Chapter 5, Solution 1.
(a) Rin = 1.5 MΩ
(b) Rout = 60 Ω
(c) A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB
Chapter 5, Solution 2.
v0 = Avd = A(v2 - v1)
= 105 (20-10) x 10-6 = 0.1V
Chapter 5, Solution 3.
v0 = Avd = A(v2 - v1)
= 2 x 105 (30 + 20) x 10-6 = 10V
Chapter 5, Solution 4.
v0 = Avd = A(v2 - v1)
v −4
v2 - v1 = 0 = = −20µV
A 2 x10 5
If v1 and v2 are in mV, then
v2 - v1 = -20 mV = 0.02
1 - v1 = -0.02
v1 = 1.02 mV
Chapter 5, Solution 5.
I
R0
-
vd Rin
+ + +
Avd v0
-
vi + -
-
-vi + Avd + (Ri - R0) I = 0 (1)
But vd = RiI,
-vi + (Ri + R0 + RiA) I = 0
vi R i
vd = (2)
R 0 + (1 + A)R i
-Avd - R0I + v0 = 0
(R 0 + R i A) v i
v0 = Avd + R0I = (R0 + RiA)I =
R 0 + (1 + A)R i
v0 R 0 + RiA 100 + 10 4 x10 5
= = ⋅ 10 4
v i R 0 + (1 + A)R i 100 + (1 + 10 )
5
10 9 100,000
≅ ⋅ 10 4 = = 0.9999990
(
1 + 10 5
) 100,001
Chapter 5, Solution 6.
vi
+ -
R0
- I
vd
Rin
+ +
+ Avd
- vo
-
(R0 + Ri)R + vi + Avd = 0
But vd = RiI,
vi + (R0 + Ri + RiA)I = 0
− vi
I= (1)
R 0 + (1 + A)R i
-Avd - R0I + vo = 0
vo = Avd + R0I = (R0 + RiA)I
Substituting for I in (1),
R 0 + R iA
v0 = −
R + (1 + A)R vi
0 i
= −
( )
50 + 2 x10 6 x 2 x10 5 ⋅ 10 −3
( )
50 + 1 + 2x10 5 x 2 x10 6
− 200,000 x 2 x10 6
≅ mV
200,001x 2 x10 6
v0 = -0.999995 mV
Chapter 5, Solution 7.
100 kΩ
Rout = 100 Ω
10 kΩ 1 2
+
VS + +
- Vd Rin + AVd
- Vout
-
-
At node 1, (VS – V1)/10 k = [V1/100 k] + [(V1 – V0)/100 k]
10 VS – 10 V1 = V1 + V1 – V0
which leads to V1 = (10VS + V0)/12
At node 2, (V1 – V0)/100 k = (V0 – AVd)/100
But Vd = V1 and A = 100,000,
V1 – V0 = 1000 (V0 – 100,000V1)
0= 1001V0 – 100,000,001[(10VS + V0)/12]
0 = -83,333,334.17 VS - 8,332,333.42 V0
which gives us (V0/ VS) = -10 (for all practical purposes)
If VS = 1 mV, then V0 = -10 mV
Since V0 = A Vd = 100,000 Vd, then Vd = (V0/105) V = -100 nV
Chapter 5, Solution 8.
(a) If va and vb are the voltages at the inverting and noninverting terminals of the op
amp.
va = v b = 0
0 − v0
1mA = v0 = -2V
2k
(b)
10 kΩ
-
2V
+
ia va
- 2V
10 kΩ
+
+-
vb + + +
2 kΩ
vo va
1V + ia vo
- - -
-
(a) (b)
Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b),
-va + 2 + v0 = 0 va = va - 2 = 1 - 2 = -1V
Chapter 5, Solution 9.
(a) Let va and vb be respectively the voltages at the inverting and noninverting
terminals of the op amp
va = vb = 4V
At the inverting terminal,
4 − v0
1mA = v0 = 2V
2k
(b) 1V
+-
+ +
vb vo
- -
Since va = vb = 3V,
-vb + 1 + vo = 0 vo = vb - 1 = 2V
Chapter 5, Solution 10.
Since no current enters the op amp, the voltage at the input of the op amp is vs.
Hence
10 v o vo
vs = v o = =2
10 + 10 2 vs
Chapter 5, Solution 11.
8 kΩ
2 kΩ
a io
−
+ +
5 kΩ b
+
3V 10 kΩ 4 kΩ vo
−
−
10
vb = (3) = 2V
10 + 5
At node a,
3 − va va − vo
= 12 = 5va – vo
2 8
But va = vb = 2V,
12 = 10 – vo vo = -2V
va − vo 0 − vo 2 + 2 2
–io = + = + = 1mA
8 4 8 4
i o = -1mA
Chapter 5, Solution 12.
4 kΩ
1 kΩ
a
−
+ 2 kΩ
b +
+
1.2V
− vo
4 kΩ
−
4 2 2
At node b, vb = vo = vo = vo
4+2 3 3
1 .2 − v a v a − v o 2
At node a, = , but va = vb = v o
1 4 3
2 2 3x 4.8
4.8 - 4 x vo = vo − vo vo = = 2.0570V
3 3 7
2 9.6
va = vb = vo =
3 7
1 .2 − v a − 1 .2
is = =
1 7
− 1.2
p = vsis = 1.2 = -205.7 mW
7
Chapter 5, Solution 13.
10 kΩ
a
+ io
− 100 kΩ i2 i1 +
b
4 kΩ vo
+ 90 kΩ
1V
− 50 kΩ −
By voltage division,
90
va = (1) = 0.9V
100
50 v
vb = vo = o
150 3
v0
But va = vb = 0 .9 vo = 2.7V
3
v v
io = i1 + i2 = o + o = 0.27mA + 0.018mA = 288 µA
10k 150k
Chapter 5, Solution 14.
Transform the current source as shown below. At node 1,
10 − v1 v1 − v 2 v1 − v o
= +
5 20 10
10 kΩ vo
10 kΩ
5 kΩ 20 kΩ
v1 v2 −
+ +
+
10V
− vo
−
But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo 40 = 7v1 - 2vo (1)
v1 − v 2 v 2 − v o
At node 2, = , v 2 = 0 or v1 = -2vo (2)
20 10
From (1) and (2), 40 = -14vo - 2vo vo = -2.5V
Chapter 5, Solution 15
(a) Let v1 be the voltage at the node where the three resistors meet. Applying
KCL at this node gives
v v −v 1 1 vo
i s = 1 + 1 o = v1 R + − (1)
R2 R3 2 R3 R3
At the inverting terminal,
0 − v1
is = → v1 = −i s R1 (2)
R1
Combining (1) and (2) leads to
R R v vo RR
i s 1 + 1 + 1 = − o → = − R1 + R3 + 1 3
R
2
R3 R3 is R2
(b) For this case,
vo 20 x 40
= − 20 + 40 + kΩ = - 92 kΩ
is 25
Chapter 5, Solution 16
10k Ω
ix
5k Ω va iy
-
vb + vo
+ 2k Ω
0.5V
- 8k Ω
Let currents be in mA and resistances be in k Ω . At node a,
0 .5 − v a v a − v o
= → 1 = 3v a − vo (1)
5 10
But
8 10
v a = vb = vo vo = v a
→ (2)
8+2 8
Substituting (2) into (1) gives
10 8
1 = 3v a − v a → v a =
8 14
Thus,
0 .5 − v a
ix = = −1 / 70 mA = − 14.28 µA
5
v − vb v o − v a 10 0 .6 8
iy = o + = 0 .6 ( v o − v a ) = 0 .6 ( v a − v a ) = x mA = 85.71 µA
2 10 8 4 14
Chapter 5, Solution 17.
vo R 12
(a) G= = − 2 = − = -2.4
vi R1 5
vo 80
(b) =− = -16
vi 5
vo 2000
(c) =− = -400
vi 5
Chapter 5, Solution 18.
Converting the voltage source to current source and back to a voltage source, we have the
circuit shown below:
20
10 20 = kΩ
3
1 MΩ
(20/3) kΩ 50 kΩ
−
+ +
+
2vi/3
− vo
−
1000 2v i vo 200
vo = − ⋅ =− = -11.764
20 3 v1 17
50 +
3
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
4
24=
3
(4/3) kΩ 4 kΩ 10 kΩ
0V
− vo
+ +
(2/3)V
−
5 kΩ
10k 2
vo = − = -1.25V
4 3
4x k
3
v v −0
io = o + o = -0.375mA
5k 10k
Chapter 5, Solution 20.
8 kΩ
2 kΩ
4 kΩ 4 kΩ
a b
−
+ +
+
9V + vo
− vs
−
−
At node a,
9 − va va − vo va − vb
= + 18 = 5va – vo - 2vb (1)
4 8 4
At node b,
va − vb vb − vo
= va = 3vb - 2vo (2)
4 2
But vb = vs = 0; (2) becomes va = –2vo and (1) becomes
-18 = -10vo – vo vo = -18/(11) = -1.6364V
Chapter 5, Solution 21.
Eqs. (1) and (2) remain the same. When vb = vs = 3V, eq. (2) becomes
va = 3 x 3 - 2v0 = 9 - 2vo
Substituting this into (1), 18 = 5 (9-2vo) – vo – 6 leads to
vo = 21/(11) = 1.909V
Chapter 5, Solution 22.
Av = -Rf/Ri = -15.
If Ri = 10kΩ, then Rf = 150 kΩ.
Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs − 0 0 0 − vo vo Rf
= + → =−
R1 R2 Rf vs R1
Chapter 5, Solution 24
v1 Rf
R1 R2
- vs + -
+
+
R4
R3 vo
v2 -
We notice that v1 = v2. Applying KCL at node 1 gives
v1 (v1 − v s ) v1 − vo 1
+ + =0
→ + 1 + 1 v1 − v s = vo (1)
R1 R2 Rf R R Rf R2 R f
1 2
Applying KCL at node 2 gives
v1 v1 − v s R3
+ =0 → v1 = vs (2)
R3 R4 R3 + R4
Substituting (2) into (1) yields
R R R R3 1
vo = R f 3 + 3 − 4 − vs
R1 R f R2 R3 + R4 R2
i.e.
R R R R3 1
k = R f 3 + 3 − 4
−
R1 R f R2 R3 + R4 R2
Chapter 5, Solution 25.
vo = 2 V
+ −
+ +
va vo
-va + 3 + vo = 0 which leads to va = vo + 3 = 5 V.
Chapter 5, Solution 26
+
vb - io
+ +
0.4V 5k Ω
- 2k Ω vo
8k Ω
-
8
vb = 0.4 = vo = 0.8vo
→ vo = 0.4 / 0.8 = 0.5 V
8+ 2
Hence,
v o 0 .5
io = = = 0.1 mA
5k 5k
Chapter 5, Solution 27.
(a) Let va be the voltage at the noninverting terminal.
va = 2/(8+2) vi = 0.2vi
1000
v 0 = 1 + v a = 10.2v i
20
G = v0/(vi) = 10.2
(b) vi = v0/(G) = 15/(10.2) cos 120πt = 1.471 cos 120πt V
Chapter 5, Solution 28.
−
+
+
−
0 − v1 v1 − v o
At node 1, =
10k 50k
But v1 = 0.4V,
-5v1 = v1 – vo, leads to vo = 6v1 = 2.4V
Alternatively, viewed as a noninverting amplifier,
vo = (1 + (50/10)) (0.4V) = 2.4V
io = vo/(20k) = 2.4/(20k) = 120 µA
Chapter 5, Solution 29
R1 va
+
vb - +
+
vi R2 R2 vo
- R1
-
R2 R1
va = vi , vb = vo
R1 + R2 R1 + R2
R2 R1
But v a = vb
→ vi = vo
R1 + R2 R1 + R2
Or
v o R2
=
vi R1
Chapter 5, Solution 30.
The output of the voltage becomes
vo = vi = 12
30 20 = 12kΩ
By voltage division,
12
vx = (1.2) = 0.2V
12 + 60
vx 0 .2
ix = = = 10µA
20k 20k
v 2 0.04
p= x
= = 2µW
R 20k
Chapter 5, Solution 31.
After converting the current source to a voltage source, the circuit is as shown below:
12 kΩ
3 kΩ 6 kΩ v
1 o
v1 + vo
2
−
+
12 V
− 6 kΩ
At node 1,
12 − v1 v1 − v o v1 − v o
= + 48 = 7v1 - 3vo (1)
3 6 12
At node 2,
v1 − v o v o − 0
= = ix v1 = 2vo (2)
6 6
From (1) and (2),
48
vo =
11
vo
ix = = 0.7272mA
6k
Chapter 5, Solution 32.
Let vx = the voltage at the output of the op amp. The given circuit is a non-inverting
amplifier.
50
v x = 1 + (4 mV) = 24 mV
10
60 30 = 20kΩ
By voltage division,
20 v
vo = v o = o = 12mV
20 + 20 2
vx 24mV
ix = = = 600nA
(20 + 20)k 40k
v o 144x10 −6
2
p= = = 204nW
R 60x10 3
Chapter 5, Solution 33.
After transforming the current source, the current is as shown below:
1 kΩ
4 kΩ vi
+ vo
va −
+
4V
− 2 kΩ 3 kΩ
This is a noninverting amplifier.
1 3
v o = 1 + v i = v i
2 2
Since the current entering the op amp is 0, the source resistor has a OV potential drop.
Hence vi = 4V.
3
vo = (4) = 6V
2
Power dissipated by the 3kΩ resistor is
2
v o 36
= = 12mW
R 3k
va − vo 4 − 6
ix = = = -2mA
R 1k
Chapter 5, Solution 34
v1 − vin v1 − vin
+ =0 (1)
R1 R2
but
R3
va = vo (2)
R3 + R 4
Combining (1) and (2),
R1 R
v1 − va + v 2 − 1 va = 0
R2 R2
R R
v a 1 + 1 = v1 + 1 v 2
R
2 R2
R 3v o R R
1 + 1 = v1 + 1 v 2
R
R3 + R 4 2 R2
R3 + R 4 R
vo = v1 + 1 v 2
R R2
R 3 1 + 1
R
2
R3 + R 4
vO = ( v1R 2 + v 2 )
R 3 ( R1 + R 2 )
Chapter 5, Solution 35.
vo R
Av = = 1 + f = 10 Rf = 9Ri
vi Ri
If Ri = 10kΩ, Rf = 90kΩ
Chapter 5, Solution 36
VTh = Vab
R1
But vs = Vab . Thus,
R1 + R2
R + R2 R
VTh = Vab = 1 v s = (1 + 2 )v s
R1 R1
To get RTh, apply a current source Io at terminals a-b as shown below.
v1
+
v2 - a
+
R2
vo io
R1
-
b
Since the noninverting terminal is connected to ground, v1 = v2 =0, i.e. no current passes
through R1 and consequently R2 . Thus, vo=0 and
v
RTh = o = 0
io
Chapter 5, Solution 37.
R R R
v o = − f v1 + f v 2 + f v 3
R1 R2 R3
30 30 30
= − (1) + (2) + (−3)
10 20 30
vo = -3V
Chapter 5, Solution 38.
R R R R
v o = − f v1 + f v 2 + f v 3 + f v 4
R1 R2 R3 R4
50 50 50 50
= − (10) + (−20) + (50) + (−100)
25 20 10 50
= -120mV
Chapter 5, Solution 39
This is a summing amplifier.
Rf Rf Rf 50 50 50
R v1 + R v 2 + R v3 = − 10 (2) + 20 v 2 + 50 (−1) = −9 − 2.5v 2
vo = −
1 2 3
Thus,
vo = −16.5 = −9 − 2.5v 2
→ v 2 = 3 V
Chapter 5, Solution 40
R1
R2 va
+
R3 vb -
+ +
v1 +
- v2 Rf vo
- +
v3 R -
-
Applying KCL at node a,
v1 − v a v 2 − v a v3 − v a v1 v 2 v3 1 1 1
+ + =0
→ + + = va ( + + ) (1)
R1 R2 R3 R1 R2 R3 R1 R2 R3
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