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Electronics Principles & Applications Eighth Edition

Electronics Principles & Applications Eighth Edition. Charles A. Schuler. Chapter 9 Operational Amplifiers. INTRODUCTION. The Differential Amplifier The Operational Amplifier Determining Gain Frequency Effects Applications Comparators. Noninverted output. Inverted output.

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Electronics Principles & Applications Eighth Edition

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  1. Electronics Principles & Applications Eighth Edition Charles A. Schuler Chapter 9 Operational Amplifiers

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

  3. Noninverted output Inverted output A differential amplifier driven at one input +VCC C C B B E E -VEE

  4. 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. Both outputs are active because Q1 drives Q2. +VCC C C B B E E Q1 Q2 -VEE

  5. Common mode input signal Reduced output Reduced output A differential amplifier driven at both inputs +VCC C C B B E E -VEE

  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. +VCC If the input signal goes positive, both transistors want to increase their current but can’t. C C B B E E Constant total current -VEE

  7. Increased output Increased output Driven at both inputs with a differential signal +VCC C C B B E E -VEE

  8. A differential amplifier driven at both inputs with a differential signal shows high gain. +VCC Here, one transistor increases its current as the other decreases so the constant total current is not a limiting factor. C C B B E E Constant total current -VEE

  9. AV(DIF) AV(CM) CMRR = 20 x log • 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.

  10. VEE - VBE 9 V - 0.7 V = 2.13 mA = IRE = RE 3.9 kW IRE IE = = 1.06 mA 2 Differential amplifier dc analysis VRL = IC x RL = 1.06 mA x 4.7 kW +9 V VCC = 4.98 V VCE = VCC - VRL - VE 4.7 kW RL RL 4.7 kW = 9 - 4.98 -(-0.7) IC = IE = 1.06 mA C C = 4.72 V B B E E RB 10 kW RB 10 kW RE 3.9 kW -9 V VEE

  11. IC IB = = b 1.06 mA 200 Differential amplifier dc analysis continued Assume b = 200 VB = VRB = IB x RB = 5.3 mA x 10 kW +9 V VCC = 53 mV = 5.3 mA 4.7 kW RL RL 4.7 kW C C B B E E RB 10 kW RB 10 kW RE 3.9 kW -9 V VEE

  12. 50 mV 50 mV = 47 W = rE = IE 1.06 mA RL RL AV(DIF) = AV(CM) = 2 x RE 2 x rE 4.7 kW 4.7 kW = 2 x 3.9 kW = = 50 2 x 47 W Differential amplifier ac analysis (50 mV is conservative) +9 V VCC 4.7 kW RL RL 4.7 kW C C = 0.6 B B E E RB 10 kW RB 10 kW RE 3.9 kW -9 V VEE

  13. = 20 x log CMRR = 20 x log = 38.4 dB 50 AV(DIF) 0.6 AV(CM) Differential amplifier ac analysis continued +9 V VCC 4.7 kW RL RL 4.7 kW C C B B E E RB 10 kW RB 10 kW RE 3.9 kW -9 V VEE

  14. RL AV(CM) = 2 x RE * 2 mA * A current source can replace RE to decrease the common mode gain. VCC Replaces this with a very high resistance value. 4.7 kW RL RL 4.7 kW C C B B E E RB 10 kW RB 10 kW NOTE: Arrow shows conventional current flow.

  15. IZ = = 10 mA IE = = 2 mA 5.1 V - 0.7 V 2.2 kW 9 V - 5.1 V 390 W A practical current source IC 390 W IC = IE = 2 mA 5.1 V 2.2 kW -9 V

  16. The common-mode signal cannot be seen in the output. A demonstration of common-mode rejection 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

  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 ________. 2rE The common-mode gain can be found by dividing the collector load by ________. 2RE

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

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

  20. Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT. With both inputs grounded through equal resistors, VOUT should be zero volts. +VCC VOUT -VEE

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

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

  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. 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

  25. 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

  26. A VOUT VIN AB +1 The feedback ratio = 1 AV(CL) = 200,000 @ 1 (200,000)(1) + 1 Op-amp follower AV(OL) = 200,000 B = 1 VOUT RL VIN

  27. 200,000 AV(CL) = = 11 (200,000)(0.091) + 1 R1 B = RF + R1 10 kW = 100 kW + 10 kW The closed-loop gain is increased by decreasing the feedback with a voltage divider. RF 100 kW R1 10 kW VOUT RL = 0.091 VIN

  28. VIN = VOUT x VOUT R1 VIN = R1 + RF 1 + RF R1 It’s possible to develop a different model for the closed loop gain by assuming VDIF = 0. RF Divide both sides by VOUT and invert: 100 kW R1 10 kW VDIF = 0 VOUT RL VIN AV(CL) = 11

  29. VIN = R1 -VOUT -RF VOUT = RF R1 VIN 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 By Ohm’s Law: 1 kW R1 VDIF = 0 VIN VOUT RL = -10 The minus sign designates an inverting amplifier.

  30. Due to the virtual ground, the input impedance of the inverting amplifier is equal to R1. Virtual ground Although op amp input currents are small, in most applications, offset error is minimized by providing equal resistance paths for the input currents. RF 10 kW R1 1 kW VDIF = 0 VIN R2 = R1|| RF = 910 W This resistor reduces offset error.

  31. 1 fB = 2pRC A typical op amp has internal frequency compensation. R Output C Break frequency:

  32. Bode plot of a typical op amp Break frequency 120 100 80 60 Gain in dB 40 20 0 1k 10 k 1M 100 k 10 100 1 Frequency in Hz

  33. AV(CL) = = 101 VOUT 100 kW = 1 + VIN 1 kW = 1 + RF R1 Op amps are typically operated with negative feedback (closed loop). This increases their useful frequency range. RF 100 kW R1 1 kW dB Gain = 20 x log 101 = 40 dB VOUT RL VIN

  34. Using the Bode plot to find closed-loop bandwidth: 120 100 Gain in dB 80 Break frequency 60 AV(CL) 40 20 0 1k 10 k 1M 100 k 10 100 1 Frequency in Hz

  35. 70 V ms 0.5 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.

  36. The Bode plot for a fast op amp shows increased small-signal bandwidth. 120 100 80 Gain in dB 60 40 fUNITY 20 0 1k 10M 10 k 1M 100 k 10 100 1 Frequency in Hz

  37. AV(CL) = = 101 VOUT 100 kW = 1 + VIN 1 kW = fUNITY fB = AV(CL) 1 + RF 10 MHz = 99 kHz R1 101 fUNITY can be used to find the small-signal bandwidth. RF 100 kW R1 1 kW VOUT RL VIN 318 Op amp =

  38. 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

  39. 1 fb = 2pRC R Amplitude response of an RC lag circuit Vout C fb 10fb 100fb 1000fb f 0 dB -20 dB Vout -40 dB -60 dB

  40. -XC R R Phase response of an RC lag circuit Vout C  = tan-1 0.1fb fb 10fb f 0o Vout -45o -90o

  41. CMiller 1 fb = 2pRCInput Interelectrode capacitance and Miller effect CBC The gain from base to collector makes CBC effectively larger in the input circuit. CBE R CBE CMiller = AVCBC CInput = CMiller + CBE

  42. 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

  43. Phase reversal Multiple lag circuits: Vout R1 R2 R3 C1 C2 C3 f 0o Vout -180o Negative feedback becomes positive!

  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. 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

  46. 5 kW 5 kHz Inverted sum of three sinusoidal signals RF 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

  47. Difference of two sinusoidal signals (V1 = V2) RF 1 kW Subtracting Amplifier (A demonstration of common-mode rejection) VOUT = V2 - V1 1 kW 1 kW 1 kW V1 V2

  48. 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.)

  49. 0 Active filter -20 Cascade RC -40 Amplitude in dB -60 100 10 Frequency in Hz

  50. Active low-pass filter with feedback VOUT C2 C1 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.

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