2. 0 Chap 2. Operational amplifiers (op-amps) Circuit symbol of an op-amp Widely used Often requires 2 power supplies + V Responds to difference between two signals

3. 1 2.2 Ideal op-amp Characteristics of an ideal op-amp R in = infinity R out = 0 A vo = infinity (A vo is the open-loop gain, sometimes A or A v of the op-amp) Bandwidth = infinity (amplifies all frequencies equally)

4. 2 Model of an ideal op-amp V+V+ V-V- V out = A(V + - V - ) + - + - Usually used with feedback Open-loop configuration not used much I-I- I+I+

5. 3 Summary of op-amp behavior V out = A(V + - V - ) V out /A = V + - V - Let A infinity then, V + -V - 0

6. 4 Summary of op-amp behavior V + = V - I + = I - = 0 Seems strange, but the input terminals to an op-amp act as a short and open at the same time

7. 5 To analyze an op-amp circuit Write node equations at + and - terminals (I + = I - = 0) Set V + = V - Solve for V out

8. 6 2.3 Inverting configuration Very popular circuit

9. 7 Analysis of inverting configuration I 1 = (V i - V - )/R 1 I 2 = (V - - V o )/R 2 set I 1 = I 2, (V i - V - )/R 1 = (V - - V o )/R 2 but V - = V + = 0 V i / R 1 = -V o / R 2 Solve for Vo V o / V i = -R 2 / R 1 Gain of circuit determined by external components I1I1 I2I2

10. 8 2.4 Applications of the inverting configuration V1 V2 V3 R1 R2 R3 Rf Current in R 1, R 2, and R 3 add to current in R f (V 1 - V - )/R 1 + (V 2 - V - )/R 2 + (V 3 - V - )/R 3 = (V - - Vo)/R f Set V - = V + = 0, V 1 /R 1 + V 2 /R 2 + V 3 /R 3 = - Vo/R f solve for V o, V o = -R f (V 1 /R 1 + V 2 /R 2 + V 3 /R 3 ) This circuit is called a weighted summer

11. 9 Integrator (2.4 Applications of the inverting configuration) I 1 = (V i - V - )/R 1 I 2 = C d(V - - V o )/dt set I 1 = I 2, (V i - V - )/R 1 = C d(V - - V o )/dt but V - = V + = 0 V i /R 1 = -C d(V o )/dt Solve for Vo V o = -(1/CR 1 )(  V i dt) Output is the integral of input signal. CR 1 is the time constant I1I1 I2I2

12. 10 Problem 2.28 Given an integrator with a time constant of 1mS, with a square wave input as shown below, find the output. V o = -(1/CR 1 )(  V i dt) CR 1 = 1mS, and V i =1V for 0 < t < 0.5mS V o = -(1/1mS)(  dt), for 0 < t < 0.5mS V o = -(t/1mS), for 0 < t < 0.5mS V o = (t/1mS), for 0.5 < t < 1.0mS t mS 1 0.5 1.5 1V -1V 0

13. 11 Problem 2.28 t mS 10.5 1V -1V 0 V o = -(t/1mS), for 0 < t < 0.5mS V o = (t/1mS), for 0.5 < t < 1.0mS When t = 0.5mS, Vo = -0.5V CR 1 = 1mS CR 1 = 0.5mS

14. 12 Exercise 2.6 Design an integrator (find RC) such that the output has a 20Vp-p amplitude given the input below.(Find the RC time constant) V o = -(1/CR)(  V i dt) V i = 10V, 0 < t < 1.0mS V o = -(1/RC) V o = -10t/RC let t = 1.0mS, and V o = -20 20 = 0.01/RC RC = 0.5mS t 2mS 10V -10V Vi Vout

15. 13 Transient response of an integrator

16. 14 2.5 Noninverting configuration (0 - V - )/R 1 = (V - - V o )/R2 But, V i = V + = V-, ( - V i )/R 1 = (V i - V o )/R 2 Solve for Vo, V o = V i (1+R 2 /R 1 ) Vi I I

17. 15 Buffer amplifier V i = V + = V - = V o V o = V i Isolates input from output

18. 16 Analyzing op-amp circuits Write node equations using: V + = V - I + = I - = 0 Solve for V out Usually easier, can solve most problems this way. Write node equations using: model, let A infinity Solve for V out Works for every op-amp circuit. OR

19. 17 Input resistance of noninverting amplifier Vout = A(V + -V - ) V- V+ R in = V in / I, from definition R in = V in / 0 R in = infinity I

20. 18 Input resistance of inverting amplifier V out = A(V + - V - ) V-V- V+V+ R in = V in / I, from definition I = (V in - V out )/R I = [V in - A (V + - V - )] / R But V + = 0 I = [V in - A( -V in )] / R R in = V in R / [V in (1+A)] As A approaches infinity, R in = 0 I

21. 19 Summary of op-amp behavior ViInverting configurationNoninverting configuration V o /V i = 1+R 2 /R 1 R in = infinity V o /V i = - R 2 /R 1 R in = R 1 Rin = 0 at this point Vi

22. 20 Fig. 2.21 A difference amplifier. Difference amplifier Use superposition, set V 1 = 0, solve for V o (noninverting amp) set V 2 = 0, solve for V o (inverting amp)

23. 21 Add the two results V o = -(R 2 /R 1 )V 1 + (1 + R 2 /R 1 ) [R 4 /(R 3 +R 4 )] V 2 Difference amplifier V o1 = -(R 2 /R 1 )V 1 V o2 = (1 + R 2 /R 1 ) [R 4 /(R 3 +R 4 )] V 2 V 2 R 4 /(R 3 +R 4 )

24. 22 Design of difference amplifiers For V o = V 2 - V 1 Set R 2 = R 1 = R, and set R 3 = R 4 = R For V o = 3V 2 - 2V 1 Set R 1 = R, R 2 = 2R, then 3[R 4 /(R 3 +R 4 )] = 3 Set R 3 = 0 V o = -(R 2 /R 1 )V 1 + (1 + R 2 /R 1 ) [R 4 /(R 3 +R 4 )]V 2

25. 23 Input resistance of difference amplifiers When measuring Rin at one input, ground all other inputs. Rin at V 1 = R 1, same as inverting amp Rin at V 2 = R 3 + R 4

26. 24 Improving the input resistance of amplifiers Add buffer amplifiers to the inputs R in = infinity at both V 1 and V 2

27. 25 Prob. 2.55 (a) Find Rin at V 1. Ground V 2, R in = R. (b) Find Rin at V 2. Ground V 1, R in = 2R. R in = V/I from definition. Write voltage loop, V = IR + IR, I = V/2R R in = V/I = V2R/V R in = 2R +V-+V- I I (c) Find Rin between V 1 and V 2.

28. 26 Rin = V/I from definition I 1 = V/2R from Ohm’s law I 2 = (V - V/2)R = V/2R because V + = V - R in = V/ (I 1 + I 2 ) = V2R/2V R in = R I +V-+V- I 1 = V/2R I2I2 V/2 (d) Find Rin between V 1 and V 2 connected together and ground. Prob. 2.55

29. 27 2.7 Frequency response of op-amps Vout = A(V + - V - ) + - + - Model of an ideal op-amp Model of an op-amp with frequency response Vout = A(V + - V - ) + - + - C We assume low-pass filter behavior

30. 28 Magnitude response of single capacitor circuit Fig. 1.23 (a) Magnitude response of (single time constant) STC networks of the low-pass type. 20 log|A| where  0 = 1/RC

31. 29 Open-loop frequency response of op-amp Unity gain frequency, occurs where Ao = 1 (A = 0dB) Break frequency(bandwidth), occurs where Ao drops 3dB below maximum Open-loop gain at low frequencies

32. 30 Gain-bandwidth product For any op-amp circuit without external capacitors: f t = |A| f b A = voltage gain of circuit f b = break frequency f t = unity gain frequency, fixed!

33. 31 Frequency response of op-amp f t = 1MHz f t = |A| f b 1MHz = (10 000) 100Hz 1MHz = (1000) 1KHz 1MHz = (10) 100KHz Tradeoff between gain and bandwidth

34. 32 Example Given A = 1 000 000 at low frequencies, at 10kHz A = 100, and f t = 1MHz, find A at 1kHz. Find break frequency f t = A f b f b = f t / A = 1MHz/1M f b = 1 A = 1 000 000 = 120 dB A = 100 = 40 dB Because 1kHz > f b, A = 20dB greater gain than at 10kHz.

35. 33 Example Given A = 106 dB at DC, f t = 2MHz, find A at f = 2kHz, 20kHz, and 200kHz. A = 106 dB (200 000) f t = 2MHz f t = A f b f b = f t / A = 2MHz/200 000 f b = 10 A at 2kHz = 60 dB A at 20kHz = 40 dB A at 200kHz = 20 dB

36. 34 Problem 2.69 Given an inverting amp circuit with A = -20, Ao of op-amp at low frequencies = 10 000, f t = 1MHz. (a) find f b, f b = f t / A f b = 1MHz / |20| f b = 50kHz (b) find A at 0.1 f b. For f < f b, A = A of circuit = -20 = 26dB. (c) find A at 10 f b. For f > f b, A = A of circuit - 20 dB, A = 26 dB - 20 dB = 6dB

37. 35 Problem 2.74 Design an circuit for A = 100 at 5MHz using op-amps with f t = 40MHz. --------- For f b = 5MHz, A = f t / f b = 8. This is the maximum gain of the op-amp! Cannot meet the specifications unless we use multiple op-amps with lower gains. If A = 100, use A = A 1 A 2 A 3 = 5 x 5 x 4. The bandwidth is now f b = f t / A = 8 MHz. R 2 = 4K  R 1 = 1K     R 2 = 5K  R 1 = 1K     R 2 = 4K  R 1 = 1K    

38. 36 Frequency response of op-amp circuits Open-loop op-amp Inverting and noninverting amplifiers Low-pass filter High-pass filter

39. 37 Frequency response of open-loop op-amp Open-loop op-amp: f t = A o f b where A o is gain of op-amp

40. 38 Constant Gain-Bandwidth product of op-amp f t = 1MHz f t = |A| f b 1MHz = (10 000) 100Hz 1MHz = (1000) 1KHz 1MHz = (10) 100KHz Tradeoff between gain and bandwidth

41. 39 Frequency response of inverting and noninverting amplifiers Inverting or noninverting amplifier: f t = |A| f b, where A = gain of circuit A = - R 2 / R 1, inverting A = 1 + R 2 /R 1, noninverting fbfb ftft |A| Vi - 20 dB/decade

42. 40 Frequency response of low-pass filter ViVi C A = - Z 2 / Z 1 Z2Z2 Z1Z1 At large frequencies A becomes zero. Passes only low frequencies.

43. 41 Frequency response of low-pass filter ViVi C Low-pass filter: C acts as a short at high frequencies, gain drops to zero at high frequencies, f t |A| f b. f b = 1/2  R 2 C Due to external capacitor Due to op-amp fbfb - 20 dB/decade

44. 42 Frequency response of high-pass filter C ViVi A = - Z 2 / Z 1 Z2Z2 Z1Z1 At large frequencies A becomes - R 2 / R 1. Passes only high frequencies.

45. 43 Frequency response of high-pass filter C ViVi High-pass filter: C acts as an open at low frequencies, gain is zero at low frequencies, f L = 1/2  R 1 C Due to external capacitor Due to op-amp bandwidth

46. 44 Design of a high-pass filter C ViVi Design the circuit to obtain: High-frequency R in = 1K  High-frequency gain = 40dB lower 3 dB frequency = 100Hz R in = R 1 + 1/sC. At high frequencies, s becomes large, R in  R 1. Let R 1 = 1K  A = - R 2 / (R 1 + 1/sC). At high frequencies, s becomes large, A  - R 2 / R 1. A = 40dB = 100, 100 = R 2 / 1K , R 2 = 100K  f L = 1/2  R 1 CC = 1/2  R 1 f L, C = 1/2  1K  F

47. 45 Design of a high-pass filter f L = 100Hz 20 dB/decade (due to capacitor) -20 dB/decade (due to op-amp)

48. 46 SPICE circuit of high-pass filter in example

49. 47 Output of high-pass filter in example

50. 48 Bandpass filter C2C2 C1C1 Vi Both C 2 and C 1 act as shorts at high frequencies. C 2 limits high-frequency gain C 1 limits low-frequency gain The gain at midrange frequencies = - R 2 / R 1 f L = 1/2  R 1 C 1 f H = 1/2  R 2 C 2 20 dB/decade (due to C 1 ) -20 dB/decade (due to C 2 ) bandwidth fLfL fHfH

51. 49 2.8 Large-signal operation of op-amps Saturation Input must be small enough so the output remains less than the supply voltage. Slew rate Maximum slope of output voltage. Response time of op-amps are described by a slew rate rather than a delay.

52. 50 Slew-rate Slew-rate (SR) = d(Vout)/dt | max = Volts/sec (V/  s) V in V out = V in SR = 2.5 V/  s

53. 51 Slew-rate SR = infinity SR = 25 V/  s SR = 5 V/  sSR = 2.5 V/  s

54. 52 Effect of slew rate on sinusoid

55. 53 Prob. 2.80 A buffer amp uses an op-amp with a SR = 10 V/  s, and the input rises between 0 and 5 V, what is the shortest pulse that can be used for full-amplitude output? Slope = 10 V/  s Shortest pulse = Vout/slope = 5V/(10V/  s) = 0.5  s 0.5  s

56. 54 Prob. 2.82 Find the highest frequency of a triangular wave of 20Vp-p that can be reproduced by an op-amp with a SR = 10 V/  s. Slope = 10 V/  s Slew rate = slope = 20/(T/2) = 10V/  s T = 4  s f = 1/T = 250 kHz T +10 -10

57. 55 Prob. 2.83 Given an op-amp with a SR = 10 V/  s, what is the highest frequency at which a 20 Vp-p sine wave can be produced at the output without distortion? Set the maximum slope of a sine wave equal to the slew rate. But  = 2  f, Eq. 2.33 f = (10 V/  s)/2  20 f = 159kHz

58. 56 2.9 DC Imperfections Offset voltage V out should be 0 If V out  0, an offset voltage is present. V os  1-5 mV nonideal op-amp

59. 57 Offset voltage Modeling of offset voltage Ideal op-amp V os +-

60. 58 Input bias currents I-I- I+I+ Input offset current I o = |I + - I - |

61. 59 Minimizing the effect of input bias currents The + and - terminals of the op-amp should “see” the same resistance Measure resistance from “ - ” terminal Resistance measured at “ - ” terminal is R 1 || R 2 Set R 3 equal to this resistance The voltage at the + and - terminals due to offset currents are now the same R3R3

62. 60 Input and output resistances of an op-amp Model of an ideal op-amp Model with resistances R i  1M  R o  100  R cm  100M 

63. 61 Prob. 2.85 Given an inverting op-amp with R 2 = 100K , and R 1 = 1K , and the input grounded, V out is measured to be -0.5V. Find V os. V os + - Nonideal op-amp V out = -0.5V V out = V os (1+R 2 /R 1 ) V os = V out / (1+R 2 /R 1 ) V os = -0.5 / (101) V os  -0.5 mV

64. 62 A noninverting amp with A = 100 uses an op-amp with V os = + 2mV. Find V out when V in = 0.01sin(  t). V out = (V in + V os )(1+R 2 /R 1 ) = (1sin(  t) + 0.2)V. For an inverting amp, V out = - V in (R 2 /R 1 ) + V os (1+R 2 /R 1 ) Prob. 2.86 V os

65. 63 Prob. 2.89 Compensate for the offset voltage in an integrator. Want to compensate for V os here 10V = iR 1 + i1K  But, we want i1K  = 1mV i = 1mV/1K  = 1  A 10V = (1  A)R 1 + (1  A) 1K  R  =  10M  i Voltage loop:

66. 64 Design of power amplifier The power output of op-amps are often limited Determine voltage needed for particular power specification Measure distortion of output

67. 65 Simulation Model source - use Thevenin equivalent circuit Model load - usually by a resistor Supply voltage is determined by power delivered to load: (V out ) 2 /R load Measure distortion by Fourier analysis and Total Harmonic Distortion

68. 66 Circuit Model of source Model of load

69. 67 Distortion measurement in simulation Measurements are for one frequency. Select Transient Analysis from Analysis Menu. Select Enable Fourier Set Center Frequency to frequency of source Set number of harmonics (5 harmonics typical). A harmonic is a multiple of the frequency of the source. Simulate Select FFT for a graphical display of frequency content. Select Examine Output in Analysis Menu for more detailed data.

70. 68 Total Harmonic Distortion THD is the ratio of the energy in the other harmonics compared to the fundamental harmonic (harmonic 1) THD should be low Medium quality amplifier has about 0.1% THD Very good quality amplifier has about 0.01% THD Energy at frequency of source Energy at frequencies not of source

71. 69 Transient output

72. 70 Fourier (frequency) analysis of output

73. 71 Total Harmonic Distortion FOURIER COMPONENTS OF TRANSIENT RESPONSE DC COMPONENT = -7.003561E-03 Average value HARMONIC FREQUENCY FOURIER NORMALIZED PHASE NORMALIZED NO (HZ) COMPONENT COMPONENT (DEG) PHASE (DEG) 1 1.000E+03 4.450E+00 1.000E+00 1.794E+02 0.000E+00 2 2.000E+03 1.115E-03 2.506E-04 -1.007E+02 -2.801E+02 3 3.000E+03 9.572E-04 2.151E-04 -9.129E+01 -2.707E+02 4 4.000E+03 1.128E-03 2.536E-04 -5.363E+01 -2.330E+02 5 5.000E+03 1.971E-03 4.430E-04 -6.763E+01 -2.470E+02 TOTAL HARMONIC DISTORTION = 6.080162E-02 PERCENT Magnitude at a particular frequency Magnitude normalized to fundamental frequency Phase at a particular frequency Phase normalized to fundamental frequency THD

74. 72 Find Vout U1A is an inverting amp U2A is a difference amp V 1 = -3(-6/2) = 9V Using superposition for second amp V out = 9(-1/2) + 2(1+1/2) V out = -4.5 + 3 = -1.5V + V out - V1V1 Op-amp problem

75. 73 Prob. 2.65 Circuit used to drive floating (not connected to ground) loads. Find V B, V C, and V out (V B -V C ) for a 1Vp-p sine wave input Find the voltage gain, V out /V in. What is the largest sine wave output? V B + V out - V C

76. 74 Voltage outputs

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