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


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10000 V
6a
v i= = 
Apply KCL at node a: 
 
0
0 0 3 5 10000 20000
a a
a
v v v v v\u2212+ = \u21d2 = = V 
 
 
 
 
 
P6.4-9 
 
 
KCL at node b: 
12 0 4
40000 20000
b b
b
v v v+ + = \u21d2 = \u2212 V 
The node voltages at the input nodes of an ideal op amp are equal, so 4 Vc bv v= = \u2212 . 
 
The node voltages at the input nodes of an ideal op amp are equal, so . 40 10 4 Vd cv v= + × = \u2212
 
KCL at node g: 3 3
20
20 10 40 10 3
f g g
g f
v v v
v v
\u2212\u239b \u239e\u2212 + = \u21d2 =\u239c \u239f× ×\u239d \u23a0
 
 
 
The node voltages at the input nodes of an ideal op amp are equal, so 2
3e g
v v v= = f . 
KCL at node d: 3 3 3 3
2
6 2430 V
20 10 20 10 20 10 20 10 5 5
d f
d f d fd e
f d
v vv v v vv v v v
\u2212\u2212 \u2212\u2212= + = + \u21d2 = = \u2212× × × × 
 
Finally, 2 16 V
3 5e g f
v v v= = = \u2212 . 
 
P6.4-10 
By voltage division (or by applying KCL at 
node a) 
0
1 0
 a s
Rv v
R R
= + 
 
Applying KCL at node b: 
 
( )
0
1 0
0
0
1
 0
 
b s b
b s b
v v v v
R R R
R R v v v v
R
\u2212 \u2212+ =+\u394
+\u394\u21d2 \u2212 + =
 
 
The node voltages at the input nodes of an 
ideal op amp are equal so . b av v=
 
 
 
 
0 0 0 0 0
1 1 0 1 1 0 1 0 0
 1 s s s
R R R R R RR Rv v v
R R R R R R R R
\u23a1 \u23a4 \u239b \u239e\u239b \u239e+\u394 +\u394 v
R
\u394 \u394= + \u2212 = \u2212 = \u2212\u23a2 \u23a5 \u239c \u239f\u239c \u239f + +\u239d \u23a0 \u239d \u23a0\u23a3 \u23a6 +
 
 
 
P6.4-11 
Node equations: 
2s s a
a s
1 2 1
0 1
Rv v v v v
R R R
\u239b \u239e\u2212+ = \u21d2 = +\u239c \u239f\u239c \u239f\u239d \u23a0
 
and 
s a a a o
2 3 4
v v v v v
R R R
\u2212 \u2212= + 
so 
4 4 4
o a
2 3 2
1
R R R
v v
R R R
\u239b \u239e= + + \u2212\u239c \u239f\u239c \u239f\u239d \u23a0 s
v 
 
 
( )( )
4 4 2 4 4 4 2 2 4
o s s
2 3 1 2 3 1 1 1 3
1 2 3 4 3 4
s
1 3
1 1 1
R R R R R R R R R
v v v
R R R R R R R R R
R R R R R R
v
R R
\u239b \u239e \u239b \u239e \u239b= + + + \u2212 = + + + +\u239c \u239f \u239c \u239f \u239c\u239c \u239f \u239c \u239f \u239c\u239d \u23a0 \u239d \u23a0 \u239d
\u239b \u239e+ + +\u239c \u239f= \u239c \u239f\u239d \u23a0
sv
\u239e\u239f\u239f\u23a0
 
 
with the given values: 
 
( )( ) ( )o s20 20 10 8 10 8 40 18 80 420 10 200v v
+ + + ×\u239b \u239e × +\u239b \u239e= =\u239c \u239f \u239c \u239f× \u239d \u23a0\u239d \u23a0 s s
v v= 
 
(checked: LNAP 5/24/04) 
 
 
P6.4-12 
Notice that the currents in resistance R1 and R2 are 
both zero, as shown. Consequently, the voltages at 
the noninverting inputs of the op amps are v1 and v2, 
as shown. The voltages at the inverting inputs of the 
ideal op amps are also v1 and v2, as shown. 
 
Apply KCL at the top node of R6 to get 
 
5 6a 2 2
a 2
5 6 6
R Rv v v v v
R R R
\u239b \u239e+\u2212 = \u21d2 = \u239c \u239f\u239c \u239f\u239d \u23a0
 
 
Apply KCL at the top node of R4 to get 
 
3 3o 1 1 a
o 1
3 4 4 4
1
R Rv v v v v v
R R R R
\u239b \u239e \u239b \u239e\u2212 \u2212= \u21d2 = + \u2212\u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0 a
v 
 
3 3 5 6
o 1
4 4 6
1
R R R R
v v
R R R
\u239b \u239e \u239b \u239e \u239b \u239e+= + \u2212\u239c \u239f \u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0 \u239d \u23a0 2
v 
When 3 6
4 5
R R
R R
= 
( )
3 3 5 3 3 4
o 1 2 1
4 4 6 4 4 3
3
1 2
4
1 1 1 1
1
R R R R R R
v v v v
R R R R R R
R
v v
R
\u239b \u239e \u239b \u239e \u239b \u239e \u239b \u239e \u239b \u239e \u239b \u239e= + \u2212 + = + \u2212 +\u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f\u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f \u239c \u239f\u239d \u23a0 \u239d \u23a0 \u239d \u23a0 \u239d \u23a0 \u239d \u23a0 \u239d \u23a0
\u239b \u239e= + \u2212\u239c \u239f\u239c \u239f\u239d \u23a0
2v
 
 
 
so vo is proportional to the difference of the inputs, v1 \u2212 v2, as required. 
Next, choose R3 and R4 so that 3
4
5 1
R
R
= + , e.g. 
R1 = 50 k\u3a9, R2 = 50 k\u3a9, R3 = 40 k\u3a9, R4 = 10 k\u3a9, R5 = 10 k\u3a9 and R6 = 40 k\u3a9. 
 
(checked: LNAP 5/24/04) 
 
 
P6.4-13 
Write a node equation at the inverting input of the 
bottom op amp: 
 
o a 4
a o
3 4 3
0
v v R
v v
R R R
+ = \u21d2 = \u2212 
 
Write a node equation at the inverting input of the top op 
amp: 
4
o
i a i 3 2 3
o i
1 2 1 2 1 4
0
R
v
v v v R R R
v v
R R R R R R
\u2212
= + = + \u21d2 = 
 
The output is proportional to the input and the constant of proportionality is 2 3
1 4
R R
R R
. We require 
 so o 20 iv v= 2 3
1 4
20
R R
R R
= . For example, 1 4 2 310 k , 40 k and 50 kR R R R= = \u3a9 = \u3a9 = \u3a9 . 
 
 
P6.4-14 
 
Represent this circuit by node equations. 
 
 
( )o a o s 2 s 1 2 o 1 a
2 1
0 R R R R
R R
v v v v
v v
\u2212 \u2212+ = \u21d2 = + \u2212 v 
o a o 4
a o
4 5 5
R
0 1
R R R
v v v
v v
\u239b \u239e\u2212 + = \u21d2 = +\u239c \u239f\u239c \u239f\u239d \u23a0
 
So ( ) ( )1 2 5 1 4 5 o1 1 4 2 5
s o o s
2 2 5 2 5 2 5 1
R R R R R RR R R R R
1 1
R R R R R R R R R
v
v v v
+ \u2212\u239b \u239e \u239b \u239e\u239b \u239e= + \u2212 + = =\u239c \u239f \u239c \u239f\u239c \u239f\u239c \u239f \u239c \u239f\u239c \u239f \u2212\u239d \u23a0 \u239d \u23a0\u239d \u23a0 4
v 
then 
2 5 1 4
2 5 1 4 2 5
R R R R1920 
R R - R R 20 R R
= \u21d2 = 
For example 
1 4 2 5 3R 19 k , R 10 k , R 20 k , R 10 k , R 10 k .= \u3a9 = \u3a9 = \u3a9 = \u3a9 = \u3a9 
 
(checked: LNAP 5/24/04) 
 
 
P6.4-15 
 
 
Writing node equations: 
s 1
1 s0 
v v
v v
R R
+ = \u21d2 = \u2212 
1 1 1 2
2 10 3 3
v v v v
v v v
R R R s
\u2212+ + = \u21d2 = = \u2212 
2 1 2 2 o
o 2 10 3 8
v v v v v
v v v
R R R
\u2212 \u2212+ + = \u21d2 = \u2212 = \u2212 sv 
The gain of this circuit, o
s
8
v
v
= \u2212 , does not depend on R. 
(checked: LNAP 6/21/04) 
 
 
 
P6.4-16 
 
Represent this circuit by node equations. 
 
s a o a a
1 3
v v v v v
2R R R
\u2212 \u2212+ = 
and 
a o 4
o a
2 4 2
0
v v R
v v
R R R
+ = \u21d2 = \u2212 
 
So 
s o 1 2 1 3 2 3 2
a o
1 3 1 2 2 1 2 3 4
1 1 1v v R R R R R R Rv v
R R R R R R R R R
\u239b \u239e \u239b \u239e\u239b+ ++ = + + = \u2212\u239c \u239f \u239c \u239f\u239c\u239c \u239f \u239c \u239f\u239c\u239d \u23a0 \u239d \u23a0\u239d
\u239e\u239f\u239f\u23a0
 
s 1 2 1 3 2 3 1 4 1 2 1 3 2 3
o o
1 3 1 3 4 1 3 4
1v R R R R R R R R R R R R R Rv v
R R R R R R R R
\u239b \u239e \u239b+ + + + += \u2212 + = \u2212\u239c \u239f \u239c\u239c \u239f \u239c\u239d \u23a0 \u239d
\u239e\u239f\u239f\u23a0
 
3 4
o s
1 4 1 2 1 3 2 3
R R
v v
R R R R R R R R
\u239b \u239e= \u2212\u239c \u239f\u239c \u239f+ + +\u239d \u23a0
 
We require 
3 4
1 4 1 2 1 3 2 3
20
R R
R R R R R R R R
= + + + 
Try 
1 2 3 4 and R R R R R a= = = = R 
Then 
2
20
3 1
a
a
= + 
So 
2 60 20 0a a\u2212 \u2212 = 
( )60 3600 4 80
60.332, 0.332
2
a
+ ± += = \u2212 
e.g. 
1 2 3 410 k and 603.32 kR R R R= = \u3a9 = = \u3a9 
(checked: LNAP 6/9/04) 
 
P6.4-17 
 
Represent this circuit by node equations. 
 
s a o
1 2 3
0
v v v
R R R
+ + = 
and 
 
a a o 4
a o
4 5 4 5
0
v v v R
v v
R R R R
\u239b \u239e\u2212+ = \u21d2 = \u239c \u239f\u239c \u239f+\u239d \u23a0
 
So 
( )s o a 4 o1 3 2 2 4 5
v v v R
v
R R R R R R
+ = \u2212 = \u2212 + 
( )
( )
( )2 4 5 4 3s 4 o o1 3 2 4 5 2 3 4 5
1 R R R R Rv R v v
R R R R R R R R R
\u239b \u239e + +\u239c \u239f= \u2212 + = \u2212\u239c \u239f +\u239d \u23a0
 
( )
( )2 3 4 5o s1 2 4 2 5 3 4
R R R R
v v
R R R R R R R
+= \u2212 + + 
We require ( )
( )2 3 4 51 2 4 2 5 3 420
R R R R
R R R R R R R
+= + + 
Try 
1 4 5 2 3 and R R R R R R a= = = = = R 
then 
2 3
3
2 220 30
3 3
a R a a
aR
= = \u21d2 = 
e.g. 
1 4 5 2 310 k and 300 kR R R R R= = = \u3a9 = = \u3a9 
 
(checked: LNAP 6/10/04) 
 
 
 
P6.4-18 
Label the node voltages as shown. Represent this 
circuit by node equations. 
 
b a a 1
a b
2 1 1 2
 
v v v R
v v
R R R R
\u2212 = \u21d2 = + 
o b
o b
3
0 
v v
i v
R
\u2212+ = \u21d2 = +3 o oR i v 
o a s a s 1 2 o
a
2 1 1 1 2
0 
v v v v v R R v
v
2R R R R R R
\u239b \u239e\u2212 \u2212 ++ = \u21d2 = \u2212\u239c \u239f\u239c \u239f\u239d \u23a0
 
So 
( )s 1 2 1 o 33 o o o
1 1 2 1 2 2 2
v R R R v R
R i v i
R R R R R R R
\u239b \u239e\u239b \u239e+= +\u239c \u239f\u239c \u239f\u239c \u239f\u239c \u239f+\u239d \u23a0\u239d \u23a0
\u2212 = 
o 2
s 1
i R
v R R
=
3
 
 
We require 2 2 1 3
1 3
0.02, e.g. 8 k , 20 k
R
R R R
R R
= = \u3a9 = = \u3a9 . 
(checked: LNAP 6/21/04) 
 
 
P6.4-19 
 
 
(a) Use units of volts, mA, and k\u3a9. Apply KCL at the inverting input of the left op amp to get 
 
s a o
a s
500 5
10 50
v v v
v v v
R R
\u239b \u239e+ + = \u21d2 = \u2212 +\u239c \u239f\u239d \u23a0o 
 
o a s o o
4 40 404 1 4
5
v v v v v
R R
\u239b \u239e= = \u2212 \u2212 \u21d2 + = \u2212\u239c \u239f\u239d \u23a0 sv 
 
o
s
4 4
40 401
v R
v R
R
= \u2212 = \u2212 ++
 
(b) o
s
0 4 0
v
R
v
\u2264 \u2264 \u221e \u21d2 \u2212 \u2264 \u2264 
(c) We require 43 120 k
40
R R
R
\u2212 = \u2212 \u21d2 = \u3a9+ 
(checked: LNAP 6/21/04) 
 
 
P6.4-20 
 
 
(a) Use units of V, mA and k\u3a9. Apply KCL at the inverting input of