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

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

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

4. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-4 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.

5. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-5 Reduced output A differential amplifier driven at both inputs C B E C B E +V CC -V EE Common mode input signal

6. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-6 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

7. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-7 Increased output Driven at both inputs with a differential signal C B E C B E +V CC -V EE

8. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-8 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

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

10. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-10 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

11. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-11 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

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

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

14. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-14 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.

15. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-15 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 

16. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-16 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.

17. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-17 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

18. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-18 Inverting input Non-inverting input Output Op amps have two inputs

19. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-19 Op-amp Characteristics High CMRR High input impedance High gain Low output impedance Available as ICs Inexpensive Reliable Widely applied

20. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-20 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.

21. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-21 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

22. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-22 Slew-rate distortion f MAX = Slew Rate 2  x V P f > f MAX VPVP

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

24. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-24 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.

25. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-25 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

26. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-26 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

27. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-27 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

28. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-28 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

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

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

31. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-31 Output A typical op amp has internal frequency compensation. Break frequency: f B = 2  RC 1 R C

32. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-32 100 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

33. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-33 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

34. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-34 100 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)

35. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-35 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

36. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-36 100 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

37. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-37 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 =

38. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-38 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

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

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

41. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-41 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

42. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-42 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

43. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-43 0 o Multiple lag circuits: -180 o R1R1 C1C1 V out f R2R2 C2C2 R3R3 C3C3 Phase reversal Negative feedback becomes positive!

44. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-44 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.

45. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-45 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

46. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-46 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

47. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-47 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)

48. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-48 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.)

49. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-49 Frequency in Hz Amplitude in dB 0 -20 -40 -60 10 100 Cascade RC Active filter

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

51. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-51 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.

52. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-52 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).

53. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-53 V IN Active high-pass filter V OUT Frequency Gain fCfC - 3 dB feedback

54. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-54 V IN Active band-pass filter (multiple feedback) V OUT Frequency Gain - 3 dB Bandwidth

55. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-55 V IN Active band-stop filter (multiple feedback) V OUT Frequency Gain - 3 dB Stopband

56. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-56 40 mV 0 V 56.6 mV 0 V - 56.6 mV Active rectifier

57. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-57 V IN V OUT Integrator R C Slope = - V IN x 1 RC V s Slope =

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

59. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-59 V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a 1 Volt reference

60. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-60 V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a noisy input signal

61. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-61 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:

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

63. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-63 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.

64. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-64 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.

65. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-65 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

66. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. McGraw-Hill 9-66 REVIEW The Differential Amplifier The Operational Amplifier Determining Gain Frequency Effects Applications Comparators