Feedback in Operational Amplifier
Lesson: Feedback in Operational Amplifier
Lesson Developer: Dr. Arun Vir Singh
College/Department: Shivaji College, University of Delhi
Institute of Lifelong Learning, University of Delhi
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Feedback in Operational Amplifier
Table of Contents
Feedback in Operational Amplifier
1.1 Introduction
1.1.1 An Amplifier Black Box with Feedback
1.1.2 Types of Feedback
1.2 Block Diagram Representation of Feedback Configuration
1.3 Closed-Loop Operational amplifier
1.4. Noninverting Amplifier
1.4.1 Closed-Loop Voltage Gain
1.4.2 Relationship between Closed-Loop and Feed-Back Circuit Gain
1.4.3 Relationship between Closed-Loop, Open-Loop and Feedback
Circuit Gain.
1.4.4 Input Resistance with Feedback
1.4.5 Output Resistance with Feedback
1.4.6 Band-Width with Feedback
1.4.7 Total Output Offset Voltage with Feedback
1.5 Inverting Amplifier
1.5.1 Closed-Loop Voltage Gain
1.5.2 Relationship between Closed-Loop, Open-Loop and Feedback
Circuit Gain.
1.5.3 Input Resistance with Feedback
1.5.4 Output Resistance with Feedback
1.5.5 Band-Width with Feedback
1.5.6 Total Output Offset Voltage with Feedback
1.6 Virtual Ground
Summary
Exercise
Glossary
References/ Further Reading Source
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1.1 Introduction
The important characteristics of amplifier are voltage gain, output impedance, input
impedance bandwidth and frequency response. These parameters are more or less constant
for an amplifier. Quite often, values of these parameters are required to change. This could
be done in number of ways. Feedback in one of them. The feedback is a process of taking a
part of output (voltage or current) and feeding it back to input signal. An amplifier that uses
feedback is called a feedback amplifier. A feedback amplifier is sometimes referred to as
closed- loop gain amplifier because the feedback forms a closed-loop between input and
output. When no connection exists between the input and output terminals then amplifier is
called open-loop amplifier.
This lesson presents how feedback is achieved and its effect on op-amp characteristics in
inverting and noninverting configurations for practical op-amp.
1.1.1 An amplifier Black Box With Feedback
A typical feedback arrangement in amplifiers will be viewed as “black box,” and is shown
figure 1. No knowledge of internal circuits is needed. In this figure an amplifier is
represented by triangular symbol of gain A.
is the input to the amplifier and
is the
output of amplifier.
A portion of the output is connected to feedback network or β network, represented by a
rectangle symbol. This provides a fraction of output ( ) as feedback signal to the input
mixer network represented by circle.
A
Σ
V
o
Vf
β
Vo
Fig:1 Simple block diagram of feedback amplifier.
Developed by: ILLL
The introduction of the β network requires:
1. That the β network does not load the output or input circuit of amplifier.
2. The input signal be transmitted forward through the amplifier only.
3. The reverse signal be transmitted from the output to the input only through the β
network.
Feedback circuit may be made up of passive components, active components or
combination of both. This lesson presents purely resistive feedback circuits.
1.1.2 Types of Feedback
Depending upon whether the feedback signal increases or decreases the input signal, there
are two basic types of feedback amplifiers
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Feedback in Operational Amplifier
Positive feedback
If the feedback signal (
is applied in such as way that it is in phase with the input signal
and thus the net input voltage
to the amplifier of gain A, becomes sum of the external
input voltage
and . That is
Since the net input to the amplifier is increased so the gain of amplifier increases. This type
of feedback is called a positive feedback. Sometimes it is known as regenerative feedback.
Positive feedback increases the gain of amplifier. However it produces excessive distortion
due to which it is seldom used in amplifiers. The positive feedback is used in oscillators.
Negative feedback
If the feedback signal is applied is such that it is out of phase with the input signal then the
voltage to the amplifier decreases.
That is
Therefore, the gain of the amplifier decreases. This type of feedback is called a negative
feedback. Sometimes, it is also called degenerative feedback.
The feedback can also be classified as voltage or current feedback. In both the cases, the
feedback signal is proportional to the output (voltage or current)
1.2 Block Diagram Representation of Feedback
Configuration
There are four ways to connect the amplifier block and feedback block. These connections
are classified according to whether the voltage or current is feedback to the input in series
or in parallel.
1. Voltage-series feedback (Fig. 2a). In this arrangement both input and output are
voltages. The circuit that uses this type of feedback is called voltage-controlled voltage
source.(VCVS)
+
+
+
Vin
Ao=Vo /Vid
Vid
-
RL
Vo
-
+
+
β=Vf /Vo
Vf=Vo β
-
-
Fig:2a.Voltage-series feedback
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Feedback in Operational Amplifier
2. Voltage-shunt feedback (Fig. 2.b). In the second type of feedback, an input current
controlling an output voltage and is called current controlled voltage source. The circuit
uses this type of feedback is called a current controlled voltage source (ICVS). It is
transresistance amplifier.
Iin
Iin
IB +
+
Ao=Vo /IB
If
RL
Vo
-
+
+
+
β=If /Vo
-
-
Voltage-shunt feedback (Fig. 2.b).
3. Current-series feedback (Fig. 2.c). In this feedback arrangement ,circuit is knows as
voltage-controlled current source(VCIS). Because an input voltage controls the output
current. VCIS is called transconductance amplifier.
Iin
+
Io=IL
+
+
Vin
Ao=Io /Vid
Vid
-
+
RL I L
+
β=Vf /Io
Current-series feedback (Fig. 2.c).
4. Current-shunt feedback (Fig. 2.d). This is the fourth type of negative feedback ,in which
both the input and output are currents and called current-controlled current source (ICIS).
ICIS is called current amplifier.
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Iin
IB
Io=IL
+
+
Iin
Ao=Io /IB
If
IL
RL
+
β=If /Io
Current-shunt feedback (Fig. 2.d).
In all these configurations the signal direction through the op-amp is from input to output.
However, the signal direction through the feedback network is exactly opposite: output to
input.
Voltage-shunt and voltage-series feedback configurations are most commonly used.
Did you know ?
Voltage dividing circuit
Body Text:
In this circuit
and
are connected in series. Vo is the voltage
applied across this combination. Let I is the current flowing
+
through it, and given by Ohm’s Law
Rf
Vo
+
Vf
By Ohm’s law ,voltage across
is
=
=
R1
0
Suggesting Reading:
Electronic Principles :A.P. Malvino ,6th Edition.
Basic Electronics: Solid State :B.L.Theraja
1.3 Closed-Loop Op-Amp
In closed-loop amplifiers a part of output (voltage or current) is fed back (mixed) to the
input signal. This is done by using feedback circuit between input and output terminals of
amplifier. This forms a closed loop, hence the name closed-loop amplifier.
The
voltage-series and voltage-shunt feedback configurations are important and most commonly
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Feedback in Operational Amplifier
used. A detailed analysis of these two configurations is presented in this lesson, computing
voltage gain, band width, input resistance and output resistance for each.
1.4.Noninverting Configuration
It is an example of voltage series feedback arrangement. Figure 3 depicts voltage series
feedback amplifier using op-amp.This circuit is commonly known as non inverting amplifier
with feedback, because it uses a feedback circuit consisting of resistors R1 and Rf.
-VEE
V1
Vin
3
+
V-
A
Vid
3
V2
2
Vo
A
V+
-
Rf
+VCC
+
V2=Vf
R1
-
0
Fig:3. Voltage series feedback amplifier for noninverting amplifier with negative feedback.
Input signal (Vin) is applied at noninverting (+) terminal (pin-3) and inverting (-) terminal
(pin-2) is connected to ground through resistor R1. Feedback voltage (Vf) is producd by
feedback circuit and is applied to inverting terminal, so we can say,
.
Differential voltage (
as shown in the diagram is equal to the input voltage (Vin) minus
feedback voltage (Vf). Op-amp amplifies the differential input. It means that feedback
voltage opposes the input voltage ( because it is out of phase 180o with respect to input
voltage) therefore feedback is said to be negative.
1.4.1 Closed-Loop Voltage Gain
Let us first define some important terms to understand the noninverting amplifier with
negative feedback
(i) Open-loop voltage gain :
[1]
Where
is the differential voltage, and given by
, where
is the voltage at
the noninverting terminal and
is the voltage at the inverting terminal . is output
voltage.
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(ii) Closed-loop Voltage (Feedback) gain
[2]
(iii) Gain of feedback circuit
[3]
Where
is feedback voltage.
Since voltage appearing at noninverting terminal
terminal
and voltage appearing at inverting
By voltage dividing rule
Substituting for
and
[4]
in expression,
[5]
Substituting Eq.[5] in Eq.[1]
Comparing with Eq.[2]
[6]
This is the actual closed-loop gain of a noninverting op-amp or voltage gain of practical
op-amp. It is determined by both the external resistors and open-loop gain of op-amp.
Since
is of the order off 105. Thus
and
. Thus
[7]
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Hence the closed-loop gain for ideal op-amp is independent of open-loop gain and it
depends only on the values of externally connected resistors,
and
Substituting for
in
Eq.[2]
[8]
[9]
Where
, is
a constant it is known as positive scalar.
Fig. 4: Output voltage as a function of input voltage for noninverting op-amp,
Source: Self
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Feedback in Operational Amplifier
Variation of output voltage as a function of input voltage is obtained using probe and is
shown in figure [4], for
,
, power supply voltage =± 15V and,
.
Output voltage is proportional to input voltage in the range -2.5 <Vin< 2.5. There after a
saturated output is obtained because the calculated output voltage
=15V, cannot be larger than the output voltage swing ±13V.
Output voltage is of same polarity as that of input voltage. Hence the name noninverting
amplifier.
Value Addition: Did you know?
Noise gain and noise minimisation
Body Text:
Noise will be amplified as the input signal. Since the gain of the noninverting voltage
amplifier is,
, so noise will be amplified by the same amount.
Minimisation:
(i) Avoid large resistor values.
(ii) Connect a small capacitor 3pf across the feedback resister. This reduces the noise gain
at high frequencies
Suggested Reading:
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition.
1.4.2 Relationship between Closed-Loop Gain and Feedback Circuit
Gain
Substituting for
from Eq.[4] in Eq. [3],
[10]
Comparing with Eq.[7]
[11]
Thus the gain of feedback circuit is reciprocal of the closed-loop voltage gain.
1.4.3 Relationship Between Closed-Loop , Open-Loop
Feedback Circuit Gain.
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and
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Feedback in Operational Amplifier
Dividing numerator and denominator of Eq. [6] by
[12]
Gain of feedback amplifier is dependent on open-loop gain and gain of feedback circuit (On
R1 and Rf).
How the negative gain stabilizes the overall voltage gain: If the open-loop voltage gain (
)
increases, the output voltage will increase and feedback loop sends more feedback voltage
to the inverting input. So Vid decreases. Therefore, even though open-loop voltage gain
increases,
decreases, and the final output (
) increases much less than it would
without the negative feedback. The overall result is very slight increase in output voltage.
Did you know?
What is loop gain and why its value is kept very large?
Body Text:
β is called loop gain because it is the voltage gain of forward and backward path. It is
defined as the ratio of open-loop gain to closed-loop gain. It determines the circuit
performance. In practice its value is kept very high to stabilize the voltage gain and
to
enhance the effects on other parameters of op-amp. It is desired that exact values should
be equal to ideal values or % error should be less. To understand this let us calculate the %
error using the relation.
(i)
,
(ii)
,
(iii)
. This means the exact gain is 0.1% less than ideal value.
. In this case exact gain is 1% less than ideal value.
,
. Exact gain is 5% less than ideal value.
Hence higher the value of
β ,lesser is the % error
Suggested Reading:
Electronic Principles :A.P. Malvino ,6th Edition.
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition.
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1.4.4 Input Resistance with Feedback
Let us define input resistance
(i) without feedback using Ohm’s law.
[13]
(ii) with feedback
[14]
Differential input is given by
, Substituting for
, Substituting for
, Substituting for
from Eq.[1]
from [Eq.13]
,
comparing with Eq.[14]
[15]
This equation reveals that series voltage feedback increases the input impedance of
amplifier (Noninverting) by a factor of
. Since
is ≈
, hence
will be very
high.
1.4.5 Output Resistance with Feedback
Output resistance of noninverting op-amp in open-loop configuration is
[16]
By applying Kirchhoff’s law to the output loop for the circuit shown in figure 5
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Fig:5. Output resistance of noninverting op-amp: Developed by :ILLL
,
, substituting for
Dividing both sides by
, substituting for
from Eq.[16]
[17]
This equation shows that output resistance of the op-amp with feedback decreases and is
very much less than the output resistance of the op-amp without feedback.
1.4.6 Bandwidth with Feedback
The bandwidth of an amplifier is defined as the range of frequencies for which gain remains
constant. Open-loop voltage gain
of 741 op-amp is 2×105 and bandwidth 5Hz. It has
one break frequency ( . Gain bandwidth product is 2×105×5=1MHz. If gain is unity
bandwidth is 1MHz .Thus, gain bandwidth product is constant.
The frequency at which the gain equals 1 is knows as unity gain-band width (UGBW).741
has unity gain-bandwidth of 1MHz.
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Since for an op-amp with single break frequency, the gain band-width product is constant
and is equal to the unity gain-bandwidth (UGBW).
[18]
For a single break frequency one can also write
[19]
where
is closed-loop gain and
[18] and [19]
, substituting for
is band-width with feedback. Therefore equating Eqs.
from Eq.[12]
[20]
This equation reveals that bandwidth increase with negative feedback. Since gain-bandwidth
product of an op-amp is constant, if bandwidth increases then gain decreases .
1.4.7 Total Output Offset Voltage with Feedback
For an op-amp amplifier without any input, output is expected to be zero. In practice it is no
so, its effect can be minimized using negative feedback. Since gain reduces
to
due to negative feedback in noninverting op-amp. So the total output offset voltage must be
decreased and is given by
[21]
Did you know?
Effect of input offset voltage and bias current on output voltage.
Body Text:
Output offset voltages are caused by input offset voltage (Vio) and input bias current (IB).
This could be positive or negative. This causes error in the output voltage because these
voltages also get amplified along with the input signal.
Maximum output offset voltage (error Voltage ) is given by equation
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This value is same for inverting and noninverting amplifier
Output offset voltage due to bias current is
Total output offset voltage is sum of these voltages
This can be minimised by taking small value of
decreases and product
Hence
. For small value of Rf ,
also decreases.
will decrease.
Suggested Reading:
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition
Electronic Devices by Thomas L. Floyd,6th Edition.
1.5 Inverting Amplifier
It is a voltage shunt fed back amplifier and is illustrated using an op-amp in figure 6. In this
configuration noninverting terminal (+) is connected to ground where as input (Vin) is
applied to inverting (-) terminal through resistor R1. Combibation of resistors R1 and Rf
forms the feedback circuit. This arrangement provides a negative feedback because
input signal after amplification and after inversion , is applied to input terminal via feedback
resistor. It results in decrease in output by decreasing the input signal.
1.5.1 Closed-Loop Voltage Gain
In circuit shown in figure 6, noninverting terminal is grounded, so
. Because of very
high value of input resistance ( ,[Fig.9, sec. 1.4.3,Lesson - Characteristics and open-loop
op-amp] of op-amp, no current flows through Ri . Or no current flows from inverting (-)
terminal to noninverting (+) terminal. So no voltage drop occurs between these input
terminals. This means that voltages of these terminals are equal.
Rf
If
R1
V2 N
Vin
Iin
2
IB
V1
A A
3
0
-VEE
-
V-
Vo
V+
+
+VCC
0
Fig:6. Voltage shunt series feedback amplifier for inverting amplifier with negative feedback.
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Currents at the node V2 by Kirchhoff’s Current Law
[22]
Where
is bias current and
is current through feedback resistance
Since no current is at inverting input (
are equal.
Or
, negligibly small), so current through R1 and Rf
(Using Ohm’s law)
Since
but
Substituting for
[23]
(being grounded). So
and
in Eq.[20]
, Substituting
for
Eq.[2]
[24]
This is the actual closed-loop gain of a practical inverting op-amp. It is determined by both
the external resistors and open-loop gain of op-amp. Minus sign indicate change of phase
between input and output is 180o .Because of this phase inversion; the configuration is
called as inverting amplifier with feedback.
Since
is very high. (105).Therefore
and
(ideal case )
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[25]
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Feedback in Operational Amplifier
This equations shows that the in case of ideal op-amp gain of the inverting amplifier is set
by selecting a ratio of feedback resistance
to input resistance
. The ratio
can be set
to any value whatsoever, even to less than 1. Because of this property, inverting
configuration lends itself to a majority of applications as against those of the noninverting
amplifier.
If the open-loop voltage gain increases, the output voltage will increase and feedback
voltage to the inverting input terminal will also increase. This negative feedback voltage
thus reduces V2 , thus the final output increases much less than it would without the
negative feedback.
Triangular wave input (green color) ( -1< Vin <1) and output wave forms ( red color )
for an inverting op-amp of gain 5, is obtained by using probe and is illustrated in figure
7. It can be seen from the figure that output is inverted or out of phase by 180o with
respect to input.
Fig. 7: Inverting amplifier input and output wave forms as a function of time,
Source:Self
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Feedback in Operational Amplifier
Did you know?
Closed-loop gain of inverting op-amp, when input resistance
resistance
of op-amp is considered
Body Text:
and output
Suggested reading:
Operational amplifier and linear integrated Circuits by Rajiv Kapadia
1.5.2 Input Resistance with Feedback
The input resistance of inverting amplifier with negative feedback can be understood with
the help of figure 8[a].Input signal and feedback signal are applied to inverting terminal
through input resistor ( ) and feedback resistor ( ). Input resistance for an inverting
negative feedback amplifier is
Fig 8(a).
[26]
Rf
V2 N
2
(a)
0
AA
V1
3
0
R1
-
Vin
Virtual
Ground: 0V
Vin
+VCC
R1
Vo
+
Virtual Ground
(b)
-VEE
0
Fig:8. Input resistance
This can be explained on the basis of concept of virtual ground. Since the inverting terminal
of inverting op-amp is at virtual ground potential, and the input signal sees
, to ground
as illustrated in figure 8(b). Hence input resistance with feedback is
.
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1.5.3 Output Resistance with Feedback:
Output resistance in this case is same as that of noninverting op-amp.
[27]
So the output resistance of the op-amp with feedback decreases and is very much less than
the output resistance of the op-amp without feedback. (For derivation sec. 1.4.5)
1.5.5 Relationship between Closed-Loop Gain ,Open-Loop Gain and
Feed Back circuit Gain.
Dividing numerator and denominator of Eq.[24] by
[28]
where
and
Comparing Eq.[28] with Eq.[12],
[Closed loop gain of noninverting op-amp ]
[29]
Closed- loop gain of the inverting op-amp is K time the closed-loop gain of noninverting
op-amp, where K<1. So closed- loop gain of the inverting op-amp is less than the closedloop gain of noninverting.
1.5.6 Total Output Offset Voltage with Feedback
For an op-amp amplifier without any input output is expected to be zero. In practice it is no
so, its effect can be minimized using negative feedback. Since gain reduces
to
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Feedback in Operational Amplifier
due to negative feedback in noninverting op-amp. So the total Output Offset Voltage must
be decreased and is given by
[30]
Did you know?
How open-loop gain affects the actual closed-loop gain of op-amp.
Body text:
Consider an inverting op-amp circuit having
(Rf=100K, R1= 1k) and
=50
(Rf=50K, R1=1k). For different values of Ao=1,10,100,1000,10000,10000, actual closedloop gain (
was calculate using Eq.[28] and plotted against the open-loop gain (
, as
shown below.
Closed-loop Gain (Af)
120
100
80
60
40
20
0
1
10
100
1000
10000 100000
Open-loop Gain (Ao)
Fig.: Variation of actual closed-loop with open-loop gain of op-amp.
The ideal closed-loop gains are shown by dotted lines. With the increase in open-loop gain
, closed-loop gain
increases. For higher values of
, actual closed-loop gain is equal
to the ideal close-loop gain or we can say at high values of
, actual closed-loop gain does
not depend on
.
So that the external resistors and not the op-amp’s
determine the actual gain.
This is also applicable in the case of noninverting amplifier.
Source:Self
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1.5.5 Bandwidth with Feedback.
For an open-loop operational amplifier, single break frequency (fo) value is 5Hz and
open-loop gain (Ao) is equal to 2×105. Gain-bandwidth product is 1MHz. On the other hand
bandwidth is 1MHz for unity gain. In other words gain-bandwidth product of a single break
frequency op-amp is always a constant quantity.
Unity gain-bandwidth (UGBW), may be written as
[31]
Alternatively ,only for one break frequency
[32]
Equating Eqs.[31] and [32]
[33]
For noninverting amplifier gain with feedback is,
refer to Eq.[12]
Substituting for Af in Eq.[33]
.
[34]
So the bandwidth op-amp with negative feedback increases by a factor of
.
Substituting for fo in Eq.[34]
. Substituting
from Eq.[28]
[35]
where
[36]
If
is known, closed-loop bandwidth of inverting amplifier can be obtained using Eq.[34]
and use equation [35] if unity gain-bandwidth (UGBW) is given.
For the closed-loop gain (
) equal to unity let us compare the bandwidth of inverting and
noninverting amplifiers.
For inverting amplifier bandwidth is,
from Eq.[35] and
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bandwidth for noninverting amplifier is
[using .Eq.19]
Thus the closed-loop bandwidth for inverting amplifier is lower than for the noninverting
amplifier by a factor of K<1.
Frequency response curve for voltage gain 5 (≈14dB) and 10 (20 dB) is illustrated in figure
9.Cut off frequencies are shown by dotted lines. For 14dB gain cut off frequency is
≈180kHz and for the amplifier having gain 20dB is 90kHz. So higher is the gain lesser is
the bandwidth.
Fig:9. Frequency response curve and bandwidth (Gain= 5 (14dB) and 10 (20dB) for an
inverting op-amp.
1.6
Virtual ground
To understand this concept, let us first understand mechanical ground and mechanical
short.
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(i) Mechanical Ground: When
a piece of wire is connected between ground and some
point in the circuit. The potential of that point becomes zero. This wire provides a path for
current to flow to ground. So we can say that it is ground for both voltage and current.
(ii) Mechanical Short: When a piece of wire is connected between two points in a circuit
The potential of both points with respect to ground is same. This wire provides a path for
current to flow between the two points. So we can say that it is short for both voltage and
current.
A virtual ground is different. Its concept is based on two characteristics of an ideal op-amp
(i) infinite open-loop voltage gain (Ao) and, (ii) infinite input resistance (Ri).
Rf
+VCC
R1
Vin
V2 N
Virtual
Ground: 0V
V1
2
-
R
Ai
A
Vo
3
+
0
-VEE
0
Fig.10: Virtual ground in inverting op-amp
Because of these characteristics we can deduce the following properties.
(i)
(Input bias current) is zero because of infinite value of
.
(ii) Maximum output voltage of op-amp is
65μV , where
where
, but
is the open-loop gain of op-amp.
=0 ,so
This value of
compared to all other voltages is very small and may be considered to be ≈
zero volt, the virtual ground shown in figure [10]. It is not exactly 0V.
Since is zero in shown in figure, the current passing through resistance
to the current through
.
must be equal
The concept of virtual ground implies that although voltage is ≈ zero, but no current can
flow through the op-amp input resistance Ri to ground. We can say that inverting input
acts like a ground for voltage but open for current. A dotted line in figure 10 between
inverting input terminal and ground shows that no current can flow to the ground.
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This concept is used in ideal inverting op-amp voltage amplifier. [ See Next Lesson
Applications of op-amp]
Figure11 shows inverting op-amp configuration, which has been used to obtain voltages (V2)
at noninverting terminal and output voltage (Vo) along with the current at node (N) using
P-spice probe. Gain of the circuit is 5.
Fig.11: Practical inverting op-amp circuit. Source: Self
It can be seen that potential at inverting input terminal is 48.9μV, it is not zero. This value
is close to zero, so we can say it is at virtual ground.
Output voltage is -4.999V which is less than the ideal voltage -5.0V. This slight difference
from the ideal op-amp is due to the actual gain and input impedance of 741 op-amps.
Output voltage is inverted. At node N input bias current
, which is very small,
and
is equal to
(1 mA) .
Virtual Short: This can be used to analyze quickly noninverting op-amp. The following
properties are used
(i) Infinite open-loop voltage gain (Ao)
Maximum output voltage of op-amp is
65μV
Where
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Feedback in Operational Amplifier
This value of
zero volt.
So,
compared to all other voltages, is very small and may be considered to be
or
(ii) Infinite input resistance (Ri). Both input currents are zero.
-VEE
Vin
V1
3
+
V-
A
Virtual
Short
V2
2
3
A
V+
-
Rf
+VCC
+
R1
Fig. 12: A virtual short in noninverting op-amp
-
0
Dotted line in figure 12 shows a virtual short between the inverting and noninverting input
terminals of noninverting op-amp. It is short for the voltage but an open for a current.
Because of virtual short, inverting input voltage follows the noninverting voltage.
Fig.13: Noninverting amplifier. Source: Self
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Figure 13 shows the circuit of a non inverting op-amp. Ideal gain =11. Voltages and
currents are obtained using probe. For an input of 1V, the resulting output will be 11V.
It can be seen from the figure that voltage of inverting terminal (V2) and noninverting
terminals (V1) are equal to input voltage Vin.
Magnitude of output voltage is 11.01 V which is more than the ideal voltage 11.0V. This
slight difference from the ideal op-amp value is due to the actual gain and input impedance
of 741 op-amp.
Do you know ?
Following facts related to op-amp amplifiers.
Body Text:
(i) All external components values should be less than 1 MΩ so that they do not
adversely affect the internal circuitry of the op-amp.
(ii) Noninverting amplifier with feedback has the characteristics of the perfect
voltage amplifier, because of its very high input resistance, very low output
resistance , stable voltage gain, large bandwidth and very little output offset
voltage.
Suggested Reading:
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition.
Electronic Principles :A.P. Malvino ,6th Edition.
SUMMARY:
Upon completing this lesson on you will be able to :
 Draw the schematic for inverting and noninverting amplifiers.
 Calculate exact closed-loop gain of op-amp in inverting and noninverting amplifier
configuration.
 Plot/Show the output waveforms of either an inverting or a noninverting amplifier for
any input voltage shape.
 To obtain the transfer curve
 Show how closed-loop gain, of inverting and noninverting amplifier depends on openloop gain.
 To measure the frequency response of inverting and noninverting amplifier.
 Explain virtual ground and virtual short in op-amp.
 Understand the effect of negative feedback on characteristics of op-amp.
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Feedback in Operational Amplifier
Question Number
Type of question
1
Multiple choice
1. An inverting amplifier has a closed-loop gain of 50. If an op-amp having an open-loop
gain of 30,000 is replaced by another op-amp with an open-loop gain of 60,000 , then
closed-loop gain
(a) doubles (b) drops to 12.5 (c) remains at 25 (d) increases slightly
2.With the use of negative feedback the gain and bandwidth product of an op-amp
(a) increases (b) decreases (c) remains same (d) fluctuates
3.Phase difference between input and output for noninverting op-amp is
(a) 180o (b) zero (c) 60o (d) (90o)
4. An op-amp has a closed-loop gain of 10 and an upper critical frequency of 10 MHz, the
gain-bandwidth product is
(a) 20 MHz (b) the unity-gain frequency MHz (c) 100 (d) answers (b) and (c)
5. Current cannot flow to ground through (a) mechanical ground (b) an ac ground (c) an
ordinary ground (d) A virtual ground.
Ans.
1.c
2.remains same
3.zero
4.d
5.d
Question Number
Type of question
2
Fill in the blanks
1. For an op-amp with negative feedback, the output is fed back to ----------- terminal.
2. A noninverting amplifier has input resistance of 20kΩ and an feedback resistance of
100kΩ. The closed-loop gain is --------.
3. Negative feed-back ------ noise.
4. The frequency at which the open-loop gain is equal to 1 is called ---------.
5. Input and output of inverting amplifier is ------ phase.
6. Negative feedback-------- the voltage gain.
7. The feedback fraction β is ---- than 1.
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`
1.inverting
2 .6
3. reduces
4 unit-gain frequency.
5. out of
6. decreases
7. less
Ans.
Question Number
3
Type of question
Subjective Questions
1. What is feedback? Which type is used in linear applications.
2.Why feedback is desirable in amplifier applications.
3. What is the difference between open-loop and closed-loop gain of op-amp.
4.What is the effect of negative on the input and output impedances of a voltage series
feedback amplifier?
5. What is band width?
6. What is the effect of negative feedback on the bandwidth of an amplifier?
7. What kind of feedback is present in a noninverting op-amp?
8. What is an noninverting amplifier?
9. Draw the Circuit diagram of an op-amp used as a noninverting amplifier and drive the
expression for its voltage gain, input resistance.
10. Draw the Circuit diagram of an op-amp in inverting configuration and drive the
expression for its voltage gain. Show that
.
β
Question Number
4
Type of question
Problems
1. An op-amp has a dc open-loop gain of 2×105 . It is used in closed-loop configuration
as a (i) an inverting voltage amplifier and (ii) noninverting voltage amplifier. Calculate
the actual and ideal closed loop voltage gain for input resistance R 1=10kΩ and feedback
resistance Rf=100kΩ. (iii) Compare the results.
2.The 741 is configured as an closed-loop inverting amplifier with R1=470 Ω and
Rf=4.7kΩ.The op-amp connected in the circuit has the following parameters: Ao=2×105,
Ri=2MΩ, Ro=75Ω, and fo=5Hz.Compute Af, Rif,Rof and ff.
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3. Determine the voltage gain, input resistance, output resistance, bandwidth and total
offset voltage for a closed-loop noninverting op-amp ,if input resistance R1=470Ω and
feedback resistance Rf=4.7kΩ. Op-amp has the following specifications. : Ao=4×105,
Ri=2MΩ, Ro=60Ω, UGBW=0.6MHz,Supply voltage=±15 V Maximum output swing
voltage=±13 V
4. For an ideal op-amp if input resistance R1=10kΩ ,feedback resistance Rf=40kΩ and
Vin= 20mVpp. Compute the closed voltage gain and bandwidth. What is the output voltage
at 250kHz and at 1MHz.
5.The unity –gain bandwidth of an op-amp is 10 MHz. It is used to design a noninverting
amplifier with an ideal closed-loop gain of 100. Determine (i) small signal bandwidth(ii) Af
at ff.
6. A noninverting op-amp has a voltage gain of 600 without feedback, and 50 with
feedback. Calculate the percentage of output which is feedback to the input.
Ans.
1. (i) (Af)Ideal = -10 (Af)Practical ==-9.9994
(ii) (Af)Ideal = 11 (Af)Practical = 10.999
(iii) In both the cases Exact and ideal gains are
2. Af = -9.999
3. Af = 10.99
Rif =470
,
ff =90.91KHz
Rif =727.3×109Ω, Rof =
4.
Solutions:
1.
Rof =
,
1: For given values of
β
,
approximately equal.
Ω , ff =
,
,
,
Ω,
Ω,
,
β
(i) For Inverting Amplifier
=-9.9994
β
(ii): For Noninverting Amplifier
β
2.
For
Ω,
Ω,
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β
,
β
(i)
Rif=470Ω
(iii)
Ω
β
(iv)
3.
= -9.9994
β
β
For
Ω,
Ω,
β
(i)
=10.99
β
Rif=Ri(1+Aoβ)=2×106×36364.63=72729.26×106=727.3×109Ω
(iii)
(iv)
Ω
β
=
4.
(i)
(ii)
(iii) Gain decreases at the rate of 20dB per decade ,so gain at 25 kHz=12dB=4
(iv)
=20mVpp
5.
, Gain at ff is the .707×A ,or
6.
Voltage gain without feedback
, and with feedback
=70.7
=60
, β=0.015 =1.5%
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GROSSARY
Amplifier: An electronic circuit used to amplify current or voltage.
Bandwidth: Characteristics of amplifier that specify the usable range of frequencies. In
this range gain /output remains constant.
Break frequency: The frequency at which the gain is 3dB down from its value at 0 Hz.
Closed- Loop gain: Gain of op-amp with feedback.
Closed- Loop: An op-amp configuration in which the output is connected back to input
through feedback resistor
Feedback: The process of sending back a small fraction of circuit output to input. It could
be in phase or out of phase.
Gain –bandwidth product: The frequency at which gain of an op-amp is 1(0dB).
Input bias current: It is the average of the currents that flow into inverting and
noninverting terminals of op-amp.
Inverting amplifier: Closed-loop op-amp in which input is applied to inverting terminal of
op-amp.
Inverting terminal: A terminal which produces output 180o out of phase with respect to
input signal .
Node:
Noninverting amplifier: Closed-loop op-amp in which input is applied to non inverting
terminal of op-amp.
Noninverting terminal: A input applied to it , output
Open Loop: An op-amp configuration in which the output is connected to input.
References/ Bibliography/ Further Reading Source:
Op-Amps and linear Integrated Circuits :Ramakant A. Gayakwad,3rd Edition.
Electronic Principles :A.P. Malvino ,6th Edition.
Electronics Devices and Circuit Theory: Robert.L Boylestad and L. Nashelsky.,2nd Edition.
Electronic Devices : Thomas L. Floyd,6th Edition.
Electronics Fundamentals and Applications: J.D.Ryder, Vth Edition.
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