© 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-1 Electronics Principles & Applications Eighth Edition Chapter 9 Operational

© 2013 The McGraw-Hill Companies, Inc. All rights reserved.McGraw-Hill

9-1

ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsEighth EditionEighth Edition

Chapter 9Operational Amplifiers

Charles A. Schuler


9-2

• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators

INTRODUCTION


9-3

Noninverted outputInverted output

A differential amplifier driven at one input

C

BE

C

BE

+VCC

-VEE


9-4

Both outputs are active because Q1 drives Q2.

C

BE

C

BE

+VCC

-VEE

Q1 Q2

Q2 serves as a common-base

amplifier in this mode. It’s driven

at its emitter.

Q1 serves as an emitter-followeramplifier in this mode to drive Q2.


9-5

Reduced outputReduced output

A differential amplifier driven at both inputs

C

BE

C

BE

+VCC

-VEE

Common mode input signal


9-6

A differential amplifier driven at both inputs with a common-mode signal shows low gain (usually a loss) because the total emitter current is fairly constant.

C

BE

C

BE

+VCC

-VEE

If the input signalgoes positive, bothtransistors want toincrease theircurrent but can’t.

Constanttotal

current


9-7

Increased outputIncreased output

Driven at both inputs with a differential signal

C

BE

C

BE

+VCC

-VEE


9-8

A differential amplifier driven at both inputs with a differential signal shows high gain.

+VCCHere, one transistorincreases its currentas the other decreasesso the constant totalcurrent is not a limitingfactor. C

BE

C

BE

-VEE

Constanttotal

current


9-9

The amplifier has two gains:

• High for differential signals• Low for common-mode signals

The ratio of the two gains is called the common-moderejection ratio (CMRR) and is perhaps the most important feature of this amplifier.

CMRR = 20 x logAV(DIF)

AV(CM)


9-10

Differential amplifier dc analysis

C

BE

C

BE

+9 V

-9 V

3.9 k

4.7 k4.7 k

10 k10 k

RE

RL

RBRB

RL

VEE

VCC

IRE =

VEE - VBE

RE

9 V - 0.7 V

3.9 k= = 2.13 mA

IE =IRE

2= 1.06 mA

IC = IE = 1.06 mA

VRL = IC x RL

= 1.06 mA x 4.7 k= 4.98 V

VCE = VCC - VRL - VE

= 9 - 4.98 -(-0.7)

= 4.72 V


9-11

Differential amplifier dc analysis continued

C

BE

C

BE

+9 V

-9 V

3.9 k

4.7 k4.7 k

10 k10 k

RE

RL

RBRB

RL

VEE

VCC

Assume = 200

IB =IC

1.06 mA

=

= 5.3 A

VB = VRB = IB x RB

= 5.3 A x 10 k

= 53 mV


9-12

Differential amplifier ac analysis

C

BE

C

BE

+9 V

-9 V

3.9 k

4.7 k4.7 k

10 k10 k

RE

RL

RBRB

RL

VEE

VCC

rE =50 mV

IE

=50 mV

1.06 mA= 47 (50 mV is conservative)

AV(DIF) = RL

2 x rE

AV(CM) = RL

2 x RE

= 504.7 k

2 x 47 =

= 0.6

4.7 k2 x 3.9 k

=


9-13

Differential amplifier ac analysis continued

C

BE

C

BE

+9 V

-9 V

3.9 k

4.7 k4.7 k

10 k10 k

RE

RL

RBRB

RL

VEE

VCC

CMRR = 20 x logAV(DIF)

AV(CM)

= 20 x log500.6

= 38.4 dB


9-14

A current source can replace RE to decrease the common mode gain.

C

BE

C

BE

4.7 k4.7 k

10 k10 k

RL

RBRB

RL

VCC

2 mA*

*NOTE: Arrow shows conventional current flow.

AV(CM) = RL

2 x RE

Replaces thiswith a very highresistance value.


9-15

A practical current source

390

5.1 V2.2 k

-9 V

IC = IE = 2 mA

IC

IZ = 9 V - 5.1 V

390 = 10 mA

IE = = 2 mA5.1 V - 0.7 V

2.2 k


9-16

6.3 V60 Hz

212 mV1 kHz

The amplitude of thecommon-mode signalis almost 30 times the

amplitude of thedifferential signal.

A demonstration of common-mode rejection

The common-mode signalcannot be seen in the output.


9-17

Differential amplifier quiz

When a diff amp is driven at one input,the number of active outputs is _____. two

When a diff amp is driven at both inputs, thereis high gain for a _____ signal. differential

When a diff amp is driven at both inputs, thereis low gain for a ______ signal. common-mode

The differential gain can be found by dividingthe collector load by ________. 2rE

The common-mode gain can be found by dividingthe collector load by ________. 2RE


9-18

Invertinginput

Non-invertinginput

Output

Op amps have two inputs


9-19

Op-amp Characteristics

• High CMRR• High input impedance• High gain• Low output impedance

• Available as ICs• Inexpensive• Reliable• Widely applied


9-20

Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT.

-VEE

+VCC

VOUT

With both inputs grounded through equal resistors, VOUT should be zero volts.


9-21

V

t

Vt

Slew rate =

The output of an op amp cannot change instantaneously.

741

0.5 Vs


9-22

Slew-rate distortion

fMAX = Slew Rate

2 x VP

f > fMAX

VP


9-23

Operational amplifier quiz

The input stage of an op amp is a__________ amplifier. differential

Op amps have two inputs: one is invertingand the other is ________. noninverting

An op amp’s CMRR is a measure of its abilityto reject a ________ signal. common-mode

The offset null terminals can be used to zeroan op amp’s __________. output

The ability of an op amp output to changerapidly is given by its _________. slew rate


9-24

RL

Op-amp follower

AV(OL) = the open loop voltage gain

AV(CL) = the closed loop voltage gain

This is a closed-loopcircuit with a voltage

gain of 1.

It has a high input impedanceand a low output impedance.


9-25

RL

Op-amp follower

AV(OL) = 200,000

AV(CL) = 1

The differential inputapproaches zero dueto the high open-loop

gain. Using this model,VOUT = VIN.

VIN

VOUT

VDIF = 0


9-26

RLVIN

VOUT

Op-amp follower

AV(OL) = 200,000

B = 1

The feedback ratio = 1

200,000

(200,000)(1) + 1 1AV(CL) =

AB +1AVIN VOUT


9-27

RLVIN

VOUT

The closed-loop gain is increased by decreasing the feedback with a voltage divider.

RF

R1

200,000

(200,000)(0.091) + 1= 11AV(CL) =

B =R1

RF + R1

100 k

10 k 10 k100 k+ 10 k

=

= 0.091


9-28

RLVIN

VOUT

RF

100 k10 k

VDIF = 0

It’s possible to develop a different model for the closed loop gain

by assuming VDIF = 0.

VIN = VOUT xR1

R1 + RF

=VOUT

VIN

1 +RF

R1

Divide both sides by VOUT and invert:

AV(CL) = 11

R1


9-29

RLVIN VOUT

RF

10 k1 k

VDIF = 0R1

In this amplifier, the assumption VDIF = 0 leads to the conclusion that the inverting op amp terminal is

also at ground potential. This is called a virtual ground.

Virtual ground We can ignore the op amp’s inputcurrent since it is so small. Thus:

IR1 = IRF

VIN

R1

=-VOUT

RF

VOUT

VIN

=-RF

R1

= -10

By Ohm’s Law:

The minus sign designates an inverting amplifier.


9-30

VIN

RF

10 k1 k

VDIF = 0

R1

Virtual ground

Due to the virtual ground, the input impedance of the inverting amplifier is equal to R1.

R2 = R1 RF = 910

Although op amp inputcurrents are small, in

most applications, offseterror is minimized by

providing equal resistance paths for the input

currents.

This resistor reduces offset error.


9-31

Output

A typical op amp has internal frequency

compensation.

Break frequency:

fB = 2RC1

R

C


9-32

100 k10 k1 10 100 1k 1M0

20

80

40

60

100

120

Frequency in Hz

Gain in dB

Bode plot of a typical op amp

Break frequency


9-33

RLVIN

VOUT

RF

100 k1 k

Op amps are typically operated with negative feedback(closed loop). This increases their useful frequency range.

R1

=VOUT

VIN

1 +RF

R1

AV(CL) =

= 1 +100 k1 k

= 101

dB Gain = 20 x log 101 = 40 dB


9-34

100 k10 k1 10 100 1k 1M0

20

80

40

60

100

120

Frequency in Hz

Gain in dB

Using the Bode plot to find closed-loop bandwidth:

Break frequency

AV(CL)


9-35

There are two frequency limitations:Slew rate determines the large-signal bandwidth.

Internal compensation sets the small-signal bandwidth.

0.5 Vs

70 Vs

A 741 op amp slews at A 318 op amp slews at


9-36

100 k10 k1 10 100 1k 1M0

20

80

40

60

100

120

Frequency in Hz

Gain in dB

The Bode plot for a fast op amp showsincreased small-signal bandwidth.

10M

fUNITY


9-37

RLVIN

VOUT

RF

100 k1 k

fUNITY can be used to find the small-signal bandwidth.

R1

=VOUT

VIN

1 +RF

R1

AV(CL) =

= 1 +100 k1 k

= 101

318 Op amp

fB = fUNITY

AV(CL)

10 MHz

101= 99 kHz=


9-38

Op amp feedback quiz

The open loop gain of an op amp is reducedwith __________ feedback negative

The ratio RF/R1 determines the gain of the___________ amplifier. inverting

1 + RF/R1 determines the gain of the___________ amplifier. noninverting

Negative feedback makes the - input of theinverting circuit a ________ ground. virtual

Negative feedback _________ small signalbandwidth. increases


9-39

R

C

Amplitude responseof an RC lag circuit

0 dB

-20 dB

-40 dB

-60 dB

10fbfb 100fb 1000fb

fb = RC1

Vout

Vout

f


9-40

0o

0.1fb fb 10fb

Phase response ofan RC lag circuit

-90o

-45o

R

C

R

-XC = tan-1

Vout

Vout

f


9-41

Interelectrode capacitance and Miller effect

CBECMiller

CBE

CBC

R

CMiller = AVCBC

CInput = CMiller + CBE

The gain frombase to collector

makes CBC

effectively largerin the input circuit.

fb = RCInput

1


9-42

10 Hz 100 Hz 1 kHz 10 kHz 100 kHz

50 dB

40 dB

30 dB

20 dB

10 dB

0 dB

Bode plot of an amplifier with two break frequencies.

20 dB/decade

40 dB/decade

fb1 fb2


9-43

0o

Multiple lag circuits:

-180o

R1C1

Vout

Vout

f

R2C2

R3C3

Phase reversal

Negative feedback becomes positive!


9-44

Op amp compensation

• Interelectrode capacitances create several break points.

• Negative feedback becomes positive at some frequency due to cumulative phase lags.

• If the gain is > 0 dB at that frequency, the amplifier is unstable.

• Frequency compensation reduces the gain to 0 dB or less.


9-45

Op amp compensation quiz

Beyond fb, an RC lag circuit’s output dropsat a rate of __________ per decade. 20 dB

The maximum phase lag for one RC networkis __________. 90o

An interelectrode capacitance can be effectivelymuch larger due to _______ effect. Miller

Op amp multiple lags cause negative feedbackto be ______ at some frequency. positive

If an op amp has gain at the frequency wherefeedback is positive, it will be ______. unstable


9-46

RF

10 k

1 k

1 kHz

3 kHz

3.3 k5 kHz

5 k

Summing Amplifier

Inverted sum of three sinusoidal signals

Amplifier scaling: 1 kHz signal gain is -103 kHz signal gain is -35 kHz signal gain is -2


9-47

RF

1 k

1 k 1 k

Subtracting Amplifier

Difference of twosinusoidal signals

(V1 = V2)

1 k

V1 V2

VOUT = V2 - V1

(A demonstration of common-mode rejection)


9-48

A cascade RC low-pass filter

An active low-pass filter

(A poor performer since later sections load the earlier ones.)

(The op amps provide isolation and better performance.)


9-49

Frequency in Hz

Am

pli

tud

e in

dB

0

-20

-40

-60

10 100

Cascade RC

Active filter


9-50

VIN

Active low-pass filterwith feedback VOUT

C1C2

At relatively low frequencies, Vout and Vin

are about the same. Thus, the signal voltageacross C1 is nearly zero. C1 has little effect

at these frequencies.

feedback


9-51

VIN

Active low-pass filterwith feedback VOUT

Frequency

Gain

fC

-3 dBFeedback canmake a filter’sperformanceeven better!

C1C2

As fIN increases and C2

loads the input, Vout

drops. This increasesthe signal voltageacross C1. This

sharpens the knee.


9-52

Frequency in Hz

Am

pli

tud

e in

dB

0

-20

-40

-60

10 100

Active filterusing feedback

(two stages)

Note the flat pass bandand the sharp knee.

The slope eventually reaches24 dB/octave or 80 db/decade

for all the filters (4 RC sections).


9-53

VIN

Active high-pass filter VOUT

Frequency

Gain

fC

-3 dB

feedback


9-54

VIN

Active band-pass filter(multiple feedback)

VOUT

Frequency

Gain

-3 dB

Bandwidth


9-55

VIN Active band-stop filter(multiple feedback)

VOUT

Frequency

Gain

-3 dB

Stopband


9-56

40 mV

0 V

56.6 mV

0 V

- 56.6 mV

Active rectifier


9-57

VIN

VOUT

Integrator

R

C

Slope = -VIN x1

RC

VsSlope =


9-58

Input Waveforms (Blue) Integrator Output Waveform (Red) Differentiator Output Waveform (Red)

Square

Triangle

Sine

A differentiator shows R and C reversed.

Vout = -(Vin/t)RC

Differentiation is the opposite of integration.


9-59

VIN

VOUT

0 V

1 V +VSAT

-VSAT

1 V

Comparator with a 1 Volt reference


9-60

VIN

VOUT

0 V

1 V +VSAT

-VSAT

1 V

Comparator with a noisy input signal


9-61

VINVOUT

+VSAT

-VSAT

Schmitt trigger with a noisy input signal

UTP

LTP

Hysteresis = UTP - LTPRF

R1

R1 + RF

R1VSAT x

Trip points:


9-62

VIN

VOUT

R2

R14.7 k

4.7 k

+5 V

3 V

1 V

Window comparator

311

311VUL

VLL VOUT is LOW (0 V) when VIN

is between 1 V and 3 V.


9-63

VIN

VOUT

+5 V

3 V

1 V

Window comparator

311

311VUL

VLL

Many comparator ICs require pull-up resistors in

applications of this type.


9-64

VIN

VOUT

R2

R14.7 k

4.7 k

+5 V

3 V

1 V

Window comparator

311

311VUL

VLL VOUT is TTL logic compatible.


9-65

Op amp applications quiz

A summing amp with different gains for theinputs uses _________. scaling

Frequency selective circuits using op ampsare called _________ filters. active

An op amp integrator uses a _________ asthe feedback element. capacitor

A Schmitt trigger is a comparator with__________ feedback. positive

A window comparator output is active whenthe input is ______ the reference points. between


9-66

REVIEW

• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators

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© 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-1 Electronics Principles & Applications Eighth Edition Chapter 9 Operational