1. 9-1 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Electronics Principles & Applications Eighth Edition Chapter 9 Operational Amplifiers (student version) Charles A. Schuler ©2013

2. 9-2 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. The Differential Amplifier The Operational Amplifier Determining Gain Frequency Effects Applications Comparators INTRODUCTION

3. 9-3 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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. 9-4 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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. 9-5 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Noninverted output Inverted output A differential amplifier driven at one input C B E C B E +V CC -V EE

6. 9-6 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Both outputs are active because Q 1 drives Q 2. C B E C B E +V CC -V EE Q1Q1 Q2Q2 Q 2 serves as a common-base amplifier in this mode. It’s driven at its emitter. Q 1 serves as an emitter-follower amplifier in this mode to drive Q 2.

7. 9-7 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Reduced output A differential amplifier driven at both inputs C B E C B E +V CC -V EE Common mode input signal

8. 9-8 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 B E C B E +V CC -V EE If the input signal goes positive, both transistors want to increase their current but can’t. Constant total current

9. 9-9 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Increased output Driven at both inputs with a differential signal C B E C B E +V CC -V EE

10. 9-10 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. A differential amplifier driven at both inputs with a differential signal shows high gain. +V CC Here, one transistor increases its current as the other decreases so the constant total current is not a limiting factor. C B E C B E -V EE Constant total current

11. 9-11 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. The amplifier has two gains: High for differential signals Low for common-mode signals The ratio of the two gains is called the common-mode rejection ratio (CMRR) and is perhaps the most important feature of this amplifier. CMRR = 20 x log A V(DIF) A V(CM)

12. 9-12 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

13. 9-13 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

14. 9-14 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Differential amplifier dc analysis C B E C B E +9 V -9 V 3.9 k  4.7 k  10 k  RERE RLRL RBRB RBRB RLRL V EE V CC I R E = V EE - V BE RERE 9 V - 0.7 V 3.9 k  = = 2.13 mA I E = IREIRE 2 = 1.06 mA I C = I E = 1.06 mA V R L = I C x R L = 1.06 mA x 4.7 k  = 4.98 V V CE = V CC - V RL - V E = 9 - 4.98 -(-0.7) = 4.72 V

15. 9-15 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Differential amplifier dc analysis continued C B E C B E +9 V -9 V 3.9 k  4.7 k  10 k  RERE RLRL RBRB RBRB RLRL V EE V CC Assume  = 200 I B = ICIC  1.06 mA  = = 5.3  A V B = V R B = I B x R B = 5.3  A x 10 k  = 53 mV

16. 9-16 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Differential amplifier ac analysis C B E C B E +9 V -9 V 3.9 k  4.7 k  10 k  RERE RLRL RBRB RBRB RLRL V EE V CC r E = 50 mV IEIE = 1.06 mA = 47  (50 mV is conservative) A V(DIF) = RLRL 2 x r E A V(CM) = RLRL 2 x R E = 50 4.7 k  2 x 47  = = 0.6 4.7 k  2 x 3.9 k  =

17. 9-17 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Differential amplifier ac analysis continued C B E C B E +9 V -9 V 3.9 k  4.7 k  10 k  RERE RLRL RBRB RBRB RLRL V EE V CC CMRR = 20 x log A V(DIF) A V(CM) = 20 x log 50 0.6 = 38.4 dB

18. 9-18 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. A current source can replace R E to decrease the common mode gain. C B E C B E 4.7 k  10 k  RLRL RBRB RBRB RLRL V CC 2 mA * * NOTE: Arrow shows conventional current flow. A V(CM) = RLRL 2 x R E Replaces this with a very high resistance value.

19. 9-19 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. A practical current source 390  5.1 V 2.2 k  -9 V I C = I E = 2 mA ICIC I Z = 9 V - 5.1 V 390  = 10 mAI E = = 2 mA 5.1 V - 0.7 V 2.2 k 

20. 9-20 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 6.3 V 60 Hz 212 mV 1 kHz The amplitude of the common-mode signal is almost 30 times the amplitude of the differential signal. A demonstration of common-mode rejection The common-mode signal cannot be seen in the output.

21. 9-21 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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 ________. 2r E The common-mode gain can be found by dividing the collector load by ________. 2R E

22. 9-22 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

23. 9-23 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

24. 9-24 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Inverting input Non-inverting input Output Op amps have two inputs

25. 9-25 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Op-amp Characteristics High CMRR High input impedance High gain Low output impedance Available as ICs Inexpensive Reliable Widely applied

26. 9-26 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Imperfections can make V OUT non-zero. The offset null terminals can be used to zero V OUT. -V EE +V CC V OUT With both inputs grounded through equal resistors, V OUT should be zero volts.

27. 9-27 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. VV tt VV tt Slew rate = The output of an op amp cannot change instantaneously. 741 0.5 V ss

28. 9-28 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Slew-rate distortion f MAX = Slew Rate 2  x V P f > f MAX VPVP

29. 9-29 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

30. 9-30 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

31. 9-31 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

32. 9-32 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL Op-amp follower A V(OL) = the open loop voltage gain A V(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.

33. 9-33 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL Op-amp follower A V(OL) = 200,000 A V(CL) = 1 The differential input approaches zero due to the high open-loop gain. Using this model, V OUT = V IN. V IN V OUT V DIF = 0

34. 9-34 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT Op-amp follower A V(OL) = 200,000 B = 1 The feedback ratio = 1 200,000 (200,000)(1) + 1  1 A V(CL) = AB +1 A V IN V OUT

35. 9-35 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT The closed-loop gain is increased by decreasing the feedback with a voltage divider. RFRF R1R1 200,000 (200,000)(0.091) + 1 = 11 A V(CL) = B = R1R1 R F + R 1 100 k  10 k  100 k  + 10 k  = = 0.091

36. 9-36 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT RFRF 100 k  10 k  V DIF = 0 It’s possible to develop a different model for the closed loop gain by assuming V DIF = 0. V IN = V OUT x R1R1 R 1 + R F = V OUT V IN 1 + RFRF R1R1 Divide both sides by V OUT and invert: A V(CL) = 11 R1R1

37. 9-37 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT RFRF 10 k  1 k  V DIF = 0 R1R1 In this amplifier, the assumption V DIF = 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 input current since it is so small. Thus: I R 1 = I R F V IN R1R1 = - V OUT RFRF V OUT V IN = -RF-RF R1R1 = - 10 By Ohm’s Law: The minus sign designates an inverting amplifier.

38. 9-38 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN RFRF 10 k  1 k  V DIF = 0 R1R1 Virtual ground Due to the virtual ground, the input impedance of the inverting amplifier is equal to R 1. R 2 = R 1  R F = 910  Although op amp input currents are small, in most applications, offset error is minimized by providing equal resistance paths for the input currents. This resistor reduces offset error.

39. 9-39 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

40. 9-40 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

41. 9-41 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Output A typical op amp has internal frequency compensation. Break frequency: f B = 2  RC 1 R C

42. 9-42 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 100 k 10 k 1 10100 1k 1M 0 20 80 40 60 100 120 Frequency in Hz Gain in dB Bode plot of a typical op amp Break frequency

43. 9-43 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT RFRF 100 k  1 k  Op amps are typically operated with negative feedback (closed loop). This increases their useful frequency range. R1R1 = V OUT V IN 1 + RFRF R1R1 A V(CL) = =1 + 100 k  1 k  = 101 dB Gain = 20 x log 101 = 40 dB

44. 9-44 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 100 k 10 k 1 10100 1k 1M 0 20 80 40 60 100 120 Frequency in Hz Gain in dB Using the Bode plot to find closed-loop bandwidth: Break frequency A V(CL)

45. 9-45 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. There are two frequency limitations: Slew rate determines the large-signal bandwidth. Internal compensation sets the small-signal bandwidth. 0.5 V ss 70 V ss A 741 op amp slews at A 318 op amp slews at

46. 9-46 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 100 k 10 k 1 10100 1k 1M 0 20 80 40 60 100 120 Frequency in Hz Gain in dB The Bode plot for a fast op amp shows increased small-signal bandwidth. 10M f UNITY

47. 9-47 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RLRL V IN V OUT RFRF 100 k  1 k  f UNITY can be used to find the small-signal bandwidth. R1R1 = V OUT V IN 1 + RFRF R1R1 A V(CL) = =1 + 100 k  1 k  = 101 318 Op amp f B = f UNITY A V(CL) 10 MHz 101 = 99 kHz =

48. 9-48 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Op amp feedback quiz The open loop gain of an op amp is reduced with __________ feedback negative The ratio R F /R 1 determines the gain of the ___________ amplifier. inverting 1 + R F /R 1 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

49. 9-49 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

50. 9-50 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

51. 9-51 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. R C Amplitude response of an RC lag circuit 0 dB -20 dB -40 dB -60 dB 10f b fbfb 100f b 1000f b f b =  RC 1 V out f

52. 9-52 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 0 o 0.1f b fbfb 10f b Phase response of an RC lag circuit -90 o -45 o R C R -X C  = tan -1 V out f

53. 9-53 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Interelectrode capacitance and Miller effect C BE C Miller C BE C BC R C Miller = A V C BC C Input = C Miller + C BE The gain from base to collector makes C BC effectively larger in the input circuit. f b =  RC Input 1

54. 9-54 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 10 Hz100 Hz1 kHz10 kHz100 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 f b1 f b2

55. 9-55 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 0 o Multiple lag circuits: -180 o R1R1 C1C1 V out f R2R2 C2C2 R3R3 C3C3 Phase reversal Negative feedback becomes positive!

56. 9-56 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

57. 9-57 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Op amp compensation quiz Beyond f b, an RC lag circuit’s output drops at a rate of __________ per decade. 20 dB The maximum phase lag for one RC network is __________. 90 o 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

58. 9-58 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

59. 9-59 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

60. 9-60 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RFRF 10 k  1 k  1 kHz 3 kHz 3.3 k  5 kHz 5 k  Summing Amplifier Inverted sum of three sinusoidal signals Amplifier scaling: 1 kHz signal gain is -10 3 kHz signal gain is -3 5 kHz signal gain is -2

61. 9-61 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. RFRF 1 k  Subtracting Amplifier Difference of two sinusoidal signals (V 1 = V 2 ) 1 k  V1V1 V2V2 V OUT = V 2 - V 1 (A demonstration of common-mode rejection)

62. 9-62 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.)

63. 9-63 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Frequency in Hz Amplitude in dB 0 -20 -40 -60 10 100 Cascade RC Active filter

64. 9-64 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN Active low-pass filter with feedback V OUT C1C1 C2C2 At relatively low frequencies, V out and V in are about the same. Thus, the signal voltage across C 1 is nearly zero. C 1 has little effect at these frequencies. feedback

65. 9-65 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN Active low-pass filter with feedback V OUT Frequency Gain fCfC - 3 dB Feedback can make a filter’s performance even better! C1C1 C2C2 As f IN increases and C 2 loads the input, V out drops. This increases the signal voltage across C 1. This sharpens the knee.

66. 9-66 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Frequency in Hz Amplitude in dB 0 -20 -40 -60 10 100 Active filter using feedback (two stages) Note the flat pass band and the sharp knee. The slope eventually reaches 24 dB/octave or 80 db/decade for all the filters (4 RC sections).

67. 9-67 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

68. 9-68 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

69. 9-69 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN Active high-pass filter V OUT Frequency Gain fCfC - 3 dB feedback

70. 9-70 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN Active band-pass filter (multiple feedback) V OUT Frequency Gain - 3 dB Bandwidth

71. 9-71 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN Active band-stop filter (multiple feedback) V OUT Frequency Gain - 3 dB Stopband

72. 9-72 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 40 mV 0 V 56.6 mV 0 V - 56.6 mV Active rectifier

73. 9-73 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT Integrator R C Slope = - V IN x 1 RC V s Slope =

74. 9-74 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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.

75. 9-75 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a 1 Volt reference

76. 9-76 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a noisy input signal

77. 9-77 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT +V SAT -V SAT Schmitt trigger with a noisy input signal UTP LTP Hysteresis = UTP - LTP RFRF R1R1 R 1 + R F R1R1 V SAT x Trip points:

78. 9-78 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT R2R2 R1R1 4.7 k  +5 V 3 V 1 V Window comparator 311 V UL V LL V OUT is LOW (0 V) when V IN is between 1 V and 3 V.

79. 9-79 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT +5 V 3 V 1 V Window comparator 311 V UL V LL Many comparator ICs require pull-up resistors in applications of this type.

80. 9-80 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. V IN V OUT R2R2 R1R1 4.7 k  +5 V 3 V 1 V Window comparator 311 V UL V LL V OUT is TTL logic compatible.

81. 9-81 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

82. 9-82 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 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

83. 9-83 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. REVIEW The Differential Amplifier The Operational Amplifier Determining Gain Frequency Effects Applications Comparators