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INTRODUCTION
An operational amplifier, or Op-Amp, is a very high gain differential amplifier with high input
impedance and low output impedance. Typical uses of the operational amplifier are to provide
voltage amplitude changes (amplitude and polarity), oscillators, fiber circuits and many types of
instrumentation circuits. An Op-Amp contains a number of differential amplifier stages to achieve
a very high voltage gain.
Figure 4.1 shows a basic Op-Amp with two inputs and one output as would result using a
differential amplifier input stage. Each input result in either the same or an opposite polarity (or
phase) output, depending on whether the signal is applied to the plus (+) or the minus (-) input.
Input 1
Output
Input 2
Figure 4.1 Basic Op-Amp
4.2.1Single-Ended Input
Single-ended input operation results when the input signal is connected to one input
with the other input connected to ground. Figure 4.2 shows the signals connected
for this operation. In Fig. 4.2a, the input is applied to the plus input (with
minus input at ground), which results in an output having the same polarity as
the applied input signal. Figure 4.2b shows an input signal applied to the
minus input, the output then being opposite in phase to the applied signal.
4.2.2 Double-Ended (Differential) Input
In addition to using only one input, it is possible to apply signals at each
input-this being a double-ended operation. Figure 4.3a shows an input,
Vd, applied between the two input terminals (recall that neither input is
at ground), with the resulting amplified output in phase with that
applied between the plus and minus inputs. Figure 4.3b shows the same
action resulting when two separate signals are applied to the inputs, the
difference signal being 𝑉𝑖1 − 𝑉𝑖2
.
(a) (b)
Figure 4.3 Double-ended (differential) operation.
4.2.3 Double-Ended Output
While the operation discussed so far had a single output, the Op-Amp can also be
operated with opposite outputs, as Shown in Fig. 4.4. An input applied to either input
will result in outputs from both output terminals, these outputs always being opposite
in polarity. Figure 4.5 shows a single-ended input with a double-ended output. Asshown, the signal applied to the plus input results in two amplified outputs of
opposite polarity. Figure 4.6 shows the same operation with a single output measured
Figure 4.4 Double-ended output
Figure 4.5 Double-ended output with single-ended input Figure 4.6 Double-ended output
between output terminals (not with to ground). This difference output signal is 𝑉𝑂1 −
𝑉𝑂2
. The difference output is also referred to as a floating signal since neither output
terminal is the ground (reference) terminal. Notice that the difference output is twice
as large as either 𝑉𝑂1
𝑜𝑟 𝑉𝑂2
since they are of opposite polarity and subtracting them
results in twice their amplitude [i.e., 10 V - (-10V) = 20 V]. Figure 4.7 shows a
differential input, differential output operation

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Electronic Devices, 9th edition
Thomas L. Floyd

The Operational Amplifier
(Op-Amp)

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Chapter Goals
•Develop understanding of linear amplification
concepts such as:
–Voltage gain, current gain, and power gain
–Gain conversion to decibel representation
–Input and output resistances
–Biasing for linear amplification
–Distortion in amplifiers
–Two-port representations of amplifiers
–Understand behavior and characteristics of ideal differential
and op amps.
–Demonstrate circuit analysis techniques for ideal op amps.
–Characterize inverting, non-inverting amplifiers.

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Amplification
Introduction
A complex periodic signal can be represented as the sum of many
individual sine waves. We consider only one component with
amplitude
V
s = 1 mV and frequency ω
s with 0 phase (signal is
used as reference):

Amplifier output is sinusoidal with same frequency but different
amplitude
V
o and phase θ:

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Amplification
Introduction (cont.)
Amplifier output power is:

Here, we desire
P
O = 100 W with R
L = 8 Ω and V
s = 1 mV


Output power also requires output current which is:



Input current is given by

Output phase is zero because circuit is purely resistive.

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Thomas L. Floyd

Amplification
Voltage Gain & Current Gain
•Voltage Gain:
Magnitude and phase of voltage gain are given by
and
For our example,
•Current Gain:
Magnitude of current gain is given by

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Thomas L. Floyd

Amplification
Power Gain
•Power Gain:


For our example,

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Thomas L. Floyd

Amplification
Expressing Gain in Decibels (dB)
The logarithmic decibel or dB scale compresses the huge numeric range of
gains encountered in real systems.

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Thomas L. Floyd

Amplification
Expressing Gain in dB - Example
For our example:

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Mismatched Source and Load
Resistances
In introductory circuit theory, the maximum power transfer
theorem is usually discussed.
Maximum power transfer occurs when the source and load
resistances are matched (equal in value).
In most amplifier applications, however, the opposite situation is
desired.
A completely mismatched condition is used at both the input and
output ports of the amplifier.

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Thomas L. Floyd

Mismatched Source and Load
Resistances
If R
in >> R
s and R
out<< R
L, then
In an ideal voltage amplifier,
and
R
out = 0
For the voltage amplifier shown

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Thomas L. Floyd

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Thomas L. Floyd

Distortion in Amplifiers
•In this graph,
different gains for
positive and
negative values of
the input cause
distortion in the
output.
•Total Harmonic
Distortion (THD) is
a measure of signal
distortion that
compares
undesired harmonic
content of a signal
to the desired
component.

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Total Harmonic Distortion
dc
desired
output
2nd harmonic
distortion
3rd harmonic
distortion
Numerator = rms amplitude of distortion terms

Denominator = desired component

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Differential Amplifier
Basic Model
v
o= A v
id
A
= open- circuit voltage gain
v
id = (v
+- v
-) = differential input
signal voltage
R
id = amplifier input resistance
R
o = amplifier output
resistance
An ideal differential amplifier produces an
output that depends on the voltage
difference between its two input terminals.
Signal developed at amplifier output is in
phase with the voltage applied at + input
(non-inverting) terminal and 180
o
out of
phase with that applied at - input (inverting)
terminal.

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Differential Amplifier Model
Impact of Source and Load
R
L = load resistance
R
S = Thevenin equivalent
resistance of signal source
v
s = Thevenin equivalent voltage
of signal source
•Op amp circuits are mostly dc-coupled amplifiers. Signals v
o and v
s may have
a dc component representing a dc shift of the input away from the Q-point.
•Op-amp amplifies both dc and ac components.

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Differential Amplifier Model
Example including Source and Load Resistances
•Problem: Calculate voltage gain for an amplifier








•Given Data: A = 100, R
id = 100kΩ, R
o = 100Ω, R
S = 10kΩ, R
L =
1000Ω

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Differential Amplifier Model
Example including Source and Load Resistances
•Analysis:




•An ideal amplifier’s output depends only on the input voltage
difference and not on the source and load resistances. This can
be achieved by using a fully mismatched resistance condition
(
R
id >> R
S or infinite R
id, and R
o << R
L or zero R
o ). Then:

•A = open-loop gain (maximum voltage gain available from the
device)

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Thomas L. Floyd

Operational Amplifers
Op-amp is an electronic device that amplify the difference of voltage
at its two inputs.
–
+
+V
–V
Very high gain dc coupled amplifiers with differential inputs.
One of the inputs is called the inverting input (−); the other is called the
non-inverting input. Usually there is a single output.
Most op-amps operate from plus and minus supply voltages, which
may or may not be shown on the schematic symbol.

1
20
1
8
DIP
1
8
DIP SMT
1
8
SMT

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The Ideal Op- Amp
Ideally, op-amps have characteristics (used in circuit analysis):
Infinite voltage gain
Infinite input impedance (does not load the driving sources)
Zero output impedance (drive any load)
Infinite bandwidth (flat magnitude response, zero phase shift)
Zero input offset voltage.

–
+
Z
in = ‘V
in V
o ut
Z
o ut = 0
A
vV
in
A
v
= ‘
The ideal op-amp has characteristics that
simplify analysis of op-amp circuits.
The concept of infinite input impedance is
particularly a valuable analysis tool for
several op-amp configurations.

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The Practical Op- Amp
Real op-amps differ from the ideal model in various respects.
In addition to finite gain, bandwidth, and input impedance,
they have other limitations.
–
+
Z
inV
in V
o ut
Z
o ut
A
vV
in
 Finite open loop gain.
 Finite input impedance.
 Non-zero output impedance.
 Input current.
 Input offset voltage.
Temperature effects.

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Internal Block Diagram of an Op-Amp
Internally, the typical op- amp has a differential input, a
voltage amplifier, and a push- pull output. Recall from the
discussion in Section 6-7 of the text that the differential
amplifier amplifies the
difference in the two inputs.
Differential
amplifier
input stage
Voltage
amplifier(s)
gain stage
Push-pull
amplifier
output
stage
V
in V
o ut
+
–

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Input Signal modes
As the input stage of an op- amp is a differential amplifier,
there are two input modes possible: differential mode
and common mode.
In differential mode any one of the two scenarios can
occur.
Either one input is applied to one input while the other
input is grounded (single-ended).
Or opposite polarity signals are applied to the inputs
(double-ended).

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Thomas L. Floyd

Differential Mode Operation
V
in
–
+
V
o ut
V
in
–
+
V
o ut
Single-ended differential amplifier
Double-ended differential amplifier

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Common Mode Operation
The same input tend to cancel
each other and the output is
zero.
This is called common- mode
rejection.
V
in
V
in
–
+
V
o ut
In common mode, two signals voltages of the same
amplitude, frequency and phase are applied to the two
inputs.
This is useful to reject unwanted signal that appears to both
inputs. It is cancelled and does not appear at the output

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Common- Mode Rejection Ratio
The ability of an amplifier to amplify differential signals and
reject common- mode signals is called the common-mode
rejection ratio (CMRR).
CMRR is defined as
ol
cm
A
A
=CMRR
where A
ol is the open-loop
differential-gain and
A
cm is
the common-mode gain.
CMRR can also be expressed in decibels as
20log
ol
cm
A
A

= 

CMRR
A
cm is zero in ideal op-amp and much less than 1
is practical op-amps.
A
ol ranges up to 200,000 (106dB)
CMRR = 100,000 means that desired signal is amplified 100 ,000 times
more than un wanted noise signal.
Op-Amp parametrs

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Common- Mode Rejection Ratio
Example
What is CMRR in decibels for a typical 741C op- amp?
The typical open-loop differential gain for the 741C is 200,000 and the
typical common-mode gain is 6.3.
90 dB
20log
ol
cm
A
A

= 

CMRR
200,000
20log
6.3
= =
(The minimum specified CMRR is 70 dB.)

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Maximum Output Voltage Swing
V
O(p-p): The maximum output voltage swing is determined
by the op- amp and the power supply voltages
With no input signal, the output of an op-amp is ideally 0 V.
This is called the quiescent output voltage.
When an input signal is applied, the ideal limits of the peak-
to-peak output signal are ± V
CC .
In practice, however, this ideal can be approached but never
reached (varies with load resistance).

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Input Offset Voltage
Ideal op- amp produces zero output voltage if the
differential input is zero
But practical op- amp produces a non- zero output voltage
when there is no differential input applied. This output
voltage is termed V
OUT(error) .
It is due to unavoidable mismatches in the differential
stage of the op amp.
The amount of differential input voltage required
between the inputs to force the output to zero volts is the
input offset voltage V
OS .
Typical value of V
OS is about 2mV.

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The input bias current is the
average of the two dc currents
required to bias the differential
amplifier
Ideally, input bias current is zero.
12
BIAS
2
II
I
+
=
Input Bias Current ( I
BIAS )

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The input impedance of an op- amp is
specified in two ways:
Differential input impedance and
common- mode input impedance.
Differential input impedance, Z
IN(d),
is the total resistance between
inverting and noninverting input.
Common- mode input impedance,
Z
IN(c), is the resistance between each
input and ground.
Input Impedance
Z
IN(d )
–
+
Z
IN(cm )
–
+

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The offset voltage developed by the input offset current is
????????????
????????????????????????=????????????
????????????????????????????????????
????????????????????????
The output error volt is
????????????
????????????????????????????????????(????????????????????????????????????????????????????????????) =????????????
????????????????????????
????????????????????????????????????
????????????????????????
Input offset Current I
OS
Ideally, the two input bias currents are equal, and thus their
difference is zero

The input offset current is
the difference of the input
bias currents ????????????
????????????????????????=
????????????
1−????????????
2

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Z
out : The output impedance is the resistance viewed from the
output of the circuit.
Output Impedance
Z
o ut
–
+

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Slew Rate
The slew rate is the maximum
rate of change of the output
voltage in response to a step input
voltage (V/µs)
Slew rate is measured with an op-
amp connected as shown and is
given as

The slew rate is dependent upon
the high-frequency response of
the amplifier stages within the
op-amp.

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Example
Determine the slew rate for the
output response to a step input
shown.
Since this response is not
ideal, the limits are taken
at the 90% points.

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Negative Feedback
Negative feedback is the process of returning a portion of
the output signal of an amplifier to the input with a phase
angle that is opposite to the input signal.
V
o ut
+
–
V
in
V
f
Internal inversion makes V
f
180° out of phase with V
in.
Negative
feedback
circuit
The advantage of negative
feedback is that precise
values of amplifier gain
can be set. In addition,
bandwidth and input and
output impedances can be
controlled.

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Importance of Negative Feedback
The usefulness of an op-
amp operated without
negative feedback is
generally limited to
comparator applications
With negative feedback the gain of op- amp (called close-
loop gain
A
cl) can be reduced and controlled so that an
op-amp can function as a linear amplifier.
The inherent open- loop voltage gain of a typical op- amp is
very high. Therefore, an extremely small input voltage (even
the input offset voltage) drives the op- amp into saturation.

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Op-Amps with Negative Feedback
An op- amp can be connected using negative feedback to
stabilize the gain and increase frequency response.
This close-loop gain ( A
cl ) is usually much less than the open-
loop gain (
A
ol).
The close-loop voltage gain is the voltage of an op-amp with
external feedback.
The amplifier circuit consists of an op-amp and an external
negative feedback circuit.
The feedback from the output is connected to the inverting input
of the op- amp.
The negative feedback is determined and controlled by external
components.

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Noninverting Amplifier
A noninverting amplifier is a configuration in which the
signal is on the noninverting input and a portion of the
output is returned to the inverting input.
The feedback circuit is formed by input resistance
R
i and
feedback resistance
R
f .
R
f
R
i
V
f
V
in
+
–
Feedback
circuit
V
o ut
This feedback creates a
voltage divider circuit
which reduces
V
out and
connects the reduced
voltage
V
f to the
inverting input and can
be expressed as:

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Noninverting Amplifier
The differential input is
amplified by the open-
loop gain and produces
the output voltage as:
The attenuation, B, of the feedback circuit is

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Noninverting Amplifier
(NI)
1
f
cl
i
R
A
R
= +
Substituting BV
out for V
f , we get
The overall voltage gain of the amplifier can be expressed
as
The product A
olB is typically much greater than 1, so the
equation simplifies to
Which means

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Thomas L. Floyd

Noninverting Amplifier
(NI)
1
f
cl
i
R
A
R
= +
Determine the gain of the noninverting amplifier shown.
R
f
82 kΩ
V
in +
–
V
o ut
R
i
3.3 kΩ
82 k
1
3.3 k
Ω
= +
Ω
= 25.8

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Voltage Follower
A special case of the inverting amplifier is when R
f =0 and
R
i = ∞. This forms a voltage follower or unity gain buffer
with a gain of 1.
R
f
82 kΩ
V
in +
–
V
o ut
R
i
3.3 kΩ
This configuration offers very
high input impedance and its very
low output impedance.
These features make it a
nearly ideal buffer amplifier for
interfacing high-impedance
sources and low-impedance loads
V
in +
–
V
o ut
It produces an excellent circuit for isolating one circuit stage from another,
which avoids "loading" effects.

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Inverting Amplifier
We have: and
Since I
in = I
f , then

The overall gain is

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Thomas L. Floyd

Example
Determine the gain of the inverting amplifier shown.
82 k
3.3 k
Ω
=−
Ω
= −24.8
–
+
R
f
V
o ut
R
i
V
in
82 kΩ
3.3 kΩ
(I)
f
cl
i
R
A
R
=−
The minus sign
indicates inversion.
Inverting Amplifier

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Oxford University Publishing
Microelectronic Circuits by Adel S. Sedra and Kenneth C. Smith (0195323033)
2.2.1.
Closed- Loop Gain
•question: how will we…
–step #4: define v
Out in terms of
current flowing across
R
2
–step #5: substitute v
in / R
1 for
i
1.

Analysis of the inverting configuration. The circled numbers indicate
the order of the analysis steps.
closed-loop
gain
G = -R
2/R
1

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Thomas L. Floyd

Effect of Finite Open- Loop Gain
•Q: How does the gain expression change if open
loop gain (
A) is not assumed to be infinite?
–A: One must employ analysis similar to the
previous, result is presented below…

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Thomas L. Floyd

Effect of Finite Open- Loop Gain

if then the previous
gain expression is
21
2
yielded
2
11
/
1(/)
1
AG
A
Out
A
In
v RR
G
RRv
A
R
R
=∞
=∞
<∞
−
= =
+
+

≠−


ideal gain non-ideal gain
Collecting terms, the closed- loop
gain
G is found as

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Thomas L. Floyd

Effect of Finite Open- Loop Gain
•Q: Under what condition can G = -R
2 / R
1 be
employed over the more complex expression?
–A: If 1 + ( R
2/R
1) << A, then simpler expression
may be used.
2 2 21
2111
if then el
/
1
1(/
1
se
)
AA
R R RR
AG G
RRRR
A
=∞ <∞
−
+ << =− =
+
+

ideal gain non-ideal gain

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Oxford University Publishing
Microelectronic Circuits by Adel S. Sedra and
Kenneth C. Smith (0195323033)

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From the Figure we get
Impedances of the Noninverting Amplifier
Input Impedance
Substituting I
inZ
in = V
d , where Z
in is open loop input impedance

( )
(NI)
1
in ol in
Z ABZ= +
The input impedance of the noninverting amplifier with negative
feedback is much greater than the internal open- loop input
impedance of the op- amp.

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Referring to the shown Figure,
after some mathematical
manipulations (Floyd 620) it
can be shown that the output
impedance of a noninverting
(NI) amplifier can be given as
Impedances of the Noninverting Amplifier
Output Impedance
The output impedance of the noninverting amplifier with
negative feedback is much less than the internal open- loop input
impedance of the op- amp.

( )
(NI)
1
out
out
ol
Z
Z
AB
=
+

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Since voltage-follower is a special case on noninverting
amplifier, the same formulas are used with
B = 1, therfore
Impedances of the Voltage-Follower
The input impedance of voltage-follower is very high, it can be
seen that it is extremely high as compared to the noninverting
amplifier as the feedback attenuation
B = 1.
The same is true for the output impedance where the factor of
is removed so it becomes even less than the output impedance of
the noninverting amplifier.

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The input impedance of the inverting (I) amplifier is
Impedances of the Inverting Amplifier
This is because the input is in series with R
i and that is connected
to virtual ground so that is the only resistance seen by input.

Input Impedance

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As with the noninverting amplifier, the output impedance of an
inverting (I) amplifier is decreased by the negative feedback.
In fact the the expression is the same as for the noninverting case.
Impedances of the Inverting Amplifier
The output impedance of the both noninverting and inverting
amplifier is low. In fact, practically it can be considered zero..

Output Impedance

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Noninverting amplifier:
Impedances
( )
(NI)
1
in ol in
Z ABZ= +
( )
(NI)
1
out
out
ol
Z
Z
AB
=
+
Generally, assumed to be ∞
Generally, assumed to be 0
(I) in i
ZR≅
( )
(I)
1
out
out
ol
Z
Z
AB
=
+
Generally, assumed to be R
i
Generally, assumed to be 0
Inverting amplifier:
Note that the output impedance has the same form for both amplifiers.

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Example

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Solution

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Example

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Determine the closed- loop gain of each amplifier in Figure.
(a) 11 (b) 101 (c) 47.8 (d) 23

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If a signal voltage of 10 mV
rms is applied to each amplifier in
Figure, what are the output voltages and what is there phase
relationship with inputs?.
(a) V
out ≅ V
in = 10 mV, in phase (b) V
out = A
clV
in = – 10 mV, 180º out of
phase (c) V
out = 233 mV, in phase (d) V
out = – 100 mV, 180º out of phase

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Effect of Input Bias Current - Inverting Amplifier
An inverting amplifier with zero input voltage is shown.
Ideally, the current through
R
i is zero because the input voltage
is zero and the voltage at the inverting terminal is zero.
The small input bias current,
I
1, is through R
f from the output
terminal. This produces an output error voltage
I
1R
f when it
should be zero

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A voltage-follower with zero input voltage and a source
resistance,
R
s is shown.
In this case, an input bias current,
I
1, produces a drop across
R
s and creates an output voltage error - I
1 R
s as shown.
Effect of Input Bias Current – Voltage Follower

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Consider a noninverting amplifier with zero input voltage.
The input bias current,
I
1, produces a voltage drop across
R
f and thus creates an output error voltage of I
1R
f
Effect of Input Bias Current - Noninverting Amplifier

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Bias Current Compensation – Voltage Follower
The output error voltage due to bias currents in a voltage-
follower can be sufficiently reduced by adding a resistor,
R
f , equal to the source resistance, R
s , in the feedback
path.
The voltage drop created by
I
1 across the added resistor
subtracts from the output error voltage (if
I
1 = I
2 ).

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Bias Current Compensation – Inverting and Noninverting
To compensate for the effect of bias current in Inverting
and Noninverting Amplifiers, a resistor
R
c is added at the
noninverting terminal.
The compensating resistor value equals the parallel
combination of
R
i and R
f.
The input current creates a voltage drop across
R
c that
offsets the voltage across the combination of
R
i and R
f ,
thus sufficiently reducing the output error voltage.

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–
+
R
f
V
o ut
R
i
V
in
R
c = R
i || R
f
–
+
R
f
V
o ut
R
i
V
in
R
c = R
i || R
f
Noninverting
amplifier
Inverting amplifier
Bias Current Compensation – Inverting and Noninverting

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Effect of Input Offset Voltage
The output of an op- amp should be zero when the
differential input is zero.
Practically this is not the case. There is always an output
error voltage present whose range is typically in
microvolts to millivolts.
This is due to the imbalances within the internal op- amp
transistors
This output error voltage is aside from the one produced
by the input bias

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Input Offset Voltage Compensation
Most ICs provide a mean of compensating for offset voltage.
An external potentiometer to the
offset null pins of IC package

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Bandwidth Limitations
Frequency response of amplifiers is shown in a plot called Bode
Plot.
In Bode plot, the frequency is on the horizontal axis and is in
logarithmic scale. It means that the frequency change is not
linear but ten-times. This ten-time change in frequency is called
a decade.
The vertical axis shows the voltage gain in decibel (dB).
The maximum gain on the plot is called the midrange gain.
The point in the frequency response of amplifiers where the
gain is 3 dB less than the midrange gain is called the critical
frequency.

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Bandwidth Limitations
–20 dB/decade roll-off
Unity-gain frequency (f
T)
Critical frequency
101 100 1k 10k 100k 1M
f
(Hz)
106
100
75
50
25
0
A
ol (dB)
Midrange

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Bandwidth Limitations
An open- loop response curve (Bode plot) for a certain op- amp
is shown.
The differential open-loop gain A
ol of an op amp is not
infinite; rather, it is finite and decreases with frequency.
Note that although the gain is quite high at dc and low
frequencies, it starts to fall off at a rather low frequency.
The process of modifying the open- loop gain is termed
frequency compensation, and its purpose is to ensure that op-
amp circuits will be stable (as opposed to oscillatory).
These are units that have a network (usually a single
capacitor) included within the same IC chip whose function is
to cause the op-amp gain to have the single- time-constant
low-pass response shown.

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Gain-Versus-Frequency Analysis
The RC lag (low- pass) circuits within an
op-amp are responsible for the roll-off in
gain as the frequency increases, just as for
the discrete amplifiers. The attenuation of
an RC lag circuit shown is expressed as:
The critical frequency of an RC circuit is:

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Gain-Versus-Frequency Analysis
If an op- amp is represented by a voltage gain element with a
gain of
A
ol(mid) plus a single RC lag circuit, as shown, it is
known as a compensated op- amp. The total open- loop gain of
the op- amp is the product of the midrange open- loop gain,
A
ol(mid), and the attenuation of the RC circuit as:

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Phase Shift
An RC circuit causes a propagation delay from input to
output, thus creating a phase shift between the input signal
and the output signal. An RC lag circuit such as found in an
op-amp stage causes the output signal voltage to lag the input.
The phase shift,
θ, is given by:

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Overall Frequency Response
Most op- amps have a constant roll-off of -20 dB/decade
above its critical frequency. The more complex IC
operational amplifier may consist of two or more
cascaded amplifier stages.
The gain of each stage is frequency dependent and rolls
off at above its critical frequency. Therefore, the total
response of an op- amp is a composite of the individual
responses of the internal stages. dB gains are added and
phase lags of the stages are added as shown in next slide
for a three stage op- amp.

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closed-loop frequency response
Op-amps are usually used in a closed- loop configuration
with negative feedback in order to achieve precise
control of gain and bandwidth.
Midrange gain of an op- amp is reduced by negative
feedback as we have already seen in previous sections.
Now we will see its effects on bandwidth.

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Effect of Negative Feedback on Bandwidth
The closed-loop critical frequency of an op-amp is given by:
????????????
????????????(????????????????????????)=????????????
????????????
????????????????????????(1+????????????????????????
????????????????????????????????????????????????????????????)
Where
B is the feedback attenuation of the closed-loop op- amp. The above expression shows that the closed-loop critical
frequency, ????????????
????????????(????????????????????????), is higher than the open-loop critical
frequency ????????????
????????????
???????????????????????? by a factor of (1+????????????????????????
????????????????????????????????????????????????????????????)
Since ????????????
????????????(????????????????????????) equals bandwidth therefore the closed-loop
bandwidth, ????????????????????????
????????????????????????, is also increased:
????????????????????????
????????????????????????=????????????????????????
????????????????????????(1+????????????????????????
????????????????????????
????????????????????????????????????)

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Summary
Bandwidth Limitations
The Figure shows the concept of closed-loop response. When the open-
loop gain is reduced due to negative feedback, the bandwidth is
increased.
This means that you can
achieve a higher
BW by
accepting less gain.
A
v
f
A
ol(mid)
0 f
c(ol) f
c(cl)
A
cl(mid)
Closed-loop gain
Open-loop gain

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Gain-Bandwidth Product
An increase in the closed-loop gain causes a decrease in the
bandwidth and vice versa, such that the product of gain and
bandwidth is constant.
If ????????????
???????????????????????? is the gain of an op- amp with ????????????
????????????(????????????????????????) bandwidth then:
????????????
????????????????????????????????????
????????????(????????????????????????)=????????????
????????????????????????????????????
????????????(????????????????????????)
The equation, A
cl f
c(cl) = A
ol f
c(ol) shows that the product of the
gain and bandwidth are constant.
The gain-bandwidth product is always equal to the frequency
at which the op-amp’s open- loop gain is unity or 0 dB (unity
gain bandwidth, ????????????
????????????)
????????????
????????????=????????????
????????????????????????????????????
????????????(????????????????????????)

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Example
Determine the bandwidth of each of the amplifiers shown.
Both op-amps have an open- loop gain of 100 dB and a unity-
gain bandwidth (
f
T ) of 3 MHz.

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Solution

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Selected Key Terms
Operational
amplifier



Differential
mode




Common mode


A type of amplifier that has very high voltage
gain, very high input impedance, very low
output impedance and good rejection of
common- mode signals.
A mode of op- amp operation in which two
opposite-polarity signals voltages are applied to
the two inputs (double-ended) or in which a
signal is applied to one input and ground to the
other input (single-ended).
A condition characterized by the presence of the same signal on both inputs

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Selected Key Terms
Open-loop
voltage gain

Negative
feedback


Closed-loop
voltage gain

Gain-
bandwidth
product

The voltage gain of an op- amp without external
feedback.
The process of returning a portion of the output
signal to the input of an amplifier such that it is
out of phase with the input.
The voltage gain of an op- amp with external
feedback.
A constant parameter which is always equal to
the frequency at which the op- amp’s open- loop
gain is unity (1).

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Quiz
1. The ideal op- amp has
a. zero input impedance and zero output impedance

b. zero input impedance and infinite output impedance
c. infinite input impedance and zero output impedance
d. infinite input impedance and infinite output
impedance

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Quiz
2. The type of signal represented in the figure is a
a. single-ended common- mode signal

b. single-ended differential signal
c. double-ended common- mode signal
d. double-ended differential signal

V
in
–
+
V
o ut

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Quiz
3. CMRR can be expressed in
a. amps

b. volts
c. ohms
d. none of the above

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Quiz
4. The difference in the two dc currents required to bias the
differential amplifier in an op- amp is called the
a. differential bias current

b. input offset current
c. input bias current
d. none of the above

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Quiz
5. To measure the slew rate of an op- amp, the input signal is a
a. pulse

b. triangle wave
c. sine wave
d. none of the above

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Quiz
6. The input impedance of a noninverting amplifier is
a. nearly 0 ohms

b. approximately equal to R
i
c. approximately equal to R
f
d. extremely large

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Quiz
7. The noninverting amplifier has a gain of 11. Assume that
V
in = 1.0 V. The approximate value of V
f is
a. 0 V

b. 100 mV

c. 1.0 V
d. 11 V

R
f
10 kΩ
V
V
in
f
+
–
V
o ut
R
i
1.0 kΩ

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Quiz
8. The inverting amplifier has a gain of −10. Assume that
V
in = 1.0 V. The approximate value of the voltage at the
inverting terminal of the op- amp is
a. 0 V

b. 100 mV

c. 1.0 V
d. 10 V

–
+
R
f
V
o ut
R
i
V
in
1.0 kΩ
10 kΩ

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Quiz
9. To compensate for bias current, the value of R
c should
be equal to
a.
R
i
b. R
f
c. R
i||R
f
d. R
i + R
f

–
+
R
f
V
o ut
R
i
V
in
R
c

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Quiz
10. Given a noninverting amplifier with a gain of 10 and a
gain-bandwidth product of 1.0 MHz, the expected high
critical frequency is
a. 100 Hz

b. 1.0 kHz

c. 10 kHz
d. 100 kHz

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Quiz
Answers:
1. c
2. d
3. d
4. b
5. a
6. d
7. c
8. a
9. c
10. d