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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.