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9-1
Electronics
Principles & Applications
Eighth Edition
Charles A. Schuler
Chapter 9
Operational Amplifiers
(student version)
©2013
McGraw-Hill
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
9-2
INTRODUCTION
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators
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9-3
Dear Student:
This presentation is arranged in segments. Each segment
is preceded by a Concept Preview slide and is followed by a
Concept Review slide. When you reach a Concept Review
slide, you can return to the beginning of that segment by
clicking on the Repeat Segment button. This will allow you
to view that segment again, if you want to.
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9-4
Concept Preview
• Differential amplifiers always have two inputs.
• Differential amplifiers can have one or two
outputs.
• Driving one input provides a difference signal.
Both outputs will be active and will be out of
phase with each other.
• Driving both inputs with the same signal results in
reduced output.
• Driving both inputs with a difference signal results
in increased output.
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9-5
A differential amplifier driven at one input
+VCC
Noninverted output
Inverted output
C
B
E
C
E
B
-VEE
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9-6
Both outputs are active because Q1 drives Q2.
Q1 serves as an
emitter-follower
amplifier in this
mode to drive Q2.
+VCC
C
B
Q1 E
Q2 serves as a
common-base
amplifier in this
mode. It’s driven
at its emitter.
C
E Q
2
B
-VEE
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9-7
A differential amplifier driven at both inputs
Common mode
input signal
+VCC
Reduced output
Reduced output
C
B
E
C
E
B
-VEE
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A differential amplifier driven at both inputs with a
common-mode signal shows low gain (usually a loss)
because the total emitter current is fairly constant.
+VCC
If the input signal
goes positive, both
transistors want to
increase their
current but can’t.
B
C
E
Constant
total
current
McGraw-Hill
9-8
C
E
B
-VEE
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Driven at both inputs with a differential signal
9-9
+VCC
Increased output
Increased output
C
B
E
C
E
B
-VEE
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A differential amplifier driven at both inputs with a
differential signal shows high gain.
Here, one transistor
increases its current
as the other decreases
so the constant total
current is not a limiting
factor.
B
+VCC
C
E
Constant
total
current
McGraw-Hill
9-10
C
E
B
-VEE
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9-11
The amplifier has two gains:
• High for differential signals
• Low for common-mode signals
The ratio of the two gains is called the common-mode
rejection ratio (CMRR) and is perhaps the most
important feature of this amplifier.
CMRR = 20 x log
McGraw-Hill
AV(DIF)
AV(CM)
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9-12
Concept Review
• Differential amplifiers always have two inputs.
• Differential amplifiers can have one or two
outputs.
• Driving one input provides a difference signal.
Both outputs will be active and will be out of
phase with each other.
• Driving both inputs with the same signal results in
reduced output.
• Driving both inputs with a difference signal results
in increased output.
Repeat Segment
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9-13
Concept Preview
• The current in the emitter resistor divides equally
between the two transistors in a differential amp.
• The differential gain is determined by the collector
load and the ac emitter resistance.
• The common mode gain is determined by the
collector load and the emitter resistor.
• The ratio of the differential gain to the common
mode gain is called the CMRR.
• The CMRR is greatly improved by using a current
source in the emitter circuit.
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9-14
Differential amplifier dc analysis
IRE =
IE =
VEE - VBE
RE
IRE
2
9 V - 0.7 V
= 2.13 mA
=
3.9 kW
VCC
+9 V
= 4.98 V
= 1.06 mA
4.7 kW
RL RL
IC = IE = 1.06 mA
C
B
10 kW
RB
E
3.9 kW
VEE
McGraw-Hill
VRL = IC x RL
= 1.06 mA x 4.7 kW
VCE = VCC - VRL - VE
4.7 kW
= 9 - 4.98 -(-0.7)
C
E
= 4.72 V
B
RB
10 kW
RE
-9 V
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9-15
Differential amplifier dc analysis continued
Assume b = 200
IB =
IC
b
VB = VRB = IB x RB
1.06 mA
=
200
= 5.3 mA
VCC
4.7 kW
10 kW
RB
E
3.9 kW
VEE
McGraw-Hill
+9 V
= 53 mV
RL RL 4.7 kW
C
B
= 5.3 mA x 10 kW
C
E
B
RB
10 kW
RE
-9 V
© 2013 The McGraw-Hill Companies, Inc. All rights reserved.
9-16
Differential amplifier ac analysis
50 mV
50 mV
(50 mV is conservative)
= 47 W
=
rE =
IE
1.06 mA
RL
VCC +9 V
AV(CM) =
RL
2 x RE
AV(DIF) =
2 x rE
4.7 kW
4.7 kW RL RL 4.7 kW
=
4.7 kW
2 x 3.9 kW
=
= 50
2 x 47 W
C
C
= 0.6
B
10 kW
RB
E
3.9 kW
VEE
McGraw-Hill
E
B
RB
10 kW
RE
-9 V
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Differential amplifier ac analysis continued
CMRR = 20 x log
AV(DIF)
AV(CM)
VCC
4.7 kW
10 kW
RB
E
3.9 kW
VEE
McGraw-Hill
50
= 38.4 dB
0.6
+9 V
RL RL 4.7 kW
C
B
= 20 x log
9-17
C
E
B
RB
10 kW
RE
-9 V
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9-18
AV(CM) =
A current source can
replace RE to decrease
the common mode gain.
RL
VCC
2 x RE
Replaces this
with a very high
resistance value.
4.7 kW
C
B
10 kW
RL RL 4.7 kW
RB
E
*
C
E
B
RB
10 kW
2 mA
*NOTE: Arrow shows conventional current flow.
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9-19
A practical current source
IC
9 V - 5.1 V
IZ =
= 10 mA
390 W
IE =
390 W
5.1 V - 0.7 V
2.2 kW
= 2 mA
IC = IE = 2 mA
5.1 V
2.2 kW
-9 V
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9-20
A demonstration of common-mode rejection
The common-mode signal
cannot be seen in the output.
The amplitude of the
common-mode signal
is almost 30 times the
amplitude of the
differential signal.
McGraw-Hill
6.3 V
60 Hz
212 mV
1 kHz
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9-21
Differential amplifier quiz
When a diff amp is driven at one input,
the number of active outputs is _____. two
When a diff amp is driven at both inputs, there
is high gain for a _____ signal. differential
When a diff amp is driven at both inputs, there
is low gain for a ______ signal. common-mode
The differential gain can be found by dividing
the collector load by ________. 2rE
The common-mode gain can be found by dividing
the collector load by ________. 2RE
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9-22
Concept Review
• The current in the emitter resistor divides equally
between the two transistors in a differential amp.
• The differential gain is determined by the collector
load and the ac emitter resistance.
• The common mode gain is determined by the
collector load and the emitter resistor.
• The ratio of the differential gain to the common
mode gain is called the CMRR.
• The CMRR is greatly improved by using a current
source in the emitter circuit.
Repeat Segment
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9-23
Concept Preview
• Operational amplifiers have one output and two
inputs: inverting and non-inverting.
• Some op amps have offset null terminals which
can be used to zero the dc output.
• The output of an op can change no faster than its
slew rate.
• Slew rate is specified in volts per microsecond.
• The slew rate and the amplitude of the output
signal determine the power bandwidth of an op
amp.
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9-24
Op amps have
two inputs
Inverting
input
Output
Non-inverting
input
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9-25
Op-amp Characteristics
•
•
•
•
McGraw-Hill
High CMRR
High input impedance
High gain
Low output impedance
•
•
•
•
Available as ICs
Inexpensive
Reliable
Widely applied
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9-26
With both inputs grounded through equal
resistors, VOUT should be zero volts.
+VCC
VOUT
-VEE
Imperfections can make VOUT non-zero. The
offset null terminals can be used to zero VOUT.
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Dt
9-27
DV
741
DV
Slew rate =
Dt
0.5 V
ms
The output of an op amp cannot change instantaneously.
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9-28
VP
f > fMAX
Slew Rate
fMAX =
2p x VP
Slew-rate distortion
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9-29
Operational amplifier quiz
The input stage of an op amp is a
__________ amplifier.
differential
Op amps have two inputs: one is inverting
and the other is ________.
noninverting
An op amp’s CMRR is a measure of its ability
to reject a ________ signal.
common-mode
The offset null terminals can be used to zero
an op amp’s __________.
output
The ability of an op amp output to change
rapidly is given by its _________. slew rate
McGraw-Hill
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9-30
Concept Review
• Operational amplifiers have one output and two
inputs: inverting and non-inverting.
• Some op amps have offset null terminals which
can be used to zero the dc output.
• The output of an op can change no faster than its
slew rate.
• Slew rate is specified in volts per microsecond.
• The slew rate and the amplitude of the output
signal determine the power bandwidth of an op
amp.
Repeat Segment
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9-31
Concept Preview
• An op amp follower has a closed loop gain of 1.
• The input and output signals are in-phase in a
follower amplifier.
• The closed loop gain can be increased by
decreasing the feedback ratio.
• The input and output signals are out of phase in an
inverting amplifier.
• The – terminal of an inverting amplifier acts as a
virtual ground.
• The input impedance of an inverting amplifier is
equal to the input resistor.
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9-32
Op-amp follower
AV(OL) = the open loop voltage gain
AV(CL) = the closed loop voltage gain
This is a closed-loop
circuit with a voltage
gain of 1.
RL
McGraw-Hill
It has a high input impedance
and a low output impedance.
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9-33
Op-amp follower
AV(OL) = 200,000
AV(CL) = 1
The differential input
approaches zero due
to the high open-loop
gain. Using this model,
VOUT = VIN.
VDIF = 0
VIN
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VOUT
RL
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9-34
Op-amp follower
AV(OL) = 200,000
B=1
A
AB +1
VIN
VOUT
The feedback ratio = 1
200,000
@1
AV(CL) =
(200,000)(1) + 1
VIN
McGraw-Hill
VOUT
RL
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9-35
The closed-loop gain is increased by decreasing
the feedback with a voltage divider.
R1
200,000
AV(CL) =
= 11
(200,000)(0.091) + 1
RF
R1
B=
100 kW
RF + R1
10 kW
=
VIN
McGraw-Hill
VOUT
RL
10 kW
100 kW + 10 kW
= 0.091
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It’s possible to develop a different
model for the closed loop gain
by assuming VDIF = 0.
R1
VIN = VOUT x
R1 + RF
RF
Divide both sides by
VOUT and invert:
100 kW
R1 10 kW
VOUT
VDIF = 0
VIN
McGraw-Hill
9-36
VOUT
VIN
RF
=1+
R1
RL
AV(CL) = 11
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9-37
In this amplifier, the assumption VDIF = 0 leads
to the conclusion that the inverting op amp terminal is
also at ground potential. This is called a virtual ground.
Virtual ground
RF
We can ignore the op amp’s input
current since it is so small. Thus:
I R 1 = IR F
10 kW
By Ohm’s Law:
1 kW
R1
VIN
VDIF = 0
VIN
R1
VOUT
RL VOUT
VIN
=
=
-VOUT
RF
-RF
R1
= -10
The minus sign designates an inverting amplifier.
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9-38
Due to the virtual ground, the input impedance
of the inverting amplifier is equal to R1.
Virtual ground
RF
10 kW
R1
1 kW
VDIF = 0
VIN
Although op amp input
currents are small, in
most applications, offset
error is minimized by
providing equal resistance
paths for the input
currents.
R2 = R1 || RF = 910 W
This resistor reduces offset error.
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9-39
Concept Review
• An op amp follower has a closed loop gain of 1.
• The input and output signals are in-phase in a
follower amplifier.
• The closed loop gain can be increased by
decreasing the feedback ratio.
• The input and output signals are out-of-phase in an
inverting amplifier.
• The – terminal of an inverting amplifier acts as a
virtual ground.
• The input impedance of an inverting amplifier is
equal to the input resistor.
Repeat Segment
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9-40
Concept Preview
• Most op amps have built-in frequency
compensation.
• The internal frequency compensation produces a
break frequency of 10 Hz or so.
• The closed loop small signal bandwidth is greater
than the break frequency.
• A Bode plot can be used to determine the small
signal bandwidth of a closed loop amplifier.
• The gain-bandwidth product can also be used to
determine the closed loop small signal bandwidth.
McGraw-Hill
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9-41
A typical op amp has
internal frequency
compensation.
R
Output
C
Break frequency:
1
fB =
2pRC
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9-42
Bode plot of a typical op amp
Break frequency
120
100
80
60
Gain in dB
40
20
0
1
10 100 1k 10 k 100 k 1M
Frequency in Hz
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9-43
Op amps are typically operated with negative feedback
(closed loop). This increases their useful frequency range.
AV(CL) =
RF
dB Gain = 20 x log 101 = 40 dB
VIN
McGraw-Hill
VIN
RF
=1+
R1
100 kW
=1+
= 101
1 kW
100 kW
R1 1 kW
VOUT
VOUT
RL
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9-44
Using the Bode plot to find closed-loop bandwidth:
120
100
Gain in dB
80
Break frequency
60
AV(CL)
40
20
0
1
10 100 1k 10 k 100 k 1M
Frequency in Hz
McGraw-Hill
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9-45
A 741 op amp slews at
0.5 V
ms
70
V
A 318 op amp slews at
ms
There are two frequency limitations:
Slew rate determines the large-signal bandwidth.
Internal compensation sets the small-signal bandwidth.
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9-46
The Bode plot for a fast op amp shows
increased small-signal bandwidth.
120
100
80
Gain in dB 60
40
fUNITY
20
0
1
10 100 1k 10 k 100 k 1M 10M
Frequency in Hz
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9-47
fUNITY can be used to find the small-signal bandwidth.
AV(CL) =
RF
R1 1 kW
McGraw-Hill
VIN
RF
=1+
R1
100 kW
=1+
= 101
1 kW
100 kW
VIN
VOUT
VOUT
318 Op amp
RL
fUNITY
fB =
AV(CL)
10 MHz
=
= 99 kHz
101
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9-48
Op amp feedback quiz
The open loop gain of an op amp is reduced
with __________ feedback
negative
The ratio RF/R1 determines the gain of the
___________ amplifier.
inverting
1 + RF/R1 determines the gain of the
___________ amplifier.
noninverting
Negative feedback makes the - input of the
inverting circuit a ________ ground. virtual
Negative feedback _________ small signal
bandwidth.
increases
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9-49
Concept Review
• Most op amps have built-in frequency
compensation.
• The internal frequency compensation produces a
break frequency of 10 Hz or so.
• The closed loop small signal bandwidth is greater
than the break frequency.
• A Bode plot can be used to determine the small
signal bandwidth of a closed loop amplifier.
• The gain-bandwidth product can also be used to
determine the closed loop small signal bandwidth.
Repeat Segment
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9-50
Concept Preview
• The amplitude response of an RC lag network is
–20 dB per decade beyond the break frequency.
• The phase response of an RC lag network is –45
degrees at the break frequency.
• The Miller effect makes some interelectrode
capacitances appear to be larger.
• Multiple lag networks inside an op amp make
negative feedback become positive at some
frequency. Frequency compensation insures that
the gain is less than 0 dB at that frequency.
McGraw-Hill
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9-51
R
Amplitude response
of an RC lag circuit
Vout
C
1
fb =
2pRC
fb
0 dB
10fb
100fb
1000fb
f
-20 dB
Vout
-40 dB
-60 dB
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9-52
R
Phase response of
an RC lag circuit
 = tan
-1
C
-XC
R
0.1fb
0o
Vout
fb
10fb
f
Vout -45o
-90o
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9-53
Interelectrode capacitance and Miller effect
The gain from
base to collector
makes CBC
effectively larger C
BE
in the input circuit.
CBC
R
CMiller = AVCBC
CInput = CMiller + CBE
McGraw-Hill
CMiller
CBE
1
fb =
2pRCInput
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9-54
Bode plot of an amplifier
with two break frequencies.
50 dB
40 dB
20 dB/decade
30 dB
20 dB
40 dB/decade
10 dB
0 dB
10 Hz
100 Hz
1 kHz
fb1
McGraw-Hill
10 kHz 100 kHz
fb2
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9-55
Multiple lag circuits:
Vout
R1
C1
R2
C2
R3
0o
C3
f
Vout
Phase reversal
-180o
Negative feedback becomes positive!
McGraw-Hill
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9-56
Op amp compensation
• Interelectrode capacitances create several
break points.
• Negative feedback becomes positive at
some frequency due to cumulative phase
lags.
• If the gain is > 0 dB at that frequency, the
amplifier is unstable.
• Frequency compensation reduces the
gain to 0 dB or less.
McGraw-Hill
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9-57
Op amp compensation quiz
Beyond fb, an RC lag circuit’s output drops
at a rate of __________ per decade. 20 dB
The maximum phase lag for one RC network
is __________.
90o
An interelectrode capacitance can be effectively
much larger due to _______ effect. Miller
Op amp multiple lags cause negative feedback
to be ______ at some frequency. positive
If an op amp has gain at the frequency where
feedback is positive, it will be ______. unstable
McGraw-Hill
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9-58
Concept Review
• The amplitude response of an RC lag network is
–20 dB per decade beyond the break frequency.
• The phase response of an RC lag network is –45
degrees at the break frequency.
• The Miller effect makes some interelectrode
capacitances appear to be larger.
• Multiple lag networks inside an op amp make
negative feedback become positive at some
frequency. Frequency compensation insures that
the gain is less than 0 dB at that frequency.
Repeat Segment
McGraw-Hill
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9-59
Concept Preview
• Op amps can be used to sum (add) two or more
signals.
• Scaling in a summing amp provides different gain
for each signal.
• Op amps can be used to subtract two signals.
• Cascade RC filters have relatively poor
performance.
• Active filters combine op amps with RC networks.
• Feedback in an op amp active filter sharpens the
knee of the frequency response curve.
McGraw-Hill
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9-60
Inverted sum of three sinusoidal signals
RF
10 kW
5 kW
5 kHz
3.3 kW
3 kHz
1 kW
1 kHz
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Summing Amplifier
Amplifier scaling:
1 kHz signal gain is -10
3 kHz signal gain is -3
5 kHz signal gain is -2
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9-61
Difference of two
sinusoidal signals
(V1 = V2)
RF
1 kW
Subtracting Amplifier
(A demonstration of
common-mode rejection)
1 kW
1 kW
1 kW
V1
V2
McGraw-Hill
VOUT = V2 - V1
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9-62
A cascade RC low-pass filter
(A poor performer since later sections load the earlier ones.)
An active low-pass filter
(The op amps provide isolation and better performance.)
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9-63
0
Active filter
Amplitude in dB
-20
Cascade RC
-40
-60
10
100
Frequency in Hz
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9-64
Active low-pass filter
with feedback
C1
VOUT
C2
VIN
feedback
At relatively low frequencies, Vout and Vin
are about the same. Thus, the signal voltage
across C1 is nearly zero. C1 has little effect
at these frequencies.
McGraw-Hill
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9-65
Active low-pass filter
with feedback
C1
VOUT
C2
VIN
As fIN increases and C2
loads the input, Vout
drops. This increases
the signal voltage
across C1. This
sharpens the knee.
-3 dB
Gain
Feedback can
make a filter’s
performance
even better!
Frequency
McGraw-Hill
fC
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9-66
0
Active filter
using feedback
(two stages)
Amplitude in dB
-20
-40
-60
10
Note the flat pass band
and the sharp knee.
The slope eventually reaches
24 dB/octave or 80 db/decade
for all the filters (4 RC sections).
100
Frequency in Hz
McGraw-Hill
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9-67
Concept Review
• Op amps can be used to sum (add) two or more
signals.
• Scaling in a summing amp provides different gain
for each signal.
• Op amps can be used to subtract two signals.
• Cascade RC filters have relatively poor
performance.
• Active filters combine op amps with RC networks.
• Feedback in an op amp active filter sharpens the
knee of the frequency response curve.
Repeat Segment
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Concept Preview
• Other active filters include high-pass, band-pass
and band-stop.
• An active rectifier will work with millivolt level
signals.
• The output slope of an op amp integrator is equal
to the dc input voltage times the reciprocal of the
time constant.
• Comparators can be used to change analog
waveforms to digital waveforms.
• A Schmitt trigger uses positive feedback to
produce hysteresis and noise immunity.
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9-69
Active high-pass filter
VOUT
VIN
feedback
-3 dB
Gain
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fC Frequency
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9-70
VIN
VOUT
Active band-pass filter
(multiple feedback)
-3 dB
Gain
Frequency
Bandwidth
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9-71
VOUT
VIN
Active band-stop filter
(multiple feedback)
-3 dB
Gain
Frequency
Stopband
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9-72
56.6 mV
Active rectifier
0V
40 mV
0V
- 56.6 mV
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9-73
Integrator
C
R
VIN
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V
Slope = s
VOUT
1
Slope = -VIN x
RC
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9-74
A differentiator shows R and C reversed.
Vout = -(Vin/t)RC
Differentiation is the opposite of integration.
Input Waveforms (Blue)
Integrator Output Waveform (Red)
Differentiator Output Waveform (Red)
Square
Triangle
Sine
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9-75
Comparator with a 1 Volt reference
+VSAT
1V
0V
-VSAT
VOUT
VIN
1V
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Comparator with a noisy input signal
+VSAT
1V
0V
-VSAT
VOUT
VIN
1V
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Schmitt trigger with a noisy input signal
+VSAT
UTP
LTP
Trip points:
R1
VSAT x
R1 + RF
-VSAT
VIN
R1
McGraw-Hill
VOUT
RF
Hysteresis = UTP - LTP
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9-78
+5 V
Window comparator
4.7 kW
VUL
3V
R1
311
VOUT
R2
4.7 kW
VIN
311
VLL
1 V VOUT is LOW (0 V) when VIN
is between 1 V and 3 V.
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Window comparator
VUL
3V
+5 V
311
VOUT
VIN
311
VLL
McGraw-Hill
1V
Many comparator ICs
require pull-up resistors in
applications of this type.
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9-80
+5 V
Window comparator
4.7 kW
VUL
3V
R1
311
VOUT
R2
4.7 kW
VIN
311
VLL
McGraw-Hill
1V
VOUT is TTL logic
compatible.
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9-81
Op amp applications quiz
A summing amp with different gains for the
inputs uses _________.
scaling
Frequency selective circuits using op amps
are called _________ filters.
active
An op amp integrator uses a _________ as
the feedback element.
capacitor
A Schmitt trigger is a comparator with
__________ feedback.
positive
A window comparator output is active when
the input is ______ the reference points. between
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9-82
Concept Review
• Other active filters include high-pass, band-pass
and band-stop.
• An active rectifier will work with millivolt level
signals.
• The output slope of an op amp integrator is equal
to the dc input voltage times the reciprocal of the
time constant.
• Comparators can be used to change analog
waveforms to digital waveforms.
• A Schmitt trigger uses positive feedback to
produce hysteresis and noise immunity.
Repeat Segment
McGraw-Hill
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9-83
REVIEW
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators
McGraw-Hill
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