A full-wave bridge rectifier circuit with a capacitor filter is designed to meet the following specifications Approximate dc output voltage: 12 V DC time averaged load current: 0.5 A Maximum ripple: 0.2 V peak-to-peak If the transformer input ac voltage is 120 Vrms, determine the value of the capacitance required
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Answer:a.Turns Ratio: 12 VPto VRMS=12√2= 8.49 V;N1N2=V1V2=120V8.49V=14:1b.Filter Capacitor:R =VLIL=12V120mA=12.12=100 Ω;VR= 0.05VP= (0.05)(12V) = 0.6 V =600 mV; VR=IfC;C =If VR; C =120mA60Hz×600mV=0.12060×0.6=.12036= .003333 ≈3300 µF2.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=2Vmπ=2×12Vπ=243.14=7.6 VPb.Ripple:ffor full-wave = 120 HzIL=VLRL=12V12Ω=1 A;VR=IfC=1120(0)=undefinedc.Efficiency:n =100×IL×RVm= 100 ×1A×12Ω12V=100%3.A half-wave rectifier using silicon diode has a secondary emf of 14.14 V (rms) with a resistanceof 0.2 Ω. The diode has a forward resistance of 0.05 Ω and a threshold voltage of 0.7 V. If loadresistance is 10 Ω, determine the following:a.dc load current:IL=VL−0.7RT=6.4V−0.7V0.2Ω+0.05Ω+10Ω=5.710.25=560 mAb.dc load voltage:V2=VRMS2× 0.45 = 14.14 V × 0.45 =6.4 Vc.voltage regulation:VR=IfC; C = 0;undefinedd.circuit efficiency:n= 100 ×IL×RVL= 100 ×0.56×10.256.4= 100 ×5.746.4≈90%e.diode PIV and current rating: PIV = VP=6.4V
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