 1 (3)
⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠
⎛ R1 ⎞ ⎛ R2 ⎞
v+ = ⎜ ⎟ vB + ⎜ ⎟ vI 2 (4)
⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠
22. Now v− = v+ ⇒ R1vA + R2 vI 1 = R1vB + R2 vI 2
R
So that v A = vB + 2 ( vI 2 − vI 1 )
R1
vI 1 ⎛ 1 1 1 ⎞ v v
= vA ⎜ + + ⎟− B − 0 (1)
R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV R2
vI 2 ⎛ 1 1 1 ⎞ v
= vB ⎜ + + ⎟− A ( 2)
R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV
Then
vI 1 ⎛ 1 1 1 ⎞ v v ⎛ R ⎞⎛ 1 1 1 ⎞
= vB ⎜ + + ⎟ − B − 0 + ⎜ 2 ⎟⎜ + + ⎟ ( vI 2 − vI 1 ) (1)
R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV R2 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠
vI 2 ⎛ 1 1 1 ⎞ 1 ⎡ R2 ⎤
= vB ⎜ + + ⎟− ⎢ vB + ( vI 2 − vI 1 ) ⎥ (2)
R1 + R2 ⎝ R1 + R2 RV R2 ⎠ RV ⎣ R1 ⎦
Subtract (2) from (1)
1 ⎛ R ⎞⎛ 1 1 1 ⎞ v 1 R2
( vI 1 − vI 2 ) = ⎜ 2 ⎟ ⎜ + + ⎟ ( vI 2 − vI 1 ) − 0 + ⋅ ( vI 2 − vI 1 )
R1 + R2 ⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R2 RV R1
v0 ⎧⎛ R ⎞ ⎛ 1
⎪ 1 1 ⎞ 1 ⎫
1 R2 ⎪
= ( vI 2 − vI 1 ) ⎨⎜ 2 ⎟ ⎜ + + ⎟+ + ⋅ ⎬
R2 ⎪⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R1 + R2 RV R1 ⎪
⎩ ⎭
⎛ R ⎞ ⎧ R2 R R1 R ⎫
v0 = ( vI 2 − vI 1 ) ⎜ 2 ⎟ ⎨ + 2 +1+ + 2⎬
⎝ R1 ⎠ ⎩ R1 + R2 RV R1 + R2 RV ⎭
2 R2 ⎛ R2 ⎞
v0 = ⎜1 + ⎟ ( vI 2 − vI 1 )
R1 ⎝ RV ⎠
9.64
23. vI 1 − vI 2 ( 0.50 − 0.030sin ω t ) − ( 0.50 + 0.030sin ω t )
i1 = =
R1 20
−0.060sin ω t
=
20
i1 = −3sin ω t ( μ A )
vO1 = i1 R2 + vI 1 = ( −0.0030sin ω t )(115 ) + 0.50 − 0.030sin ω t
vO1 = 0.50 − 0.375sin ω t
vO 2 = vI 2 − i1 R2 = 0.50 + 0.030sin ω t − ( −0.003sin ω t )(115 )
vO 2 = 0.50 + 0.375sin ω t
R4 200
vO = ( vO 2 − vO1 ) = ⎡ 0.50 + 0.375sin ω t − ( 0.50 − 0.375sin ω t ) ⎤
R3 50 ⎣ ⎦
vO = 3sin ω t ( V )
vO 2 0.50 + 0.375sin ω t
i3 = =
R3 + R4 50 + 200
i3 = 2 + 1.5sin ω t ( μ A )
vO1 − vO ( 0.5 − 0.375sin ω t ) − ( 3sin ω t )
i2 = =
R3 + R4 250
i2 = 2 − 13.5sin ω t ( μ A )
9.65
⎛ 40 ⎞
(a) vOB = ⎜1 + ⎟ vI = 2.1667 sin ω t
⎝ 12 ⎠
30
(b) vOC = − vI = −1.25sin ω t
12
(c)
vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t )
vO = 3.417 sin ω t
vO 3.417
(d) = = 6.83
vI 0.5
9.66
vI
iO =
R
9.67
vO R ⎛ 2R ⎞
Ad = = 4 ⎜1 + 2 ⎟
vI 2 − vI 1 R3 ⎝ R1 ⎠
200 ⎛ 2 (115 ) ⎞
vO = ⎜1 + ⎟ ( 0.06sin ω t )
50 ⎝ R1 ⎠
230
For vO = 0.5 = 1.0833 ⇒ R1 = 212.3 K
R1
230
vO = 8 V = 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K
R1
9.68
24. ⎛ 2 R2 ⎞
R4
vO = ⎜1 + ⎟ ( vI 2 − vI 1 )
⎝
R3 R1 ⎠
Set R2 = 15 K, Set R1 = 2 K + 100 k ( Rot )
R4
Want ≈8 Set R3 = 10 K
R3
R 4 = 75 K
Now
75 ⎛ 2 (15 ) ⎞
Gain (min) = ⎜1 + ⎟ = 9.71
10 ⎝ 102 ⎠
75 ⎛ 2 (15 ) ⎞
Gain ( max ) = ⎜1 + ⎟ = 120
10 ⎝ 2 ⎠
9.69
For a common-mode gain, vcm = vI 1 = vI 2
Then
⎛ R ⎞ R
v01 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm
⎝ R1 ⎠ R1
⎛ R ⎞ R
v02 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm
⎝ R1 ⎠ R1
From Problem 9.62 we can write
R4 R4
−
R3 R3 ′
Acm =
⎛ R4 ⎞
⎜1 + ⎟
⎝ R3 ⎠
′
R3 = R4 = 20 kΩ, R3 = 20 kΩ ± 5%
20
1−
R3 1 ⎛ 20 ⎞
′
Acm = = ⎜1 − ⎟
(1 + 1) 2 ⎝ ′
R3 ⎠
′
For R3 = 20 kΩ − 5% = 19 kΩ
1 ⎛ 20 ⎞
Acm = ⎜ 1 − ⎟ = −0.0263
2 ⎝ 19 ⎠
′
For R3 = 20 kΩ + 5% = 21 kΩ
1 ⎛ 20 ⎞
Acm = ⎜ 1 − ⎟ = 0.0238
2⎝ 21 ⎠
So Acm max = 0.0263
9.70
a.
1
⋅ vI ( t ′ ) dt ′
R1C2 ∫
v0 =
0.5
∫ 0.5sin ω t dt = − ω cos ω t
1 ( 0.5 ) 0.5
v0 = 0.5 = ⋅ =
R1C2 ω 2π R1C2 f
1 1
f = = ⇒ f = 31.8 Hz
2π R1C2 2π ( 50 × 103 )( 0.1× 10−6 )
Output signal lags input signal by 90°
25. b.
0.5
i. f = ⇒ f = 15.9 Hz
2π ( 50 × 103 )( 0.1× 10−6 )
0.5
ii. f = ⇒ f = 159 Hz
( 0.1)( 2π ) ( 50 ×103 )( 0.1×10−6 )
9.71
1 − vI ⋅ t
vO = −
RC ∫ vI ( t ) dt = RC
vI = −0.2
Now
− ( −0.2 )( 2 )
8=
RC
(a) RC = 0.05 s
( 0.2 ) t
(b) 14 = ⇒ t = 3.5 s
0.05
9.72
a.
1 1
− R2 R2 ⋅
v0 jω C2 jω C2
= =−
vI R1 ⎛ 1 ⎞
R1 ⎜ R2 + ⎟
⎝ jω C2 ⎠
v0 R 1
=− 2⋅
vI R1 1 + jω R2 C2
v0 R
b. =− 2
vI R1
1
c. f =
2π R2 C2
9.73
a.
v0 − R2 R ( jω C1 )
= =− 2
vI R + 1 1 + jω R1C1
jω C1
1
v0 R jω R1C1
=− 2⋅
vI R1 1 + jω R1C1
v0 R
b. =− 2
vI R1
1
c. f =
2π R1C1
9.74
Assuming the Zener diode is in breakdown,
26. R2 1
vO = − ⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V
R1 1
0 − vO 0 − ( −6.8 )
i2 = = ⇒ i2 = 6.8 mA
R2 1
10 − Vz 10 − 6.8
iz = − i2 = − 6.8 ⇒ iz = −6.2 mA!!!
Rs 5.6
Circuit is not in breakdown. Now
10 − 0 10
= i2 = ⇒ i2 = 1.52 mA
Rs + R1 5.6 + 1
vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V
iz = 0
9.75
⎛ v ⎞ ⎡ v ⎤ ⎛ vI ⎞
vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −10 ⎟
⎝ I s R1 ⎠ ⎢ (10 )(10 ) ⎥
⎣ ⎦ ⎝ 10 ⎠
For vI = 20 mV , vO = 0.497 V
For vI = 2 V , vO = 0.617 V
9.76
27. ⎛ 333 ⎞
v0 = ⎜ ⎟ ( v01 − v02 ) = 16.65 ( v01 − v02 )
⎝ 20 ⎠
⎛i ⎞
v01 = −vBE1 = −VT ln ⎜ C1 ⎟
⎝ IS ⎠
⎛i ⎞
v02 = −vBE 2 = −VT ln ⎜ C 2 ⎟
⎝ IS ⎠
⎛i ⎞ ⎛i ⎞
v01 − v02 = −VT ln ⎜ C1 ⎟ = VT ln ⎜ C 2 ⎟
⎝ iC 2 ⎠ ⎝ iC1 ⎠
v v
iC 2 = 2 , iC1 = 1
R2 R1
⎛v R ⎞
So v01 − v02 = VT ln ⎜ 2 ⋅ 1 ⎟
⎝ R2 v1 ⎠
Then
⎛v R ⎞
v0 = (16.65 )( 0.026 ) ln ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
⎛v R ⎞
v0 = 0.4329 ln ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
ln ( x ) = log e ( x ) = ⎡ log10 ( x ) ⎤ ⋅ ⎡log e (10 ) ⎤
⎣ ⎦ ⎣ ⎦
= 2.3026 log10 ( x )
⎛v R ⎞
Then v0 ≅ (1.0 ) log10 ⎜ 2 ⋅ 1 ⎟
⎝ v1 R2 ⎠
9.77
( )
vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT
vO = (10 −10
)e vI / 0.026
For vI = 0.30 V , vo = 1.03 × 10−5 V
For vI = 0.60 V , vo = 1.05 V
![⎛R R R ⎞
vO = − ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟
⎝ R1 R2 R3 ⎠
⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤
= − ⎢⎜ ⎟ ( 0.5 ) + ⎜ ⎟ ( 0.75 ) + ⎜ ⎟ ( 2.5 ) ⎥
⎣ ⎝ 50 ⎠ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦
= − [1 + 3.75 + 2.5]
vO = −7.25 V
(b)
⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤
−2 = − ⎢⎜ ⎟ (1) + ⎜ ⎟ ( 0.8 ) + ⎜ ⎟ vI 3 ⎥
⎣⎝ 50 ⎠ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦
2 = 2 + 4 + vI 3
vI 3 = −4 V
9.28
− RF R R
vo = ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1 R2 R3
= −4vI 1 − 8vI 2 − 2vI 3
RF RF RF
=4 =8 =2
R1 R2 R3
Largest resistance = RF = 250 K ⇒ R1 = 62.5 K R2 = 31.25 K R3 = 125 K
9.29
RF R
v0 = −4vI 1 − 0.5vI 2 = − vI 1 − F vI 2
R1 R2
RF RF
=4 = 0.5 ⇒ R1 is the smallest resistor
R1 R2
vI 2
i = 100 μ A = = ⇒ R1 = 20 kΩ
R1 R1
⇒ RF = 80 kΩ
⇒ R2 = 160 kΩ
9.30
vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft )
1 1
f = 1 kHz ⇒ T = 3 ⇒ 1 ms vI 2 ⇒ T2 = ⇒ 10 ms
10 100
R R 10 10
vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2
R1 R2 1 5
vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V )
vO = −0.707 sin ( 2π ft ) − ( ±2 V )](https://image.slidesharecdn.com/ch09s-120608121927-phpapp01/75/ch09s-9-2048.jpg?cb=1670018920)
![9.31
RF R R
vO = − ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3
R1 R2 R3
20 20 20
vO = − ⋅ vI 1 − ⋅ vI 2 − ⋅ vI 3
10 5 2
K sin ω t = −2vI 1 − 4 [ 2 + 100sin ω t ] − 0
Set vI 1 = −4 mV
9.32
Only two inputs.
⎡R R ⎤
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
⎣ R1 R2 ⎦
⎡ 1 ⎤
= − ⎢3vI 1 + ⋅ vI 2 ⎥
⎣ 4 ⎦
RF RF 1
=3 =
R1 R2 4
Smallest resistor = 10 K = R1
RF = 30 K R2 = 120 K
9.33
⎡R R ⎤
vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥
⎣ R1 R2 ⎦
− RF −R R RF
−5 − 5sin ω t = ( 2.5sin ω t ) F ⋅ ( 2 ) ⇒ F = 2 = 2.5
R1 R2 R1 R2
RF = largest resistor ⇒ RF = 200 K
R1 = 100 K R2 = 80 K
9.34
a.
RF R R R
v0 = − ⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 )
R3 R2 R1 R0
RF ⎡ a3 a2 a1 a0 ⎤
So v0 = + + + ( 5)
10 ⎢ 2 4 8 16 ⎥
⎣ ⎦
R 1
b. v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ
10 2
c.
10 1
i. v0 = ⋅ ⋅ 5 ⇒ v0 = 0.3125 V
10 16](https://image.slidesharecdn.com/ch09s-120608121927-phpapp01/75/ch09s-10-2048.jpg?cb=1670018920)
