ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition

Chapter 9Operational Amplifiers

(student version)

©2003 Glencoe/McGraw-Hill

Charles A. Schuler

• The Differential Amplifier

• The Operational Amplifier

• Determining Gain

• Frequency Effects

• Applications

• Comparators

INTRODUCTION

Dear Student:

This presentation is arranged in segments. Each segment is preceded by a Concept Preview slide and is followed by a Concept Review slide. When you reach a Concept Review slide, you can return to the beginning of that segment by clicking on the Repeat Segment button. This will allow youto view that segment again, if you want to.

Concept Preview

• Differential amplifiers always have two inputs.

• Differential amplifiers can have one or two outputs.

• Driving one input provides a difference signal. Both outputs will be active and will be out of phase with each other.

• Driving both inputs with the same signal results in reduced output.

• Driving both inputs with a difference signal results in increased output.

Noninverted outputInverted output

A differential amplifier driven at one input

C

BE

C

BE

+VCC

-VEE

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.

Reduced outputReduced output

A differential amplifier driven at both inputs

C

BE

C

BE

+VCC

-VEE

Common mode input signal

Increased outputIncreased output

A differential amplifier driven at both inputs

C

BE

C

BE

+VCC

-VEE

Differential mode input signal

Concept Review• Differential amplifiers always have two inputs.

• Differential amplifiers can have one or two outputs.

• Driving one input provides a difference signal. Both outputs will be active and will be out of phase with each other.

• Driving both inputs with the same signal results in reduced output.

• Driving both inputs with a difference signal results in increased output.

Repeat Segment

Concept Preview• The current in the emitter resistor divides

equally between the two transistors in a differential amp.

• The differential gain is determined by the collector load and the ac emitter resistance.

• The common mode gain is determined by the collector load and the emitter resistor.

• The ratio of the differential gain to the common mode gain is called the CMRR.

• The CMRR is greatly improved by using a current source in the emitter circuit.

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

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

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

=

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

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.

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

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.

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, there is high gain for a _____ signal. differential

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

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

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

Concept Review• The current in the emitter resistor divides

equally between the two transistors in a differential amp.

• The differential gain is determined by the collector load and the ac emitter resistance.

• The common mode gain is determined by the collector load and the emitter resistor.

• The ratio of the differential gain to the common mode gain is called the CMRR.

• The CMRR is greatly improved by using a current source in the emitter circuit.

Repeat Segment

Concept Preview

• Operational amplifiers have one output and two inputs: inverting and non-inverting.

• Some op amps have offset null terminals which can be used to zero the dc output.

• The output of an op can change no faster than its slew rate.

• Slew rate is specified in volts per microsecond.

• The slew rate and the amplitude of the output signal determine the power bandwidth of an op amp.

Invertinginput

Non-invertinginput

Output

Op amps have two inputs

Op-amp Characteristics

• High CMRR

• High input impedance

• High gain

• Low output impedance

• Available as ICs

• Inexpensive

• Reliable

• Widely applied

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.

V

t

Vt

Slew rate =

The output of an op amp cannot change instantaneously.

741

0.5 Vs

Slew-rate distortion

fMAX = Slew Rate

2 x VP

f > fMAX

VP

fMAX is known as the power bandwidth.

Operational Amplifier Quiz

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

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

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

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

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

Concept Review• Operational amplifiers have one output and two

inputs: inverting and non-inverting.

• Some op amps have offset null terminals which can be used to zero the dc output.

• The output of an op can change no faster than its slew rate.

• Slew rate is specified in volts per microsecond.

• The slew rate and the amplitude of the output signal determine the power bandwidth of an op amp.

Repeat Segment

Concept Preview• An op amp follower has a closed loop gain of 1.

• The input and output signals are in-phase in a follower amplifier.

• The closed loop gain can be increased by decreasing the feedback ratio.

• The input and output signals are out of phase in an inverting amplifier.

• The – terminal of an inverting amplifier acts as a virtual ground.

• The input impedance of an inverting amplifier is equal to the input resistor.

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.

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

RLVIN

VOUT

AV(OL) = 200,000

B = 1

The feedback ratio = 1

200,000

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

AB +1AVIN VOUT

Op-amp Follower

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

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

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.

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

some applications, offseterror is minimized by

providing equal paths forthe input currents.

This resistor reduces offset error.

Concept Review• An op amp follower has a closed loop gain of 1.• The input and output signals are in-phase in a

follower amplifier.• The closed loop gain can be increased by

decreasing the feedback ratio.• The input and output signals are out-of-phase

in an inverting amplifier.• The – terminal of an inverting amplifier acts

as a virtual ground.• The input impedance of an inverting amplifier

is equal to the input resistor.

Repeat Segment

Concept Preview• Most op amps have built-in frequency

compensation.

• The internal frequency compensation produces a break frequency of 10 Hz or so.

• The closed loop small signal bandwidth is greater than the break frequency.

• A Bode plot can be used to determine the small signal bandwidth of a closed loop amplifier.

• The gain-bandwidth product can also be used to determine the closed loop small signal bandwidth.

Output

A typical op amp has internal frequency

compensation.

Break frequency:

fB = 2RC1

R

C

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 AmpBreak frequency

RLVIN

VOUT

RF

100 k1 k

Op amps are usually 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

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)

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

100 k10 k1 10 100 1k 1M0

20

80

40

60

100

120

Frequency in Hz

Gai

n in

dB

The Bode plot for a fast op ampshows increased bandwidth.

10M

fUNITY

fUNITY is alsocalled the

gain-bandwidthproduct.

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=

Op Amp Feedback Quiz

The open loop gain of an op amp is reduced with __________ 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 the inverting circuit a ________ ground. virtual

Negative feedback _________ small signal bandwidth.

increases

Concept Review• Most op amps have built-in frequency

compensation.

• The internal frequency compensation produces a break frequency of 10 Hz or so.

• The closed loop small signal bandwidth is greater than the break frequency.

• A Bode plot can be used to determine the small signal bandwidth of a closed loop amplifier.

• The gain-bandwidth product can also be used to determine the closed loop small signal bandwidth.

Repeat Segment

Concept Preview

• The amplitude response of an RC lag network is –20 dB per decade beyond the break frequency.

• The phase response of an RC lag network is –45 degrees at the break frequency.

• The Miller effect makes some interelectrode capacitances appear to be larger.

• Multiple lag networks inside an op amp make negative feedback become positive at some frequency. Frequency compensation insures that the gain is less than 0 dB at that frequency.

R

C

Amplitude Responseof RC Lag Circuit

0 dB

-20 dB

-40 dB

-60 dB

10fbfb 100fb 1000fb

fb = RC1

Vout

Vout

f

0o

0.1fb fb 10fb

Phase Responseof RC Lag Circuit

-90o

-45o

R

C

R

-XC = tan-1

Vout

Vout

f

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

10 Hz 100 Hz 1 kHz 10 kHz 100 kHz

50 dB

40 dB

30 dB

20 dB

10 dB

0 dB

20 dB/decade

40 dB/decade

fb1 fb2

Bode Plot of an Amplifier with Two Break Frequencies

0o

Multiple Lag Circuits:

-180o

R1C1

Vout

Vout

f

R2C2

R3C3

Phase reversal

Negative feedback becomes positive!

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.

Op Amp Compensation Quiz

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

The maximum phase lag for one RC network is __________. 90o

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

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

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

Concept Review• The amplitude response of an RC lag network is

–20 dB per decade beyond the break frequency.

• The phase response of an RC lag network is –45 degrees at the break frequency.

• The Miller effect makes some interelectrode capacitances appear to be larger.

• Multiple lag networks inside an op amp make negative feedback become positive at some frequency. Frequency compensation insures that the gain is less than 0 dB at that frequency.

Repeat Segment

Concept Preview• Op amps can be used to sum (add) two or

more signals.

• Scaling in a summing amp provides different gain for each signal.

• Op amps can be used to subtract two signals.

• Cascade RC filters have relatively poor performance.

• Active filters combine op amps with RC networks.

• Feedback in an op amp active filter sharpens the knee of the frequency response curve.

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

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)

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.)

Frequency in Hz

Am

pli

tud

e in

dB

0

-20

-40

-60

10 100

Cascade RC

Active filter

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 effectat these frequencies.

feedback

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.

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).

Concept Review• Op amps can be used to sum (add) two or more

signals.

• Scaling in a summing amp provides different gain for each signal.

• Op amps can be used to subtract two signals.

• Cascade RC filters have relatively poor performance.

• Active filters combine op amps with RC networks.

• Feedback in an op amp active filter sharpens the knee of the frequency response curve.

Repeat Segment

Concept Preview• Other active filters include high-pass, band-

pass and band-stop.

• An active rectifier will work with millivolt level signals.

• The output slope of an op amp integrator is equal to the dc input voltage times the reciprocal of the time constant.

• Comparators can be used to change analog waveforms to digital waveforms.

• A Schmitt trigger uses positive feedback to produce hysteresis and noise immunity.

VIN

Active high-pass filter VOUT

Frequency

Gain

fC

-3 dB

feedback

VIN

Active band-pass filter(multiple feedback)

VOUT

Frequency

Gain

-3 dB

Bandwidth

VIN Active band-stop filter(multiple feedback)

VOUT

Frequency

Gain

-3 dB

Stopband

40 mV

0 V

56.6 mV

0 V

- 56.6 mV

Active rectifier

VIN

VOUT

Integrator

R

C

Slope = -VIN x1

RC

VsSlope =

VIN

VOUT

0 V

1 V +VSAT

-VSAT

1 V

Comparator with a 1 Volt Reference

VIN

VOUT

0 V

1 V +VSAT

-VSAT

1 V

Comparator with a Noisy Input Signal

VINVOUT

+VSAT

-VSAT

Schmitt Trigger with a Noisy Input Signal

UTP

LTP

Hysteresis = UTP - LTPRF

R1

R1 + RF

R1VSAT x

Trip points:

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.

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.

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.

Op Amp Applications Quiz

A summing amp with different gains for the inputs uses _________. scaling

Frequency selective circuits using op amps are called _________ filters. active

An op amp integrator uses a _________ as the 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

Concept Review• Other active filters include high-pass, band-

pass and band-stop.

• An active rectifier will work with millivolt level signals.

• The output slope of an op amp integrator is equal to the dc input voltage times the reciprocal of the time constant.

• Comparators can be used to change analog waveforms to digital waveforms.

• A Schmitt trigger uses positive feedback to produce hysteresis and noise immunity.

Repeat Segment

REVIEW

• The Differential Amplifier

• The Operational Amplifier

• Determining Gain

• Frequency Effects

• Applications

• Comparators