electronics principles & applications sixth edition chapter 9 operational amplifiers (student...
TRANSCRIPT
![Page 1: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/1.jpg)
ElectronicsElectronics
Principles & ApplicationsPrinciples & ApplicationsSixth EditionSixth Edition
Chapter 9Operational Amplifiers
(student version)
©2003 Glencoe/McGraw-Hill
Charles A. Schuler
![Page 2: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/2.jpg)
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators
INTRODUCTION
![Page 3: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/3.jpg)
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 youto view that segment again, if you want to.
![Page 4: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/4.jpg)
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.
![Page 5: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/5.jpg)
Noninverted outputInverted output
A differential amplifier driven at one input
C
BE
C
BE
+VCC
-VEE
![Page 6: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/6.jpg)
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.
![Page 7: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/7.jpg)
Reduced outputReduced output
A differential amplifier driven at both inputs
C
BE
C
BE
+VCC
-VEE
Common mode input signal
![Page 8: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/8.jpg)
Increased outputIncreased output
A differential amplifier driven at both inputs
C
BE
C
BE
+VCC
-VEE
Differential mode input signal
![Page 9: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/9.jpg)
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
![Page 10: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/10.jpg)
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.
![Page 11: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/11.jpg)
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
![Page 12: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/12.jpg)
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
![Page 13: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/13.jpg)
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
=
![Page 14: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/14.jpg)
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
![Page 15: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/15.jpg)
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.
![Page 16: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/16.jpg)
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
![Page 17: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/17.jpg)
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.
![Page 18: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/18.jpg)
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
![Page 19: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/19.jpg)
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
![Page 20: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/20.jpg)
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.
![Page 21: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/21.jpg)
Invertinginput
Non-invertinginput
Output
Op amps have two inputs
![Page 22: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/22.jpg)
Op-amp Characteristics
• High CMRR
• High input impedance
• High gain
• Low output impedance
• Available as ICs
• Inexpensive
• Reliable
• Widely applied
![Page 23: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/23.jpg)
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.
![Page 24: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/24.jpg)
V
t
Vt
Slew rate =
The output of an op amp cannot change instantaneously.
741
0.5 Vs
![Page 25: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/25.jpg)
Slew-rate distortion
fMAX = Slew Rate
2 x VP
f > fMAX
VP
fMAX is known as the power bandwidth.
![Page 26: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/26.jpg)
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
![Page 27: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/27.jpg)
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
![Page 28: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/28.jpg)
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.
![Page 29: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/29.jpg)
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.
![Page 30: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/30.jpg)
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
![Page 31: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/31.jpg)
RLVIN
VOUT
AV(OL) = 200,000
B = 1
The feedback ratio = 1
200,000
(200,000)(1) + 1 1AV(CL) =
AB +1AVIN VOUT
Op-amp Follower
![Page 32: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/32.jpg)
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
![Page 33: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/33.jpg)
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
![Page 34: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/34.jpg)
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.
![Page 35: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/35.jpg)
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.
![Page 36: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/36.jpg)
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
![Page 37: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/37.jpg)
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.
![Page 38: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/38.jpg)
Output
A typical op amp has internal frequency
compensation.
Break frequency:
fB = 2RC1
R
C
![Page 39: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/39.jpg)
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 AmpBreak frequency
![Page 40: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/40.jpg)
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
![Page 41: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/41.jpg)
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)
![Page 42: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/42.jpg)
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
![Page 43: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/43.jpg)
100 k10 k1 10 100 1k 1M0
20
80
40
60
100
120
Frequency in Hz
Gai
n in
dB
The Bode plot for a fast op ampshows increased bandwidth.
10M
fUNITY
fUNITY is alsocalled the
gain-bandwidthproduct.
![Page 44: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/44.jpg)
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=
![Page 45: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/45.jpg)
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
![Page 46: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/46.jpg)
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
![Page 47: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/47.jpg)
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.
![Page 48: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/48.jpg)
R
C
Amplitude Responseof RC Lag Circuit
0 dB
-20 dB
-40 dB
-60 dB
10fbfb 100fb 1000fb
fb = RC1
Vout
Vout
f
![Page 49: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/49.jpg)
0o
0.1fb fb 10fb
Phase Responseof RC Lag Circuit
-90o
-45o
R
C
R
-XC = tan-1
Vout
Vout
f
![Page 50: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/50.jpg)
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
![Page 51: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/51.jpg)
10 Hz 100 Hz 1 kHz 10 kHz 100 kHz
50 dB
40 dB
30 dB
20 dB
10 dB
0 dB
20 dB/decade
40 dB/decade
fb1 fb2
Bode Plot of an Amplifier with Two Break Frequencies
![Page 52: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/52.jpg)
0o
Multiple Lag Circuits:
-180o
R1C1
Vout
Vout
f
R2C2
R3C3
Phase reversal
Negative feedback becomes positive!
![Page 53: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/53.jpg)
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.
![Page 54: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/54.jpg)
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
![Page 55: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/55.jpg)
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
![Page 56: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/56.jpg)
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.
![Page 57: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/57.jpg)
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
![Page 58: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/58.jpg)
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)
![Page 59: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/59.jpg)
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.)
![Page 60: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/60.jpg)
Frequency in Hz
Am
pli
tud
e in
dB
0
-20
-40
-60
10 100
Cascade RC
Active filter
![Page 61: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/61.jpg)
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 effectat these frequencies.
feedback
![Page 62: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/62.jpg)
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.
![Page 63: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/63.jpg)
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).
![Page 64: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/64.jpg)
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
![Page 65: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/65.jpg)
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.
![Page 66: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/66.jpg)
VIN
Active high-pass filter VOUT
Frequency
Gain
fC
-3 dB
feedback
![Page 67: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/67.jpg)
VIN
Active band-pass filter(multiple feedback)
VOUT
Frequency
Gain
-3 dB
Bandwidth
![Page 68: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/68.jpg)
VIN Active band-stop filter(multiple feedback)
VOUT
Frequency
Gain
-3 dB
Stopband
![Page 69: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/69.jpg)
40 mV
0 V
56.6 mV
0 V
- 56.6 mV
Active rectifier
![Page 70: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/70.jpg)
VIN
VOUT
Integrator
R
C
Slope = -VIN x1
RC
VsSlope =
![Page 71: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/71.jpg)
VIN
VOUT
0 V
1 V +VSAT
-VSAT
1 V
Comparator with a 1 Volt Reference
![Page 72: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/72.jpg)
VIN
VOUT
0 V
1 V +VSAT
-VSAT
1 V
Comparator with a Noisy Input Signal
![Page 73: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/73.jpg)
VINVOUT
+VSAT
-VSAT
Schmitt Trigger with a Noisy Input Signal
UTP
LTP
Hysteresis = UTP - LTPRF
R1
R1 + RF
R1VSAT x
Trip points:
![Page 74: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/74.jpg)
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.
![Page 75: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/75.jpg)
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.
![Page 76: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/76.jpg)
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.
![Page 77: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/77.jpg)
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 whenthe input is ______ the reference points. between
![Page 78: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/78.jpg)
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
![Page 79: Electronics Principles & Applications Sixth Edition Chapter 9 Operational Amplifiers (student version) ©2003 Glencoe/McGraw-Hill Charles A. Schuler](https://reader031.vdocument.in/reader031/viewer/2022013108/56649cd95503460f949a2440/html5/thumbnails/79.jpg)
REVIEW
• The Differential Amplifier
• The Operational Amplifier
• Determining Gain
• Frequency Effects
• Applications
• Comparators