Chapter 2 Operational Amplifier.pptx

Oxford University Publishing
Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)
Chapter #2: Operational Amplifier
from Microelectronic Circuits Text
by Sedra and Smith
Oxford Publishing
Karama Jeribi Friha

Introduction
 IN THIS CHAPTER YOU WILL LEARN
 The terminal characteristics of the ideal op-amp.
 How to analyze circuits containing op-amps, resistors,
and capacitors.
 How to use op-amps to design amplifiers having
precise characteristics.

 1. Introduction
 2. The Ideal Op Amp
 3. The Inverting Configuration
 4. The Noninverting Configuration
 5. Difference Amplifiers
 Problem set
3 Outline

Introduction
The Operational Amplifiers
4
 Invented in 1941 by Bell Labs engineer Karl D. Swartzel Jr. using
vacuum tubes.
 First monolithic IC op-amp was designed by Bob Widlar at
Fairchild Semiconductor.
 The 741 op-amp is perhaps the best known op-amp in the world.
Many other op-amps use the same pin configuration as the 741.
 It is an active circuit element designed to perform amplification
and mathematical operations of addition, subtraction,
multiplication, division, differentiation and integration.
 Op-amps are ubiquitous low cost components used in countless
applications for analog signal processing (gain, filtering, signal
conditioning).

The 741 op-amp
741
5
pin configuration circuit symbol
A typical op amp

Classic 741 Schematic
6
The op amp is an electronic device consisting of a complex arrangement of resistors, transistors, capacitors,

The Op Amp Terminals
 terminal #1
 inverting input
 terminal #2
 non-inverting input
 terminal #3
 output
 terminal #4
 positive supply VCC
 terminal #5
 negative supply VEE

Function and
Characteristics
of the Ideal Op Amp
 The op amp is designed to:
1. sense the difference between the voltage signals applied
at its two input terminals (i.e., the quantity V2 – V1),
2. multiply this by a number A, and cause the resulting
voltage A( V2 – V1) to appear at output terminal 3 .
3. input terminal 1 is called the inverting input terminal and
is distinguished by a "-" sign,
4. input terminal 2 is called the noninverting input terminal
and is distinguished by a "+" sign
8

Cont’d
9
A is called the differential gain = open-loop gain

 The op amp responds only to the difference signal v2 − v1 and
hence ignores any signal common to both inputs.
 That is, if v1 = v2 = 1 V, then the output will (ideally) be zero.
This property is called common-mode rejection
  we conclude that an ideal op amp has zero common-mode
gain or, equivalently, infinite common-mode rejection
 An important characteristic of op amps is that they are direct-
coupled or dc amplifiers, where dc stands for direct-coupled (it
could equally well stand for direct current, since a direct-
coupled amplifier is one that amplifies signals whose frequency
is as low as zero).
10 Characteristic of Op Amps

Differential &
Common-Mode
Signals
 Q: How is common-mode input (vIcm) defined in
terms of v1 and v2?
𝑣𝐼𝑐𝑚 =
1
2
(𝑣1 + 𝑣2)
common−mode input
but also...
𝑣1 = 𝑣𝐼𝑐𝑚 − 𝑣𝐼𝑑
diff
/2
inverting input
𝑣2 = 𝑣𝑐𝑚𝑖 + 𝑣𝐼𝑑/2
non−inverting input
𝑣𝐼𝑑= 𝑣2 − 𝑣1

2.1.3. Differential &
Common-Mode
Signals
𝑣𝐼𝑐𝑚 =
1
2
(𝑣1 + 𝑣2)
common−mode input
but also...
Representation of the signal sources 𝑣1and 𝑣2 in terms of their differential and common mode

 1. Infinite input impedance (Due to this almost any
source can drive it)
 2. Zero output impedance (So that there is no change in
output due to change in load current)
 3. Zero common-mode gain or, equivalently, infinite
common-mode rejection
 4. Infinite open-loop gain A (Huge Gain!!!!!)
 5. Infinite bandwidth
13 Characteristics of the Ideal Op Amps

Typical Op Amp Parameters
Parameter Variable Typical Ranges Ideal Values
Open-Loop
Voltage Gain
A 105 to 108
∞
Input Resistance Ri 105 to 1013 W
∞W
Output Resistance Ro 10 to 100 W 0 W
Supply Voltage Vcc/V+
-Vcc/V-
5 to 30 V
-30V to 0V
N/A
N/A

2. For close loop configuration, the OP-AMP tries to
keep the inputs the same voltage
Two Rules to remember !!!
1. No Current flows In or Out of the Inputs
15

The Inverting
Configuration
 Q: What are two basic closed-loop op-amp
configurations which employ op-amp and resistors
alone?
 A: 1) inverting and 2) non-inverting op amp

The Inverting Configuration
 It consists of one op amp and two resistors 𝑅1and 𝑅2.
 Resistor 𝑅2 closes the loop around the op amp
18
negative
feedback

The Inverting
Configuration
 question: what are two basic closed-loop op amp configurations which
employ op-amp and resistors alone?
 answer: inverting and non-inverting op amp
 note: here we examine the inverting type
source is applied to
inverting input
non-inverting input is
grounded
R2 facilitates “negative
feedback”
R1 regulates level of
this feedback
Figure 2.5: The inverting closed-loop configuration.

The Closed-Loop Gain
 Aim: to determine the closed-loop gain G
 Assumption: the op amp to be ideal
 the equivalent circuit 
20

The Inverting
Configuration
 question: what are two basic closed-loop op amp configurations which
employ op-amp and resistors alone?
 answer: inverting and non-inverting op-amp
 note: here we examine the inverting type
Figure 2.5: The inverting closed-loop configuration.
i1 i = 0
i1

Result:
 We started out with an amplifier having very large gain
A, and through applying negative feedback we have
obtained a closed-loop gain R2/R1 that is much smaller
than A but is stable and predictable.
 That is, we are trading gain for accuracy.
23

Effect of Finite Open-Loop Gain
 Assumption: the op-amp open-loop gain A is
finite.
24
The current 𝑖1 through 𝑅1 can now be found from

Con’t…
 The infinite input impedance of the op amp forces the current
𝑖1 to flow entirely through 𝑅2.
 The output voltage Vo can thus be determined from
 the closed-loop gain G
25

Effect of Finite
Open-Loop Gain
 Q: How does the gain expression change if open loop
gain (A) is not assumed to be infinite?
if then the previous
gain expression is
2 1
2
yielded
2
1
1
/
1 ( / )
1
A
G
A
Out
A
In
v R R
G
R R
v
A
R
R




 

 
 

 

ideal gain
non-ideal gain

2.2.1. Effect of
Finite
Open-Loop Gain
 Q: Under what condition can G = -R2 / R1 be
employed over the more complex expression?
 A: If 1 + (R2/R1) << A, then simpler expression may
be used.
2 2 2 1
2 1
1 1
if then el
/
1
1 ( /
1
se
)
A A
R R R R
A G G
R R
R R
A
 

    

 
 
 
ideal gain non-ideal gain

The Non-inverting
Configuration
 The second closed-loop configuration we shall
study is
28

The Non-Inverting
Configuration
source is applied to
non-inverting input
inverting input is
grounded
through R1
R1 and R2 act as voltage divider,
regulating negative feedback to the
inverting input
node #1
node #2

Example 3
 2.9 Use the superposition principle to find the output
voltage of the circuit
32

The Voltage Follower = Buffer amplifier
 We will explore the property of high input impedance
 to connect a source with a high impedance to a low-impedance load.
 It is used mainly as an impedance transformer or a power amplifier.
 voltage follower, since the output “follows” the input. In the ideal case,
Vo = Vi, Rin = ∞, Rout = 0,
33

Difference
Amplifiers
 difference amplifier – is a closed-loop configuration
which responds to the difference between two
signals applied at its input and ideally rejects signals
that are common to the two.
 Ideally, the amp will amplify only the differential
signal (vId) and reject completely the common-
mode input signal (vIcm). However, a practical
circuit will behave as below…
𝑣𝑂𝑢𝑡 = 𝐴𝑑𝑣𝐼𝑑 + 𝐴𝑐𝑚𝑣𝐼𝑐𝑚

Difference
Amplifiers
differential gain
differential input
common-mode input
common-mode gain
𝑣𝑂𝑢𝑡 = 𝐴𝑑𝑣𝐼𝑑 + 𝐴𝑐𝑚𝑣𝐼𝑐𝑚

Difference
Amplifiers
 The efficacy of a differential amplifier is measured by
the degree of its rejection of common-mode signals
in preference to differential signals.
 common-mode rejection ratio (CMRR) – is the
degree to which a differential amplifier “rejects” the
common-mode input.
 Ideally, CMRR = infinity…
𝐶𝑀𝑅𝑅 = 20 𝐥𝐨𝐠10
𝐴𝑑
𝐴𝐶𝑚

2.4. Difference
Amplifiers
 10
20 Di
Cm
A
CMMR
A
log
Figure : Representing the input signals to a differential amplifier in
terms of their differential and common-mode components.

Single-Op-Amp Difference Amplifier
 The gain of the noninverting amplifier configuration is
positive, (1+R2/R1) while that of the inverting
configuration is negative, (-R2/R1)
 Combine the two configurations together
 we have to make the two gain magnitudes equal in
order to reject common-mode signals
38

A Single
Op-Amp
Difference Amp
 Q: What are the characteristics of the difference
amplifier?
 A: Refer to following equations…
 
2 1 4 2
2 1
4 3 1 1
1 3 2
2 1
2 4 1
but i
(
f
)
( )
then
Out In In
Out In In
R R R R
v v v
R R R R
R R R
v v v
R R R

 


 
 
 

 

A Superior Circuit—The
Instrumentation Amplifier
40

2.4.2. The
Instrumentation
Amplifier
 question: however, can we get “more” from these amps than simply
impedance matching?
 answer: yes, maybe additional voltage gain???
A popular circuit for an instrumentation amplifier.
stage #1
non-inverting
op amp (A1)
non-inverting
op amp (A2)
stage #2
difference op
amp (A3)
vOut = (1 + R2/R1)vIn
vOut = (R4/R3)vId

2.4.2. The
Instrumentation
Amplifier
A popular circuit for an instrumentation amplifier. (b) The circuit in
(a) with the connection between node X and ground removed and
the two resistors R1 and R1 lumped together. This simple wiring
change dramatically improves performance.

The
Instrumentation
Amplifier
 Q: How can one analyze this circuit?

Conclusion
 The IC op-amp is a versatile circuit building block. It is
easy to apply, and the performance of op-amp circuits
closely matches theoretical predictions.
 The op-amp terminals are the inverting terminal (1), the
non-inverting input terminal (2), the output terminal (3),
the positive-supply terminal (4) to be connected to the
positive power supply (VCC), and the negative-supply
terminal (5) to be connected to the negative supply (-
VEE).

Conclusion (2)
 The ideal op-amp responds only to the difference input
signal, that is (v2 - v1). It yields an output between
terminals 3 and ground of A(v2 - v1). The open-loop gain
(A) is assumed to be infinite. The input resistance (Rin) is
infinite. The output resistance (Rout) is assumed to be
zero.
 Negative feedback is applied to an op-amp by
connecting a passive component between its output
terminal and its inverting (negative) input terminal.

Conclusion (3)
 Negative feedback causes the voltage between the two
input terminals to become very small, and ideally zero.
Correspondingly, a virtual short is said to exist between
the two input terminals. If the positive input terminal is
connected to ground, a virtual ground appears on the
negative terminal.

Conclusion (4)
 The two most important assumptions in the analysis of
op-amp circuits, assuming negative feedback exists, are:
 the two input terminals of the op-amp are at the
same voltage potential.
 zero current flows into the op-amp input terminals.
 With negative feedback applied and the loop closed, the
gain is almost entirely determined by external
components: Vo/Vi = -R2/R1 or 1+R2/R1.

Conclusion (5)
 The non-inverting closed-loop
configuration features a very high input
resistance. A special case is the unity-
gain follower, frequently employed as
a buffer amplifier to connect a high-
resistance source to a low-resistance
load.
 The difference amplifier of the Figure is
designed with R4/R3 = R2/R1, resulting
in vo = (R2/R1)(vI2 - vI1).

Conclusion (6)
 The instrumentation amplifier of the
Figure is a very popular circuit. It
provides vo = (1+R2/R1)(R4/R3)(vI2 - vI1).
It is usually designed with R3 = R4 and
R1 and R2 selected to provide the
required gain. If an adjustable gain is
needed, part of R1 can be made
variable.

Problem Set
 Exercise 1
Give the expression of V0.

Problem Set
 Exercise 2
Determine the voltage gain of the op-amp circuit below.

Problem Set
 Exercise 3
Determine V0 in the following configurations.
Configuration (a) Configuration (b)

Chapter 2 Operational Amplifier.pptx

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