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 012 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 ii iv i

= == − = −

= + = −

Apply Ohm’s law to the 4 kΩ resistor

( )( )( )

o

3

4000

2.4 2.4 10 4000 12 Vo 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 20 3

8000 4000o

o

vv

− − − −+ = ⇒ = − 0 V

Apply Ohm’s law to the 8 kΩ resistor

2 3.5 mA8000

oo

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 Va10000av v−−⎛ ⎞− − × − ⇒⎜ ⎟

⎝ ⎠ Apply Ohm’s law to the 20 kΩ resistor

1 mA20000 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 03000 4000

15 V

o

o

v

v

−− −⎛ ⎞ ⎛ ⎞− − − ⋅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⇒ = −

=

Apply KCL at the output node of the op amp:

0 7.5 mA6000 3000

o oo 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 mA4000 4000

aa

vi = = =

Apply KCL at node a:

0.625 mAb ai i= = Apply KVL:

( )( )3 3

8000 4000

12 10 0.625 10 7.5 Vo 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 aa s

a ao a

ao a

Rv v v vR R R

R R R Rv vi vR R R R R R

R R Rvv v vR 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 mA4000oi = =

(checked using LNAP 8/16/02) P6.3-9 KCL at node a:

( )180 0 12 V

4000 8000a 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 V4000 8000o bv 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 3s s

2 2

1v R

i iR R

⎛ ⎞= + = +⎜ ⎟⎜ ⎟

⎝ ⎠i

Using Ohm’s law gives

o a 3 1 3s o 1 3

1 2

1v v R R R

i i v R RR R R

⎛ ⎞ ⎛ ⎞−= = + ⇒ = + +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠s

2

i

We require

1 31 3

2

20R R

R RR

+ + =

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 2a s

1 2 1 2

0v v v R

v vR 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 s

3 4 4 4 1 2 1 2 4

0R R Rv v v R R R R R

v v vR R R R R R R R R

+⎛ ⎞ ⎛ ⎞ ⎛ ⎞− + += + ⇒ = = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠ ⎝ ⎠

sv

We require

( )( )

2 3 4

1 2 4

5R 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 4b b b

bv v v v− +

+ + = ⇒ = −

The node voltages at the input nodes of an ideal op amp are equal so 1 V4e bv v= = − .

KCL at node e: 100 101000 9000 4

e e dd e

v v v v v V−+ = ⇒ = = −

(checked using LNAP 8/16/02)

P6.4-2 Apply KCL at node a:

12 00 4 V6000 6000 6000a a a

av 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 mA3000

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 vR R

Rv vR

⎛ ⎞ ⎛ ⎞− −− − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ = −

Apply KCL at node a:

0 2 3 2 4 3 4 0 44 3 2 4 3 2 2 3

2 3 2 4 3 4

1 3

0 1 1 1 0

a a aa a

s

v v v v R R R R R Rv R v vR R R R R R R R

R R R R R R vR R

⎛ ⎞− − + ++ + = ⇒ = + + =⎜ ⎟

⎝ ⎠+ +

= −

o 30 900 30Plug in values yields 200 V/V4.8s

vv

+ +⇒ = − = −

P6.4-4

Ohm’s law:

1 2

2

v viR−=

KVL:

( ) ( )1 2 30 1 2 3 1

2

R R Rv R R R i v vR

+ += + + = − 2

P6.4-5

1 1 2 1 11 2

1 7 7 7

2 1 2 2 22 1

2 7 7 7

0 0 1

0 0 1

aa

bb

v v v v R Rv vR R R R

v v v v R Rv v vR R R R

⎛ ⎞− −+ + = ⇒ = + −⎜ ⎟

⎝ ⎠⎛ ⎞− −

− + = ⇒ = + −⎜ ⎟⎝ ⎠

v

6

4 6 4 6

0 50

3 5 3

0 0 0

0 0 (1 )

b c cc b

a c ca c

v v v Rv vR R R R

v v v v R Rv vR R R R

⎛ ⎞− −− + + = ⇒ =⎜ ⎟ +⎝ ⎠⎛ ⎞ ⎛ ⎞− −

− + + = ⇒ = − + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

5

3

v

( )( )

( )( )

6 3 5 6 3 5 25 1 52 10 2

3 7 3 4 6 7 3 7 3 4 6 7

00

(1 ) (1 )

5

c

R R R R R R RR R RR Rv vR R R R R R R R R R R R

v viR

⎡ ⎤ ⎡+ += + + − + +⎢ ⎥ ⎢+ +⎣ ⎦ ⎣

−= =

1v⎤⎥⎦

P6.4-6

KCL at node b: 3 3

5020 10 25 10 4

a cc a

v v v v+ = ⇒ = −× ×

KCL at node a: ( )3 3 3 3

512 0 124 0 V

40 10 40 10 20 10 10 10 13

a aa a a

a

v vv v v v

⎛ ⎞− −⎜ ⎟− − + ⎝ ⎠+ + + = ⇒ = −× × × ×

So 5 14 1c av 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 010000 30000

1.5 V

aa

a

vv

v

+ −⎛ ⎞−⎛ ⎞− + − =⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

⇒ = −

Apply KCL to the super node corresponding the voltage source:

( )

( )

0 6 010000 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 00 0 0 0.7 mA

30000 30000bv v vi i−

+ + = ⇒ = −

P6.4-8 Apply KVL to the bottom mesh:

0 0

0

(10000) (20000) 5 01 mA6

i i

i

− − +

⇒ =

=

The node voltages at the input nodes of an ideal op amp are equal. Consequently

010 10000 V6av i= =

Apply KCL at node a:

00 0 3 5

10000 20000a a

av v v v v−

+ = ⇒ = = V

P6.4-9

KCL at node b: 12 0 4

40000 20000b b

bv 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

2020 10 40 10 3

f g gg f

v v vv v

−⎛ ⎞− + = ⇒ =⎜ ⎟× ×⎝ ⎠

The node voltages at the input nodes of an ideal op amp are equal, so 23e gv v v= = f .

KCL at node d: 3 3 3 3

26 2430 V

20 10 20 10 20 10 20 10 5 5

d fd f d fd e

f d

v vv v v vv v v v−− −−

= + = + ⇒ = = −× × × ×

Finally, 2 16 V3 5e g fv v v= = = − .

P6.4-10 By voltage division (or by applying KCL at node a)

0

1 0

a sRv v

R R=

+

Applying KCL at node b:

( )

0

1 0

00

1

0

b s b

b s b

v v v vR R R

R R v v v vR

− −+ =

+Δ+Δ

⇒ − + =

The node voltages at the input nodes of an ideal op amp are equal so . b av v=

0 0 0 0 01 1 0 1 1 0 1 0 0

1 s s sR R R R R RR Rv v v

R R R R R R R R⎡ ⎤ ⎛ ⎞⎛ ⎞+Δ +Δ v

RΔ Δ= + − = − = −⎢ ⎥ ⎜ ⎟⎜ ⎟ + +⎝ ⎠ ⎝ ⎠⎣ ⎦ +

P6.4-11 Node equations:

2s s aa s

1 2 1

0 1Rv v v v v

R R R⎛ ⎞−

+ = ⇒ = +⎜ ⎟⎜ ⎟⎝ ⎠

and s a a a o

2 3 4

v v v v vR R R− −

= +

so

4 4 4o a

2 3 2

1R R R

v vR R R

⎛ ⎞= + + −⎜ ⎟⎜ ⎟⎝ ⎠

sv

( )( )

4 4 2 4 4 4 2 2 4o s s

2 3 1 2 3 1 1 1 3

1 2 3 4 3 4s

1 3

1 1 1R R R R R R R R R

v v vR R R R R R R R R

R R R R R Rv

R R

⎛ ⎞ ⎛ ⎞ ⎛= + + + − = + + + +⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠ ⎝ ⎠ ⎝

⎛ ⎞+ + +⎜ ⎟=⎜ ⎟⎝ ⎠

sv⎞⎟⎟⎠

with the given values:

( )( ) ( )o s

20 20 10 8 10 8 40 18 80 420 10 200

v v+ + + ×⎛ ⎞ × +⎛ ⎞= =⎜ ⎟ ⎜ ⎟× ⎝ ⎠⎝ ⎠

s sv 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 2a 2

5 6 6

R Rv v v v vR R R

⎛ ⎞+−= ⇒ = ⎜ ⎟⎜ ⎟

⎝ ⎠

Apply KCL at the top node of R4 to get

3 3o 1 1 ao 1

3 4 4 4

1R Rv v v v v v

R R R R⎛ ⎞ ⎛ ⎞− −

= ⇒ = + −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

av

3 3 5 6o 1

4 4 6

1R R R R

v vR R R

⎛ ⎞ ⎛ ⎞ ⎛ ⎞+= + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2v

When 3 6

4 5

R RR R

=

( )

3 3 5 3 3 4o 1 2 1

4 4 6 4 4 3

31 2

4

1 1 1 1

1

R R R R R Rv v v v

R R R R R R

Rv 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 1RR

= + , 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 4a o

3 4 30

v v Rv v

R R R+ = ⇒ = −

Write a node equation at the inverting input of the top op amp:

4o

i a i 3 2 3o i

1 2 1 2 1 40

Rv

v v v R R Rv 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 RR R

. We require

so o 20 iv v= 2 3

1 420

R RR 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 s2 s 1 2 o 1 a

2 1

0 R R R RR R

v v v vv v

− −+ = ⇒ = + − v

o a o 4a o

4 5 5

R0 1

R R Rv v v

v v⎛ ⎞−

+ = ⇒ = +⎜ ⎟⎜ ⎟⎝ ⎠

So ( ) ( )1 2 5 1 4 5 o1 1 4 2 5

s o o s2 2 5 2 5 2 5 1

R R R R R RR R R R R1 1

R R R R R R R R Rv

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 vR R+ = ⇒ = −

1 1 1 22 10 3 3

v v v vv v v

R R R s

−+ + = ⇒ = = −

2 1 2 2 oo 2 10 3 8

v v v v vv v v

R R R− −

+ + = ⇒ = − = − sv

The gain of this circuit, o

s

8vv

= − , 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 a2 4 2

0v v R

v vR R R

+ = ⇒ = −

So

s o 1 2 1 3 2 3 2a 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 3o 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 4o s

1 4 1 2 1 3 2 3

R Rv v

R R R R R R R R⎛ ⎞

= −⎜ ⎟⎜ ⎟+ + +⎝ ⎠

We require 3 4

1 4 1 2 1 3 2 3

20R R

R R R R R R R R=

+ + +

Try 1 2 3 4 and R R R R R a= = = = R

Then 2

203 1

aa

=+

So 2 60 20 0a a− − =

( )60 3600 4 8060.332, 0.332

2a

+ ± += = −

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

0v v vR R R

+ + =

and

a a o 4a o

4 5 4 5

0v v v R

v vR R R R

⎛ ⎞−+ = ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠

So

( )s o a 4

o1 3 2 2 4 5

v v v Rv

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 Rv v

R R R R R R R R R

⎛ ⎞ + +⎜ ⎟= − + = −⎜ ⎟ +⎝ ⎠

( )( )

2 3 4 5o s

1 2 4 2 5 3 4

R R R Rv v

R R R R R R R

+= −

+ +

We require ( )

( )2 3 4 5

1 2 4 2 5 3 4

20R 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

2 220 303 3a R a aaR

= = ⇒ =

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 1a b

2 1 1 2

v v v R

v vR R R R−

= ⇒ =+

o bo b

3

0 v v

i vR−

+ = ⇒ = +3 o oR i v

o a s a s 1 2 oa

2 1 1 1 2

0 v v v v v R R v

v2R 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 RR i v i

R R R R R R R⎛ ⎞⎛ ⎞+

= +⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠− =

o 2

s 1

i Rv R R

=3

We require 22 1 3

1 3

0.02, e.g. 8 k , 20 kR

R R RR 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 oa s

500 510 50v v v

v v vR R

⎛ ⎞+ + = ⇒ = − +⎜ ⎟⎝ ⎠

o

o a s o o4 40 404 1 45

v v v v vR R

⎛ ⎞= = − − ⇒ + = −⎜ ⎟⎝ ⎠

sv

o

s

4 440 401

v Rv R

R

= − = −++

(b) o

s

0 4 0v

Rv

≤ ≤ ∞ ⇒ − ≤ ≤

(c) We require 43 120 k40

R RR

− = − ⇒ = Ω+

(checked: LNAP 6/21/04) P6.4-20

(a) Use units of V, mA and kΩ. Apply KCL at the inverting input of the left op amp to get

1 a oa s

300 310 30v v v

v vR R

⎛ ⎞+ + = ⇒ = − +⎜ ⎟⎝ ⎠

ov

o a s o o90 903 9 1 9v v v v vR R

⎛ ⎞= = − − ⇒ + = −⎜ ⎟⎝ ⎠

sv

o

s

9 990 901

v Rv R

R

= − = −++

(b) o

s

0 9 0v

Rv

≤ ≤ ∞ ⇒ − ≤ ≤

(c) We require 95 112.5 k90R R

R−

− = ⇒ = Ω+

(checked: LNAP 7/8/04)

P6.4-21 Use units of V, mA and kΩ.

1 2 3 1o120 20 120 20 120 201 3 0.5 840 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 45 3 2 1

1 10 240 20 3 3 3

v v v vv v v v

− −+ = ⇒ = + = − 2

2 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 21 2

Summing Amplifier: 6 26 2

Inverting Amplifier:a

oo a

v v vv 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 4b a

3 3 3

0v v v R R

v vR R R

⎛ ⎞− ++ = ⇒ = ⎜ ⎟⎜ ⎟

⎝ ⎠

Apply KCL at the noninverting input of the op amp to get

a in a b

out1 2

0v v v v

iR R− −

+ + =

Solving gives

1 2 in b 1 2 3 4 ina out a out

1 2 1 2 1 2 2 3 1

0 0R R v v R R R R v

v i vR R R R R R R R R

⎛ ⎞ ⎛ ⎞+ + +− − + = ⇒ − − + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠i

When 2 3 1 4R R R R= the quantity in parenthesis vanishes leaving

out in1

1i vR

=

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 4b t

3 4 3

0v v v R R

v vR R R

⎛ ⎞− ++ = ⇒ = ⎜ ⎟⎜ ⎟

⎝ ⎠

Apply KCL at the noninverting input of the op amp to get

3 4t

3t t b t 4 2 3 1 4t t t

1 2 1 2 1 2 3 1 2 3

1

R Rv

Rv v v v R R R R Ri v

R R R R R R R R R R

⎛ ⎞+⎜ ⎟⎜ ⎟ ⎛ ⎞− −⎝ ⎠= + = + = − =⎜ ⎟⎜ ⎟

⎝ ⎠v

t 1 2 3 2

o2 4t 2 3 1 4

1 3

v R R R RR 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 5a o

5 4 4 5

v v v Rv v

R R R R⎛ ⎞−

= ⇒ = ⎜ ⎟⎜ ⎟+⎝ ⎠

Solving these equations gives

s b 1 2 2 3 1 3 5a 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 5o s

1 2 2 3 1 3 5 1 2 2 3 1 3 5

R R R R R R R Rv v

R R R R R R R R R R R R R R⎛ ⎞+ +

= × + ×⎜ ⎟⎜ ⎟+ + + +⎝ ⎠bv×

So

2 3 4 5 1 3 4 5s 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 bR R R R R R R R R R R R R R

⎛ ⎞+ += × = ×⎜ ⎟⎜ ⎟+ + + +⎝ ⎠

v×

(b) The equation of the straight line is

o s5 54

v v= +

We require 2 3 4 5

1 2 2 3 1 3 5

54

R R R RR 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 5b b

1 2 2 3 1 3 5

554

R R R Rv 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 3o s

1 2 3 1 2

0v v v R R

v vR R R R R

⎛ ⎞+ + = ⇒ = − −⎜ ⎟⎜ ⎟

⎝ ⎠bv

So

3 3b

1 2

and R R

a bR R

= − = − v

(b) The equation of the straight line is

o s5 52

v v= − +

We require 3

1

52

RR

− = −

e.g. . Next, we require 1 320 k and 50 kR R= Ω = Ω

3 b

2

5R vR

= −

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

teq

t

vR

i=

First, use KVL to get ( )( ) (

( ))t p

pt

p

0 1

1

i R a R i R aR

R a Ri i

R aR

= + + + +

⎛ ⎞+ −⇒ = −⎜ ⎟⎜ ⎟+⎝ ⎠

p

Use KVL to get

( ) ( )p pt t t

p p

1 21

R a R a Rv Ri Ri R i R i

R aR R aR⎛ ⎞ ⎛+ − −

= + = − =⎜ ⎟ ⎜⎜ ⎟ ⎜+ +⎝ ⎠ ⎝t

1 ⎞⎟⎟⎠

so

( ) ( )t peq

pt p

2 1 2 1

1

v R a R a RR p

Ri R aR aR

− −= = =

+ +

(checked: LNAPDC 7/24/04)

P6.5-16

(a)

( )4 1 2 2 6 1 4 2 3 6

o 13 4 1 1 5 51 3 4

1 1R R R R R R R R R R

v vR R R R R RR R R

⎡ ⎤⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ −= − + =⎢ ⎥⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟+ +⎢ ⎥⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦

1v+

So the gain is

( )o 1 4 p 3 6

1 51 3 4

1v R R aR R Rv RR R R

⎛ ⎞−= +⎜ ⎟⎜ ⎟+ ⎝ ⎠

(b) When 1 3 412 pR R R R= = = the gain becomes

o 6

i 5

1 12

v Ra

v R⎛ ⎞⎡ ⎤= − +⎜ ⎟⎢ ⎥ ⎜ ⎟⎣ ⎦ ⎝ ⎠

so

6 o 6

5 i 5

1 11 12 2

R v RR v R

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞− + ≤ ≤⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠+

We require

66 5

5

110 1 192

RR 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 osb

v v v i−= +

× ×

Solving for vout:

( ) ( )

( ) ( )( )

33 3

1 13

3 3 9

100 10 1 100 10 11 100 1010 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: 0

1 10000 90000

os osb

v vi v−+ =

Solving for vo:

( ) ( )

( )

33 3

o 1 os b13

3 3 9 3

90 101 90 10 10 90 10 i10 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 vvR R R v R R AR

v Av v v v R R R R R R AR R

− − ⎫+ + = ⎪ −⎪ ⇒ =⎬+ − + + + +⎪+ =⎪⎭

)

P6.7-4 a) 3

0 23

1

49 10 = 9.60785.1 10in

v Rv R

×= − = − −

×

b) ( ) ( )( )( )6 3

03 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

vv

× − ×= =

× + × + × + × × +−

c) 6 3

03 6 3 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

vv

× − ×= −

× × × × ×

P6.7-5

Apply KCL at node b: 3

2 3

( )b cmRv v

R R= −

+ pv

Apply KCL at node a: 0

4 1

( ) 0a a cm nv v v v vR R− − +

+ =

The voltages at the input nodes of an ideal op amp are equal so a bv v= .

4 40

1 1

( )cm n aR Rv v vR R

1R v+= − + +

40

1

4 1 3

1 2 3

( )

( ) ( )( )

cm n

cm p

Rv v vR

R R R v vR R R

= − + +

+−

+

4

3 4 1 3 34 41

31 2 1 2 3 2 1

2

4 4 40

1 1 1

1( )when then

( ) 1

so

( ) ( ) (cm n cm p n p

RR R R R RR RR

RR R R R R R RR

R R Rv v v v v vR R R

++

= = ×+ +

= − + + − = − + )v

=

Section 6-9 How Can We Check…?

P6.9-1

Apply KCL at the output node of the op amp

( )ooo

50

10000 4000vvi− −

= + =

Try the given values: io =−1 mA and vo = 7 V

( )3 3 7 571 10 3.7 1010000 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 V10 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:

( ) ( )o6 4 43 1 2 18 6 12 V2 2 2

v ⎛ ⎞⎛ ⎞ ⎛ ⎞= − − − + + = − + = −⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

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 010000 40000 4000− − −

+ + =

node c: 0 2 2 5 04000 6000− −

= +

node d: 2 5 5 5 116000 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 020000 40000 40000

− − −− − −+ + =

node e: ( )2.5 0.25 0.25 09000 1000

− − − −≠ +

Therefore, the analysis is not correct.

Notice that ( )2.5 0.25 0.25 09000 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.