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(answered) - 1. A full-wave rectifier is to be designed to produce a peak

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(answered) – 1. A full-wave rectifier is to be designed to produce a peakDescriptionSolution downloadThe QuestionMy feedback for this assignment was simply “review 9-11 (all parts)” — can anybody explain how to do these correctly?1. A full-wave rectifier is to be designed to produce a peak output voltage of 12 V, deliver 120 mAto the load, and produce an output with a ripple of not more than 5 percent. An input line voltageof 120 V (rms), 60 Hz is available. Consider a bridge type rectifier. Specify the transformer ratioand the size of the required filter capacitor. Show all work.= 8.49 V;N1N212V120 mA =12.1212?2a. Turns Ratio: 12 VP to VRMS ==V1V2120 V8.49V ==14 : 1b. Filter Capacitor: R =VLIL=IfC ; C =V) = 0.6 V = 600 mV; VR =0.12060 ? 0.6=.12036= 100 ?; VR = 0.05VP = (0.05)(12If VR ; C =120 mA60 Hz ? 600 mV== .003333 ? 3300 ?F2. Silicon diodes are used in a two-diode full-wave rectifier circuit to supply a load with 12 voltsD.C. Assuming ideal diodes and that the load resistance is 12 ohms, compute the secondarytransformer voltage, the load ripple voltage, and the efficiency of the rectifier. Show all work.a. Secondary: VO =2Vm?2 ?12 V?=b. Ripple: f for full-wave = 120 Hz IL =VLRL243.14=== 7.6 VP12 V12 ?= 1 A; VR =IfC=1120 (0) = undefinedI L?RVmc. Efficiency: n =100 ?= 100 ?1 A ?12 ?12 V= 100%3. A half-wave rectifier using silicon diode has a secondary emf of 14.14 V (rms) with a resistanceof 0.2 ?. The diode has a forward resistance of 0.05 ? and a threshold voltage of 0.7 V. If loadresistance is 10 ?, determine the following:a. dc load current: IL =V L ?0.7RTb. dc load voltage: V2 =V RMS ? 0.45 = 14.14 V ? 0.45 = 6.4 Vc. voltage regulation: VR ==6.4 V ?0.7 V0.2 ?+0.05 ?+10 ?5.7= 10.25= 560 mA2IfC ; C = 0; undefinedd. circuit efficiency: n = 100 ?I L?RVL= 100 ?90%e. diode PIV and current rating: PIV = VP = 6.4V0.56? 10.256.4= 100 ?5.746.4?

(answered) – 1. A full-wave rectifier is to be designed to produce a peak

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