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9-1 Electronics Principles & Applications Eighth Edition Charles A. Schuler Chapter 9 Operational Amplifiers (student version) ©2013 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-2 INTRODUCTION • The Differential Amplifier • The Operational Amplifier • Determining Gain • Frequency Effects • Applications • Comparators McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-3 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-4 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-5 A differential amplifier driven at one input +VCC Noninverted output Inverted output C B E C E B -VEE McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-6 Both outputs are active because Q1 drives Q2. Q1 serves as an emitter-follower amplifier in this mode to drive Q2. +VCC C B Q1 E Q2 serves as a common-base amplifier in this mode. It’s driven at its emitter. C E Q 2 B -VEE McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-7 A differential amplifier driven at both inputs Common mode input signal +VCC Reduced output Reduced output C B E C E B -VEE 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. +VCC If the input signal goes positive, both transistors want to increase their current but can’t. B C E Constant total current McGraw-Hill 9-8 C E B -VEE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Driven at both inputs with a differential signal 9-9 +VCC Increased output Increased output C B E C E B -VEE 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. Here, one transistor increases its current as the other decreases so the constant total current is not a limiting factor. B +VCC C E Constant total current McGraw-Hill 9-10 C E B -VEE © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-11 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 McGraw-Hill AV(DIF) AV(CM) © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-12 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-13 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-14 Differential amplifier dc analysis IRE = IE = VEE - VBE RE IRE 2 9 V - 0.7 V = 2.13 mA = 3.9 kW VCC +9 V = 4.98 V = 1.06 mA 4.7 kW RL RL IC = IE = 1.06 mA C B 10 kW RB E 3.9 kW VEE McGraw-Hill VRL = IC x RL = 1.06 mA x 4.7 kW VCE = VCC - VRL - VE 4.7 kW = 9 - 4.98 -(-0.7) C E = 4.72 V B RB 10 kW RE -9 V © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-15 Differential amplifier dc analysis continued Assume b = 200 IB = IC b VB = VRB = IB x RB 1.06 mA = 200 = 5.3 mA VCC 4.7 kW 10 kW RB E 3.9 kW VEE McGraw-Hill +9 V = 53 mV RL RL 4.7 kW C B = 5.3 mA x 10 kW C E B RB 10 kW RE -9 V © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-16 Differential amplifier ac analysis 50 mV 50 mV (50 mV is conservative) = 47 W = rE = IE 1.06 mA RL VCC +9 V AV(CM) = RL 2 x RE AV(DIF) = 2 x rE 4.7 kW 4.7 kW RL RL 4.7 kW = 4.7 kW 2 x 3.9 kW = = 50 2 x 47 W C C = 0.6 B 10 kW RB E 3.9 kW VEE McGraw-Hill E B RB 10 kW RE -9 V © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Differential amplifier ac analysis continued CMRR = 20 x log AV(DIF) AV(CM) VCC 4.7 kW 10 kW RB E 3.9 kW VEE McGraw-Hill 50 = 38.4 dB 0.6 +9 V RL RL 4.7 kW C B = 20 x log 9-17 C E B RB 10 kW RE -9 V © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-18 AV(CM) = A current source can replace RE to decrease the common mode gain. RL VCC 2 x RE Replaces this with a very high resistance value. 4.7 kW C B 10 kW RL RL 4.7 kW RB E * C E B RB 10 kW 2 mA *NOTE: Arrow shows conventional current flow. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-19 A practical current source IC 9 V - 5.1 V IZ = = 10 mA 390 W IE = 390 W 5.1 V - 0.7 V 2.2 kW = 2 mA IC = IE = 2 mA 5.1 V 2.2 kW -9 V McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-20 A demonstration of common-mode rejection The common-mode signal cannot be seen in the output. The amplitude of the common-mode signal is almost 30 times the amplitude of the differential signal. McGraw-Hill 6.3 V 60 Hz 212 mV 1 kHz © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-21 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-22 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-23 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-24 Op amps have two inputs Inverting input Output Non-inverting input McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-25 Op-amp Characteristics • • • • McGraw-Hill High CMRR High input impedance High gain Low output impedance • • • • Available as ICs Inexpensive Reliable Widely applied © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-26 With both inputs grounded through equal resistors, VOUT should be zero volts. +VCC VOUT -VEE Imperfections can make VOUT non-zero. The offset null terminals can be used to zero VOUT. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. Dt 9-27 DV 741 DV Slew rate = Dt 0.5 V ms The output of an op amp cannot change instantaneously. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-28 VP f > fMAX Slew Rate fMAX = 2p x VP Slew-rate distortion McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-29 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-30 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-31 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-32 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. RL McGraw-Hill It has a high input impedance and a low output impedance. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-33 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 VIN McGraw-Hill VOUT RL © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-34 Op-amp follower AV(OL) = 200,000 B=1 A AB +1 VIN VOUT The feedback ratio = 1 200,000 @1 AV(CL) = (200,000)(1) + 1 VIN McGraw-Hill VOUT RL © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-35 The closed-loop gain is increased by decreasing the feedback with a voltage divider. R1 200,000 AV(CL) = = 11 (200,000)(0.091) + 1 RF R1 B= 100 kW RF + R1 10 kW = VIN McGraw-Hill VOUT RL 10 kW 100 kW + 10 kW = 0.091 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. It’s possible to develop a different model for the closed loop gain by assuming VDIF = 0. R1 VIN = VOUT x R1 + RF RF Divide both sides by VOUT and invert: 100 kW R1 10 kW VOUT VDIF = 0 VIN McGraw-Hill 9-36 VOUT VIN RF =1+ R1 RL AV(CL) = 11 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-37 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. Virtual ground RF We can ignore the op amp’s input current since it is so small. Thus: I R 1 = IR F 10 kW By Ohm’s Law: 1 kW R1 VIN VDIF = 0 VIN R1 VOUT RL VOUT VIN = = -VOUT RF -RF R1 = -10 The minus sign designates an inverting amplifier. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-38 Due to the virtual ground, the input impedance of the inverting amplifier is equal to R1. Virtual ground RF 10 kW R1 1 kW VDIF = 0 VIN Although op amp input currents are small, in most applications, offset error is minimized by providing equal resistance paths for the input currents. R2 = R1 || RF = 910 W This resistor reduces offset error. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-39 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-40 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-41 A typical op amp has internal frequency compensation. R Output C Break frequency: 1 fB = 2pRC McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-42 Bode plot of a typical op amp Break frequency 120 100 80 60 Gain in dB 40 20 0 1 10 100 1k 10 k 100 k 1M Frequency in Hz McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-43 Op amps are typically operated with negative feedback (closed loop). This increases their useful frequency range. AV(CL) = RF dB Gain = 20 x log 101 = 40 dB VIN McGraw-Hill VIN RF =1+ R1 100 kW =1+ = 101 1 kW 100 kW R1 1 kW VOUT VOUT RL © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-44 Using the Bode plot to find closed-loop bandwidth: 120 100 Gain in dB 80 Break frequency 60 AV(CL) 40 20 0 1 10 100 1k 10 k 100 k 1M Frequency in Hz McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-45 A 741 op amp slews at 0.5 V ms 70 V A 318 op amp slews at ms There are two frequency limitations: Slew rate determines the large-signal bandwidth. Internal compensation sets the small-signal bandwidth. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-46 The Bode plot for a fast op amp shows increased small-signal bandwidth. 120 100 80 Gain in dB 60 40 fUNITY 20 0 1 10 100 1k 10 k 100 k 1M 10M Frequency in Hz McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-47 fUNITY can be used to find the small-signal bandwidth. AV(CL) = RF R1 1 kW McGraw-Hill VIN RF =1+ R1 100 kW =1+ = 101 1 kW 100 kW VIN VOUT VOUT 318 Op amp RL fUNITY fB = AV(CL) 10 MHz = = 99 kHz 101 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-48 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-49 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-50 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-51 R Amplitude response of an RC lag circuit Vout C 1 fb = 2pRC fb 0 dB 10fb 100fb 1000fb f -20 dB Vout -40 dB -60 dB McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-52 R Phase response of an RC lag circuit = tan -1 C -XC R 0.1fb 0o Vout fb 10fb f Vout -45o -90o McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-53 Interelectrode capacitance and Miller effect The gain from base to collector makes CBC effectively larger C BE in the input circuit. CBC R CMiller = AVCBC CInput = CMiller + CBE McGraw-Hill CMiller CBE 1 fb = 2pRCInput © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-54 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 fb1 McGraw-Hill 10 kHz 100 kHz fb2 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-55 Multiple lag circuits: Vout R1 C1 R2 C2 R3 0o C3 f Vout Phase reversal -180o Negative feedback becomes positive! McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-56 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-57 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-58 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-59 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-60 Inverted sum of three sinusoidal signals RF 10 kW 5 kW 5 kHz 3.3 kW 3 kHz 1 kW 1 kHz McGraw-Hill Summing Amplifier Amplifier scaling: 1 kHz signal gain is -10 3 kHz signal gain is -3 5 kHz signal gain is -2 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-61 Difference of two sinusoidal signals (V1 = V2) RF 1 kW Subtracting Amplifier (A demonstration of common-mode rejection) 1 kW 1 kW 1 kW V1 V2 McGraw-Hill VOUT = V2 - V1 © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-62 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.) McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-63 0 Active filter Amplitude in dB -20 Cascade RC -40 -60 10 100 Frequency in Hz McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-64 Active low-pass filter with feedback C1 VOUT C2 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-65 Active low-pass filter with feedback C1 VOUT C2 VIN As fIN increases and C2 loads the input, Vout drops. This increases the signal voltage across C1. This sharpens the knee. -3 dB Gain Feedback can make a filter’s performance even better! Frequency McGraw-Hill fC © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-66 0 Active filter using feedback (two stages) Amplitude in dB -20 -40 -60 10 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). 100 Frequency in Hz McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-67 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-68 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. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-69 Active high-pass filter VOUT VIN feedback -3 dB Gain McGraw-Hill fC Frequency © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-70 VIN VOUT Active band-pass filter (multiple feedback) -3 dB Gain Frequency Bandwidth McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-71 VOUT VIN Active band-stop filter (multiple feedback) -3 dB Gain Frequency Stopband McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-72 56.6 mV Active rectifier 0V 40 mV 0V - 56.6 mV McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-73 Integrator C R VIN McGraw-Hill V Slope = s VOUT 1 Slope = -VIN x RC © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-74 A differentiator shows R and C reversed. Vout = -(Vin/t)RC Differentiation is the opposite of integration. Input Waveforms (Blue) Integrator Output Waveform (Red) Differentiator Output Waveform (Red) Square Triangle Sine McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-75 Comparator with a 1 Volt reference +VSAT 1V 0V -VSAT VOUT VIN 1V McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-76 Comparator with a noisy input signal +VSAT 1V 0V -VSAT VOUT VIN 1V McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-77 Schmitt trigger with a noisy input signal +VSAT UTP LTP Trip points: R1 VSAT x R1 + RF -VSAT VIN R1 McGraw-Hill VOUT RF Hysteresis = UTP - LTP © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-78 +5 V Window comparator 4.7 kW VUL 3V R1 311 VOUT R2 4.7 kW VIN 311 VLL 1 V VOUT is LOW (0 V) when VIN is between 1 V and 3 V. McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-79 Window comparator VUL 3V +5 V 311 VOUT VIN 311 VLL McGraw-Hill 1V Many comparator ICs require pull-up resistors in applications of this type. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-80 +5 V Window comparator 4.7 kW VUL 3V R1 311 VOUT R2 4.7 kW VIN 311 VLL McGraw-Hill 1V VOUT is TTL logic compatible. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-81 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-82 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 McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved. 9-83 REVIEW • The Differential Amplifier • The Operational Amplifier • Determining Gain • Frequency Effects • Applications • Comparators McGraw-Hill © 2013 The McGraw-Hill Companies, Inc. All rights reserved.