40 pág.

# Chapter 6 - The Operational Amplifier

DisciplinaCircuitos Elétricos I8.487 materiais104.862 seguidores
Pré-visualização4 páginas
```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```