Solution manual fundamentals of electric circuits 3rd edition chapter05

Chapter 5, Problem 1.
The equivalent model of a certain op amp is shown in Fig. 5.43. Determine:
(a) the input resistance.
(b) the output resistance.
(c) the voltage gain in dB.

8x104vd

Figure 5.43 for Prob. 5.1
Chapter 5, Solution 1.
(a)
(b)
(c)

Rin = 1.5 MΩ
Rout = 60 Ω
A = 8x104
Therefore AdB = 20 log 8x104 = 98.0 dB

Chapter 5, Problem 2
The open-loop gain of an op amp is 100,000. Calculate the output voltage when there are
inputs of +10 µV on the inverting terminal and + 20 µV on the noninverting terminal.
Chapter 5, Solution 2.
v0 = Avd = A(v2 - v1)
= 105 (20-10) x 10-6 = 1V

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Chapter 5, Problem 3
Determine the output voltage when .20 µV is applied to the inverting terminal of an op
amp and +30 µV to its noninverting terminal. Assume that the op amp has an open-loop
gain of 200,000.

Chapter 5, Solution 3.
v0 = Avd = A(v2 - v1)
= 2 x 105 (30 + 20) x 10-6 = 10V

Chapter 5, Problem 4
The output voltage of an op amp is .4 V when the noninverting input is 1 mV. If the
open-loop gain of the op amp is 2 × 106, what is the inverting input?

Chapter 5, Solution 4.
v0 = Avd = A(v2 - v1)
v
−4
= −2μV
v2 - v1 = 0 =
A 2x10 6
v2 - v1 = -2 µV = –0.002 mV
1 mV - v1 = -0.002 mV
v1 = 1.002 mV

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Chapter 5, Problem 5.
For the op amp circuit of Fig. 5.44, the op amp has an open-loop gain of 100,000, an
input resistance of 10 kΩ, and an output resistance of 100 Ω. Find the voltage gain vo/vi
using the nonideal model of the op amp.

Figure 5.44 for Prob. 5.5

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Chapter 5, Solution 5.

I

R0

Rin

vd
+
vi

+

-

Avd

+

v0

-

-

-vi + Avd + (Ri + R0) I = 0
But

+

(1)

vd = RiI,
-vi + (Ri + R0 + RiA) I = 0
I=

vi
R 0 + (1 + A)R i

(2)

-Avd - R0I + v0 = 0
v0 = Avd + R0I = (R0 + RiA)I =

(R 0 + R i A) v 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
5
100,001
1 + 10

(

)

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Chapter 5, Problem 6
Using the same parameters for the 741 op amp in Example 5.1, find vo in the op amp
circuit of Fig. 5.45.

Figure 5.45 for Prob. 5.6

Example 5.1
A 741 op amp has an open-loop voltage gain of 2×105, input resistance of 2 MΩ, and
output resistance of 50Ω. The op amp is used in the circuit of Fig. 5.6(a). Find the closedloop gain vo/vs . Determine current i when vs = 2 V.

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Chapter 5, Solution 6.
vi
+ -

R0
I

Rin

vd

+

-

+

Avd

+
vo

-

(R0 + Ri)R + vi + Avd = 0
But

vd = RiI,
vi + (R0 + Ri + RiA)I = 0
I=

− vi
R 0 + (1 + A)R i

(1)

-Avd - R0I + vo = 0
vo = Avd + R0I = (R0 + RiA)I
Substituting for I in (1),

⎛ R 0 + RiA ⎞
⎟⎟ vi
v0 = − ⎜⎜

⎝ R 0 + (1 + A)R i ⎠
50 + 2x10 6 x 2 x10 5 ⋅ 10 −3
= −
50 + 1 + 2 x10 5 x 2x10 6

(

≅

(

)

)

− 200,000x 2 x10 6
mV
200,001x 2 x10 6
v0 = -0.999995 mV

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Chapter 5, Problem 7
The op amp in Fig. 5.46 has Ri = 100 kΩ, Ro = 100 Ω, A = 100,000. Find the differential
voltage vd and the output voltage vo.

+
–

Figure 5.46 for Prob. 5.7

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Chapter 5, Solution 7.
100 kΩ

10 kΩ

VS

+

–

Rout = 100 Ω

1

2

+
Vd

Rin

–
+

AVd

–

At node 1,

+
Vout

–

(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 + 99,999,999[(10VS + V0)/12]

0 = 83,333,332.5 VS + 8,334,334.25 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

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Chapter 5, Problem 8
Obtain vo for each of the op amp circuits in Fig. 5.47.

Figure 5.47 for Prob. 5.8
Chapter 5, Solution 8.
(a)

If va and vb are the voltages at the inverting and noninverting terminals of the op
amp.
va = vb = 0
1mA =

0 − v0
2k

v0 = -2V

(b)

10 kΩ

2V

-

+

ia

va
vb

1V

+

2V

+

+
vo

2 kΩ

-

+

va

10 kΩ

+-

+
i

vo

(b)

(a)

Since va = vb = 1V and ia = 0, no current flows through the 10 kΩ resistor. From Fig. (b),
-va + 2 + v0 = 0

v0 = va - 2 = 1 - 2 = -1V

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Chapter 5, Problem 9
Determine vo for each of the op amp circuits in Fig. 5.48.

+
–

Figure 5.48 for Prob. 5.9
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 =
2k

v0 = 2V

(b)

1V
+-

+

+

vb

vo

-

-

Since va = vb = 3V,
-vb + 1 + vo = 0

vo = vb - 1 = 2V

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Chapter 5, Problem 10
Find the gain vo/vs of the circuit in Fig. 5.49.

Figure 5.49 for Prob. 5.10

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
vs = vo ⎜
⎟=
⎝ 10 + 10 ⎠ 2

vo
=2
vs

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Chapter 5, Problem 11
Find vo and io in the circuit in Fig. 5.50.

Figure 5.50 for Prob. 5.11
Chapter 5, Solution 11.

−
+
+
−

+
vo

vb =

10
(3) = 2V
10 + 5

At node a,
3 − va va − vo

=
2
8

12 = 5va – vo

But va = vb = 2V,
12 = 10 – vo
–io =

vo = –2V

va − vo 0 − vo 2 + 2 2
+
=
+ = 1mA
8
4
8
4
i o = –1mA

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Chapter 5, Problem 12.

Calculate the voltage ratio vo/vs for the op amp circuit of Fig. 5.51. Assume that the op
amp is ideal.
25 kΩ
5 kΩ
–
+
vs

+
vo

+
_
10 kΩ

–

Figure 5.51

For Prob. 5.12.

Chapter 5, Solution 12.
This is an inverting amplifier.
25
vo = −
vs
⎯⎯
→
5

vo
= −5
vs

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Chapter 5, Problem 13
Find vo and io in the circuit of Fig. 5.52.

Figure 5.52 for Prob. 5.13
Chapter 5, Solution 13.

+
−

+
+
−

vo

By voltage division,
va =

90

(1) = 0.9V
100

vb =

v
50
vo = o
150
3

But va = vb

io = i1 + i2 =

v0
= 0.9
3

vo = 2.7V

vo
v
+ o = 0.27mA + 0.018mA = 288 μA
10k 150k

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Chapter 5, Problem 14
Determine the output voltage vo in the circuit of Fig. 5.53.

Figure 5.53 for Prob. 5.14
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

−
+

+
−

+
vo

But v2 = 0. Hence 40 - 4v1 = v1 + 2v1 - 2vo
At node 2,

v1 − v 2 v 2 − v o

=
,
20
10

40 = 7v1 - 2vo

v 2 = 0 or v1 = -2vo

From (1) and (2), 40 = -14vo - 2vo

(1)
(2)

vo = -2.5V

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Chapter 5, Problem 15
(a). Determine the ratio vo/is in the op amp circuit of Fig. 5.54.
(b). Evaluate the ratio for R1 = 20 kΩ, R2 = 25 kΩ, R3 = 40 2kOmega$.

Figure 5.54
Chapter 5, Solution 15
(a) Let v1 be the voltage at the node where the three resistors meet. Applying

KCL at this node gives

⎛ 1
v1 v1 − vo
1 ⎞ vo
⎟⎟ −
+
= v1 ⎜⎜
+
R2
R3
⎝ R2 R3 ⎠ R3
At the inverting terminal,
is =

0 − v1
⎯
⎯→ v1 = −i s R1
R1
Combining (1) and (2) leads to
⎛
v
R
R ⎞
i s ⎜⎜1 + 1 + 1 ⎟⎟ = − o
⎯
⎯→
R2 R3 ⎠
R3
⎝

is =

(1)

(2)
⎛
vo
RR ⎞
= −⎜⎜ R1 + R3 + 1 3 ⎟⎟
is
R2 ⎠
⎝

(b) For this case,

vo
20 x 40 ⎞
⎛
= −⎜ 20 + 40 +
⎟ kΩ = - 92 kΩ
25 ⎠
is
⎝
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Chapter 5, Problem 16
Obtain ix and iy in the op amp circuit in Fig. 5.55.

Figure 5.55

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Chapter 5, Solution 16
10k Ω

5k Ω

ix

va
vb

+
0.5V
-

iy

+

vo

2k Ω
8k Ω

Let currents be in mA and resistances be in k Ω . At node a,
0.5 − v a v a − vo
=
⎯
⎯→ 1 = 3v a − vo
5
10

(1)

But
8
10
(2)
vo
⎯
⎯→ vo = v a
8+2
8
Substituting (2) into (1) gives
10
8
1 = 3v a − v a
⎯
⎯→

va =
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(vo − v a ) = 0.6( v a − v a ) =
x mA = 85.71 μA
2
10
8
4 14
v a = vb =

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Chapter 5, Problem 17
Calculate the gain vo/vi when the switch in Fig. 5.56 is in:

(a) position 1 (b) position 2 (c) position 3

Figure 5.56

Chapter 5, Solution 17.
(a)
(b)
(c)

G=

vo
R
12
= − 2 = − = -2.4
vi
R1
5

vo
80
=−
= -16
vi
5
vo
2000
=−
= -400
vi

5

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* Chapter 5, Problem 18.
For the circuit in Fig. 5.57, find the Thevenin equivalent to the left of terminals a-b.
Then calculate the power absorbed by the 20-kΩ resistor. Assume that the op amp is
ideal.
10 kΩ
2 kΩ
12 kΩ
a

–
2 mV

+
_

8 kΩ
20 kΩ
b
Figure 5.57

For Prob. 5.18.

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Chapter 5, Solution 18.
We temporarily remove the 20-kΩ resistor. To find VTh, we consider the circuit below.
10 kΩ
2 kΩ
12 kΩ

–
+
2 mV

+

+
_

8Ω

VTh
–

This is an inverting amplifier.
10k

VTh = −
(2mV) = −10mV
2k
To find RTh, we note that the 8-kΩ resistor is across the output of the op amp which is
acting like a voltage source so the only resistance seen looking in is the 12-kΩ resistor.
The Thevenin equivalent with the 20-kΩ resistor is shown below.
12 kΩ

a
I

–10 mV

+
_

20 k
b

I = –10m/(12k + 20k) = 0.3125x10–6 A
p = I2R = (0.3125x10–6)2x20x103 = 1.9531 nW

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Chapter 5, Problem 19

Determine io in the circuit of Fig. 5.58.

Figure 5.58
Chapter 5, Solution 19.
We convert the current source and back to a voltage source.
24=

+
−

4
3

−
+

10k ⎛ 2 ⎞
⎜ ⎟ = -1.25V
4⎞ ⎝ 3⎠
⎛
⎜ 4 + ⎟k
3⎠
⎝
v
v −0
= -0.375mA
io = o + o
5k
10k
vo = −

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Chapter 5, Problem 20
In the circuit in Fig. 5.59, calculate vo if vs = 0.

Figure 5.59
Chapter 5, Solution 20.

−
+
+
−

+
−

+
vo

At node a,

9 − va va − vo va − vb
=
+

4
8
4

18 = 5va – vo - 2vb

(1)

At node b,

va − vb vb − vo
=
4
2

va = 3vb - 2vo

(2)

But vb = vs = 0; (2) becomes va = –2vo and (1) becomes
-18 = -10vo – vo

vo = -18/(11) = -1.6364V

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Chapter 5, Problem 21.
Calculate vo in the op amp circuit of Fig. 5.60.
10 kΩ
4 kΩ
–
+
3V

+

+
_

1V

Figure 5.60

vo
–

+
_

For Prob. 5.21.

Chapter 5, Solution 21.
Let the voltage at the input of the op amp be va.

va = 1 V,

3-v a va − vo
=
4k
10k

⎯⎯
→

3-1 1− vo
=
4
10

vo = –4 V.
Chapter 5, Problem 22
Design an inverting amplifier with a gain of -15.
Chapter 5, Solution 22.
Av = -Rf/Ri = -15.
If Ri = 10kΩ, then Rf = 150 kΩ.

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Chapter 5, Problem 23
For the op amp circuit in Fig. 5.61, find the voltage gain vo/vs.

Figure 5.61

Chapter 5, Solution 23
At the inverting terminal, v=0 so that KCL gives
vs − 0
0 0 − vo
=
+
R1
R2
Rf

⎯⎯⎯→

vo
vs

=−

Rf
R1

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