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CHAPTER 5 - OPERATIONAL AMPLIFIER

List of topics for this chapter :

Operational Amplifiers

Ideal Operational Amplifier

Inverting AmplifierNoninverting Amplifier

Summing Amplifier

Difference Amplifier

Cascaded Operational Amplifier Circuits

Operational Amplifier Circuits with PSpice

Applications

OPERATIONAL AMPLIFIERS

Problem 5.1 Calculate outV for = k100,k10,k1,100,1R L , given the

circuit in Figure 5.1.

Figure 5.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.The value of the load resistor, LR , is varied over a wide range of values.

The dependent voltage source (a voltage-controlled voltage source) has a large gain (100k).

Obviously, the goal of the problem is to determine the output voltage in terms of the input (or

source) voltage.

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.

100 k 100k Vin

1 k

+

Vs

+

Vin

+

50

2 k

RL

+

Vout

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outV can be determined using nodal analysis, mesh analysis, or circuit analysis. Because we

want to find a voltage, rather than a current, we will use nodal analysis.

ATTEMPT a problem solution.Use nodal analysis to find outV in terms of SV and LR .

At the left node or inV :

0k2

VV

k100

0V

k1

VV outininsin =

+

+

At the right node or outV :

0R

0V

50

Vk100V

k2

VV

L

outinoutinout =

++

+

Simplifying the left node equation,

0)VV)(50(V)VV)(100( outininsin =++0V50V100V151 outsin =

Simplifying the right node equation,

0VR

k2)Vk100V)(40()VV( out

L

inoutinout =+++

0V)1M4(VR

k241 inout

L

=+

+

outL

in V1M4R

k241-

V

+

=

Substituting the simplified right node equation into the simplified left node equation,

0V50V100V1M4

R

k241(151)-

outsout

L =

+

sout

LV100V

)1M4(

)1M4)(50(R

k241(151)-

=

+

s

L

out V

)1M4)(50(R

k241(151)-

)1M4)(100(V

+

=

Substitute each value for LR into this equation to find outV in terms of sV .

For = 1R L , =outV sV99692283.1-

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1.5 M 8 104 vd

60

vd+

For = 100R L , =outV sV99990789.1-For = k1R L , =outV sV99993507.1-For = k10R L , =outV sV99993779.1-For = k100R L , =outV sV99993806.1-

EVALUATE the solution and check for accuracy.First, the answers appear reasonable with the gain of the entire circuit approaching 2 as LRincreases in size.

In addition, even for = 1R L , sout V2V = is a good approximation

Clearly, using an ideal op amp is reasonable.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.

This problem has been solved satisfactorily.

For = 1R L , =outV sV99692283.1-

For = 100R L , =outV sV99990789.1-

For = k1R L , =outV sV99993507.1-

For = k10R L , =outV sV99993779.1-

For = k100R L , =outV sV99993806.1-

Problem 5.2 [5.1] The equivalent model of a certain op amp is shown in Figure 5.1.Determine:

(a) the input resistance,(b) the output resistance,(c) the voltage gain in dB.

Figure 5.1

(a) =inR M5.1(b) =outR 60(c) 4108A =

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)108(log20A 410dB ==dBA 06.98

Problem 5.3 Solve for outV when = M1,k100,k1,1R L , given the circuit inFigure 5.1.

Figure 5.1

Use nodal analysis to find outV in terms of SV and LR .

At the left node or inV :

0k100

0V

k1

VV inSin =

+

At the right node or outV :

0R

0V

50

Vk100V

L

outinout =++

Simplifying the left nodal equation,

0V)VV)(100( inSin =+

Sin V100V101 =

Sin V101

100V =

Simplifying the right node equation, 0Vk2VR

1

50

1inout

L=+

+

inout

L

LVk2-V

R50

50R=

+

in

L

L

out V50R

Rk100-V

+=

100 k100k Vin

1 k

+

Vs

+

Vin

+

50

RL

+

Vout

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Substituting the simplified left node equation into the simplified right node equation,

S

L

L

S

L

L

out V)50R)(101(

RM10-V

101

100

50R

Rk100-V

+=

+=

Substitute each value for LR into this equation to find outV in terms of SV .

For = 1R L , =outV SV3706.941,1-

For = k1R L , =outV SV1438.295,94-

For = k100R L , =outV SV4208.960,98-

For = M1R L , =outV SV9507.004,99-

IDEAL OPERATIONAL AMPLIFIER

An ideal op amp has infinite open-loop gain, infinite input resistance, and zero output resistance.

Problem 5.4 Looking at the circuit in Figure 5.1, what effect does LR have on the valueof outV ?

Figure 5.1

Carefully DEFINE the problem.Each component is labeled completely. The problem is clear.

PRESENT everything you know about the problem.Essentially, we are to determine if the value of LR affects the output voltage in any way.

Thus, the goal of the problem is to solve for oV in terms of the other variables.

RF

R1

+

Vs

+

+

Vo

RL

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Treat the operational amplifier as ideal. Due to infinite input resistance, we know that the

currents into both input terminals are zero. The voltage across the terminals is negligibly

small or ba VV = .

Establish a set of ALTERNATIVE solutions and determine the one that promises thegreatest likelihood of success.Because this is a circuit problem, we can use nodal analysis, mesh analysis, or basic circuit

analysis. Nodal analysis typically works best for op amp circuits.

ATTEMPT a problem solution.Referring to the circuit below, there are three unknown nodes.

Write a node equation at node a. The node voltage at node b is already known, 0Vb = .Writing a node equation at node c will only introduce an additional unknown. This gives two

equations with four unknowns. Solving for oV in terms of sV (and the resistances) implies

that we need one equation with two unknowns. Hence, we need a constraint equation to

reduce the number of unknowns.

At node a, 00R

VV

R

VV

F

ca

1

sa =+

+

At node b, 0Vb =

The constraint equation comes from a characteristic of the ideal op amp.

ba VV =

Thus, 0VVba ==

Substitute the constraint into the node equation for node a to solve for oV .

0R

V-

R

V-

F

c

1

s =+

RF

R1

+

Vs

Va

Vb

+

+

Vo

RL

Vc

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Hence, s1

F

c VR

R-V =

Clearly, s1

F

co VR

R-VV ==

We have shown that the value of LR has no effect on the value of oV , assuming that LR is

finite and not equal to zero.

EVALUATE the solution and check for accuracy.Consider the following circuit.

Using Ohm's law,1

as

R

VVI

=

From a characteristic of the ideal op amp, ba VV =

But 0V0V ab ==

So,1

s

R

VI =

Also,F

c

F

ca

R

V-

R

VVI =

=

But oc VV =

So,F

o

R

V-I =

Thus,

s

1

F

o

F

o

1

sV

R

R-V

R

V-

R

V==

RF

R1

+

Vs

Va

Vb

+

+

Vo

RL

VcI

I

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Our check for accuracy was successful.

Has the problem been solved SATISFACTORILY? If so, present the solution; if not,then return to ALTERNATIVE solutions and continue through the process again.This problem has been solved satisfactorily.

The value of RL has no effect on the value of Vo, assuming that RL is finite and not zero.

Problem 5.5 [5.11] Find ov and oi in the circuit in Figure 5.1.

Figure 5.1

At node b, V2)3(510

10vb =

+=

At node a, oaoaa

vv5128

vv

2

v3 ==

But V2vv ba ==

So, ov)2)(5(12 ==ov V2-

m5.0-0.5m4k

2-

k8

22-

k4

0v

k8

vvi

oao

o =+

=

+

=

=oi mA1-

+

vo

8 k

5 k +

2 k

10 k+3 V 4 k

iova

vb

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INVERTING AMPLIFIER

An inverting amplifier reverses the polarity of the input signal while amplifying it.

Problem 5.6 [5.19] Using the circuit in Figure 5.1, calculate ov if 0vs = .

Figure 5.1

At node a,

k4

vv

k8

vv

k4

v9 baoaa +

=

boa v2vv518 = (1)

At node b,

k2

vv

k4

vv obba =

oba v2v3v = (2)

But 0vv sb ==

Hence, (2) becomes oa v2-v =

and (1) becomes ooo v-11vv10-18 == and =ov V1.6364-

8 k

4 k

va

vb+

+

+

vo

vs

4 k2 k

+

9 V

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Problem 5.7 Express oV in terms of sV for the circuit shown in Figure 5.1.

Figure 5.1

Using nodal analysis,

0R

VV

R

VV

F

oa

1

sa =

+

where 0VV ba == is the constraint equation.

0R

V-

R

V-

F

o

1

s =+

=oV s1

FV

R

R-

NONINVERTING AMPLIFIER

Problem 5.8 How does the circuit in Figure 5.1 differ from the circuit in Figure 5.1?

Figure 5.1

RF

R1 Va

Vb

+

+

+

Vout

Vs

RL

RF

R1

+

Vs

Va

Vb

+ +

Vo

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0R

VV

R

0V

F

outa

1

a =

+

where sba VVV == is the constraint equation.

Simplifying,

0R

VV

R

V

F

outs

1

s =

+

s

F1

out

F

VR

1

R

1V

R

1

+=

s

1

F

out V1R

RV

+=

The significant difference between the two circuits is that the voltage gain for this circuit ispositive. In addition it should be noted that in the circuit ofFigure 5.1, the relationship

between the output voltage, Vo, and the input voltage, Vs, is a simple ratio of RF and R1. For

the circuit in Figure 5.1, however, the gain can never be less than one. Since there is rarely

a case where the gain is less than one, this is not normally a problem.

SUMMING AMPLIFIER

A summing amplifier combines several inputs and produces an output that is the weighted sum of

the inputs.

Problem 5.9 [5.33] A four-input summing amplifier has ==== 4321 RRRRk12 . What value of feedback resistor is needed to make it an averaging amplifier?

In order for )vvvv(k12

Rv

R

Rv

R

Rv

R

Rv

R

Rv 4321

f4

4

f3

3

f2

2

f1

1

fo +++=+++=

to becomek12

R

4

1)vvvv(

4

1-v f4321o =+++=

==4

k12R f 3 k

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Problem 5.10 Express outV in terms of 1V and 2V for the circuit shown in Figure 5.1.What have we done here?

Figure 5.1


0k10

VV

k10

VV

k10

VV outa2a1a =

+

+

where 0VV ba == is the constraint equation.

0k10

V-

k10

V-

k10

V- out21 =++

21out VV-V ==outV )VV(- 21 ++++

We have constructed an inverting, summing amplifier.

DIFFERENCE AMPLIFIER

A difference amplifier amplifies the difference between two inputs but rejects any signals

common to the two inputs.

Problem 5.11 Using an operational amplifier, can we construct a circuit where12out VVV = ?

Yes, we want to construct what is called a difference amplifier. We can do this using the circuit

shown in Figure 5.1.

10 k

10 k

+

V2

Va

Vb

+ +

Vout

RL

10 k

+

V1

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Figure 5.1

Now, verify that this circuit will amplify the difference of the two inputs.


0R

VV

R

VV

2

outa

1

1a =

+

and 0R

0V

R

VV

4

b

3

2b =

+

where ba VV = is the constraint equation.

Simplifying,

out1

1

2a

1

2 VVR

RV1

R

R=

+ and 2

43

4b V

RR

RV

+=

Using the constraint equation to combine the two equations yields

1

1

22

43

4

1

2out V

R

RV

RR

R1

R

RV

+

+=

or

1

1

22

43

21

1

2out V

R

RV

RR1

RR1

R

RV

++

=

When4

3

2

1

R

R

R

R= , )VV(

R

RV 12

1

2

out =

If 21 RR = and 43 RR = ,

12out VVV =

This was the desired case.

+

Vout

R2

R3

+

V2

Va

Vb

+

RL

R1

+

V1

R4

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Problem 5.12 [5.39] Design a difference amplifier to have a gain of 2 and a commonmode input resistance of 10 k at each input.

The input resistances are

== k10RR 31

For a gain of 2,

=== k20R2R2R

R12

1

2

A property of difference amplifiers is

4

3

2

1

R

R

R

R=

Thus,

== k20RR 24

Now, verify the results,

1

1

2

2

43

21

1

2

o vR

Rv

)RR1(

)RR1(

R

Rv

++

=

12o vk10

k20v

5.01

5.01

k10

k20v

++

=

)vv(2v 12o =

This is the desired result. Therefore,== 31 RR k10 == 42 RR k20

CASCADED OPERATIONAL AMPLIFIER CIRCUITS

A cascade connection is a head-to-tail arrangement of two or more op amp circuits such that the

output of one op amp circuit is the input to the next op amp circuit.

Problem 5.13 [5.45] Refer to the circuit in Figure 5.1. Calculate oi if:(a) mV12vs =(b) mV)t377cos(10vs =

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Figure 5.1

This is a cascading system of two inverting amplifiers.

sso v6v6

12-

4

12-v =

=

s3-

3

s

o v103102

vi =

=

(a) When mV12vs = ,=oi A36

(b) When mV)t377cos(10vs ==oi A)t377cos(30

OPERATIONAL AMPLIFIER CIRCUITS WITH PSPICE

Problem 5.14 Solve Problem 5.1 using PSpice.PSpice does not perform symbolic simulations. So, let V1Vs = . Add a VIEWPOINT to thecircuit to indicate the output voltage. Set the load resistor to the desired value, save the schematic

and simulate. With the repetition of setting the load resistor and simulating the circuit, the output

voltage for each load resistor in Problem 5.1 can be verified.

12 k

6 k

+

+

vs 2 k

io

12 k

4 k +

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From the schematic,

for = 1R L , =outV V99692283.1-

By changing the load resistor and simulating the circuit, it can be shown that

for = 100R L , =outV V99990789.1-

for = k1R L , =outV V99993507.1-

for = k10R L , =outV V99993779.1-

for = k100R L , =outV V99993806.1-

This matches the answers obtained in Problem 5.1 when V1Vs = .

Problem 5.15 Solve Problem 5.7 using PSpice.Consider the following schematic.

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Because PSpice does not perform symbolic simulations, let V1Vs = . We also need to choosevalues for the resistors.

For the circuit above, let

=sV V1 , =1R k1 , and =FR k10

which produces,

=== )1(1k

10k-V

R

R-V s

1

F

o V10-

This verifies the answer obtained in Problem 5.7.

One must realize that Problem 5.7 was performed assuming an ideal op amp; PSpice does not

simulate an ideal op amp. Thus, the output voltage may not be an integer value even though thecalculations from Problem 5.7 would predict an integer value.

Also, the output voltage cannot be greater than V+ or less than V, where V+ and V are thepower supply voltages of the op amp.

There are three modes in which real op amps can operate. The most desirable is to have them

give the desired output. The second mode is when the op amp goes into saturation, reaching its

maximum output voltage and remaining there. The third mode is that the op amp can act like anoscillator; its output voltage can be some type of periodic signal such as a sine wave.

Problem 5.16 Solve Problem 5.8 using PSpice.Consider the following schematic.

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For the circuit above,

=sV V1 , =1R k1 , and =FR k10and

=

+=

+= )1(1

1k

10kV1

R

RV s

1

F

o V11

This verifies the answer obtained in Problem 5.8

Also, see the comments made concerning ideal versus real op amps in Problem 5.15.

APPLICATIONS

Problem 5.17 Use an operational amplifier to change an ideal voltage source to an idealcurrent source.

Consider the following circuit.

Clearly,

1

in

1 R

Vi =

In the circuit above, the op amp will maintain the current through the black box in the

feedback path at i1. Thus, the op amp is working like an ideal current source.

R1

+

Vin

+ +

Vout

?

i1

i1

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