1. Principles & Applications Operational Amplifiers
Electronics
Principles & Applications
Sixth Edition
Charles A. Schuler
Chapter 9
Operational Amplifiers
(student version)
© Glencoe/McGraw-Hill

2. INTRODUCTION The Differential Amplifier The Operational Amplifier
Determining Gain
Frequency Effects
Applications
Comparators

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.

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.

5. C C B B E E A differential amplifier driven at one input +VCC -VEE
Inverted output
Noninverted output
+VCC C C B B E E
-VEE

6. Both outputs are active because Q1 drives Q2.
Q1 serves as an
emitter-follower
amplifier in this
mode to drive Q2.
Q2 serves as a
common-base
amplifier in this
mode. It’s driven
at its emitter.
+VCC C C B B E E Q1 Q2
-VEE

7. C C B B E E A differential amplifier driven at both inputs +VCC -VEE
Common mode
input signal
Reduced output
Reduced output
+VCC C C B B E E
-VEE

8. C C B B E E A differential amplifier driven at both inputs +VCC -VEE
Differential mode
input signal
Increased output
Increased output
+VCC C C B B E E
-VEE

9. 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

10. 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.

11. Differential Amplifier dc Analysis
IRE =
VEE - VBE RE
9 V V
3.9 kW =
= 2.13 mA
VRL = IC x RL
= 1.06 mA x 4.7 kW
IE =
IRE 2
= 1.06 mA
VCC
+9 V
= 4.98 V
VCE = VCC - VRL - VE
4.7 kW RL RL
4.7 kW
= (-0.7)
IC = IE = 1.06 mA C C
= 4.72 V B B E E
10 kW RB RB
10 kW
3.9 kW RE
VEE
-9 V

12. Differential Amplifier dc Analysis (continued)
Assume b = 200
VB = VRB = IB x RB
= 5.3 mA x 10 kW
IB = IC b
1.06 mA
200 =
VCC
+9 V
= 53 mV
= 5.3 mA
4.7 kW RL RL
4.7 kW C C B B E E
10 kW RB RB
10 kW
3.9 kW RE
VEE
-9 V

13. Differential Amplifier ac Analysis
50 mV IE =
1.06 mA
= 47 W
(50 mV is conservative)
VCC
+9 V
AV(CM) = RL
2 x RE
AV(DIF) = RL
2 x rE
4.7 kW
2 x 3.9 kW =
4.7 kW RL RL
4.7 kW
= 50
4.7 kW
2 x 47 W = C C
= 0.6 B B E E
10 kW RB RB
10 kW
3.9 kW RE
VEE
-9 V

14. Differential Amplifier ac Analysis (continued)
CMRR = 20 x log
AV(DIF)
AV(CM)
= 20 x log 50
0.6
= 38.4 dB
VCC
+9 V
4.7 kW RL RL
4.7 kW C C B B E E
10 kW RB RB
10 kW
3.9 kW RE
VEE
-9 V

15. * C C B B E E A current source can replace RE to decrease
the common mode gain.
AV(CM) = RL
2 x RE
VCC
Replaces this
with a very high
resistance value.
4.7 kW RL RL
4.7 kW C C B B E E
10 kW RB RB
10 kW *
2 mA
NOTE: Arrow shows conventional current flow.

16. A Practical Current Source
IZ =
9 V V
390 W
= 10 mA IC
IE =
= 2 mA
5.1 V V
2.2 kW
390 W
IC = IE = 2 mA
5.1 V
2.2 kW
-9 V

17. The common-mode signal cannot be seen in the output.
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.
6.3 V
60 Hz
212 mV
1 kHz

18. 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

19. 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

20. 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.

21. Op amps have
two inputs
Inverting
input
Output
Non-inverting
input

22. Op-amp Characteristics
High CMRR
High input impedance
High gain
Low output impedance
Available as ICs
Inexpensive
Reliable
Widely applied

23. With both inputs grounded through equal
resistors, VOUT should be zero volts.
+VCC
VOUT
Imperfections can make VOUT non-zero. The
offset null terminals can be used to zero VOUT.
-VEE

24. Dt DV DV Dt
Slew rate =
741
0.5 V ms
The output of an op amp cannot change instantaneously.

25. Slew-rate distortion VP
f > fMAX
fMAX =
Slew Rate
2p x VP
fMAX is known as the power bandwidth.

26. 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

27. 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

28. 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.

29. It has a high input impedance and a low output impedance.
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.
It has a high input impedance
and a low output impedance. RL

30. The differential input
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
VOUT RL
VIN

31. Op-amp Follower AV(OL) = 200,000 B = 1 AB +1 A VIN VOUT
The feedback ratio = 1
200,000
(200,000)(1) + 1
@ 1
AV(CL) =
VOUT RL
VIN

32. The closed-loop gain is increased by decreasing
the feedback with a voltage divider.
200,000
(200,000)(0.091) + 1
= 11
AV(CL) = RF
B = R1
RF + R1
100 kW R1
10 kW
10 kW
100 kW + 10 kW =
VOUT RL
VIN
= 0.091

33. It’s possible to develop a different model for the closed loop gain
by assuming VDIF = 0.
VIN = VOUT x R1
R1 + RF RF
Divide both sides by
VOUT and invert:
100 kW R1
10 kW =
VOUT
VIN
1 + RF R1
VDIF = 0
VOUT RL
VIN
AV(CL) = 11

34. -VOUT -RF 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.
We can ignore the op amp’s input
current since it is so small. Thus:
Virtual ground RF
IR1 = IRF
10 kW
1 kW
By Ohm’s Law: R1
VIN R1 =
-VOUT RF
VDIF = 0
VIN
VOUT RL
VOUT
VIN =
-RF R1
= -10
The minus sign designates an inverting amplifier.

35. Due to the virtual ground, the input impedance
of the inverting amplifier is equal to R1.
Virtual ground RF
Although op amp input
currents are small, in
some applications, offset
error is minimized by
providing equal paths for
the input currents.
10 kW R1
1 kW
VDIF = 0
VIN
R2 = R1 || RF = 910 W
This resistor reduces offset error.

36. 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

37. 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.

38. A typical op amp has internal frequency compensation.
Output C
Break frequency:
fB =
2pRC 1

39. Bode Plot of a Typical Op Amp
Break frequency
120
100 80 60
Gain in dB 40 20 1 10
100 1k
10 k
100 k 1M
Frequency in Hz

40. Op amps are usually operated with negative feedback
(closed loop). This increases their useful frequency range. =
VOUT
VIN
1 + RF R1
AV(CL) = RF =
1 +
100 kW
1 kW
= 101
100 kW R1
1 kW
dB Gain = 20 x log 101 = 40 dB
VOUT RL
VIN

41. Using the Bode plot to find closed-loop bandwidth:
120
100
Gain in dB 80
Break frequency 60
AV(CL) 40 20 1 10
100 1k
10 k
100 k 1M
Frequency in Hz

42. There are two frequency limitations:
0.5 V ms
70 V ms
A 741 op amp slews at
A 318 op amp slews at
There are two frequency limitations:
Slew rate determines the large-signal bandwidth.
Internal compensation sets the small-signal bandwidth.

43. The Bode plot for a fast op amp shows increased bandwidth.
120
100
fUNITY is also
called the
gain-bandwidth
product. 80
Gain in dB 60 40
fUNITY 20 1 10
100 1k
10 k
100 k 1M
10M
Frequency in Hz

44. fUNITY can be used to find the small-signal bandwidth.=
VOUT
VIN
1 + RF R1
AV(CL) = RF =
1 +
100 kW
1 kW
= 101
100 kW R1
1 kW
fB =
fUNITY
AV(CL)
VOUT RL
VIN
318 Op amp
10 MHz
101
= 99 kHz =

45. 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

46. 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

47. 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.

48. Amplitude Response of RC Lag Circuit
Vout C
fb =
2pRC 1 fb
10fb
100fb
1000fb f
0 dB
-20 dB
Vout
-40 dB
-60 dB

49. Phase Response of RC Lag Circuit
Vout C R
-XC
 = tan-1
0.1fb fb
10fb 0o f
Vout
-45o
-90o

50. Interelectrode Capacitance and Miller Effect
The gain from
base to collector
makes CBC
effectively larger
in the input circuit.
CBC
CBE R
CMiller
CBE
CMiller = AVCBC
fb =
2pRCInput 1
CInput = CMiller + CBE

51. 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
10 kHz
100 kHz
fb1
fb2

52. Multiple Lag Circuits:
Vout R1 R2 R3 C1 C2 C3 0o f
Vout
Phase reversal
-180o
Negative feedback becomes positive!

53. 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.

54. 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

55. 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

56. 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.

57. RF Summing Amplifier Amplifier scaling: 1 kHz signal gain is -10
Inverted sum of three sinusoidal signals RF
5 kHz
5 kW
10 kW
Summing Amplifier
3.3 kW
Amplifier scaling:
1 kHz signal gain is -10
3 kHz signal gain is -3
5 kHz signal gain is -2
3 kHz
1 kW
1 kHz

58. common-mode rejection)
Difference of two
sinusoidal signals
(V1 = V2) RF
1 kW
Subtracting Amplifier
(A demonstration of
common-mode rejection)
1 kW
VOUT = V2 - V1
1 kW
1 kW V1 V2

59. 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.)

60. Active filter
-20
Cascade RC
-40
Amplitude in dB
-60 10
100
Frequency in Hz

61. Active low-pass filter
with feedback
VOUT C1 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.

62. Active low-pass filter
with feedback
VOUT C1 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
Feedback can
make a filter’s
performance
even better!
Gain
Frequency fC

63. The slope eventually reaches 24 dB/octave or 80 db/decade
Note the flat pass band
and the sharp knee.
Active filter
using feedback
(two stages)
-20
-40
Amplitude in dB
The slope eventually reaches
24 dB/octave or 80 db/decade
for all the filters (4 RC sections).
-60 10
100
Frequency in Hz

64. 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

65. 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.

66. Active high-pass filter
VOUT
VIN
feedback
-3 dB
Gain fC
Frequency

67. Active band-pass filter
VIN
VOUT
Active band-pass filter
(multiple feedback)
-3 dB
Gain
Frequency
Bandwidth

68. Active band-stop filter
VOUT
VIN
Active band-stop filter
(multiple feedback)
-3 dB
Gain
Frequency
Stopband

69. 0 V
56.6 mV
Active rectifier
40 mV
0 V mV

70. Integrator C V
Slope = s R
VOUT
VIN
Slope =
-VIN x 1 RC

71. Comparator with a 1 Volt Reference
+VSAT
1 V
0 V
-VSAT
VOUT
VIN
1 V

72. Comparator with a Noisy Input Signal
+VSAT
1 V
0 V
-VSAT
VOUT
VIN
1 V

73. Schmitt Trigger with a Noisy Input Signal
+VSAT
UTP
LTP
-VSAT
Trip points:
R1 + RF R1
VSAT x
VOUT
VIN RF R1
Hysteresis = UTP - LTP

74. VOUT is LOW (0 V) when VIN is between 1 V and 3 V.
Window Comparator
4.7 kW R1
311
VOUT
VUL
3 V R2
4.7 kW
VIN
311
VLL
VOUT is LOW (0 V) when VIN
is between 1 V and 3 V.
1 V

75. require pull-up resistors in applications of this type.
+5 V
Window Comparator
311
VOUT
VUL
3 V
VIN
311
Many comparator ICs
require pull-up resistors in
applications of this type.
VLL
1 V

76. VOUT is TTL logic compatible.
Window Comparator
4.7 kW R1
311
VOUT
VUL
3 V R2
4.7 kW
VIN
311
VLL
VOUT is TTL logic
compatible.
1 V

77. 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

78. 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

79. REVIEW The Differential Amplifier The Operational Amplifier
Determining Gain
Frequency Effects
Applications
Comparators