BASIC CONCEPTS
Amplifier: Electronic circuit that produces an output
(voltage/current) in linear proportion to the input quantity.
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.
1
HISTORY OF OP-AMP
Harold S. Black develops the feedback amplifier for
the Western Electric Company
19201930
The First Op-Amp: Designed by Karl Swartzel
at Bell Labs
1930–
1940
Vacuum-tube
based
electronic
circuits
Loebe Julie then develops an Op-Amp with
two inputs: Inverting and Non-inverting
19401950
19501960
Advent
of
solid-state
(semiconductor) electronics
Bipolar junction transistors
2
HISTORY OF OP-AMP
Beginning of the Solid State Op-Amp, GAP/R P45
19601961
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
Integrated
circuits (ICs)
National Semiconductors: The LM101
and then the LM101A (both by Widlar)
19671968
19681969
Fairchild
Semiconductors:
The “famous” μA741 (by Dave
Fullager) and then the μA748
3
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
4
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)
5
OP-AMP INTRODUCTION
Op-amp Basic Circuit
6
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.
7
OP-AMP CHARACTERISTICS
Differential mode operation:
Vd
Vo = AdVi
Ad typically very large
Vi1
Vi2
Vo
Common mode operation:
Vo = AcVi
Vo
Vi
Ac << Ad
8
OP-AMP CHARACTERISTICS
Output voltage Vo = AdVd + AcVc
Vd = (Vi1 – Vi2) , Vc = (Vi1 + Vi2)/2
Vd
Vo
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.
10
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 Vi2  (150140)V  10V
Vi1 Vi2  150140V 145V
Common voltage is given by Vc 
2
2
11
The output voltage is given by


AcVc 
1 Vc 
Vo  AdVd  AV
=> Vo  AdVd  1
c c  Ad Vd  1


A
V

d d 
 CMRR Vd 
(a) CMRR = 100


145 
1 Vc 
Vo  AdVd 1
  4000*10 1
  45.8mV
 100*10 
 CMRR Vd 
(b) CMRR = 105

1 Vc 

145 
Vo  AdVd 1
  4000*10 1 5
  40.006mV
 10 *10 
 CMRR Vd 
12
NUMERICAL EXAMPLE 2
Problem: Calculate the CMRR in dB for the op-amp below
Differential
Mode
Common
Mode
13
NUMERICAL EXAMPLE 2
Vo
Soln: The differential gain is given by Ad   8  8000
Vd 1m
Vo 12m
 12
Common mode gain is given as Ac  
Vc 1m
Ad 8000

 666.7
CMRR:
Ac
12
Ad
 56.48dB
CMRR (dB): CMRR  20log10
Ac
14
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
 2 fK cos2 ft
The rate of change of signal w.r.t time is
dt
dv
 2 fK
Max. rate of change
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.
15
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
A
Soln: For voltage gain A, Vo  AVi =>
t
t
=>
Vo
2
SR
t
A


 40
Vi Vi 0.5
10
t
t
16
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
17
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.
I I
I IB 
2

IB

IB
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.
18
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
I IB  I IB  IO  30  5 / 2  32.5nA
2
I
I IB  I IB  IO  30  5 / 2  27.5nA
2
19
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.
20
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
IO
os

Vdrift
TAnoise
TR f
v 
T
T
μV/oC
Effective
Voltage gain
nA/oC
“Feedback”
Resistance
from output
to inptut
21
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
22
NUMERICAL EXAMPLE
Soln:
VosIO
I
IO
os
Vdrift 
TAnoise

TR f
v
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
23
EQUIVALENT CIRCUIT
An op-amp is an active circuit
element that can be used to perform
linear mathematical operations like
addition, subtraction,
differentiation, and integration.
+VCC
VN
vVx
D
Rroout
Rr in
v
Vo o
d
Avx
AVD
VP
-VEE
RL 
 Rin 

vout  vs 
 A 


R
R

R

R
 in
s 
 L
out 
24
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)
25
OP-AMP INPUT-OUTPUT CHARACTERISTICS
Vo
Positive Saturation
VCC
Vin,max
Vin,min
Negative Saturation
0
Vd
-VEE
Linear region
(slope = voltage gain)
26
IDEAL OP-AMP
iN =0
VN
Vo
VD
AVD
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)
27
PRACTICAL OP-AMP
out
Rroout
VD
Rrd in
Vo
AV D
in
x
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.
29
+VCC
VIRTUAL GROUND CONCEPT
• Vo ≤ |VCC| ~ 5 - 15 V
• e.g. for Vo = 10 V & A = 105,
ro
Vo
rd
VD
AVD
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.
30
VIRTUAL GROUND CONCEPT
Rf
i1
R1
if
i
v
vi
v’
vo
By the concept of virtual ground,
i = 0 => i1 = −if
and, v = v’ = 0
31
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
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
32
NEGATIVE FEEDBACK CONCEPT
Input voltage Vi = Ve + Vf
…(1)
Feedback voltage Vf = βVo = βAolVe
…(3)
=> Vi = (1 + βAol)Ve
…(2)
…(4)
Vi
Closed loop gain:
Acl = Vo/Vi = Aol /(1 + βAol)
Output voltage Vo = AolVe
Vi
For βAol >> 1, Acl ~ 1/β
Ve V
e
Vf
Vf
Sacrifice factor S = Aol / Acl ~ βAol
Aol
Vo
iload
Vo
Feedback
β
33
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.
34
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
=>
Soln: Acl 
1  Aol 
200
Acl 
 22.22
1 0.04*200
200
Aol
Sacrifice factor S 
9

Acl 22.22
f2cl  f2ol S
=>
f2cl  10k *9  90kHz
35
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
vi
R1
v
v = v’ = 0 and i1 = -if
v’
 vi / R1 = -vo / Rf
vo
 Av = vo /vi = -(Rf / R1)
36
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
37
NUMERICAL EXAMPLE
Now for the second op-amp, the circuit reduces to
As v = v’ = 0, KCL at (−):
v1  0 -vv23  0
vo0


R1
R2
R1
v0
1
=>
v1  v2   
R1
R2
v1
R2
R1
v’
-vv23
R1
v
vo
R2
 vo 
v2  v1 
R1
38
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
vi
v
v = v’ = vi and i1 = -if
v’
 -vi / R1 = (vi – vo) / Rf
vo
 Av = vo /vi = 1 + (Rf / R1)
39
NUMERICAL EXAMPLE 1
Problem: Given vi = 1 V in the circuit below.
Find the output voltage vo and output current io
v’
v
i’ =0
i=0
io
vi
R
52 kΩ
1
40 RkΩ
vo
R20
3 kΩ
Soln: v = v’= vi = 1 V
As i=0,
vi  0 vo vi

5k
40k
 vo  9V
vo vo  vi

io 
 io  0.65mA
20k 40k
40
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
41
Solved Problems
42
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

fmax  6631Hz
f max 
2Vo
2 *12
For this application, the 741 is ~2x as fast as it needs to be.
Therefore slewing doesn’t takes place.
43
NUMERICAL 2
Problem: Determine Vo and Io in the circuit below
44
Soln: KCL at
inverting (-) terminal:
2 V1
 0
10 20
=> V1  4V
V1 V1 V1 V0
 
 0 => V0  2V1  8V
KCL at node V1:
5 20
4
8
8(4)
Output current is given by I 0 

 2mA
8
4
45
NUMERICAL 3
Problem: Determine vo in the op-amp circuit below.
46
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
47
NUMERICAL 4
Problem: Find the output voltage Vo for the circuit below.
2
4
8
6
Vo
12
48
2
Soln: KCL at the noninverting (+) input,
V V Vo V 6


0
12
8
6
=> Vo  3V  8
4
8
6
Vo
12
By voltage division,
4
2
V  V  
Vo  Vo
42
3
2 

=> Vo  3  Vo  8 => Vo  8V
3 
49
NUMERICAL 5
Problem: Determine the input impedance and output voltage for the
op-amp circuit shown below. RL is the load resistance.
50
NUMERICAL 5
Soln: Since V− = 0, the input impedance is Zin = Vin / Iin = 5 kΩ
Vo  AVi
Iin
V−
Vin
 Rf 
20k
A  

 4

5k
 R1 
Vo 100m4  400mV
51
Unsolved Problems
52
PRACTICE NUMERICAL 1
Problem:
Determine Vo.
Ans. Vo = -1.95V
53
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
54
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
55
PRACTICE NUMERICAL 4
Problem: Calculate vo
Ans. 21V
56
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
57
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-CircuitTheory”, 7th-Edition.
4. Ramakant A. Gayakwad, “Op-Amps and Linear Integrated Circuits”,
4th edition, 2008.
58