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

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

3. 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

4. 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

5. 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

6. 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

7. 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

8. 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

9. 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

10. 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

11. * 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.

12. 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

13. 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

14. 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

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

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

17. 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

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

19. VP
f > fMAX
fMAX =
Slew Rate
2p x VP
Slew-rate distortion

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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

26. -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.

27. 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.

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

29. 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

30. 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

31. 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

32. 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.

33. The Bode plot for a fast op amp shows
increased small-signal bandwidth.
120
100 80
Gain in dB 60 40
fUNITY 20 1 10
100 1k
10 k
100 k 1M
10M
Frequency in Hz

34. 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 =

35. 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

36. 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

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

38. 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

39. 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

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

41. 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.

42. 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

43. 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

44. 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

45. 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.)

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

47. 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.

48. 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

49. 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

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

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

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

53. 0 V
56.6 mV
Active rectifier
40 mV
0 V mV

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

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

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

57. 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

58. 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

59. 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

60. 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

61. 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

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