Electronics Principles & Applications Fifth Edition Chapter 9 Operational Amplifiers ©1999...

ElectronicsElectronics

Principles & ApplicationsPrinciples & ApplicationsFifth EditionFifth Edition

Chapter 9Operational Amplifiers

©1999 Glencoe/McGraw-Hill

Charles A. Schuler

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

INTRODUCTION

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

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

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

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

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

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

Op-amp follower

AV(OL) = 200,000

B = 1

The feedback ratio = 1

200,000

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

AB +1AVIN VOUT

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.

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 amp

Break 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

Gain in dB

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

10M

fUNITY

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 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

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

Bode plot of an amplifier with two break frequencies.

20 dB/decade

40 dB/decade

fb1 fb2

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 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

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)

VIN

Active low-pass filter VOUT

Frequency

Gain

fC

-3 dB

VIN

Active high-pass filter VOUT

Frequency

Gain

fC

-3 dB

VIN

Active band-pass filter

VOUT

Frequency

Gain

-3 dB

Bandwidth

VIN Active band-stop filter

VOUT

Frequency

Gain

-3 dB

Stopband

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 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

REVIEW

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