The Operational Amplifier
 The operational amplifier is a building block of modern electronic
instrumentation.
 Therefore, mastery of operational amplifier fundamentals is
paramount to any practical application of electronic circuits.
 Specialists in electronic instruments find employment in medical
schools, hospitals, research laboratories, aircraft industries, and
thousands of other industries where electronic instruments are
routinely used.
 The operational amplifier is often called the op amp for short.
1
Op Amp
 The op amp is an electronic device consisting of a complex
arrangement of resistors, transistors, capacitors, and diodes. A full
discussion of what is inside the op amp is the subject of another
course.
 In this course, it will suffice to treat the op amp as a circuit
building block and simply study what takes place at its terminals.
2
Op Amp
 The op amp is an active circuit element that behaves like a
voltage-controlled voltage source.
 It can also be used in making a voltage- or current-controlled
current source.
 An op amp is designed to perform mathematical operations of
addition, subtraction, multiplication, division, differentiation,
and integration.
 The ability of the op amp to perform these mathematical
operations is the reason it is called an operational amplifier.
3
Op Amp Pin Configuration
 Op amps are commercially available in integrated circuit packages
in several forms. The figure shows a typical op amp package.
The eight-lead DIP package of an op amp.
4
Op Amp Circuit Symbol
A typical op amp: pin configuration and circuit symbol.
5
Powering The Op Amp
 As an active element, the op amp must be powered by a voltage
supply. Although the power supplies are often ignored in op amp
circuit diagrams for the sake of simplicity, the power supply
currents must not be overlooked.
i p + in + i0 + ic+ + ic- = 0
Powering the op amp.
6
Practical Limitation
 Positive saturation, vo = VCC
 Linear region, −VCC ≤ vo = Avd ≤ VCC
 Negative saturation, vo = −VCC
Op amp output voltage vo as a function of the
differential input voltage vd= vp – vn
A is the so-called open-loop gain
Note: If A is large, then vd = vp – vn must be very small or the Op Amp will
immediately go into positive or negative saturation.
7
Op Amp Equivalent Circuit
 The input resistance Ri is
the Thevenin equivalent
resistance seen at the input
terminals
 While the output resistance
Ro is the Thevenin
equivalent resistance seen
at the output
Equivalent circuit of the non-ideal op amp.
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Open Loop Gain
 The differential input voltage vd is given by vd= vp – vn
 The op amp senses the difference between the two inputs,
multiplies it by the gain A, and causes the resulting voltage to
appear at the output. Thus, the output vo is given by
vo= Avd = A(vp – vn)
 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.
 How to stabilize an amplifier with a very large gain ?
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Closed Loop Gain
 The concept of feedback is crucial to our understanding of op amp
circuits.
 A negative feedback is achieved when the output is fed back to the
inverting terminal of the op amp.
 When there is a feedback path from output to input, the ratio of the
output voltage to the input voltage is called the closed-loop gain.
 The closed-loop gain is almost insensitive to the open-loop gain A
of the op amp.
 For this reason, op amps are used in circuits with feedback paths.
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Negative Feedback
vF+

+
v1

vF
+ 
+
vi

F
+
A
vF
vo
vo
Open loop gain A =
vi
Feedback factor F =
vo

Closed loop gain K =
vF = F vo
vo
v1
vo
vo
vo vi
A
A
Closed loop gain K = =
=
=
=
v1 vF + vi vF vi + vi vi
Fvo vi +1 FA +1
1
1
K=
»
F +1 A
F
if A  F
Assume a passive feedback network with F £ 1, then A = 105  F and
K=
1
³1
F
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Ideal Op Amp
 An op amp is ideal if it has the following characteristics:
1. Infinite open-loop gain, A ≈ ∞
2. Infinite input resistance, Ri ≈ ∞
3. Zero output resistance, Ro ≈ 0
 The currents into both input terminals are zero
 The voltage across the input terminals is negligibly small
Ri » ¥  i p = 0,
vd = v p - vn = 0 
in = 0
v p = vn
Ideal op amp model
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Inverting Amplifier
in = 0  is = -i f
vs - vn
v -v
=- o n
Rs
Rf
But vn = v p = 0 for an ideal op amp
vs
vo
=Rs
Rf
vo = -
Rf
Rs
vs
Rf
vo
=The voltage gain Av is
vs
Rs
An inverting amplifier reverses the polarity of the input signal while amplifying it.
13
Example 5.1
Assume that the op amp is ideal. Calculate the output voltage vo for the following
values of vs : 0.4V, 2.0V, 3.5V, -0.6V, -2.4V.
80k
16k
10V
-15V
14
Example 1
Find the output voltage of the op amp circuit. Calculate the current through
the feedback resistor.
i
15
Example 2
Determine vo.
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Example 5.2
a) Design an inverting amplifier with a voltage gain of -12. Use 15V power supplies.
b) What range of input voltages allows the op amp to operate in its linear region?
15V
-15V
a) Since vo = -
Rf
Rs
vs ,
choose Rs = 1k  R f = 12k
b) 15V = (-12)(vs )
 vs = 15V 12 = 1.25V
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Noninverting Amplifier
Since i p = 0  v p = vg = vn
Use voltage divider to get vn :
Rs
vn =
vo = vg
R f + Rs
 vo =
R f + Rs
Rs
æ Rf
vg = ççç1+
R
è
ö÷
÷÷ vg
÷
s ø
Rf
vo
= 1+
The voltage gain Av is
vg
Rs
An noninverting amplifier provides a positive gain which is greater or equal unity.
The series resistance for vg is often omitted. Why is it useful in practice?
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Voltage Follower
æ Rf
vo = ççç1 +
R
è
ö÷
÷÷ vi
÷
s ø
R f = 0, Rs ¥
vo = vi
The voltage follower.
Such a circuit has a very high input impedance and is therefore useful as an
intermediate-stage (or buffer) amplifier to isolate one circuit from another. The
voltage follower minimizes interaction between the two stages and eliminates inter
stage loading.
A voltage follower used to isolate two cascaded stages of a circuit.
19
Example 3
For the op amp circuit, calculate the output voltage vo.
Method 1: Using superposition
Method 2: Applying KCL
 vo  1V, as before
20
Example 4
Calculate vo.
v1
21
Summing Amplifier
ia
KCL: ia + ib + ic = in
va - vn vb - vn vc - vn vn - vo
+
+
=
Ra
Rb
Rc
Rf
vn = v p = 0
A summer can have more than
three inputs.
æ Rf
Rf
R f ö÷
ç
vo = -çç va + vb + vc ÷÷
Rb
Rc ÷ø
è Ra
An summing amplifier combines several inputs and produces an output that is
the weighted sum of the inputs.
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Example 5
Find vo and io.
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Digital-to-Analog Converter
Four-bit DAC:
(a) block diagram
(b) binary weighted ladder type
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Example 6
In the op amp circuit, let Rf = 10 kΩ, R1 = 10 kΩ, R2 = 20 kΩ, R3 = 40 kΩ, and R4 = 80
kΩ. Obtain the analog output for binary inputs [0000], [0001], [0010], . . . , [1111].
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Difference Amplifier
va - vn vn - vo
=
Ra
Rb
Rb va - Rb vn = Ra vn - Ra v0
( Ra + Rb )vn = Rb va + Ra v0
 vn =
Rd
vp =
vb
Rc + Rd
vn = v p 
Rb va + Ra v0
Ra + Rb
Rd ( Ra + Rb )
Rb
vo =
vb - va
Ra ( Rc + Rd )
Ra
A difference amplifier amplifies the difference between two inputs but rejects any
signals common to the two inputs.
Note: If
Ra
R
R
= c  v0 = b (vb - va )
Rb
Rd
Ra
If Ra = Rb and Rc = Rd  v0 = vb - va
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Example 7
Design an op amp circuit with inputs
v1 and v2 such that vo = −5v1+3v2.
vo =
R4 ( R1 + R2 )
R
v2 - 2 v1
R1 ( R3 + R4 )
R1
R2 (1 + R1 R2 )
R2
vo =
v2 - v1
R1 (1 + R3 R4 )
R1
Design 1
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Example 7 cont’d
Design a different op amp circuit with inputs v1 and v2 such that vo = −5v1+3v2.
Design 2
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Example 8
Design a difference amplifier with gain of 4.
Figure 5.24
If
R
R1
R
= 3  v0 = 2 (v2 - v1 )
R2
R4
R1
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Instrumentation Amplifier
Show that:
An instrumentation amplifier is an amplifier of low-level signals used in process
control or measurement applications and is commercially available in single-package
30
units.
Example 9
Obtain io in the instrumentation amplifier circuit of Fig. A.
Recall:
Figure A
compare
31
Summary
32
Summary
33
Op Amp Circuit Analysis With PSPICE
 PSpice does not have a model for an ideal op amp.
 PSpice has four nonideal, commercially available op amps in its eval.slb
library.
 Note that each of them requires dc supplies, without which the op amp will not
work.

Nonideal op amp models available in PSpice.
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Example 10
Use PSpice to solve for vo / vs.
vo -3.9983V
=
= -1.99915
2V
vs
35
Controlled Sources
 Voltage Controlled Voltage Source (VCVS)
Non-inverting amplifier:
vo
R2
= 1+
vi
R1
independent of RL
36
Controlled Sources
 Voltage Controlled Current Source (VCCS)
Redraw the circuit:
vo = v+ = v- =
 v2 = 2vo
R
v2
R+R
KCL: ii + i2 - io = 0
vi - vo v2 - vo
+
- io = 0
R
R
vi vo
vo vo vi
io = - + 2 - =
R R
R R
R
1
io = vi independent of RL
R
37
Controlled Sources
 Current Controlled Voltage Source (CCVS)
æ R2 ö÷
vo = ççç1 + ÷÷ vi ,
çè
R ÷ø
vi = R3ii
since i+ = 0
1
æ R2 ÷ö
 v0 = R3 ççç1 + ÷÷ ii
çè
R ÷ø
independent of RL
1
38
Controlled Sources
 Current Controlled Current Source (CCCS)
Redraw the circuit:
KVL: R1i1 - vD - R2i2 = 0
i2 =
, vD = 0
R1
i1 , i1 = ii , i2 = io
R2
R1
io =
ii
R2
independent of RL
39