9-1
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ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsEighth EditionEighth Edition
Chapter 9Operational Amplifiers
(student version)
Charles A. Schuler
©2013
9-2
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• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators
INTRODUCTION
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Dear Student:
This presentation is arranged in segments. Each segmentis preceded by a Concept Preview slide and is followed by aConcept Review slide. When you reach a Concept Reviewslide, you can return to the beginning of that segment byclicking on the Repeat Segment button. This will allow youto view that segment again, if you want to.
9-4
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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.
9-5
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Noninverted outputInverted output
A differential amplifier driven at one input
C
BE
C
BE
+VCC
-VEE
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Both outputs are active because Q1 drives Q2.
C
BE
C
BE
+VCC
-VEE
Q1 Q2
Q2 serves as a common-base
amplifier in this mode. It’s driven
at its emitter.
Q1 serves as an emitter-followeramplifier in this mode to drive Q2.
9-7
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Reduced outputReduced output
A differential amplifier driven at both inputs
C
BE
C
BE
+VCC
-VEE
Common mode input signal
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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.
C
BE
C
BE
+VCC
-VEE
If the input signalgoes positive, bothtransistors want toincrease theircurrent but can’t.
Constanttotal
current
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Increased outputIncreased output
Driven at both inputs with a differential signal
C
BE
C
BE
+VCC
-VEE
9-10
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A differential amplifier driven at both inputs with a differential signal shows high gain.
+VCCHere, one transistorincreases its currentas the other decreasesso the constant totalcurrent is not a limitingfactor. C
BE
C
BE
-VEE
Constanttotal
current
9-11
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The amplifier has two gains:
• High for differential signals• Low for common-mode signals
The ratio of the two gains is called the common-moderejection ratio (CMRR) and is perhaps the most important feature of this amplifier.
CMRR = 20 x logAV(DIF)
AV(CM)
9-12
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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
9-13
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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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Differential amplifier dc analysis
C
BE
C
BE
+9 V
-9 V
3.9 k
4.7 k4.7 k
10 k10 k
RE
RL
RBRB
RL
VEE
VCC
IRE =
VEE - VBE
RE
9 V - 0.7 V
3.9 k= = 2.13 mA
IE =IRE
2= 1.06 mA
IC = IE = 1.06 mA
VRL = IC x RL
= 1.06 mA x 4.7 k= 4.98 V
VCE = VCC - VRL - VE
= 9 - 4.98 -(-0.7)
= 4.72 V
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Differential amplifier dc analysis continued
C
BE
C
BE
+9 V
-9 V
3.9 k
4.7 k4.7 k
10 k10 k
RE
RL
RBRB
RL
VEE
VCC
Assume = 200
IB =IC
1.06 mA
=
= 5.3 A
VB = VRB = IB x RB
= 5.3 A x 10 k
= 53 mV
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Differential amplifier ac analysis
C
BE
C
BE
+9 V
-9 V
3.9 k
4.7 k4.7 k
10 k10 k
RE
RL
RBRB
RL
VEE
VCC
rE =50 mV
IE
=50 mV
1.06 mA= 47 (50 mV is conservative)
AV(DIF) = RL
2 x rE
AV(CM) = RL
2 x RE
= 504.7 k
2 x 47 =
= 0.6
4.7 k2 x 3.9 k
=
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Differential amplifier ac analysis continued
C
BE
C
BE
+9 V
-9 V
3.9 k
4.7 k4.7 k
10 k10 k
RE
RL
RBRB
RL
VEE
VCC
CMRR = 20 x logAV(DIF)
AV(CM)
= 20 x log500.6
= 38.4 dB
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A current source can replace RE to decrease the common mode gain.
C
BE
C
BE
4.7 k4.7 k
10 k10 k
RL
RBRB
RL
VCC
2 mA*
*NOTE: Arrow shows conventional current flow.
AV(CM) = RL
2 x RE
Replaces thiswith a very highresistance value.
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A practical current source
390
5.1 V2.2 k
-9 V
IC = IE = 2 mA
IC
IZ = 9 V - 5.1 V
390 = 10 mA
IE = = 2 mA5.1 V - 0.7 V
2.2 k
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6.3 V60 Hz
212 mV1 kHz
The amplitude of thecommon-mode signalis almost 30 times the
amplitude of thedifferential signal.
A demonstration of common-mode rejection
The common-mode signalcannot be seen in the output.
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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, thereis high gain for a _____ signal. differential
When a diff amp is driven at both inputs, thereis low gain for a ______ signal. common-mode
The differential gain can be found by dividingthe collector load by ________. 2rE
The common-mode gain can be found by dividingthe collector load by ________. 2RE
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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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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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Invertinginput
Non-invertinginput
Output
Op amps have two inputs
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Op-amp Characteristics
• High CMRR• High input impedance• High gain• Low output impedance
• Available as ICs• Inexpensive• Reliable• Widely applied
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Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT.
-VEE
+VCC
VOUT
With both inputs grounded through equal resistors, VOUT should be zero volts.
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V
t
Vt
Slew rate =
The output of an op amp cannot change instantaneously.
741
0.5 Vs
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Slew-rate distortion
fMAX = Slew Rate
2 x VP
f > fMAX
VP
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Operational amplifier quiz
The input stage of an op amp is a__________ amplifier. differential
Op amps have two inputs: one is invertingand the other is ________. noninverting
An op amp’s CMRR is a measure of its abilityto reject a ________ signal. common-mode
The offset null terminals can be used to zeroan op amp’s __________. output
The ability of an op amp output to changerapidly is given by its _________. slew rate
9-30
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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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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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RL
Op-amp follower
AV(OL) = the open loop voltage gain
AV(CL) = the closed loop voltage gain
This is a closed-loopcircuit with a voltage
gain of 1.
It has a high input impedanceand a low output impedance.
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RL
Op-amp follower
AV(OL) = 200,000
AV(CL) = 1
The differential inputapproaches zero dueto the high open-loop
gain. Using this model,VOUT = VIN.
VIN
VOUT
VDIF = 0
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RLVIN
VOUT
Op-amp follower
AV(OL) = 200,000
B = 1
The feedback ratio = 1
200,000
(200,000)(1) + 1 1AV(CL) =
AB +1AVIN VOUT
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RLVIN
VOUT
The closed-loop gain is increased by decreasing the feedback with a voltage divider.
RF
R1
200,000
(200,000)(0.091) + 1= 11AV(CL) =
B =R1
RF + R1
100 k
10 k 10 k100 k+ 10 k
=
= 0.091
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RLVIN
VOUT
RF
100 k10 k
VDIF = 0
It’s possible to develop a different model for the closed loop gain
by assuming VDIF = 0.
VIN = VOUT xR1
R1 + RF
=VOUT
VIN
1 +RF
R1
Divide both sides by VOUT and invert:
AV(CL) = 11
R1
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RLVIN VOUT
RF
10 k1 k
VDIF = 0R1
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 We can ignore the op amp’s inputcurrent since it is so small. Thus:
IR1 = IRF
VIN
R1
=-VOUT
RF
VOUT
VIN
=-RF
R1
= -10
By Ohm’s Law:
The minus sign designates an inverting amplifier.
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VIN
RF
10 k1 k
VDIF = 0
R1
Virtual ground
Due to the virtual ground, the input impedance of the inverting amplifier is equal to R1.
R2 = R1 RF = 910
Although op amp inputcurrents are small, in
most applications, offseterror is minimized by
providing equal resistance paths for the input
currents.
This resistor reduces offset error.
9-39
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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
9-40
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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.
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Output
A typical op amp has internal frequency
compensation.
Break frequency:
fB = 2RC1
R
C
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100 k10 k1 10 100 1k 1M0
20
80
40
60
100
120
Frequency in Hz
Gain in dB
Bode plot of a typical op amp
Break frequency
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RLVIN
VOUT
RF
100 k1 k
Op amps are typically operated with negative feedback(closed loop). This increases their useful frequency range.
R1
=VOUT
VIN
1 +RF
R1
AV(CL) =
= 1 +100 k1 k
= 101
dB Gain = 20 x log 101 = 40 dB
9-44
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100 k10 k1 10 100 1k 1M0
20
80
40
60
100
120
Frequency in Hz
Gain in dB
Using the Bode plot to find closed-loop bandwidth:
Break frequency
AV(CL)
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There are two frequency limitations:Slew rate determines the large-signal bandwidth.
Internal compensation sets the small-signal bandwidth.
0.5 Vs
70 Vs
A 741 op amp slews at A 318 op amp slews at
9-46
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100 k10 k1 10 100 1k 1M0
20
80
40
60
100
120
Frequency in Hz
Gain in dB
The Bode plot for a fast op amp showsincreased small-signal bandwidth.
10M
fUNITY
9-47
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RLVIN
VOUT
RF
100 k1 k
fUNITY can be used to find the small-signal bandwidth.
R1
=VOUT
VIN
1 +RF
R1
AV(CL) =
= 1 +100 k1 k
= 101
318 Op amp
fB = fUNITY
AV(CL)
10 MHz
101= 99 kHz=
9-48
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Op amp feedback quiz
The open loop gain of an op amp is reducedwith __________ 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 theinverting circuit a ________ ground. virtual
Negative feedback _________ small signalbandwidth. increases
9-49
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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
9-50
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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.
9-51
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R
C
Amplitude responseof an RC lag circuit
0 dB
-20 dB
-40 dB
-60 dB
10fbfb 100fb 1000fb
fb = RC1
Vout
Vout
f
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0o
0.1fb fb 10fb
Phase response ofan RC lag circuit
-90o
-45o
R
C
R
-XC = tan-1
Vout
Vout
f
9-53
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Interelectrode capacitance and Miller effect
CBECMiller
CBE
CBC
R
CMiller = AVCBC
CInput = CMiller + CBE
The gain frombase to collector
makes CBC
effectively largerin the input circuit.
fb = RCInput
1
9-54
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10 Hz 100 Hz 1 kHz 10 kHz 100 kHz
50 dB
40 dB
30 dB
20 dB
10 dB
0 dB
Bode plot of an amplifier with two break frequencies.
20 dB/decade
40 dB/decade
fb1 fb2
9-55
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0o
Multiple lag circuits:
-180o
R1C1
Vout
Vout
f
R2C2
R3C3
Phase reversal
Negative feedback becomes positive!
9-56
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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.
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Op amp compensation quiz
Beyond fb, an RC lag circuit’s output dropsat a rate of __________ per decade. 20 dB
The maximum phase lag for one RC networkis __________. 90o
An interelectrode capacitance can be effectivelymuch larger due to _______ effect. Miller
Op amp multiple lags cause negative feedbackto be ______ at some frequency. positive
If an op amp has gain at the frequency wherefeedback is positive, it will be ______. unstable
9-58
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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
9-59
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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.
9-60
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RF
10 k
1 k
1 kHz
3 kHz
3.3 k5 kHz
5 k
Summing Amplifier
Inverted sum of three sinusoidal signals
Amplifier scaling: 1 kHz signal gain is -103 kHz signal gain is -35 kHz signal gain is -2
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RF
1 k
1 k 1 k
Subtracting Amplifier
Difference of twosinusoidal signals
(V1 = V2)
1 k
V1 V2
VOUT = V2 - V1
(A demonstration of common-mode rejection)
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A cascade RC low-pass filter
An active low-pass filter
(A poor performer since later sections load the earlier ones.)
(The op amps provide isolation and better performance.)
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Frequency in Hz
Am
pli
tud
e in
dB
0
-20
-40
-60
10 100
Cascade RC
Active filter
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VIN
Active low-pass filterwith feedback VOUT
C1C2
At relatively low frequencies, Vout and Vin
are about the same. Thus, the signal voltageacross C1 is nearly zero. C1 has little effect
at these frequencies.
feedback
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VIN
Active low-pass filterwith feedback VOUT
Frequency
Gain
fC
-3 dBFeedback canmake a filter’sperformanceeven better!
C1C2
As fIN increases and C2
loads the input, Vout
drops. This increasesthe signal voltageacross C1. This
sharpens the knee.
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Frequency in Hz
Am
pli
tud
e in
dB
0
-20
-40
-60
10 100
Active filterusing feedback
(two stages)
Note the flat pass bandand the sharp knee.
The slope eventually reaches24 dB/octave or 80 db/decade
for all the filters (4 RC sections).
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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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VIN
Active high-pass filter VOUT
Frequency
Gain
fC
-3 dB
feedback
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VIN
Active band-pass filter(multiple feedback)
VOUT
Frequency
Gain
-3 dB
Bandwidth
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VIN Active band-stop filter(multiple feedback)
VOUT
Frequency
Gain
-3 dB
Stopband
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40 mV
0 V
56.6 mV
0 V
- 56.6 mV
Active rectifier
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VIN
VOUT
Integrator
R
C
Slope = -VIN x1
RC
VsSlope =
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Input Waveforms (Blue) Integrator Output Waveform (Red) Differentiator Output Waveform (Red)
Square
Triangle
Sine
A differentiator shows R and C reversed.
Vout = -(Vin/t)RC
Differentiation is the opposite of integration.
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VIN
VOUT
0 V
1 V +VSAT
-VSAT
1 V
Comparator with a 1 Volt reference
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VIN
VOUT
0 V
1 V +VSAT
-VSAT
1 V
Comparator with a noisy input signal
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VINVOUT
+VSAT
-VSAT
Schmitt trigger with a noisy input signal
UTP
LTP
Hysteresis = UTP - LTPRF
R1
R1 + RF
R1VSAT x
Trip points:
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VIN
VOUT
R2
R14.7 k
4.7 k
+5 V
3 V
1 V
Window comparator
311
311VUL
VLL VOUT is LOW (0 V) when VIN
is between 1 V and 3 V.
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VIN
VOUT
+5 V
3 V
1 V
Window comparator
311
311VUL
VLL
Many comparator ICs require pull-up resistors in
applications of this type.
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VIN
VOUT
R2
R14.7 k
4.7 k
+5 V
3 V
1 V
Window comparator
311
311VUL
VLL VOUT is TTL logic compatible.
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Op amp applications quiz
A summing amp with different gains for theinputs uses _________. scaling
Frequency selective circuits using op ampsare called _________ filters. active
An op amp integrator uses a _________ asthe feedback element. capacitor
A Schmitt trigger is a comparator with__________ feedback. positive
A window comparator output is active whenthe input is ______ the reference points. between
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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
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REVIEW
• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators