ELL 100 - Introduction to Electrical Engineering

LECTURE 26: OPERATIONAL AMPLIFIERS

Introduction

Important Specifications/Characteristics

Equivalent Circuit

Ideal and Practical/Non-Ideal Op-amp

Common ICs and Pin Configurations

Inverting and Non-Inverting Amplifier

2

Outline

BASIC CONCEPTS

3

Amplifier: Electronic circuit that produces an output quantity

(voltage/current) in linear proportion to the input quantity.

Op-amp: Operational amplifier, a high-gain amplifier with an output

that corresponds to the difference between two input signals.

Vout = A(V+ - V-), A ~ 105

Integrated Circuit (IC): Collection of semiconductor electronic devices

(diodes, transistors) combined with other circuit elements (R, L, C) printed

in a single chip.

A

4

REAL LIFE APPLICATIONS

Microphone Amplifier

Digital to Analog Converter

5

REAL LIFE APPLICATIONS

Sensors e.g. Electronic Thermometer

6

REAL LIFE APPLICATIONS

Automatic Light Operated Switch DC Volt Polarity Meter

7

REAL LIFE APPLICATIONS

Control of Motors

8

REAL LIFE APPLICATIONS

Music Players

9

REAL LIFE APPLICATIONS

Analog Computer

10

REAL LIFE APPLICATIONS

Current and Voltage Regulator

11

REAL LIFE APPLICATIONS

Waveform Generator

HISTORY OF OP-AMP

12

Harold S. Black develops the feedback amplifier for

the Western Electric Company

The First Op-Amp: Designed by Karl Swartzel

at Bell Labs

Loebe Julie then develops an Op-Amp with

two inputs: Inverting and Non-inverting

Advent of solid-state

(semiconductor) electronics

Bipolar junction transistors

1920-

1930

1930–

1940

1950-

1960

1940-

1950

Vacuum-tube

based

electronic

circuits

HISTORY OF OP-AMP

13

Beginning of the Solid State Op-Amp, GAP/R P45

The GAP/R PP65 makes the Op-Amp into a circuit

component as a potted module

Robert J.Widlar develops the μA702 Monolithic

IC Op-Amp and shortly after the μA709

National Semiconductors: The LM101

and then the LM101A (both by Widlar)

Fairchild Semiconductors:

The “famous” μA741 (by Dave

Fullager) and then the μA748

1960-

1961

1962

1963

1967-

1968

1968-

1969

Integrated

circuits (ICs)

OP-AMP INTRODUCTION

14

• Multi-stage high-gain amplifier having a differential input and a

single-ended output that draws power from an external supply voltage.

• Contains a number of transistor-based differential amplifier stages to

achieve a very high voltage gain (~105).

• Contains several transistors, resistors, a few capacitors and diodes in

it’s internal circuitry

OP-AMP INTRODUCTION

15

Differential Amplifier: Basic unit of the op-amp is a differential amplifier.

A number of differential amplifiers are connected in cascade to form op-amp.

Vout = Gv(V1 – V2)

OP-AMP INTRODUCTION

16

Op-amp Basic Circuit

OP-AMP INTERNAL CIRCUIT

17

The op-amp internal circuit can be divided into 3 stages:

(a) Input Stage

The function of the input stage is to amplify the input difference, Vp− Vn,

and convert it to a single-ended signal.

(b) Second Stage

It further amplifies the signal and provides frequency compensation via

the capacitor, CC

(c) Output Stage

The output stage provides output current drive capability.

OP-AMP CHARACTERISTICS

18

Differential mode operation:

Vo = AdVi

Ad typically very large

Common mode operation:

Vo = AcVi

Ac << Ad

Vo

Vi1 Vi2

Vd

Vo

Vi

OP-AMP CHARACTERISTICS

19

Output voltage Vo = AdVd + AcVc

Vd = (Vi1 – Vi2) , Vc = (Vi1 + Vi2)/2

Ad >> Ac

Common mode rejection

• The common signal is rejected while the difference of the signals is amplified.

• Noise (any unwanted input signal) is common to both inputs, and hence

is attenuated via the differential connection.

• This feature is known as common mode rejection ratio (CMRR).

Vo

Vi1 Vi2

Vd

OP-AMP CHARACTERISTICS

20

Common Mode Rejection Ratio (CMRR)

The ratio of the differential gain to the common mode gain yields the

common mode rejection ratio. Ideally CMRR should be infinite.

CMRR = Ad / Ac

CMRR (dB) = 20 log10 (Ad / Ac)

It is a measure of how well the op-amp suppresses identical

signals on the inputs relative to differential input signals.

NUMERICAL EXAMPLE 1

21

Problem: An op-amp with a differential gain of Ad = 4000 is supplied with

input voltages of Vi1 = 150 µV and Vi2 = 140 µV.

Determine the output voltage if the value of CMRR is: (a) 100 (b) 105

Soln: Differential voltage is given by

Common voltage is given by

1 2 (150 140) 10d i iV V V V V

1 2150 140

1452 2

i ic

VV VV V

22

The output voltage is given by

=>

(a) CMRR = 100

(b) CMRR = 105

1 c co d d c c d d

d d

A VV A V A V A V

A V

11 c

o d d

d

VV A V

CMRR V

1 1451 4000*10 1 45.8

100*10

co d d

d

VV A V mV

CMRR V

5

1 1451 4000*10 1 40.006

10 *10

co d d

d

VV A V mV

CMRR V

Problem: Calculate the CMRR in dB for the op-amp below

Differential

Mode

Common

Mode

NUMERICAL EXAMPLE 2

23

NUMERICAL EXAMPLE 2

24

Soln: The differential gain is given by

Common mode gain is given as

CMRR:

CMRR (dB):

88000

1

od

d

VA

V m

1212

1

oc

c

V mA

V m

8000666.7

12

d

c

A

A

1020log 56.48d

c

ACMRR dB

A

OP-AMP CHARACTERISTICS

25

Slew Rate: Maximum rate of change of output voltage vs time

Let the signal be a sine wave

The rate of change of signal w.r.t time is

Max. rate of change

Slew rate required =

fmax is the highest signal frequency and Vp is the maximum output voltage

required to be supported by the op-amp.

( ) sin 2v t K ft

2 cos2dv

fK ftdt

2dv

fKdt

max2 pf V

NUMERICAL EXAMPLE 1

26

Problem: For an op-amp having a slew rate of SR = 2 V/s, what is the

maximum closed-loop voltage gain that can be used when the input signal

varies by 0.5 V in 10 s?

Soln: For voltage gain A, =>

=>

o iV AV o iV VA

t t

240

0.510

o

i i

V

SRtAV V

t t

NUMERICAL EXAMPLE 2

27

Problem: Determine the maximum frequency for an input a.c. signal

of 0.02 V peak that may be amplified without any distortion using an

op-amp with slew rate SR = 0.5 V/s and closed-loop voltage gain of 24.

Soln: Peak output voltage is given by

The max signal frequency is given by

24(0.02) 0.48VoV

3

maxf 175 102 o

SRHz

V

OP-AMP CHARACTERISTICS

28

Input Bias Current: The average magnitude of the two base currents at

the input terminals with the output at a specified level.

2

IB IBIB

I II

Input bias current is a

problem as it flows into

external impedances and

produces d.c. offset voltages,

which add to system errors.

Typically, IIB ~ 50 fA – 10 μA

for low - high speed op amps.

OP-AMP CHARACTERISTICS

29

Input Offset Current: The difference between the base currents into the

two input terminals with the output at a specified level.

It is because of an imbalance between the two input terminals e.g. due to

slight differences in transistor characteristics or biasing elements.

IIO = IIB+ − IIB

−

e.g. For an input offset current IIO = 5 nA and input bias current IIB = 30 nA,

the base currents at the two input terminals will be

30 5 / 2 32.52

IOIB IB

II I nA 30 5 / 2 27.5

2

IOIB IB

II I nA

OP-AMP CHARACTERISTICS

30

Input Offset Voltage: DC voltage that must be applied between the input

terminals to provide a DC output voltage of zero. A direct consequence of

a finite input offset current.

If both inputs are grounded,

the output voltage is not zero,

but there is a small offset.

VIO is normally depicted as a

voltage source driving the

non-inverting (+) input.

OP-AMP CHARACTERISTICS

31

Drift: Variation in the output offset voltage due to change in temperature.

It depends on the IIO (input offset current) and VIO (input offset voltage)

sensitivities w.r.t temperature

μV/oC Effective

Voltage gain

os osdrift noise f

V IV TA TR

T T

IO IO

v

nA/oC

“Feedback”

Resistance

from output

to inptut

NUMERICAL EXAMPLE

32

Problem: Determine the output voltage drift for the circuit shown below

at a target temperature of 80°C. Assume that the circuit has been nulled

at 25°C and the closed-loop voltage gain is 100. The input offset voltage

and current for the op-amp vary with temperature as ΔVIO/ΔT = 5 μV/°C

and ΔIIO/ΔT = 1 nA/°C.

Rf = 100 kΩ

NUMERICAL EXAMPLE

33

Soln:

Given ΔVIO/ΔT = 5 μV/°C, ΔIIO/ΔT = 1 nA/°C,

ΔT = 80 – 25 = 55°C, Av = 100, Rf = 105 Ω

=> Vdrift = (5×10-6 × 55 × 100) V + (1×10-9 × 55 ×105) V

=> Vdrift = (0.0275 + 0.0055) V = 0.033 V = 33 mV

os osdrift noise f

V IV TA TR

T T

IO IO

v

EQUIVALENT CIRCUIT

34

An op-amp is an active circuit

element that can be used to perform

linear mathematical operations like

addition, subtraction,

differentiation, and integration.

Vo

VD rd

ro

AVD

VN

VP

vx

Avx

voRout

Rin

+VCC

-VEE

in Lout s

in s L out

R Rv v A

R R R R

EQUIVALENT CIRCUIT

35

• Op-amps do not have a 0-V ground terminal. Ground reference is

established externally via the power-supply common terminal.

• A is called the open-loop voltage gain because it is the gain of the

op-amp without any external feedback from output to input.

• A practical limitation of the op-amp is that the magnitude of its

output voltage cannot exceed supply voltages |VCC| or |VEE|

• In the linear region, the curve of output vs input voltage is

approximately a straight line and its slope represents the voltage gain.

• In the saturation region, the amplifier produces a clipped output a.c.

waveform (Vout clipped at +VCC or –VEE)

OP-AMP INPUT-OUTPUT CHARACTERISTICS

36

VCC

Vo

-VEE

0

Negative Saturation

Positive Saturation

Vd

Linear region

(slope = voltage gain)

Vin,maxVin,min

IDEAL OP-AMP

37

Vo

VD

AVD

VN

VP

iN =0

iP =0

Infinite open-loop gain (A= ∞), Infinite input impedance (Zin= ∞)

Zero output impedance (Zout = 0), Zero common-mode gain (CMRR = ∞)

Infinite bandwidth & slew rate, Zero input offsets (VIO = 0, IIO = 0) & drift (Vdrift = 0)

PRACTICAL OP-AMP

38

A is large but finite (~20,000 - 200,000), Rin is large but finite (~0.3 - 2 MΩ)

Rout is small but non-zero (~75 Ω), Bandwidth is finite (Capacitances take effect)

CMRR ~70-90 dB (~3000 - 30,000), Slew Rate <~0.5 V/μs

VIO ~2-5 mV, IIO ~20-200 nA, IIB ~80-500 nA

Vo

VD rd

ro

AVD

out

in x

Rout

Rin

39

COMMONLY USED ICS & PIN CONFIGURATIONS

741: General purpose op-amp IC

Used in general purpose amplifiers, active filters, arithmetic circuits,

voltage comparators, waveform generators, regulated power supplies etc.

40

COMMONLY USED ICS & PIN CONFIGURATIONS

LM358: Low power, dual channel op-amp IC

Used in transducer amplifiers based on sensing weak external signals like

temperature, force/pressure, sound, light etc.

41

COMMONLY USED ICS & PIN CONFIGURATIONS

UA747: 14-pin dual op-amp device

Used in analog signal processing circuits such as peak/envelope detectors etc.

42

COMMONLY USED ICS & PIN CONFIGURATIONS

LM339: 14-pin 4-channel op-amp device

Used in low-level sensing and memory applications in automotive/industrial

settings, measuring instruments, timing & oscillators etc.

43

COMMONLY USED ICS & PIN CONFIGURATIONS

TL082: 8-pin 2-channel op-amp device

Offers high slew rate, low input bias & offset current, and low output-drift.

Used in high speed integrators, fast D/A converters, sample & hold circuits etc.

VIRTUAL GROUND CONCEPT

44

• Vo ≤ |VCC| ~ 5 - 15 V

• e.g. for Vo = 10 V & A = 105,

VD = 0.1 mV

• VD ~ 0 is a very good approximation in most cases (“virtual ground”).

• Thus, at the op-amp input terminals, there exists a virtual short circuit.

• Also, there is no current through the input terminals to a very good

approximation i.e. Iin ~ 0.

Vo

VD rd

ro

AVD

Iin

+VCC

-VCC

VIRTUAL GROUND CONCEPT

45

By the concept of virtual ground, i = 0 => i1 = −if and, v = v’ = 0

vo

Rf

R1

vi

v

v

if

i1 i

NEGATIVE FEEDBACK CONCEPT

46

Definition: A negative feedback is achieved when a part of the output

is fed back to the inverting (−) input terminal of the op amp.

Why Negative Feedback?

When device's gain is simply

too large (unknown) and its

bandwidth too narrow,

negative feedback is used to

set the gain to a specific

precise value (irrespective of

internal gain) and increase

the bandwidth of operation.

A → ∞

β < 1

NEGATIVE FEEDBACK CONCEPT

47

Input voltage Vi = Ve + Vf …(1) Output voltage Vo = AolVe …(2)

Feedback voltage Vf = βVo = βAolVe …(3) => Vi = (1 + βAol)Ve …(4)

Closed loop gain:

Acl = Vo/Vi = Aol /(1 + βAol)

For βAol >> 1, Acl ~ 1/β

Sacrifice factor S = Aol / Acl ~ βAol

Vo

Vi

iload

Feedback

Ve

Vf

Aol

β

Ve

Vf

Vi

Vo

48

Effects of Negative Feedback

• Fixes the gain at a precise value using external circuit elements, thus

becoming immune to variations of op-amp open-loop gain.

• Tends to stabilize operations and reduce fluctuations.

• Reduces the effect of device nonlinearities.

• Increases the bandwidth of the system by factor of S.

• Exercises control over the input and output impedances of the circuit.

• The system gain decreases by factor of S. Thus, there’s a tradeoff

between bandwidth and gain.

NUMERICAL EXAMPLE

49

Problem: The open loop gain (Aol) of an amplifier is 200, operating from

DC (f1 ~ 0) to an upper cutoff frequency (f2-ol) of 10 kHz. If the feedback

factor (β) is 0.04, what are the closed loop gain (Acl) and new upper cutoff

frequency (f2-cl)?

Soln: =>

Sacrifice factor

=>

1

olcl

ol

AA

A

20022.22

1 0.04*200clA

2009

22.22

ol

cl

AS

A

2 2cl olf f S 2 10 *9 90clf k kHz

INVERTING AMPLIFIER

50

Input is applied to inverting (−) terminal.

Reverses the polarity (180o phase shift) of input signal while amplifying it.

vo

Rf

R1

vi

v

v

if

i1

By the virtual ground concept,

v = v’ = 0 and i1 = -if

vi / R1 = -vo / Rf

Av = vo /vi = -(Rf / R1)

NUMERICAL EXAMPLE

51

Problem: Find vo for the circuit shown below

Soln: Consider the inverting amplifier at the first op-amp 33 2

2

v Rv v

v R

NUMERICAL EXAMPLE

52

Now for the second op-amp, the circuit reduces to

As v = v’ = 0, KCL at (−):

=>

31

1 1 2

0 00 ov vv

R R R

01 2

1 2

1 vv v

R R

22 1

1

o

Rv v v

R

vo

R2R1

R1

v1

v3

v

v-v2

-v2

NON-INVERTING AMPLIFIER

53

Input is applied to non-inverting (+) terminal.

Output has same polarity/phase as input signal.

vo

Rf

R1

v

v

vi

if

i1

By the virtual ground concept,

v = v’ = vi and i1 = -if

-vi / R1 = (vi – vo) / Rf

Av = vo /vi = 1 + (Rf / R1)

NUMERICAL EXAMPLE 1

54

Problem: Given vi = 1 V in the circuit below.

Find the output voltage vo and output current io

vo

R1

vi

v

v

R2 R3

io

i=0

i =0

0

5 40

i o iv v v

k k

9ov V

20 40

o o io

v v vi

k k

0.65oi mA

Soln: v = v’= vi = 1 V

As i=0,

40 kΩ5 kΩ 20 kΩ

NUMERICAL EXAMPLE 2

55

Problem: Design a non-inverting amplifier with a gain of 26-dB and an

input impedance of 47 kΩ.

Soln: First turn 26-dB

into ordinary form.

26 = 20 log10 Av

Av = 101.3 = 20

1 + (Rf / Ri) = 20

Rf / Ri = 19 => can choose Ri = 1 kΩ and Rf = 19 kΩ

56

Solved Problems

Problem: A 741 op-amp with slew rate SR = 0.5 V/μs is used as part of a

motor control system. If the highest reproducible frequency is 3 kHz and

the maximum output level is 12 V peak, does slewing ever occur?

Soln: The maximum frequency supported by the op-amp is given by

For this application, the 741 is ~2x as fast as it needs to be.

Therefore slewing doesn’t takes place.

maxf 6631Hz

NUMERICAL 1

57

max

0.5 /f

2 2 *12o

SR V s

V

NUMERICAL 2

58

Problem: Determine Vo and Io in the circuit below

59

Soln: KCL at

inverting (-) terminal:

=>

120

10 20

V

1 01 1 05 20 4

V VV V

1 4V V

0 12 8V V V KCL at node V1: =>

Output current is given by 0

8 8 ( 4)2

8 4I mA

NUMERICAL 3

60

Problem: Determine vo in the op-amp circuit below.

61

Soln: KCL at node a:

=>

=>

Since, va = vb = 2 V =>

6

40 20

a o av v v

k k

12 2a o av v v

3 12o av v

6 12 6ov V

NUMERICAL 4

62

Problem: Find the output voltage Vo for the circuit below.

2

12

86

4

Vo

2

12

86

4

Vo

63

Soln: KCL at the non-

inverting (+) input,

=>

60

12 8 6

oV VV V

3 8oV V

4 2

4 2 3o oV V V V

23 8

3o oV V

8oV V

By voltage division,

=> =>

NUMERICAL 5

64

Problem: Determine the input impedance and output voltage for the

op-amp circuit shown below. RL is the load resistance.

NUMERICAL 5

65

Soln: Since V− = 0, the input impedance is Zin = Vin / Iin = 5 kΩ

o iV AV

1

204

5

fR kA

R k

100 4 400oV m mV

Vin

Iin V−

66

Unsolved Problems

PRACTICE NUMERICAL 1

67

Problem:

Determine Vo.

Ans. Vo = -1.95V

PRACTICE NUMERICAL 2

68

Problem: (a) A differential amplifier has an open-circuit voltage gain of

100. The input signals are 3.25 and 3.15 V. Determine the output voltage.

(b) The differential amplifier has a common input signal of 3.20 V to both

terminals. This results in an output signal of 26 mV. Determine the

common-mode gain and the CMRR.

Ans. (a) 10V (b) 0.0081, 81.8dB

PRACTICE NUMERICAL 3

69

Problem: Find the output of the op amp circuit. Calculate the current

through the feedback resistor.

Ans. -3.15 V, 11.25 μA

PRACTICE NUMERICAL 4

70

Problem: Calculate vo

Ans. 21V

PRACTICE NUMERICAL 5

71

Problem: Design an inverting amplifier with a gain of 10 and an input

impedance of 15 kΩ.

Ans. R1 = 15 kΩ , Rf = 150 kΩ

vo

Rf

R1

vi

v

v

if

i1

REFERENCES

72

1. Edward Hughes; John Hiley, Keith Brown, Ian McKenzie Smith,

“Electrical and Electronic Technology”, 10th edition, Pearson

Education Limited, Year: 2008.

2. Alexander, Charles K., and Sadiku, Matthew N. O., “Fundamentals of

Electric Circuits”, 5th Ed, McGraw Hill, Indian Edition, 2013.

3. Robert-Boylestad, Louis-Nashelsky, “Electronic-Devices-and-Circuit-

Theory”, 7th-Edition.

4. Ramakant A. Gayakwad, “Op-Amps and Linear Integrated Circuits”,

4th edition, 2008.