ELL 100 - Introduction to Electrical Engineering

LECTURE 26: OPERATIONAL AMPLIFIERS

Outline
 Introduction
 Important Specifications/Characteristics
 Equivalent Circuit
 Ideal and Practical/Non-Ideal Op-amp
 Common ICs and Pin Configurations
 Inverting and Non-Inverting Amplifier

2
 BASIC CONCEPTS
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.

A 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. 3
 REAL LIFE APPLICATIONS

Microphone Amplifier

Digital to Analog Converter

4
REAL LIFE APPLICATIONS

Sensors e.g. Electronic Thermometer

5
 REAL LIFE APPLICATIONS

Automatic Light Operated Switch DC Volt Polarity Meter

6
REAL LIFE APPLICATIONS

Control of Motors
7
REAL LIFE APPLICATIONS

Music Players
8
REAL LIFE APPLICATIONS

Analog Computer
9
REAL LIFE APPLICATIONS

Current and Voltage Regulator

10
REAL LIFE APPLICATIONS

Waveform Generator
11
 HISTORY OF OP-AMP
Harold S. Black develops the feedback amplifier for
1920- the Western Electric Company
1930 The First Op-Amp: Designed by Karl Swartzel
at Bell Labs
1930–
1940 Loebe Julie then develops an Op-Amp with
Vacuum-tube two inputs: Inverting and Non-inverting
based 1940-
electronic 1950 Advent of solid-state
circuits (semiconductor) electronics
1950- Bipolar junction transistors
1960
12
 HISTORY OF OP-AMP
Beginning of the Solid State Op-Amp, GAP/R P45
1960-
1961 The GAP/R PP65 makes the Op-Amp into a circuit
component as a potted module
1962 Robert J.Widlar develops the μA702 Monolithic
IC Op-Amp and shortly after the μA709
1963
National Semiconductors: The LM101
and then the LM101A (both by Widlar)
1967-
Integrated 1968 Fairchild Semiconductors:
circuits (ICs) The “famous” μA741 (by Dave
1968-
1969 Fullager) and then the μA748

13
 OP-AMP INTRODUCTION
• 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

14
 OP-AMP INTRODUCTION
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)

15
 OP-AMP INTRODUCTION
Op-amp Basic Circuit

16
 OP-AMP INTERNAL CIRCUIT
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.
17
 OP-AMP CHARACTERISTICS
Differential mode operation:
Vo
Vd
Vo = AdVi
Ad typically very large Vi1 Vi2

Common mode operation: Vo

Vo = AcVi
Vi
Ac << Ad
18
 OP-AMP CHARACTERISTICS
Output voltage Vo = AdVd + AcVc
Vo
Vd = (Vi1 – Vi2) , Vc = (Vi1 + Vi2)/2 Vd
Ad >> Ac
Vi1 Vi2

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). 19
 OP-AMP CHARACTERISTICS
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.
20
 NUMERICAL EXAMPLE 1

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 Vd  Vi1  Vi 2  (150  140)V  10V

Vi1  Vi 2 150  140 V

Common voltage is given by Vc    145V
2 2
21
The output voltage is given by
 AcVc   1 Vc 
Vo  Ad Vd  AcVc  Ad Vd 1   => Vo  AdVd 1  
 Ad Vd   CMRR Vd 

(a) CMRR = 100

 1 Vc   145 
Vo  AdVd 1    4000*10 1    45.8mV
 CMRR Vd   100*10 

(b) CMRR = 105

 1 Vc   145 
Vo  AdVd 1    4000*10 1  5   40.006mV
 CMRR Vd   10 *10  22
 NUMERICAL EXAMPLE 2
Problem: Calculate the CMRR in dB for the op-amp below

Differential
Mode

Common
Mode

23
 NUMERICAL EXAMPLE 2
Vo 8
Soln: The differential gain is given by Ad    8000
Vd 1m

Vo 12m
Common mode gain is given as Ac    12
Vc 1m

Ad 8000
CMRR:   666.7
Ac 12

Ad
CMRR (dB): CMRR  20log10  56.48dB
Ac
24
 OP-AMP CHARACTERISTICS
Slew Rate: Maximum rate of change of output voltage vs time

Let the signal be a sine wave v(t )  K sin 2 ft

dv
The rate of change of signal w.r.t time is  2 fK cos 2 ft
dt
dv
Max. rate of change  2 fK
dt
Slew rate required = 2 fmaxVp

fmax is the highest signal frequency and Vp is the maximum output voltage
required to be supported by the op-amp.
25
 NUMERICAL EXAMPLE 1
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?

Vo Vi
Soln: For voltage gain A, Vo  AVi => A
t t

Vo
=> A t 
SR

2
 40
Vi Vi 0.5
10
t t
26
 NUMERICAL EXAMPLE 2
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 Vo  24(0.02)  0.48V

SR
The max signal frequency is given by f max   175 103 Hz
2 Vo

27
 OP-AMP CHARACTERISTICS
Input Bias Current: The average magnitude of the two base currents at
the input terminals with the output at a specified level.

  Input bias current is a

I I
I IB IB IB
problem as it flows into
2 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.

28
 OP-AMP CHARACTERISTICS
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
 I IO  I IO
I  I IB 
IB  30  5 / 2  32.5nA I  I IB 
IB  30  5 / 2  27.5nA
2 2 29
 OP-AMP CHARACTERISTICS
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.

30
 OP-AMP CHARACTERISTICS
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

VosIO I osIO
Vdrift  TAnoise
v  TR f
T T
“Feedback”
nA/oC Resistance
μV/oC Effective from output
Voltage gain to inptut
31
 NUMERICAL EXAMPLE
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

32
 NUMERICAL EXAMPLE
Soln:

VosIO I osIO
Vdrift  TAnoise
v  TR f
T T
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

33
 EQUIVALENT CIRCUIT
+VCC
An op-amp is an active circuit VN
element that can be used to perform
linear mathematical operations like Rroout v
Vo o
vVxD Rrdin
addition, subtraction, AV DAvx
differentiation, and integration.
VP
-VEE

 Rin   RL 
vout  vs    A  
 Rin  Rs   L
R  Rout 
34
 EQUIVALENT CIRCUIT
• 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)
35
OP-AMP INPUT-OUTPUT CHARACTERISTICS
Vo

VCC Positive Saturation

Vin,min Vin,max
0 Vd

Linear region
Negative Saturation -VEE (slope = voltage gain)
36
 IDEAL OP-AMP
iN =0
VN

Vo
VD
AV D

VP
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)
37
 PRACTICAL OP-AMP
out

Rroout Vo
VD Rrdin x
AV D in

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 38
 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.
39
 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.
40
 COMMONLY USED ICS & PIN CONFIGURATIONS
UA747: 14-pin dual op-amp device

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

41
 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. 42
 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.
43
 +VCC
VIRTUAL GROUND CONCEPT
• Vo ≤ |VCC| ~ 5 - 15 V
ro Vo
• e.g. for Vo = 10 V & A = 105, VD rd
AV D
VD = 0.1 mV Iin

-VCC

• 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.
44
 VIRTUAL GROUND CONCEPT
Rf if

i1 R1 i
v

vi v
vo

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

45
 NEGATIVE FEEDBACK CONCEPT
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 A→∞
too large (unknown) and its
bandwidth too narrow,
negative feedback is used to
set the gain to a specific β<1
precise value (irrespective of
internal gain) and increase
the bandwidth of operation.
46
 NEGATIVE FEEDBACK CONCEPT
Input voltage Vi = Ve + Vf …(1) Output voltage Vo = AolVe …(2)
Feedback voltage Vf = βVo = βAolVe …(3) => Vi = (1 + βAol)Ve …(4)

Vi
Closed loop gain: Ve V
Vi e Aol
Acl = Vo/Vi = Aol /(1 + βAol) Vo
Vf iload
For βAol >> 1, Acl ~ 1/β
Vo
Vf Feedback
Sacrifice factor S = Aol / Acl ~ βAol β
47
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.

48
 NUMERICAL EXAMPLE
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)?

Aol 200
Soln: Acl  Acl   22.22
1   Aol  =>
1  0.04*200
Aol 200
Sacrifice factor S   9
Acl 22.22

f 2cl  f 2ol S => f 2cl  10k *9  90kHz

49
 INVERTING AMPLIFIER
Input is applied to inverting (−) terminal.
Reverses the polarity (180o phase shift) of input signal while amplifying it.
Rf if
By the virtual ground concept,
i1 R1
v = v’ = 0 and i1 = -if
v

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

50
 NUMERICAL EXAMPLE
Problem: Find vo for the circuit shown below

Soln: Consider the inverting amplifier at the first op-amp v3   R  v  v

3 2
v2 R
51
 NUMERICAL EXAMPLE
Now for the second op-amp, the circuit reduces to

As v = v’ = 0, KCL at (−): R1
R2
v1
v1  0 -vv32  0 vo  0
 
R1 R1 R2
1 v0 v
=>  v1  v2    -vv23
R1 v
R1 R2
vo
R2
 vo   v2  v1 
R1
52
 NON-INVERTING AMPLIFIER
Input is applied to non-inverting (+) terminal.
Output has same polarity/phase as input signal.
Rf if
By the virtual ground concept,
i1 R1
v = v’ = vi and i1 = -if
v
v  -vi / R1 = (vi – vo) / Rf

vi vo  Av = vo /vi = 1 + (Rf / R1)

53
 NUMERICAL EXAMPLE 1
Problem: Given vi = 1 V in the circuit below.
Find the output voltage vo and output current io
v i =0

Soln: v = v’= vi = 1 V
v i=0
As i=0,
io
vi  0 vo  vi
vi
40 RkΩ
1

52kΩ
R vo R20
3 kΩ 5k 40k
 vo  9V
vo vo  vi
io    io  0.65mA
20k 40k 54
 NUMERICAL EXAMPLE 2
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
55
Solved Problems

56
 NUMERICAL 1
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

SR 0.5V /  s
f max   fmax  6631Hz
2 Vo 2 *12
For this application, the 741 is ~2x as fast as it needs to be.
Therefore slewing doesn’t takes place.
57
 NUMERICAL 2
Problem: Determine Vo and Io in the circuit below

58
 Soln: KCL at
inverting (-) terminal:
2 V1
 0
10 20
=> V1  4V

V1 V1 V1  V0
KCL at node V1:    0 => V0  2V1  8V
5 20 4
8 8  (4)
Output current is given by I 0    2mA
8 4 59
 NUMERICAL 3
Problem: Determine vo in the op-amp circuit below.

60
Soln: KCL at node a:
va  vo 6  va

40k 20k
=> va  vo  12  2va

=> vo  3va  12

Since, va = vb = 2 V => vo  6  12  6V

61
 NUMERICAL 4
Problem: Find the output voltage Vo for the circuit below.
2

6 8
Vo
12

62
 2
Soln: KCL at the non-
4
inverting (+) input,
V V  Vo V  6
  0
12 8 6 6 8
Vo
=> Vo  3V  8 12

By voltage division,
4 2 2 
V  V 
42
Vo  Vo => Vo  3  Vo   8 => Vo  8V
3 3 
63
 NUMERICAL 5
Problem: Determine the input impedance and output voltage for the
op-amp circuit shown below. RL is the load resistance.

64
 NUMERICAL 5
Soln: Since V− = 0, the input impedance is Zin = Vin / Iin = 5 kΩ

Vo  AVi Iin V−
Vin
 Rf  20k
A     4
 R1  5k

Vo  100m   4  400mV

65
Unsolved Problems

66
 PRACTICE NUMERICAL 1
Problem:
Determine Vo.

Ans. Vo = -1.95V
67
 PRACTICE NUMERICAL 2
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

68
 PRACTICE NUMERICAL 3
Problem: Find the output of the op amp circuit. Calculate the current
through the feedback resistor.

Ans. -3.15 V, 11.25 μA

69
 PRACTICE NUMERICAL 4
Problem: Calculate vo

Ans. 21V
70
 PRACTICE NUMERICAL 5
Problem: Design an inverting amplifier with a gain of 10 and an input
impedance of 15 kΩ. Rf if

i1 R1
v

vi v
vo

Ans. R1 = 15 kΩ , Rf = 150 kΩ
71
 REFERENCES
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. 72