Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
The Operational Amplifier
Introduction
In this chapter we will introduce a general purpose integrated circuit (IC), the Operational
Amplifier (Op-Amp), which is a most versatile and widely used linear integrated circuit. The
Op-Amp is a direct-coupled high-gain amplifier to which feedback is added to control its
overall response characteristics.
The standard Op-Amp symbol is shown in left-hand figure below. It has two input terminals,
the inverting (-) input and the non-inverting (+) input. The typical Op-Amp operates with
two dc supply voltages, one positive and the other negative, as shown in the right-hand figure
below. Usually these dc voltage terminals are left off the schematic symbol for simplicity but
are always understood to be there.
SCHEMATIC SYMBOL
Inverting
Input
Inverting
Input
V2
Output
A
V+
VIN
A
+
+
V-
V1
Non-inverting
Input
VOUT
Non-inverting
Input
The equivalent circuit for an Op-Amp is as follows:
ROUT
+
VIN
RIN
Aol VIN
VOUT
-
Equivalent Circuit
Where
RIN
is the total resistance between the inverting and the non-inverting
inputs
ROUT is the resistance viewed from the output terminal
Aol
is the Open-Loop Voltage Gain i.e. when there are no external
components
The Op-Amp responds to a difference between the input terminals
1
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Op-amp Characteristics
Open-Loop Voltage Gain
The open-loop voltage gain, AOL, of an op-amp is the internal voltage gain of the device and
represents the ratio of output voltage to input voltage when there are no external components.
The open-loop voltage gain is set entirely by the internal design. Open-loop voltage gain can
range up to 200,00 and is not a well-controlled parameter. Data sheets often refer to the
open-loop voltage gain as the large-signal voltage gain.
Signal-Ended Input
Signal-Ended input => one input is grounded and the signal voltage is applied only to the
other input as shown below.
VIN
A
VOUT = -AVIN
+
+
VIN
A
VOUT = AVIN
-
Differential Input (Double-Ended Input)
In this mode, two signals are applied to the inputs, as shown below. The output is the sum of
the respective single ended outputs.
V1
A
Vdiff
VOUT = -AV(d)Vdiff
+
V2
2
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Common-Mode Input
In this mode, two identical polarity signals (same phase) are applied to the inputs, as shown
in the left-hand figure below.
Note that there is no output due to VCM, the common mode voltage because
VCM = V1 - V2 = 0, as shown in the right-hand figure below.
V1
-
A
VCM
A
VOUT = -ACMVCM
VOUT = 0V
+
+
V2
E
This is very useful as unwanted signals (noise) appearing with the same polarity on both
input lines are essentially cancelled out and do not appear on the output.
The common-mode input voltage range is the range of input voltages which, when applied to
both inputs, will not cause clipping or other output distortion. Many op-amps have commonmode rages of no more than +/-10 V with dc supply voltages of +/- 15 V, while in others the
output can go as high as the supply voltages (this is called rail-to-rail).
Common-Mode-Rejection-ratio (CMRR).
The ability of an Op-Amp to suppress common signals is expressed in terms of its CommonMode-Rejection-ratio (CMRR).
CMRR =
AV ( d )
ACM
The higher the CMRR, the better. A very high value of CMRR means that the differential
gain AV(d) is high and the common-mode gain ACM is low. The CMRR is often expressed in
decibels (dB) as:
 AV ( d ) 

CMRR’ = 20 log 
 ACM 
Ideally, an Op-Amp provides infinite gain for desired signals (single ended or differential)
and zero gain for common-mode signals,
Input and Output Impedance
Two basic ways of specifying the input impedance of an op-amp are the differential and the
common mode. The differential input impedance is the total resistance between the inverting
and the non-inverting inputs. The common-mode input impedance is the resistance between
each input and ground.
The output impedance is the resistance viewed from the output terminal of the op-amp.
3
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Characteristics of an Ideal Op-Amp:
ROUT
RIN
VIN
+
-
AolVIN
+
VOUT
The ideal op-Amp has the following characteristics

Input Resistance RIN = .
(i.e. no current demanded at amplifier
input terminals)

Output Resistance ROUT = 0.
(i.e. Output Voltage unaffected by load)

Voltage Gain Aol = 
(i.e. Open Loop Voltage gain is
infinite)
The ideal Op-Amp also has the following characteristics:

Bandwidth= 

Infinite CMRR
(i.e. the response extends from dc to )
(i.e. ideally VOUT = 0 for common mode inputs
where the Common-Mode Rejection Ratio (CMRR) is a measure of an
Op-Amp’s ability to reject common-mode signals )
dVOUT

i.e. the maximum rate of change of VOUT is 
 Slew rate= 
dt
where slew rate is the maximum rate of change of the output voltage in
response to a step input voltage and is dependent upon the high
frequency response of the amplifier stages within the Op- Amp.
Characteristics of a Practical Op-Amp
In practice the IC Op-Amp falls short of the ideal characteristics, however the following
applies
 Very HIGH input resistance
 Very LOW output resistance
 Very large Open Loop Voltage Gain
For example, a popular 741 Op-Amp has the following characteristics:
 Open-Loop voltage gain  200,000
 Input impedance  2 M
 Output impedance  75 
 Bandwidth for unit gain  1 MHz
 CMRR  90dB
 Slew rate  0.5V/s
4
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Need for Negative Feedback
As the Open-Loop Voltage Gain of the Op-Amp is very large, an extremely small difference
in the two input voltages drives the Op-Amp into its saturated output states, i.e. it will cause
the output voltage to go all the way to its extreme positive or negative voltage limit. As this
is seldom desirable the full gain of the Op-Amp is not usually applied to an input, instead
negative feedback (using external resistors) is applied to reduce the overall gain through
signal feedback.
Negative
Feedback
Circuit
Vf
VIN
VOUT =AclVIN
+
Where:
VIN
Vf
Acl
is the input signal
is a portion of the output signal fed back to the inverting input
the closed loop gain is the voltage gain with negative feedback
The Op-Amp responds to the voltage VIN at its non-inverting input, which moves the output
towards saturation. However, a fraction of this output is returned to the inverting terminal
through the feedback path. As the feedback signal approaches the value of V IN  there is
nothing left for the Op-Amp to amplify as
Vf – VIN  0
Thus the feedback signal tries (but never quite succeeds) in matching the input signal and
thus the gain is controlled by the amount of feedback used.
5
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Op-Amp Configurations with Negative Feedback
The Basic Inverting Amplifier:
A signal VIN is applied through a series resistor R1 to the inverting input as shown below.
The ouput is fed back through R2 to the inverting input. The non-inverting input is grounded.
i2
R1
i-
i1
V
VIN
R2
A
+
VOUT
The Closed Loop Gain (ACl) is defined as the gain with feedback applied
Basic Inverting Amplifer
 Closed Loop Gain (ACl) is:

R2
R1
RIN = R1
ACl = 
Input Resistance is ideally
Derive the Closed Loop Gain the Basic Inverting Amplifier:
We call the voltage at the inverting terminal V- and the voltage at the non-inverting terminal
V+.
Then V =
and
V- - V+
VOUT = -AV= -A(V- - V+)
From the diagram above:

=
A(V+ - V-)
V+ = 0 and V- = V
i1 
VIN  V
R1
and:
i2 
V  VOUT
R2
Kirchhoffs Current Law at the inverting terminal gives us:
i1 = i2 + i -
6
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
As the ideal amplifier has infinite input resistance then i- must be 0.
i1 = i2

VIN  V V  VOUT

R1
R2
VIN V
V VOUT



R1 R1 R 2
R2


But:
VOUT = -AV
=>
VOUT
=>

VIN
VOUT

R1
R2
VINR2 = -VOUTR1
R2
VOUT

R1
VIN
As A->  then V  



V 
VOUT
A
V -> 0
VOUT
R2

VIN
R1
ACl =
Where ACl is the Closed Loop Gain and is defined as the gain with feedback applied
Note that A, the gain of the op-amp without feedback, is called the Open Loop Gain
We see that, as expected, the gain is negative and that the gain depends only on the ratio of
the resistor values and not on the amplifier itself.
V=

V- - V+ and as shown above if A ->  then V -> 0
V- = V+
In the configuration above, we have V+ grounded so therefore
V- = 0
We cannot actually ground the inverting terminal but since its potential is V- = 0, we say that
a “virtual ground” exists at the inverting input terminal.
Since V- is at virtual ground the input impedance seen by the signal source generating VIN
is (ideally) R1 ohms.
7
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Example 1:
i2
R1
VIN
R2
i-
i1
V
A
+
VOUT
If R1 = 2k, R2 = 10 k and VOUT = 2V,
(a)
What is VIN?
(b)
What is the input resistance seen by VIN?
VIN?
ACl



VOUT
R2

VIN
R1
=
VOUT
10,000

 5
VIN
2,000
VOUT
2
VIN 
   0.4V
5
5
VIN = -400mV
Input Resistance seen by VIN
Since V- is at virtual ground the input impedance seen by the signal source generating V IN is
R1 ohms.
RIN = R1 = 2k
Note on the behaviour of the Op-Amp circuit:
Since VOUT is held by the supply rails to a maximum of 15V and since A-> , then the
action of the Op-Amp is to “look” at its input terminals and swing its output terminal around
so that the external feedback resistors bring the input differential, V1 to zero if possible
i.e. V =
VOUT  15

0
A

8
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
The Basic Non-Inverting Amplifier:
The input signal is applied to the non-inverting (+) input. A portion of the output signal is
applied back to the inverting (-) input through the feedback network. This constitutes
negative feedback.
i2
R1
R2
i-
i1
V
A
+
VOUT
VIN
Basic Non-inverting Amplifier
 Closed Loop Gain (ACl) is:

ACl = 1
R2
R1
RIN = 
Input Resistance is ideally
Derive the Closed Loop Gain of the Basic Non-inverting Amplifier:
Assuming again infinite input resistance, then i- =0 and i1 = i2
VOUT = -AV= -A(V- - V+)
=
A(V+ - V-)
VOUT
=>
V+ - V- =
A
VOUT
As A->  then
=> 0 and V+ - V- = 0
A

V+ = VV+ = VIN
But:
V IN
VOUT  VIN
and
i2 =
R1
R2
VIN VOUT  VIN

=>
R1
R2
=>
VINR2 = VOUTR1 – VINR1
=>
VIN(R2 + R1)= VOUTR1
VOUT R1  R 2
R2

1
=>
VIN
R1
R1
The Closed Loop Gain (ACl) is:
R2
ACl = 1
R1
=>
i1 =
9
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
The assumption of large voltage gain in the system implies that the output voltage takes on a
value such that the negative feedback forces the potential of the inverting input to equal
(approximately) that of the non-inverting input.
The input impedance seen by the signal source generating VIN is infinite (ideally)
Example 2:
What is the Closed Loop Gain (ACl ) and the input resistance seen by the applied signal VIN in
the following system:
R1 = 10k
R2 = 90 k
i2
R1
R2
i-
i1
V
A
+
VOUT
VIN
Closed Loop Gain
ACl = 1
R2
90,000
 1
 10
R1
10,000
Input Resistance
Since the input resistance to an Op-Amp is infinite (ideally) the signal VIN sees infinite input
resistance
RIN = 
Note also that if the output resistance of the Op-Amp is assumed to be zero, then the output
resistance of the Non-Inverting Amplifier is zero.
If in the previous example, VIN is made 1V what happens?
VOUT is initially 0V. the Op-Amp sees an enormous input imbalance (the + input is at 1V and
the – input is at 0V) and the output VOUT is forced to go positive. When VOUT reaches 10V
then the potential divider of R1 and R2 make the voltage at the inverting input equal to 1V. If
VOUT tried to go above 10V then the voltage across R1 would also rise. The result of a higher
voltage at the inverting terminal is to force VOUT back down to 10V
10
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
The Voltage Follower: (Unity Gain Buffer Amplifier)
The Voltage Follower is a special case of the non-inverting Amplifier, where all of the output
voltage is fed back to the inverting terminal by a straight connection as shown below.
V
A
+
VOUT
VIN
Voltage Follower Amplifier
 Closed Loop Gain (ACl) is:
 Input Resistance is ideally
ACl = 1
RIN = 
Derive the Closed Loop Gain of the voltage follower Amplifier:
VOUT = V + VIN
VOUT
V 
VOUT = -AV
=>
A
VOUT
VOUT = VIN 
A
VOUT
As A ->  then
-> 0
A

VOUT  VIN
R2
The Closed Loop Gain (ACl) is:
ACl = 1
R1
In this case R2 goes to zero and R1 goes to infinity

ACl = 1
It Unity Gain Buffer Amplifier is called a Voltage Follower since V IN = VOUT. The most
important features of the voltage-follower configuration are its very high input impedance
and its very low output impedance. Therefore, it may be used to allow a signal from a high
impedance source to be coupled to a low impedance load.
11
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Example 3:
An inverting amplifier is driven by a signal source whose output impedance is 40k. The
source voltage is 2.2Vr.m.s and R1=15k and R2=39k.
What is the peak value of the output voltage?
R2
RS
R1
VS
VIN
V
A
+
VOUT
R2
VIN
R1
Since the inverting terminal is a virtual ground, the input circuit can be drawn as follows:
VOUT  

RS

VS
R1
VIN

 R1 
VIN  Vs

 R1  RS 
R2
R 2  R1 
VOUT   VIN   Vs

R1
R1  R1  RS 
 R2 
VOUT  Vs

 R1  RS 


39 x10 3
 R2 

  1.56Vrms
VOUT  Vs
  2.2
3
3 
15
x
10

40
x
10
 R1  RS 


VOUTRMS 
VOUTPEAK
2
 VOUTPEAK  2.VOUTRMS  2.(1.56) = -2.2VPEAK
Therefore:
|VOUTPEAK| = 2.2V
Example 4:
An inverting voltage amplifier is driven by a signal source with internal resistance of 1 k.
The source voltage is 1.2 V, the feedback resistor R2 = 10 k and R1 = 5 k.
What is the output voltage?
Gain = - R2/R1 = - 10/5 = -2
Vin = Vs ( R1/(R1 + Rs) ) = 1.2 ( 5000/6000 ) = 1 V
Vout = 1*(-2) = -2 V
12
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Exercises:
Attempt the following exercises.
Example 5:
An inverting voltage amplifier is driven by a signal source with internal resistance of 2 k.
The source voltage is 1.4 V, the feedback resistor R2 = 10 k and R1 = 5 k.
What is the output voltage?
Example 6:
An operational amplifier is configured as a non-inverting amplifier having a gain of 10. A
d.c. voltage of 1 V is applied as an input. Draw the circuit diagram. What is the voltage at
each of the amplifier terminals.
Example 7:
An operational amplifier is configured as an inverting amplifier and driven by a signal source
whose output resistance is 5 k. The source voltage is 2.2V, the feedback resistor R2 = 60
k and R1 = 15 k. What is the output voltage.
Example 8:
It is required to construct an inverting amplifier to have a voltage gain of 20 and an input
resistance of 20 k. Determine the value of the resistive components required. Indicate the
virtual earth point.
Example 9:
 Draw a circuit diagram showing how an operational amplifier may be configured as a
non-inverting voltage amplifier.

Determine an expression for the closed loop voltage gain of this amplifier in terms of the
component values in the circuit. State any approximations used.

Give an example of resistor values that would give a voltage gain of 10.
Example 10:
Draw a circuit diagram showing how an operational amplifier may be configured as a
unity gain voltage amplifier and state why such an amplifier might be used.
Example 11:
Two IC op-amps are available to you. Their characteristics are listed below. Choose the one
you think is more desirable.
RIN
ROUT
AOL
50,000
Op-amp 1
5 M
100 
150,000
Op-amp2
10 M
75 
13
Analogue ELEK1289 - Electronic systems and practice II - Unit 4 – Operational Amplifiers
Summary:
OP-AMP Schematic Symbol And Equivalent Circuit
Inverting
Input
ROUT
Output
A
+
VIN
RIN
+
Aol VIN
VOUT
-
Non-inverting
Input
Common-Mode-Rejection-ratio (CMRR)
The ability of an Op-Amp to suppress common signals is expressed in terms of.
CMRR =
 AV ( d ) 

CMRR’ = 20 log 
 ACM 
AV ( d )
ACM
The ideal op-Amp
has the following characteristics

Input Resistance RIN = .
(i.e. no current demanded at amplifier
input terminals)

Output Resistance ROUT = 0.
(i.e. Output Voltage unaffected by load)

Voltage Gain Aol = 
(i.e. Open Loop Voltage gain is
infinite)
The Basic Inverting Amplifier:
Where ACl is the Closed Loop Gain and is defined as the gain with feedback applied
VOUT
R2

ACl =
VIN
R1
The input impedance seen by the signal source generating VIN is R1
The Basic Non-Inverting Amplifier:
The Closed Loop Gain (ACl) is:
R2
R1
The input impedance seen by the signal source generating VIN is infinite (ideally)
ACl = 1
The Voltage Follower: (Unity Gain Buffer Amplifier)
R2
The Closed Loop Gain (ACl) is: ACl = 1
R1
In this case R2 goes to zero and R1 goes to infinity =>
ACl = 1
It Unity Gain Buffer Amplifier is called a Voltage Follower since VIN = VOUT. It has
high input impedance and low output impedance. Therefore, it may be used to allow
a signal from a high impedance source to be coupled to a low impedance load.
14