# ECE 2006 Fall 2011 Homework #04 Solution

```ECE 2006 Fall 2011
Homework #04 Solution
Problems in the textbook:
5.9, 5.10, 5.21, 5.29, 5.39, 5.47
Chapter 5, Problem 9
Determine vo for each of the op amp circuits in Fig. 5.48.
2 mA
6V
+
–
5V
2V
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 = 6V
At the inverting terminal,
2mA =
6 − v0
2k
v0 = 2V
2V
(b)
+-
Since va = vb = 5V,
–vb + 2 + vo = 0
+
+
vb
vo
-
-
vo = vb – 2 = 3V
Chapter 5, Problem 10
Find the gain vo/vs of the circuit in Fig. 5.49.
37kΩ
20kΩ
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  vo
vs = vo 
=
 10 + 20  3
vo
=3
vs
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.
3-v a va − vo
=
va 1
=
V,
4k
10k
3-1 1− vo

=
→
4
10
vo = –4 V.
Chapter 5, Problem 29
Determine the voltage gain vo/vi of the op amp circuit in Fig. 5.67.
Chapter 5, Solution 29
va
R1
vb
+
vi
-
+
-
R2
+
R2
vo
R1
-
va =
R2
vi ,
R1 + R2
But v a = vb
vb =

→
R1
vo
R1 + R2
R2
R1
vi =
vo
R1 + R2
R1 + R2
Or
v o R2
=
vi
R1
Chapter 5, Problem 39
For the op amp circuit in Fig. 5.76, determine the value of v2 in order to make
vo = –16.5 V.
Figure 5.76
Chapter 5, Solution 39
This is a summing amplifier.
Rf
Rf 
 Rf
50
50

 50
vo = −
v1 +
v2 +
v3  = − (2) + v 2 + (−1)  = −9 − 2.5v 2
R2
R3 
20
50

 10
 R1
Thus,
v o = −16.5 = −9 − 2.5v 2

→
v2 = 3 V
Chapter 5, Problem 47.
The circuit in Fig. 5.79 is for a difference amplifier. Find vo given that v1 =1V and v2 =
2V.
30 kΩ
2 kΩ
–
+
2 kΩ
v1
v
+
+
_
v2
+
_
vo
20 kΩ
–
Figure 5.79
For Prob. 5.47.
Chapter 5, Solution 47.
Using eq. (5.18), R1 =
2kΩ, R2 =
30kΩ, R3 =
2kΩ, R4 =
20kΩ
vo=
30(1+ 2 / 30)
30
32
(2) − 15(1)= 14.09 V
v2 −
V1=
2(1+ 2 / 20)
2
2.2
```