2. Electronics Principles & Applications Fifth Edition Chapter 9 Operational Amplifiers ©1999 Glencoe/McGraw-Hill Charles A. Schuler

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

4. Noninverted output Inverted output A differential amplifier driven at one input C B E C B E +V CC -V EE

5. 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.

6. Reduced output A differential amplifier driven at both inputs C B E C B E +V CC -V EE Common mode input signal

7. Increased output A differential amplifier driven at both inputs C B E C B E +V CC -V EE Differential mode input signal

8. 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 RE = V EE - V BE RERE 9 V - 0.7 V 3.9 k  = = 2.13 mA I E = I RE 2 = 1.06 mA I C = I E = 1.06 mA V RL = 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

9. 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 RB = I B x R B = 5.3  A x 10 k  = 53 mV

10. 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  =

11. 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

12. 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.

13. 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 

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 ________. 2r E The common-mode gain can be found by dividing the collector load by ________. 2R E

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

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

17. 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.

18. 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

19. Slew-rate distortion f MAX = Slew Rate 2  x V P f > f MAX VPVP

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. 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.

22. 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

23. 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

24. 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

25. 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

26. 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 R1 = I RF 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.

27. 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 some applications, offset error is minimized by providing equal paths for the input currents. This resistor reduces offset error.

28. Output A typical op amp has internal frequency compensation. Break frequency: f B = 2  RC 1 R C

29. 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

30. RLRL V IN V OUT RFRF 100 k  1 k  Op amps are usually 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

31. 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)

32. 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

33. 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

34. 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 =

35. 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 + 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. R C Amplitude response of 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

37. 0 o 0.1f b fbfb 10f b Phase response of RC lag circuit -90 o -45 o R C R -X C  = tan -1 V out f

38. 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

39. 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

40. 0 o Multiple lag circuits: -180 o R1R1 C1C1 V out f R2R2 C2C2 R3R3 C3C3 Phase reversal 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 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

43. 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

44. 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)

45. V IN Active low-pass filter V OUT Frequency Gain fCfC - 3 dB

46. V IN Active high-pass filter V OUT Frequency Gain fCfC - 3 dB

47. V IN Active band-pass filter V OUT Frequency Gain - 3 dB Bandwidth

48. V IN Active band-stop filter V OUT Frequency Gain - 3 dB Stopband

49. V IN V OUT Integrator R C Slope = - V IN x 1 RC V s Slope =

50. V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a 1 Volt reference

51. V IN V OUT 0 V 1 V +V SAT -V SAT 1 V Comparator with a noisy input signal

52. 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:

53. 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.

54. 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.

55. 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.

56. 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

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