Chapter 6 - The Operational Amplifier
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Chapter 6 - The Operational Amplifier


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the left op amp to get 
 
1 a o
a s
300 3
10 30
v v v
v v
R R
\u239b \u239e+ + = \u21d2 = \u2212 +\u239c \u239f\u239d \u23a0ov 
 
o a s o o
90 903 9 1 9v v v v v
R R
\u239b \u239e= = \u2212 \u2212 \u21d2 + = \u2212\u239c \u239f\u239d \u23a0 sv 
 
o
s
9 9
90 901
v R
v R
R
= \u2212 = \u2212 ++
 
(b) o
s
0 9 0
v
R
v
\u2264 \u2264 \u221e \u21d2 \u2212 \u2264 \u2264 
(c) We require 95 112.5 k
90
R R
R
\u2212\u2212 = \u21d2 = \u3a9+ 
(checked: LNAP 7/8/04) 
 
 
 
P6.4-21 
Use units of V, mA and k\u3a9. 
 
1 2 3 1o
120 20 120 20 120 201 3 0.5 8
40 20 120 20 20 30 20
v v v v v\u23a1 \u23a4\u239b \u239e\u239b \u239e \u239b \u239e\u239b \u239e \u239b \u239e\u239b \u239e= \u2212 \u2212 + + + = \u2212 \u2212\u239c \u239f\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f\u239c \u239f\u23a2 \u23a5+\u239d \u23a0\u239d \u23a0 \u239d \u23a0\u239d \u23a0 \u239d \u23a0\u239d \u23a0\u23a3 \u23a6 2 3v v 
so 
a = 3, b = \u22120.5 and c = \u22128 
 
(checked: LNAP 6/21/04) 
 
 
P6.4-22 
 
Label the node voltages as shown. Use units of V, mA and k\u3a9. 
3 1 4 and v v v v2= = \u2212 
( )5 3 5 4 5 3 2 11 10 240 20 3 3 3v v v v v v v v\u2212 \u2212+ = \u21d2 = + = \u2212 22 v 
so 
1 2 and 
3 3
a b= \u2212 = \u2212 
(checked: LNAP 6/21/04) 
 
 
 
 
Section 6-5: Design Using Operational Amplifier 
 
P6.5-1 
Use the current-to-voltage converter, entry (g) in Figure 6.6-1. 
 
 
 
P6.5-2 
Use the voltage \u2013controlled current source, entry (i) in Figure 6.6-1. 
 
 
 
P6.5-3 
Use the noninverting summing amplifier, entry (e) in Figure 6.6-1. 
 
 
 
 
 
 
P6.5-4 
Use the difference amplifier, entry (f) in Figure 6.6-1. 
 
 
 
P6.5-5 
Use the inverting amplifier and the summing amplifier, entries (a) and (d) in Figure 6.6-1. 
 
 
 
P6.5-6 
Use the negative resistance converter, entry (h) in Figure 6.6-1. 
 
 
 
 
 
 
 
 
 
 
 
P6.5-7 
Use the noninverting amplifier, entry (b) in Figure 6.6-1. Notice that the ideal op amp forces the 
current iin to be zero. 
 
 
 
 
 
 
P6.5-8 
 
 
( )1 2
1 2
Summing Amplifier: 6 2
6 2
Inverting Amplifier:
a
o
o a
v v v
v v
v v
= \u2212 + \u23ab \u21d2 = +\u23ac= \u2212 \u23ad
v 
 
 
 
P6.5-9 
 
 
 
 
 
 
 
 
 
 
Using superposition, 1 2 3 9 16 32 7 Vov v v v= + + = \u2212 \u2212 + = 
 
 
P6.5-10 
 
R1 6 12 24 6||12 6||24 
R2 12||12||24 6||12||24 6||12||12 12||24 12||12 
-vo/vs 0.8 0.286 0.125 2 1.25 
 
R1 12||12 12||24 6||12||12 6||12||24 12||12||24 
R2 6||24 6||12 24 12 6 
-vo/vs 0.8 0.5 8 3.5 1.25 
 
 
 
 
P6.5-11 
Label the node voltages as shown. Apply KCL at the 
inverting input of the op amp to get 
 
a a b 3 4
b a
3 3 3
0
v v v R R
v v
R R R
\u239b \u239e\u2212 ++ = \u21d2 = \u239c \u239f\u239c \u239f\u239d \u23a0
 
 
Apply KCL at the noninverting input of the op amp to get 
 
a in a b
out
1 2
0
v v v v
i
R R
\u2212 \u2212+ + = 
Solving gives 
 
1 2 in b 1 2 3 4 in
a out a out
1 2 1 2 1 2 2 3 1
0 0
R R v v R R R R v
v i v
R R R R R R R R R
\u239b \u239e \u239b \u239e+ + +\u2212 \u2212 + = \u21d2 \u2212 \u2212 + =\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0
i 
 
When 2 3 1 4R R R R= the quantity in parenthesis vanishes leaving 
 
out in
1
1i v
R
= 
P6.5-12 
 
Label the node voltages as shown. Apply KCL at the 
inverting input of the op amp to get 
 
t t b 3 4
b t
3 4 3
0
v v v R R
v v
R R R
\u239b \u239e\u2212 ++ = \u21d2 = \u239c \u239f\u239c \u239f\u239d \u23a0
 
 
Apply KCL at the noninverting input of the op amp to get 
 
3 4
t
3t t b t 4 2 3 1 4
t t t
1 2 1 2 1 2 3 1 2 3
1
R R
v
Rv v v v R R R R R
i v
R R R R R R R R R R
\u239b \u239e+\u239c \u239f\u239c \u239f \u239b \u239e\u2212 \u2212\u239d \u23a0= + = + = \u2212 =\u239c \u239f\u239c \u239f\u239d \u23a0
v 
 
t 1 2 3 2
o
2 4t 2 3 1 4
1 3
v R R R R
R R Ri R R R R
R R
= = =\u2212 \u2212
 
 
 
P6.5-13 
(a) Label the node voltages as shown. The 
node equations are 
 
s a b a a
1 2
v v v v v
3R R R
\u2212 \u2212+ = 
and 
a o a 5
a o
5 4 4 5
v v v R
v v
R R R R
\u239b \u239e\u2212= \u21d2 = \u239c \u239f\u239c \u239f+\u239d \u23a0
 
 
Solving these equations gives 
s b 1 2 2 3 1 3 5
a o
1 1 2 3 2 1 2 3 4 5
1 1 1v v R R R R R R R vv v b
2R R R R R R R R R R R
\u239b \u239e \u239b \u239e+ += + + \u2212 = × \u2212\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f+\u239d \u23a0 \u239d \u23a0
 
So 
2 3 4 5 1 3 4 5
o s
1 2 2 3 1 3 5 1 2 2 3 1 3 5
R R R R R R R R
v v
R R R R R R R R R R R R R R
\u239b \u239e+ += × + ×\u239c \u239f\u239c \u239f+ + + +\u239d \u23a0 b
v× 
So 
2 3 4 5 1 3 4 5
s b
1 2 2 3 1 3 5 1 2 2 3 1 3 5
 and 
R R R R R R R R
a v b
R R R R R R R R R R R R R R
\u239b \u239e+ += × = ×\u239c \u239f\u239c \u239f+ + + +\u239d \u23a0
v× 
 
(b) The equation of the straight line is 
o s
5 5
4
v v= + 
We require 
2 3 4 5
1 2 2 3 1 3 5
5
4
R R R R
R R R R R R R
+× =+ + 
 
For example, let . Next we require 1 2 3 4 510 k , 55 k and 20 kR R R R R= = = \u3a9 = \u3a9 = \u3a9
 
1 3 4 5
b b
1 2 2 3 1 3 5
55
4
R R R R
v v
R R R R R R R
+= × ×+ + = 
i.e. 
b 4 Vv = 
(checked: LNAP 6/20/04) 
 
 
 
 
P6.5-14 
(a) Apply KCL at the inverting input of the op 
amp to get: 
s b o 3 3
o s
1 2 3 1 2
0
v v v R R
v v
R R R R R
\u239b \u239e+ + = \u21d2 = \u2212 \u2212\u239c \u239f\u239c \u239f\u239d \u23a0 b
v 
 
So 
3 3
b
1 2
 and 
R R
a b
R R
= \u2212 = \u2212 v 
 
(b) The equation of the straight line is 
o s
5 5
2
v v= \u2212 + 
We require 
3
1
5
2
R
R
\u2212 = \u2212 
e.g. . Next, we require 1 320 k and 50 kR R= \u3a9 = \u3a9
3 b
2
5
R v
R
= \u2212 
e.g. . 2 3 b50 k and 5 VR R v= = \u3a9 = \u2212
(checked: LNAP 6/20/04) 
 
 
P6.5-15 
 
Here\u2019s the circuit used to determine the equivalent 
resistance , given by 
t
eq
t
v
R
i
= 
First, use KVL to get 
( )( ) (
( )
)t p
p
t
p
0 1
1
 
i R a R i R aR
R a R
i i
R aR
= + + + +
\u239b \u239e+ \u2212\u21d2 = \u2212\u239c \u239f\u239c \u239f+\u239d \u23a0
p
 
 
Use KVL to get 
( ) ( )p p
t t t
p p
1 2
1
R a R a R
v Ri Ri R i R i
R aR R aR
\u239b \u239e \u239b+ \u2212 \u2212= + = \u2212 =\u239c \u239f \u239c\u239c \u239f \u239c+ +\u239d \u23a0 \u239d t
1 \u239e\u239f\u239f\u23a0
 
so 
 
( ) ( )t p
eq
pt p
2 1 2 1
1
v R a R a R
R pRi R aR a
R
\u2212 \u2212= = =+ +
 
 
(checked: LNAPDC 7/24/04) 
 
 
P6.5-16 
 
 
(a) 
( )4 1 2 2 6 1 4 2 3 6o 13 4 1 1 5 51 3 41 1
R R R R R R R R R R
v v
R R R R R RR R R
\u23a1 \u23a4\u239b \u239e\u239b \u239e \u239b \u239e \u239b \u239e+ \u2212= \u2212 + =\u23a2 \u23a5\u239c \u239f\u239c \u239f \u239c \u239f \u239c \u239f\u239c \u239f\u239c \u239f \u239c \u239f \u239c \u239f+ +\u23a2 \u23a5\u239d \u23a0\u239d \u23a0 \u239d \u23a0 \u239d \u23a0\u23a3 \u23a6 1
v+ 
So the gain is 
( )o 1 4 p 3 61 51 3 4 1
v R R aR R R
v RR R R
\u239b \u239e\u2212= +\u239c \u239f\u239c \u239f+ \u239d \u23a0
 
(b) When 1 3 4
1
2 p
R R R R= = = the gain becomes 
o 6
i 5
1 1
2
v R
a
v R
\u239b \u239e\u23a1 \u23a4= \u2212 +\u239c \u239f\u23a2 \u23a5 \u239c \u239f\u23a3 \u23a6 \u239d \u23a0
 
so 
6 o 6
5 i 5
1 11 1
2 2
R v R
R v R
\u239b \u239e \u239b \u239e\u239b \u239e \u239b \u239e\u2212 + \u2264 \u2264\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0\u239d \u23a0 \u239d \u23a0
+ 
We require 
6
6 5
5
110 1 19
2
R
R R
R
\u239b \u239e= + \u21d2 =\u239c \u239f\u239c \u239f\u239d \u23a0
 
 
e.g. Any convenient value of Rp will do, e.g. 5 610 k and 190 k .R R= \u3a9 = \u3a9 p 100 kR = \u3a9
 
 
 
 
Section 6-6: Operational Amplifier Circuits and Linear Algebraic Equations 
 
P6.6-1 
 
 
 
P6.6-2 
 
 
 
 
 
 
 
Section 6-7: Characteristics of the Practical Operational Amplifier 
 
P6.7-1 
 
The node equation at node a is: 13 3 100 10 10 10
out os os
b
v v v i\u2212 = +× × 
Solving for vout: 
( ) ( )
( ) ( )( )
3
3 3
1 13
3 3 9
100 10 1 100 10 11 100 10
10 10
 11 0.03 10 100 10 1.2 10 0.45 mV
out os b os bv v i v
\u2212 \u2212
\u239b \u239e×= + + × = + ×\u239c \u239f×\u239d \u23a0
= × + × × =
i
 
 
P6.7-2 
 
The node equation at node a is: 01 10000 90000
os os
b
v vi v\u2212+ = 
Solving for vo: 
 
( ) ( )
( )
3
3 3
o 1 os b13
3 3 9 3
90 101 90 10 10 90 10 i
10 10
10(5 10 )+ 90 10 (.05 10 ) 50.0045 10 50 mV
os bv v i v
\u2212 \u2212 \u2212
\u239b \u239e×= + + × = + ×\u239c \u239f×\u239d \u23a0
= × × × = × \u2212\ufffd
 
 
 
P6.7-3 
 
 
1 1 01
1 2 0 0 2
 0 1 0 1 1 0 2 1
0 2
 0
( ) 
( )( ) (1 0
in
in in
in in in
v v v vv
R R R v R R AR
v Av v v v R R R R R R A
R R
\u2212 \u2212 \u23ab+ + = \u23aa \u2212\u23aa \u21d2 =\u23ac+ \u2212 + + + +\u23aa+ = \u23aa\u23ad
)
 
 
 
 
P6.7-4 
a) 3
0 2
3
1
49 10 = 9.6078
5.1 10in
v R
v R
×= \u2212 = \u2212 \u2212× 
 
b) ( ) ( )( )( )6 30
3 6 3 3 6
2 10 75 200,000 50 10
 9.9957
(5 10 2 10 )(75 50 10 ) (5 10 )(2 10 )(1 200,000)in
v
v
× \u2212 ×= =× + × + × + × × + \u2212 
 
c) 6 3
0
3 6