CHAPTER 5 - OPERATIONAL AMPLIFIER

List of topics for this chapter :Operational AmplifiersIdeal Operational AmplifierInverting AmplifierNoninverting AmplifierSumming AmplifierDifference AmplifierCascaded Operational Amplifier CircuitsOperational Amplifier Circuits with PSpiceApplications

OPERATIONAL AMPLIFIERS

Problem 5.1 Calculate outV for ΩΩΩΩΩ= 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 (orsource) voltage.

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

100 kΩΩΩΩ 100k Vin

1 kΩΩΩΩ

+

Vs

−−−−

+

Vin

−−−−

−−−−+

50 ΩΩΩΩ

2 kΩΩΩΩ

RL

+

Vout

−−−−

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

Linoutinout =+++−

0V)1M4(VR

k241 inout

L

=−+

+

out

L

in V1M4

R

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 Ω= 1R L , =outV sV99692283.1-

1.5 MΩΩΩΩ 8 ×××× 104 vd

60 ΩΩΩΩ

−−−−

vd

+

+−−−−

For Ω= 100R L , =outV sV99990789.1-For Ω= k1R L , =outV sV99993507.1-For Ω= k10R L , =outV sV99993779.1-For Ω= 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 Ω= 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 Ω= 1R L , =outV sV99692283.1-

For Ω= 100R L , =outV sV99990789.1-

For Ω= k1R L , =outV sV99993507.1-

For Ω= k10R L , =outV sV99993779.1-

For Ω= 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 ΩΩΩΩM5.1

(b) =outR ΩΩΩΩ60

(c) 4108A ×=

)108(log20A 410dB ×==dBA 06.98

Problem 5.3 Solve for outV when ΩΩΩΩ= 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

=+

+

inoutL

L Vk2-VR50

50R=

+

inL

Lout V50R

Rk100-V

+

=

100 kΩΩΩΩ 100k Vin

1 kΩΩΩΩ

+

Vs

−−−−

+

Vin

−−−−

−−−−+

50 ΩΩΩΩ

RL

+

Vout

−−−−

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

SL

LS

L

Lout 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 Ω= 1R L , =outV SV3706.941,1-

For Ω= k1R L , =outV SV1438.295,94-

For Ω= k100R L , =outV SV4208.960,98-

For Ω= 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 value

of 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

Treat the operational amplifier as ideal. Due to infinite input resistance, we know that thecurrents into both input terminals are zero. The voltage across the terminals is negligiblysmall 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 circuitanalysis. 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 twoequations with four unknowns. Solving for oV in terms of sV (and the resistances) impliesthat we need one equation with two unknowns. Hence, we need a constraint equation toreduce 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, 0VV ba ==

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

Hence, s1

Fc VR

R-V =

Clearly, s1

Fco VR

R-VV ==

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

s1

Fo

F

o

1

s VR

R-V

R

V-

R

V=→=

RF

R1

+

Vs

−−−−

Va

Vb

−+ +

Vo

−−−−

RL

VcI

I

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 vv512

8

vv

2

v3−=→

−=

−

But V2vv ba ==

So, ov)2)(5(12 −==ov V2-

m5.0-0.5m4k

2-

k8

22-

k4

0v

k8

vvi oaoo −=+

−=

−+

−=

=oi mA1-

+

vo

−−−−

8 kΩΩΩΩ

5 kΩΩΩΩ −+

2 kΩΩΩΩ

10 kΩΩΩΩ+−

3 V 4 kΩΩΩΩ

iova

vb

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 kΩΩΩΩ

4 kΩΩΩΩ

va

vb−+

+−

+

vo

−−−−

vs

4 kΩΩΩΩ2 kΩΩΩΩ

+−

9 V

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

F VRR-

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−−

Using nodal analysis,

0R

VV

R

0V

F

outa

1

a =−+−

where sba VVV == is the constraint equation.

Simplifying,

0R

VV

R

V

F

outs

1

s =−

+

sF1

outF

VR

1

R

1V

R

1

+=

s1

Fout 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 of Figure 5.1, the relationshipbetween the output voltage, Vo, and the input voltage, Vs, is a simple ratio of RF and R1. Forthe circuit in Figure 5.1, however, the gain can never be less than one. Since there is rarelya 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 ofthe inputs.

Problem 5.9 [5.33] A four-input summing amplifier has ==== 4321 RRRRΩk12 . 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 kΩΩΩΩ

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

Using nodal analysis,

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 signalscommon to the two inputs.

Problem 5.11 Using an operational amplifier, can we construct a circuit where

12out VVV −= ?

Yes, we want to construct what is called a difference amplifier. We can do this using the circuitshown in Figure 5.1.

10 kΩΩΩΩ

10 kΩΩΩΩ

+

V2

−−−−

Va

Vb

−+ +

Vout

−−−−

RL

10 kΩΩΩΩ

+

V1

−−−−

Figure 5.1

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

Using nodal analysis,

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,

out11

2a

1

2 VVR

RV1

R

R =−

+ and 2

43

4b VRR

RV

+=

Using the constraint equation to combine the two equations yields

11

22

43

4

1

2out VR

RV

RR

R1

R

RV −

+

+=

or

11

22

43

21

1

2out VR

RV

RR1

RR1

R

RV −

++=

When 4

3

2

1

R

R

R

R = , )VV(R

RV 12

1

2out −=

If 21 RR = and 43 RR = ,

12out VVV −=

This was the desired case.

+

Vout

−−−−

R2

R3

+

V2

−−−−

Va

Vb

−+

RL

R1

+

V1

−−−−

R4

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

The input resistances areΩ== k10RR 31

For a gain of 2,

Ω==→= k20R2R2R

R12

1

2

A property of difference amplifiers is

4

3

2

1

R

R

R

R=

Thus,Ω== k20RR 24

Now, verify the results,

11

22

43

21

1

2o 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 ΩΩΩΩk10 == 42 RR ΩΩΩΩk20

CASCADED OPERATIONAL AMPLIFIER CIRCUITS

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

Figure 5.1

This is a cascading system of two inverting amplifiers.

sso v6v6

12-

4

12-v =

=

s3-

3s

o v103102

vi ×=

×=

(a) When mV12vs = ,=oi A36 µµµµ

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

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 schematicand simulate. With the repetition of setting the load resistor and simulating the circuit, the outputvoltage for each load resistor in Problem 5.1 can be verified.

12 kΩΩΩΩ

6 kΩΩΩΩ −+

+−

vs 2 kΩΩΩΩ

io

12 kΩΩΩΩ

4 kΩΩΩΩ −+

From the schematic,for Ω= 1R L , =outV V99692283.1-

By changing the load resistor and simulating the circuit, it can be shown thatfor Ω= 100R L , =outV V99990789.1-for Ω= k1R L , =outV V99993507.1-for Ω= k10R L , =outV V99993779.1-for Ω= 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.

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 ΩΩΩΩk1 , and =FR ΩΩΩΩk10which produces,

=== )1(1k

10k-V

R

R-V s

1

Fo 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 notsimulate 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 themgive the desired output. The second mode is when the op amp goes into saturation, reaching itsmaximum 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.

For the circuit above,=sV V1 , =1R ΩΩΩΩk1 , and =FR ΩΩΩΩk10

and

=

+=

+= )1(1

1k

10kV1

R

RV s

1

Fo 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

in1 R

Vi =

In the circuit above, the op amp will maintain the current through the black box in thefeedback path at i1. Thus, the op amp is working like an ideal current source.

R1

+

Vin

−−−−

−+ +

Vout

?

i1

i1

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PrefaceChapter 1 Basic ConceptsChapter 2 Basic LawsChapter 3 Methods of AnalysisChapter 4 Circuit TheoremsChapter 5 Operational AmplifierOperational AmplifiersIdeal Operational AmplifierInverting AmplifierNoninverting AmplifierSumming AmplifierDifference AmplifierCascaded Operational Amplifier CircuitsOperational Amplifier Circuit Analysis With PSpiceApplications

Chapter 6 Capacitors and InductorsChapter 7 First-Order CircuitsChapter 8 Second-Order CircuitsChapter 9 Sinusoids and PhasorsChapter 10 Sinusoidal Steady-State AnalysisChapter 11 AC Power AnalysisChapter 12 Three-Phase CircuitsChapter 13 Magnetically Coupled CircuitsChapter 14 Frequency ResponseChapter 15 Laplace TransformChapter 16 Fourier SeriesChapter 17 Fourier TransformChapter 18 Two-Port Networks

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