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electronics fundamentals circuits, devices, and applications THOMAS L. FLOYD DAVID M. BUCHLA chapter 19 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Comparators Op-amps can be used to compare the amplitude of one voltage with another. Although general-purpose op-amps can be used as comparators, special op-amps are available to optimize speed and add features. +V An example of a comparison circuit is shown. The input is compared with a reference set by the voltagedivider. Notice that there is no feedback; the op-amp is operated in open-loop, so the output will be in saturation. Electronics Fundamentals 8th edition Floyd/Buchla R1 Vin Vout + R2 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Sketch the output of the comparator in relationship to the input; assume the maximum output is ±13 V. +10 V The threshold is +4.2 V. The output is in positive saturation when Vin > +4.2 V +4.2 V Vin 0V V = +15 V R1 10 kW Vin -10 V +13 V + Vout 0V R2 3.9 kW -13 V Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Show the output of the comparator for the last example if the inputs to the op-amp are reversed. The threshold is still +4.2 V but now the output is in negative saturation when Vin > +4.2 V. +10 V +4.2 V Vin 0V V = +15 V -10 V +13 V R1 10 kW + - Vin Vout 0V R2 3.9 kW -13 V Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Summing amplifier There are a number of useful applications for the basic inverting amplifier configuration. One is the summing amplifier that uses two or more inputs and one output. The virtual ground isolates the inputs from each other. Input current from each input is passed to Rf, which develops an output voltage that is proportional to the algebraic sum of the inputs. VIN1 VIN2 VIN3 VINn Rf R1 R2 - R3 VOUT + Rn Virtual ground Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Averaging amplifier An averaging amplifier is a variation of the summing amplifier in which all input resistors are equal. The feedback resistor is the reciprocal of the number of inputs times the input resistor value. For example, if there are three input resistors, each with a value of 10 kW, then Rf = 3.3 kW to form an averaging amplifier. Electronics Fundamentals 8th edition Floyd/Buchla VIN1 VIN2 VIN3 R1 Rf 10 kW R2 3.3 kW 10 kW R3 + VOUT 10 kW © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Scaling adder A scaling adder is another variation of the summing amplifier in which the input resistors are adjusted to weight inputs differently. The input “weight” is proportional to the current from that input. Larger resistors will allow less current for a given input voltage, so they have less “weight” than smaller resistors. In the case shown, VIN3 is “weighted” 2 times more than VIN2, which is 2 times more than VIN1. Electronics Fundamentals 8th edition Floyd/Buchla VIN1 VIN2 VIN3 R1 Rf 10 kW R2 10 kW 5.0 kW R3 + VOUT 2.5 kW © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Scaling adder What is VOUT for the scaling adder if all inputs are + 1.0 V? By Ohm’s law, the currents into Rf are I1 = 0.1 mA, I2 = 0.2 mA and I3 = 0.4 mA. Using the superposition theorem, the current in Rf is 0.7 mA. From Ohm’s law, VOUT = 7 V VIN1 VIN2 VIN3 R1 Rf 10 kW R2 10 kW 5.0 kW R3 + VOUT 2.5 kW Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Integrators Mathematical integration is basically a summing process. Within certain limitations, an integrator circuit simulates this process. The ideal integrator is essentially a summing amplifier with a capacitor in place of the feedback resistor. Vin Rf C In practical circuits, a large value resistor is usually in parallel with the capacitor to prevent the output from drifting into saturation. R - Vout + Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Integrators For the ideal integrator, the rate of change of the output is given by Vout - Vin t Ri C The minus sign in the equation is due to the inverting amplifier. If the input is a square wave centered about 0 V, the output is a negative triangular wave (provided saturation is not reached). C Vin Vin 0V R + Electronics Fundamentals 8th edition Floyd/Buchla Vout 0V Vout © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary A 5 kHz square wave with 10 Vpp is applied to a practical integrator. Show the output waveform voltages. During the positive input (½ the period), the change in the output is V 5V Vout - in t 100 μs = 5.6 V Ri C 2.7 kW 33 nF Rf 270 kW C Vin R 2.7 kW 33 nF + Electronics Fundamentals 8th edition Floyd/Buchla Vout The feedback resistor (Rf) is large compared to R, so has little effect on the shape of the waveform. In a practical circuit, it will cause the output waveform to center on zero as shown on the following slide. © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary continued… The results of a computer simulation on Multisim confirm the calculated change (5.6 V) in output voltage (blue line). Rf 270 kW C Vin R 2.7 kW 33 nF + Electronics Fundamentals 8th edition Floyd/Buchla Vout © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Differentiators In mathematics, differentiation is the process of finding the rate of change. An idea differentiator circuit is shown. It produces an inverted output that is proportional to the rate of change of the input. In practical circuits, a small value resistor is added in series with the input to prevent high frequency ringing. Vin Rin Vin Rf C - Vout + Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Differentiators The output voltage for the ideal differentiator is given by VC Vout - R f C t The minus sign in the equation is due to the inverting amplifier. If the input is a ramp, the output is a negative dc level for the positive slope and a positive dc level for the negative slope. Rf Vin Vin Vout Electronics Fundamentals 8th edition Floyd/Buchla C - Vout + © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary A 1.0 kHz, 10 Vpp triangular wave is applied to a practical differentiator as shown. Show the output in relationship to the input. When the input has a positive slope, the output is V Vout - C t 10 V R C 2.7 kW 100 nF -5.4 V f 0.5 ms By symmetry, when the input has a negative slope, the Rf output will be +5.4 V. Rin Vin +5.0 V Vin 2.7 kW C 120 W 100 nF - Vout + 0V -5.0 V 0 Electronics Fundamentals 8th edition Floyd/Buchla 1 ms 2 ms See next slide for waveforms… © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary continued… The results of a computer simulation on Multisim confirm the calculated output voltages (±5.4 V). The output voltage is the blue line. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Oscillators The feedback oscillators introduced in Chapter 17 and other types of feedback oscillators can be implemented with op-amps. One type of feedback oscillator is called the Wien-bridge oscillator. This circuit is useful for generating low distortion sine waves. Lead-lag circuit Negative feedback with JFET gain control JFET bias circuit Electronics Fundamentals 8th edition Floyd/Buchla Rf C1 - R1 Vout + D1 Q1 R2 C2 R3 R4 C3 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Oscillators The lead-lag circuit in the Wien-bridge oscillator has a maximum response at the resonant frequency given by 1 Vin The lead-lag circuit fr response is… 2πRC Rf C 1 Vout This equation is valid when R’s and C’s in the lead-lag circuit are equal. Because the attenuation is ⅓ at fr, the gain of the Wien bridge must set for 3. Electronics Fundamentals 8th edition Floyd/Buchla - R1 Vout Vout ⅓Vin + D1 Q1 R2 C2 R3 R4 fr f C3 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Wien-bridge oscillator What is the frequency of the bridge? The frequency is given by fr 1 2πRC 1 2π 6.8 kW 47 nF 498 Hz Electronics Fundamentals 8th edition Floyd/Buchla Rf 10 kW C1 47 nF R1 6.8 kW - Vout + D1 Q1 R2 6.8 kW C2 47 nF R3 1.0 kW R4 10 kW C3 1.0 mF © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Triangular-wave oscillator A triangular-wave oscillator can be made from a comparator and an integrator. The integrator produces a ramp due to the constant current charging of the capacitor. When the ramp reaches a trip point, the comparator suddenly switches to opposite level and the ramp changes direction. C Vout (square) + Comparator Electronics Fundamentals 8th edition Floyd/Buchla R1 Vout (triangle) R2 R3 + Integrator © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Square-wave relaxation oscillator The square-wave relaxation oscillator uses a comparator to switch the output based on the charging and discharging of a capacitor. R1 Vout C Vout + R2 R3 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Active filters A selectsthe certain frequencies and excludes others. Byfilter reversing resistors and capacitors in the low-pass Active use op-amps optimize thefilter frequency circuit, filters a high-pass filter is to created. This has a gain response. A 2-pole where low-pass of 1 at frequencies f > filter fc. and its response is shown. The gain for this filter is 1 (0 dB) for f < fc. Gain (dB) C1 R1 V Vin R1 -- R2 ++ in C1 C2 R2 Vout 00 -3 -3 -40 dB/decade -40 dB/decade C2 f fc fc Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Voltage regulators Voltage regulators are made from integrated circuits. A basic series IC regulator has four blocks: Control element VIN Reference voltage Electronics Fundamentals 8th edition Floyd/Buchla Error detector VOUT Sample circuit © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Voltage regulators A series regulator uses a comparator to compare the output voltage with a reference voltage. The series transistor drops more or less voltage to keep the output constant. Q1 VIN The voltage at the inverting input is forced to be the same as the reference voltage by feedback action. Therefore, R2 VOUT 1 VREF R3 Electronics Fundamentals 8th edition Floyd/Buchla VOUT R1 + - R2 R3 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Voltage regulators What is the output voltage of the series regulator? VIN = R +24 V VOUT 1 2 VREF R1 R3 4.7 kW 6.8 kW 1 5.1 V 3.3 k W = 15.6 V Q1 VOUT + 5.1 V Electronics Fundamentals 8th edition Floyd/Buchla R2 6.8 kW R3 3.3 kW © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Voltage regulators A shunt regulator also has four blocks; it controls the current in the parallel control element. A series resistor drops more or less voltage to keep the output constant. R1 VIN VOUT Reference voltage Error detector Control element Sample circuit Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Summary Voltage regulators Shunt regulators are not as efficient as series regulators, but have the advantage of short circuit protection. R1 VIN Can you identify each element in this circuit? Reference voltage Electronics Fundamentals 8th edition Floyd/Buchla VOUT R2 - Error detector Q1 + Control element R3 Sample circuit R4 © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Selected Key Terms Summing An amplifier with several inputs that produces amplifier an output voltage proportional to the algebraic sum of the inputs. Averaging An amplifier with several inputs that produces amplifier an output voltage that is the mathematical average of the input voltages. Scaling adder A special type of summing amplifier with weighed inputs. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Selected Key Terms Integrator A circuit that produces an inverted output that approaches the mathematical integral of the input. Differentiator A circuit that produces an inverted output that approaches the mathematical derivative of the input, which is the rate of change. Active filter A frequency selective circuit consisting of active devices such as transistors or op-amps combined with reactive (RC) circuits. Series A type of voltage regulator with the control regulator element in series between the input and output. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 1. When an op-amp is configured as a comparator, the gain is equal to a. 0 b. 1 c. a ratio of two resistors. d. the open-loop gain. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 2. The approximate voltage at the inverting input of the op-amp shown is equal to a. the average of the input voltages. b. the sum of the input voltages. c. 0 V d. VOUT 3 VIN1 VIN2 VIN3 R1 Rf 10 kW R2 3.3 kW 10 kW R3 - VOUT + 10 kW Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 3. For the scaling adder shown, the input with the greatest weight is R1 R f VIN1 a. VIN1 b. VIN2 c. VIN3 VIN2 VIN3 10 kW R2 5.0 kW R3 10 kW + VOUT 2.5 kW d. they are all equal. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 4. In a practical integrator, the purpose of the feedback resistor (Rf) is to a. limit the gain. Rf b. prevent drift. C c. prevent oscillations. d. all of the above. Vin R + Electronics Fundamentals 8th edition Floyd/Buchla Vout © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 5. Assume the top waveform represents the input to a differentiator circuit. Which represents the expected output? Vin a. b. c. d. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 6. The lead-lag network in a Wien bridge with equal value R’s and C’s attenuates the signal by a factor of a. 2 b. 3 c. 5 d. 10 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 7. A Wien-bridge is used to produce a. sine waves. b. square waves. c. triangle waves. d. all of the above. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 8. For the circuit shown, the two outputs (in red) produce a. sine and square waves. b. triangle and square waves. c. sine and triangle waves. d. sawtooth and triangle waves. C Vout + Comparator Electronics Fundamentals 8th edition Floyd/Buchla R1 Vout R2 R3 + Integrator © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 9. The purpose of the op-amp in the series regulator is a. to sample the output. b. to establish a reference. c. as a control element. d. error detection. Q1 VIN VOUT R1 + - R2 R3 Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz 10. An advantage of a shunt regulator is a. short circuit protection. b. efficiency. c. no need for a reference voltage. d. all of the above. Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved. Chapter 19 1 Quiz Answers: Electronics Fundamentals 8th edition Floyd/Buchla 1. d 6. b 2. c 7. a 3. c 8. b 4. b 9. d 5. c 10. a © 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.