LA SIERRA PHYS 486 - Electronics fundamentals

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34Slide 35Slide 36Slide 37Slide 38Slide 39Slide 40Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.chapter 19electronics fundamentalscircuits, devices, and applicationsTHOMAS L. FLOYDDAVID M. BUCHLAChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.ComparatorsOp-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.An example of a comparison circuit is shown. The input is compared with a reference set by the voltage-divider. Notice that there is no feedback; the op-amp is operated in open-loop, so the output will be in saturation.VinR1Vout+R2+VChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Sketch the output of the comparator in relationship to the input; assume the maximum output is ±13 V.VinR1Vout+R2V = +15 V10 k3.9 kThe threshold is +4.2 V. The output is in positive saturation when Vin > +4.2 VVin+10 V10 V0 V+4.2 V+13 V13 V0 VChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Show the output of the comparator for the last example if the inputs to the op-amp are reversed.VinR1Vout+R2V = +15 V10 k3.9 kThe threshold is still +4.2 V but now the output is in negative saturation when Vin > +4.2 V.Vin+10 V10 V0 V+4.2 V+13 V13 V0 VChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Summing amplifierThere 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.R1VIN1VIN2VIN3VINnR2R3RnRf+VOUTThe 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.Virtual groundChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Averaging amplifierAn 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.R1VIN1VIN2VIN3R2R3Rf+VOUTFor example, if there are three input resistors, each with a value of 10 k, then Rf = 3.3 k to form an averaging amplifier.10 k10 k10 k3.3 kChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Scaling adderA 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.R1VIN1VIN2VIN3R2R3Rf+VOUTLarger 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 VIN1.10 k5.0 k2.5 k10 kChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.Scaling adderR1VIN1VIN2VIN3R2R3Rf+VOUT10 k5.0 k2.5 k10 k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 VChapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.IntegratorsMathematical 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. RCVinVoutIn practical circuits, a large value resistor is usually in parallel with the capacitor to prevent the output from drifting into saturation. Rf+Chapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.IntegratorsFor the ideal integrator, the rate of change of the output is given byRCVinVoutout iniV Vt R CD=-DThe 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).VinVout0 V0 V+Chapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.RCVinVout( ) ( )5 V 100 μs =2.7 k 33 nFinoutiVV tR CD =- D =-W A 5 kHz square wave with 10 Vpp is applied to a practical integrator. Show the output waveform voltages.33 nF2.7 k270 kRfDuring the positive input (½ the period), the change in the output is5.6 VThe 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.+Chapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights Reserved.RCVinVout33 nF2.7 k270 kRfThe results of a computer simulation on Multisim confirm the calculated change (5.6 V) in output voltage (blue line).continued…+Chapter 1Chapter 1Chapter 19Chapter 19Electronics Fundamentals 8th edition Floyd/Buchla© 2010 Pearson Education, Upper Saddle River, NJ 07458. All Rights


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