Chapter 6 - The Operational Amplifier

Prévia do material em texto

Problems 
Section 6-3: The Ideal Operational Amplifier 
 
P6.3-1 
 
 
(checked using LNAP 8/16/02) 
 
P6.3-2 
 
 
Apply KVL to loop 1: 
 
1 1
1
12 3000 0 2000 0
12 2.4 mA
5000
i i
i
− + + + =
⇒ = = 
The currents into the inputs of an ideal op 
amp are zero so 
( )
o 1
2 1
2
 2.4 mA
 2.4 mA
 1000 0 2.4 Va
i i
i i
v i
= =
= − = −
= + = −
 
 
Apply Ohm’s law to the 4 kΩ resistor 
( )
( )( )
o
3
4000
2.4 2.4 10 4000 12 V
o av v i
−
= −
= − − × = − 
 
(checked using LNAP 8/16/02) 
 
P6.3-3 
 
The voltages at the input nodes of an ideal op amp 
are equal so 2 Vav = − . 
 
Apply KCL at node a: 
 
( ) ( )2 12 2 0 3
8000 4000
o
o
v
v
− − − −+ = ⇒ = − 0 V 
 
Apply Ohm’s law to the 8 kΩ resistor 
 
2 3.5 mA
8000
o
o
vi − −= = 
 
(checked using LNAP 8/16/02) 
P6.3-4 
 The voltages at the input nodes of an ideal 
op amp are equal so . 5 Vv =
 
Apply KCL at the inverting input node of the 
op amp: 
35 0.1 10 0 = 0 = 4 Va10000
av v−−⎛ ⎞− − × − ⇒⎜ ⎟⎝ ⎠
 
Apply Ohm’s law to the 20 kΩ resistor 
1 mA
20000 5
avi = = 
 
(checked using LNAP 8/16/02)
 
P6.3-5 
The voltages at the input nodes of 
an ideal op amp are equal, so 
. Apply KCL at node a: 0 Vav =
 
30 12 0 2 10 0
3000 4000
 15 V
o
o
v
v
−− −⎛ ⎞ ⎛ ⎞− − − ⋅⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠
⇒ = −
=
 
Apply KCL at the output node of 
the op amp: 
0 7.5 mA
6000 3000
o o
o o
v vi i+ + = ⇒ = 
(checked using LNAP 8/16/02)
 
 
P6.3-6 
The currents into the inputs of an ideal op amp 
are zero and the voltages at the input nodes of 
an ideal op amp are equal so . 
Apply Ohm’s law to the 4 kΩ resistor: 
2.5 Vav =
2.5 0.625 mA
4000 4000
a
a
vi = = = 
 
Apply KCL at node a: 
0.625 mAb ai i= = 
Apply KVL: 
( )( )3 3
8000 4000
12 10 0.625 10 7.5 V
o b av i i
−
= +
= × × = 
 
 
(checked using LNAP 8/16/02) 
 
P6.3-7 
 
 
2
1 2 1
2 3 2 3
2 3 2 3 1 3
4 2 4
4 3 3 1
0 0 0 0 
0 0
00 0 0 o
s a
a s
a a
o a
a
o a
Rv v v v
R R R
R R R Rv vi v
R R R R R R
R R Rvv v v
R R R R R
⎛ ⎞⎛ ⎞− −− − + = ⇒ = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⎛ ⎞+ +− −= + = − = ⎜ ⎟⎜ ⎟⎝ ⎠
⎛ ⎞ ⎛ ⎞−−− − + = ⇒ = − =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ 3
s
s
v
v
 
 
 
 
 
 
 
 
 
P6.3-8 
 
 
The node voltages have been 
labeled using: 
 
1. The currents into the 
inputs of an ideal op amp are 
zero and the voltages at the 
input nodes of an ideal op 
amp are equal. 
 
2. KCL 
 
3. Ohm’s law 
 
Then 
 
0 11.8 1.8 10 Vv = − = 
and 
10 2.5 mA
4000o
i = = 
 
 
(checked using LNAP 8/16/02) 
P6.3-9 
KCL at node a: 
 
( )18 0 0 12 V
4000 8000
a a
a
v v v
− − + + = ⇒ = −
 
The node voltages at the input nodes of 
ideal op amps are equal, so . b av v=
 
Voltage division: 
 
8000 8 V
4000 8000o b
v v= =+ − 
 
 
 
 
(check using LNAP 8/16/02) 
 
 
P6.3-10 
Label the circuit as shown. The current in resistor R 3 is 
. Consequently: si
a sv i R= 3 
 
Apply KCL at the top node of R 2 to get 
a 3
s s
2 2
1
v R
i i
R R
⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠
i 
Using Ohm’s law gives 
 
o a 3 1 3
s o 1 3
1 2
1
v v R R R
i i v R R
R R R
⎛ ⎞ ⎛ ⎞− = = + ⇒ = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ s2
i
 
We require 
1 3
1 3
2
20
R R
R R
R
+ + = 
e.g. and . 1 5 kR = Ω 2 3 10 kR R= = Ω
(checked: LNAP 6/2/04) 
 
 
P6.3-11 
Label the circuit as shown. Apply KCL at the top 
node of R 2 to get 
 
s a a 2
a s
1 2 1 2
0
v v v R
v v
R R R R
⎛ ⎞− = + ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠
 
 
Apply KCL at the inverting node of the op amp to 
get ( )
( )2 3 4o a a 3 4 3 4 2o a s3 4 4 4 1 2 1 2 40
R R Rv v v R R R R R
v v v
R R R R R R R R R
+⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + += + ⇒ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠ s
v 
 
We require ( )
( )2 3 41 2 4 5
R R R
R R R
+ =+ 
 
e.g. , and 1 2 10 kR R= = Ω 3 90 kR = Ω 4 10 kR = Ω . 
(checked: LNAP 6/2/04) 
 
 
 
Section 6-4: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers 
 
P6.4-1 
 
 
 KCL at node b: 
2 5 10 V
20000 40000 40000 4
b b b
b
v v v v− ++ + = ⇒ = − 
 The node voltages at the input nodes of an ideal op amp are equal so 1 V
4e b
v v= = − . 
 KCL at node e: 100 10
1000 9000 4
e e d
d e
v v v v v V−+ = ⇒ = = − 
 
(checked using LNAP 8/16/02) 
P6.4-2 
Apply KCL at node a: 
 
12 00 4 V
6000 6000 6000
a a a
a
v v v v− −= + + ⇒ = 
 
Apply KCL at the inverting input of the op 
amp: 
 
o
o
0 00 
6000 6000
 4 V
a
a
v v
v v
− −⎛ ⎞ ⎛ ⎞− + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ = − = −
0=
 
 
Apply KCL at the output of the op amp: 
 
o
o
o o
o
0 
6000 6000
 1.33 mA
3000
v vi
vi
−⎛ ⎞− + =⎜ ⎟⎝ ⎠
⇒ = − =
0
 
 
 
 
 
 
 
(checked using LNAP 8/16/02)
 
P6.4-3 
 
 
Apply KCL at the inverting input of 
the op amp: 
 
2 1
2
1
0 0 0a s
a s
v v
R R
Rv v
R
⎛ ⎞ ⎛ ⎞− −− − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ = −
 
 
Apply KCL at node a: 
0 2 3 2 4 3 4 0 4
4 3 2 4 3 2 2 3
2 3 2 4 3 4
1 3
0 1 1 1 0 
 
a a a
a a
s
v v v v R R R R R Rv R v v
R R R R R R R R
R R R R R R v
R R
⎛ ⎞− − + ++ + = ⇒ = + + =⎜ ⎟⎝ ⎠
+ += −
 
o 30 900 30Plug in values yields 200 V/V
4.8s
v
v
+ +⇒ = − = − 
 
 
P6.4-4 
 
Ohm’s law: 
1 2
2
v vi
R
−= 
 
KVL: 
 
( ) ( )1 2 30 1 2 3 1
2
R R Rv R R R i v v
R
+ += + + = − 2 
 
 
 
P6.4-5 
 
 
1 1 2 1 1
1 2
1 7 7 7
2 1 2 2 2
2 1
2 7 7 7
0 0 1
 0 0 1
a
a
b
b
v v v v R Rv v
R R R R
v v v v R Rv v v
R R R R
⎛ ⎞− −+ + = ⇒ = + −⎜ ⎟⎝ ⎠
⎛ ⎞− −− + = ⇒ = + −⎜ ⎟⎝ ⎠
v
 
6
4 6 4 6
0 5
0
3 5 3
0 0 0 
0 0 (1 )
b c c
c b
a c c
a c
v v v Rv v
R R R R
v v v v R Rv v
R R R R
⎛ ⎞− −− + + = ⇒ =⎜ ⎟ +⎝ ⎠
⎛ ⎞ ⎛ ⎞− −− + + = ⇒ = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
5
3
v
 
 
( )
( )
( )
( )6 3 5 6 3 5 25 1 52 10 23 7 3 4 6 7 3 7 3 4 6 7
0
0
 
 (1 ) (1 )
 
 
5
c
R R R R R R RR R RR Rv v
R R R R R R R R R R R R
v vi
R
⎡ ⎤ ⎡+ += + + − + +⎢ ⎥ ⎢+ +⎣ ⎦ ⎣
−= = "
1v
⎤⎥⎦ 
 
P6.4-6 
 
 
KCL at node b: 3 3
50
20 10 25 10 4
a c
c a
v v v v+ = ⇒ = −× × 
KCL at node a: ( )3 3 3 3
5
12 0 124 0 V
40 10 40 10 20 10 10 10 13
a a
a a a
a
v vv v v v
⎛ ⎞− −⎜ ⎟− − + ⎝ ⎠+ + + = ⇒ = −× × × × 
 
So 5 1
4 1c a
v v 5
3
= − = − . 
 
(checked using LNAP 6/21/05) 
 
 
P6.4-7 
Apply KCL at the inverting input node of 
the op amp 
 
( )6 00 0 0
10000 30000
 1.5 V
aa
a
vv
v
+ −⎛ ⎞−⎛ ⎞− + − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ = −
 
 
Apply KCL to the super node 
corresponding the voltage source: 
 
( )
( )
0 6 0
10000 30000
6
 0
30000 10000
 3 6
3 6 0
 2 6 3 V
a a
a ba b
a a a b
a b
b a
v v
v vv v
v v v v
v v
v v
− + −+
+ −−+ + =
⇒ + + + −
+ + − =⎡ ⎤⎣ ⎦
⇒ = + =
 
 
Apply KCL at node b: 
 
( )
( ) ( ) ( )
0
0
0
6
 0
10000 30000 30000 10000
3 3 6 
 8 4 18 12 V
a bb b a b
b b a b a b
b a
v vv v v v v
v v v v v v v
v v v
+ −⎛ ⎞− −⎛ ⎞+ − − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠
⇒ + − − − − + − =⎡ ⎤⎣ ⎦
⇒ = − − =
0 
 
Apply KCL at the output node of the op amp: 
 
0 0
0 0 0 0.7 mA30000 30000
bv v vi i−+ + = ⇒ = −
 
 
 
 
 
P6.4-8 
Apply KVL to the bottom mesh: 
 
0 0
0
(10000) (20000) 5 0
1 mA
6
i i
i
− − +
⇒ =
=
 
 
The node voltages at the input nodes of an 
ideal op amp are equal. Consequently 
 
0
1010000 V
6a
v i= = 
Apply KCL at node a: 
 
0
0 0 3 5 10000 20000
a a
a
v v v v v−+ = ⇒ = = V 
 
 
 
 
 
P6.4-9 
 
 
KCL at node b: 
12 0 4
40000 20000
b b
b
v v v+ + = ⇒ = − V 
The node voltages at the input nodes of an ideal op amp are equal, so 4 Vc bv v= = − . 
 
The node voltages at the input nodes of an ideal op amp are equal, so . 40 10 4 Vd cv v= + × = −
 
KCL at node g: 3 3
20
20 10 40 10 3
f g g
g f
v v v
v v
−⎛ ⎞− + = ⇒ =⎜ ⎟× ×⎝ ⎠
 
 
 
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
−− −−= + = + ⇒ = = −× × × × 
 
Finally, 2 16 V
3 5e g f
v v v= = = − . 
 
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
− −+ =+Δ
+Δ⇒ − + =
 
 
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
⎡ ⎤ ⎛ ⎞⎛ ⎞+Δ +Δ v
R
Δ Δ= + − = − = −⎢ ⎥ ⎜ ⎟⎜ ⎟ + +⎝ ⎠ ⎝ ⎠⎣ ⎦ +
 
 
 
P6.4-11 
Node equations: 
2s s a
a s
1 2 1
0 1
Rv v v v v
R R R
⎛ ⎞−+ = ⇒ = +⎜ ⎟⎜ ⎟⎝ ⎠
 
and 
s a a a o
2 3 4
v v v v v
R R R
− −= + 
so 
4 4 4
o a
2 3 2
1
R R R
v v
R R R
⎛ ⎞= + + −⎜ ⎟⎜ ⎟⎝ ⎠ 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
⎛ ⎞ ⎛ ⎞ ⎛= + + + − = + + + +⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝
⎛ ⎞+ + +⎜ ⎟= ⎜ ⎟⎝ ⎠
sv
⎞⎟⎟⎠
 
 
with the given values: 
 
( )( ) ( )o s20 20 10 8 10 8 40 18 80 420 10 200v v
+ + + ×⎛ ⎞ × +⎛ ⎞= =⎜ ⎟ ⎜ ⎟× ⎝ ⎠⎝ ⎠ 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
⎛ ⎞+− = ⇒ = ⎜ ⎟⎜ ⎟⎝ ⎠
 
 
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
⎛ ⎞ ⎛ ⎞− −= ⇒ = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ a
v 
 
3 3 5 6
o 1
4 4 6
1
R R R R
v v
R R R
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 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
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= + − + = + − +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎛ ⎞= + −⎜ ⎟⎜ ⎟⎝ ⎠
2v
 
 
 
so vo is proportional to the difference of the inputs, v1 − v2, as required. 
Next, choose R3 and R4 so that 3
4
5 1
R
R
= + , e.g. 
R1 = 50 kΩ, R2 = 50 kΩ, R3 = 40 kΩ, R4 = 10 kΩ, R5 = 10 kΩ and R6 = 40 kΩ. 
 
(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
+ = ⇒ = − 
 
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
−
= + = + ⇒ = 
 
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= = Ω = Ω = Ω . 
 
 
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
− −+ = ⇒ = + − v 
o a o 4
a o
4 5 5
R
0 1
R R R
v v v
v v
⎛ ⎞− + = ⇒ = +⎜ ⎟⎜ ⎟⎝ ⎠
 
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
+ −⎛ ⎞ ⎛ ⎞⎛ ⎞= + − + = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ −⎝ ⎠ ⎝ ⎠⎝ ⎠ 4
v 
then 
2 5 1 4
2 5 1 4 2 5
R R R R1920 
R R - R R 20 R R
= ⇒ = 
For example 
1 4 2 5 3R 19 k , R 10 k , R 20 k , R 10 k , R 10 k .= Ω = Ω = Ω = Ω = Ω 
 
(checked: LNAP 5/24/04) 
 
 
P6.4-15 
 
 
Writing node equations: 
s 1
1 s0 
v v
v v
R R
+ = ⇒ = − 
1 1 1 2
2 10 3 3
v v v v
v v v
R R R s
−+ + = ⇒ = = − 
2 1 2 2 o
o 2 10 3 8
v v v v v
v v v
R R R
− −+ + = ⇒ = − = − sv 
The gain of this circuit, o
s
8
v
v
= − , 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
− −+ = 
and 
a o 4
o a
2 4 2
0
v v R
v v
R R R
+ = ⇒ = − 
 
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
⎛ ⎞ ⎛ ⎞⎛+ ++ = + + = −⎜ ⎟ ⎜ ⎟⎜⎜ ⎟ ⎜ ⎟⎜⎝ ⎠ ⎝ ⎠⎝
⎞⎟⎟⎠
 
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
⎛ ⎞ ⎛+ + + + += − + = −⎜ ⎟ ⎜⎜ ⎟ ⎜⎝ ⎠ ⎝
⎞⎟⎟⎠
 
3 4
o s
1 4 1 2 1 3 2 3
R R
v v
R R R R R R R R
⎛ ⎞= −⎜ ⎟⎜ ⎟+ + +⎝ ⎠
 
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− − = 
( )60 3600 4 80
60.332, 0.332
2
a
+ ± += = − 
e.g. 
1 2 3 410 k and 603.32 kR R R R= = Ω = = Ω 
(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
⎛ ⎞−+ = ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠
 
So 
( )s o a 4 o1 3 2 2 4 5
v v v R
v
R R R R R R
+ = − = − + 
( )
( )
( )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
⎛ ⎞ + +⎜ ⎟= − + = −⎜ ⎟ +⎝ ⎠
 
( )
( )2 3 4 5o s1 2 4 2 5 3 4
R R R R
v v
R R R R R R R
+= − + + 
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
= = ⇒ = 
e.g. 
1 4 5 2 310 k and 300 kR R R R R= = = Ω = = Ω 
 
(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
− = ⇒ = + 
o b
o b
3
0 
v v
i v
R
−+ = ⇒ = +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
⎛ ⎞− − ++ = ⇒ = −⎜ ⎟⎜ ⎟⎝ ⎠
 
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
⎛ ⎞⎛ ⎞+= +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠
− = 
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
= = Ω = = Ω . 
(checked: LNAP 6/21/04) 
 
 
P6.4-19 
 
 
(a) Use units of volts, mA, and kΩ. 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
⎛ ⎞+ + = ⇒ = − +⎜ ⎟⎝ ⎠o 
 
o a s o o
4 40 404 1 4
5
v v v v v
R R
⎛ ⎞= = − − ⇒ + = −⎜ ⎟⎝ ⎠ sv 
 
o
s
4 4
40 401
v R
v R
R
= − = − ++
 
(b) o
s
0 4 0
v
R
v
≤ ≤ ∞ ⇒ − ≤ ≤ 
(c) We require 43 120 k
40
R R
R
− = − ⇒ = Ω+ 
(checked: LNAP 6/21/04) 
 
 
P6.4-20 
 
 
(a) Use units of V, mA and kΩ. Apply KCL at the inverting input ofthe left op amp to get 
 
1 a o
a s
300 3
10 30
v v v
v v
R R
⎛ ⎞+ + = ⇒ = − +⎜ ⎟⎝ ⎠ov 
 
o a s o o
90 903 9 1 9v v v v v
R R
⎛ ⎞= = − − ⇒ + = −⎜ ⎟⎝ ⎠ sv 
 
o
s
9 9
90 901
v R
v R
R
= − = − ++
 
(b) o
s
0 9 0
v
R
v
≤ ≤ ∞ ⇒ − ≤ ≤ 
(c) We require 95 112.5 k
90
R R
R
−− = ⇒ = Ω+ 
(checked: LNAP 7/8/04) 
 
 
 
P6.4-21 
Use units of V, mA and kΩ. 
 
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⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= − − + + + = − −⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥+⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠⎣ ⎦ 2 3v v 
so 
a = 3, b = −0.5 and c = −8 
 
(checked: LNAP 6/21/04) 
 
 
P6.4-22 
 
Label the node voltages as shown. Use units of V, mA and kΩ. 
3 1 4 and v v v v2= = − 
( )5 3 5 4 5 3 2 11 10 240 20 3 3 3v v v v v v v v− −+ = ⇒ = + = − 22 v 
so 
1 2 and 
3 3
a b= − = − 
(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 –controlled 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
= − + ⎫ ⇒ = +⎬= − ⎭
v 
 
 
 
P6.5-9 
 
 
 
 
 
 
 
 
 
 
Using superposition, 1 2 3 9 16 32 7 Vov v v v= + + = − − + = 
 
 
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
⎛ ⎞− ++ = ⇒ = ⎜ ⎟⎜ ⎟⎝ ⎠
 
 
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
− −+ + = 
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
⎛ ⎞ ⎛ ⎞+ + +− − + = ⇒ − − + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
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
⎛ ⎞− ++ = ⇒ = ⎜ ⎟⎜ ⎟⎝ ⎠
 
 
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
⎛ ⎞+⎜ ⎟⎜ ⎟ ⎛ ⎞− −⎝ ⎠= + = + = − =⎜ ⎟⎜ ⎟⎝ ⎠
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
= = =− −
 
 
 
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
− −+ = 
and 
a o a 5
a o
5 4 4 5
v v v R
v v
R R R R
⎛ ⎞−= ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠
 
 
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
⎛ ⎞ ⎛ ⎞+ += + + − = × −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠
 
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
⎛ ⎞+ += × + ×⎜ ⎟⎜ ⎟+ + + +⎝ ⎠ 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
⎛ ⎞+ += × = ×⎜ ⎟⎜ ⎟+ + + +⎝ ⎠
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= = = Ω = Ω = Ω
 
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
⎛ ⎞+ + = ⇒ = − −⎜ ⎟⎜ ⎟⎝ ⎠ b
v 
 
So 
3 3
b
1 2
 and 
R R
a b
R R
= − = − v 
 
(b) The equation of the straight line is 
o s
5 5
2
v v= − + 
We require 
3
1
5
2
R
R
− = − 
e.g. . Next, we require 1 320 k and 50 kR R= Ω = Ω
3 b
2
5
R v
R
= − 
e.g. . 2 3 b50 k and 5 VR R v= = Ω = −
(checked: LNAP 6/20/04) 
 
 
P6.5-15 
 
Here’s 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
= + + + +
⎛ ⎞+ −⇒ = −⎜ ⎟⎜ ⎟+⎝ ⎠
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
⎛ ⎞ ⎛+ − −= + = − =⎜ ⎟ ⎜⎜ ⎟ ⎜+ +⎝ ⎠ ⎝ t
1 ⎞⎟⎟⎠
 
so 
 
( ) ( )t p
eq
pt p
2 1 2 1
1
v R a R a R
R pRi R aR a
R
− −= = =+ +
 
 
(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
⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −= − + =⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ 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
⎛ ⎞−= +⎜ ⎟⎜ ⎟+ ⎝ ⎠
 
(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
⎛ ⎞⎡ ⎤= − +⎜ ⎟⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠
 
so 
6 o 6
5 i 5
1 11 1
2 2
R v R
R v R
⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− + ≤ ≤⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠
+ 
We require 
6
6 5
5
110 1 19
2
R
R R
R
⎛ ⎞= + ⇒ =⎜ ⎟⎜ ⎟⎝ ⎠
 
 
e.g. Any convenient value of Rp will do, e.g. 5 610 k and 190 k .R R= Ω = Ω p 100 kR = Ω
 
 
 
 
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− = +× × 
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
− −
⎛ ⎞×= + + × = + ×⎜ ⎟×⎝ ⎠
= × + × × =
i
 
 
P6.7-2 
 
The node equation at node a is: 01 10000 90000
os os
b
v vi v−+ = 
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
− − −
⎛ ⎞×= + + × = + ×⎜ ⎟×⎝ ⎠
= × × × = × −�
 
 
 
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
− − ⎫+ + = ⎪ −⎪ ⇒ =⎬+ − + + + +⎪+ = ⎪⎭
)
 
 
 
 
P6.7-4 
a) 3
0 2
3
1
49 10 = 9.6078
5.1 10in
v R
v R
×= − = − −× 
 
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
× − ×= =× + × + × + × × + − 
 
c) 6 3
0
3 63 3 6
2 10 (75 (200,000)(49 10 )) = 9.6037
(5.1 10 +2 10 )(75+49 10 )+(5.1 10 )(2 10 )(1+200,000)in
v
v
× − ×= −× × × × × 
 
 
 
P6.7-5 
 
 
Apply KCL at node b: 
3
2 3
 ( )b cm
Rv v
R R
= −+ pv 
Apply KCL at node a: 
0
4 1
( ) 0a a cm nv v v v v
R R
− − ++ = 
The voltages at the input nodes of 
an ideal op amp are equal so 
 a bv v= . 
4 4
0
1 1
 ( )cm n a
R Rv v v
R R
1R v+= − + + 
4
0
1
4 1 3
1 2 3
 ( )
( ) ( )
( )
cm n
cm p
Rv v v
R
R R R v v
R R R
= − + +
+ −+
 
4
3 4 1 3 34 41
31 2 1 2 3 2 1
2
4 4 4
0
1 1 1
1
( )when then 
( ) 1
so
 ( ) ( ) (cm n cm p n p
R
R R R R RR RR
RR R R R R R R
R
R R Rv v v v v v
R R R
++= = ×+ +
= − + + − = − + )v
=
 
 
 
 
 
Section 6-9 How Can We Check…? 
 
P6.9-1 
 Apply KCL at the output node of the op 
amp 
 
( )oo
o
5
0
10000 4000
vvi
− −= + = 
 
Try the given values: io =−1 mA and 
vo = 7 V 
 
( )3 3 7 571 10 3.7 10
10000 4000
− − − −− × ≠ × = + 
 
KCL is not satisfied. These cannot be the 
correct values of io and vo. 
 
 
 
P6.9-2 
 
 ( )( )
( )
3 3
3
 o 3
o
4 10 2 10 8 V
12 10 1.2 8 9.6 V
10 10
 So 9.6 V instead of 9.6 V.
a
a
v
v v
v
−= × × =
×= − = − = −×
= −
 
 
 
P6.9-3 
 First, redraw the circuit as: 
 
 
 
Then using superposition, and recognizing of the inverting and noninverting amplifiers: 
 
( ) ( )o 6 4 43 1 2 18 6 12 V2 2 2v
⎛ ⎞⎛ ⎞ ⎛ ⎞= − − − + + = − + = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ 
 
The given answer is correct. 
 
 
P6.9-4 
First notice that is required by the ideal op amp. (There is zero current into the input 
lead of an ideal op amp so there is zero current in the 10 kΩ connected between nodes e and f, 
hence zero volts across this resistor. Also, the node voltages at the input nodes of an ideal op 
amp are equal.) 
e fv v v= = c
 
The given voltages satisfy all the node equations at nodes b, c and d: 
 
node b: 0 ( 5) 0 0 2 0
10000 40000 4000
− − −+ + = 
 
node c: 0 2 2 5 0
4000 6000
− −= + 
 
node d: 2 5 5 5 11
6000 5000 4000
− −= + 
 
Therefore, the analysis is correct. 
 
 
P6.9-5 
The given voltages satisfy the node equations at nodes b and e: 
 
node b: ( ).25 5.25 2 .25 0
20000 40000 40000
− − −− − −+ + = 
 
node e: ( )2.5 0.25 0.25 0
9000 1000
− − − −≠ + 
 
Therefore, the analysis is not correct. 
 
Notice that ( )2.5 0.25 0.25 0
9000 1000
− − + += + 
 
So it appears that instead of e 0.25 Vv = + e 0.25 V.v = − 
 
 Also, the circuit is an noninverting summer with Ra = 10 kΩ and Rb = 1 kΩ, K1 =1/ 2, K2 
= 1/ 4 and K4 = 9. The given node voltages satisfy the equation 
 
( ) ( ) ( )1 24 1 12.5 10 2 52 4d a cv K K v K v ⎛ ⎞− = = + = + −⎜ ⎟⎝ ⎠ 
None-the-less, the analysis is not correct. 
 
 
 
 
 
	CH6sec3
	Problems 
	Section 6-3: The Ideal Operational Amplifier 
	P6.3-1 
	P6.3-2
	P6.3-3
	P6.3-4
	P6.3-5
	P6.3-6
	P6.3-7 
	P6.3-8
	P6.3-9
	P6.3-10
	P6.3-11
	CH6sec4
	Section 6-4: Nodal Analysis of Circuits Containing Ideal Operational Amplifiers 
	P6.4-1 
	P6.4-2
	P6.4-3
	P6.4-4
	P6.4-5 
	P6.4-6
	P6.4-7
	P6.4-8
	P6.4-9 
	P6.4-10
	P6.4-11
	P6.4-13
	P6.4-14 
	P6.4-15 
	P6.4-16
	P6.4-17
	P6.4-18
	P6.4-19 
	P6.4-20 
	P6.4-21 
	P6.4-22 
	CH6sec5
	Section 6-5: Design Using Operational Amplifier 
	P6.5-1
	P6.5-2
	P6.5-3
	P6.5-4
	P6.5-5
	P6.5-6
	P6.5-7
	P6.5-8
	P6.5-9
	P6.5-10
	P6.5-11
	P6.5-12
	P6.5-13
	P6.5-14
	P6.5-15
	P6.5-16 
	CH6sec6
	Section 6-6: Operational Amplifier Circuits and Linear Algebraic Equations 
	P6.6-1 
	P6.6-2 
	Ch6sec7
	Section 6-7: Characteristics of the Practical Operational Amplifier 
	P6.7-1 
	P6.7-2 
	P6.7-3 
	P6.7-4
	 P6.7-5
	CH6sec9
	Section 6-9 How Can We Check…? 
	P6.9-1 
	P6.9-2 
	P6.9-3 
	P6.9-4 
	P6.9-5