electronics principles & applications fifth edition chapter 9 operational amplifiers ©1999...
TRANSCRIPT
![Page 1: Electronics Principles & Applications Fifth Edition Chapter 9 Operational Amplifiers ©1999 Glencoe/McGraw-Hill Charles A. Schuler](https://reader034.vdocuments.net/reader034/viewer/2022051205/56649ea95503460f94bae20a/html5/thumbnails/1.jpg)
ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsFifth EditionFifth Edition
Chapter 9Operational Amplifiers
©1999 Glencoe/McGraw-Hill
Charles A. Schuler
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• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators
INTRODUCTION
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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.
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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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Increased outputIncreased output
A differential amplifier driven at both inputs
C
BE
C
BE
+VCC
-VEE
Differential mode input signal
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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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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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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
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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
some applications, offseterror is minimized by
providing equal paths forthe input currents.
This resistor reduces offset error.
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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 usually 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
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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
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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
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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=
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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
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R
C
Amplitude responseof 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 responseof RC lag circuit
-90o
-45o
R
C
R
-XC = tan-1
Vout
Vout
f
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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
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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
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0o
Multiple lag circuits:
-180o
R1C1
Vout
Vout
f
R2C2
R3C3
Phase reversal
Negative feedback becomes positive
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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
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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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VIN
Active low-pass filter VOUT
Frequency
Gain
fC
-3 dB
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VIN
Active high-pass filter VOUT
Frequency
Gain
fC
-3 dB
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VIN
Active band-pass filter
VOUT
Frequency
Gain
-3 dB
Bandwidth
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VIN Active band-stop filter
VOUT
Frequency
Gain
-3 dB
Stopband
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VIN
VOUT
Integrator
R
C
Slope = -VIN x1
RC
VsSlope =
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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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REVIEW
• The Differential Amplifier• The Operational Amplifier• Determining Gain• Frequency Effects• Applications• Comparators