Question

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Draw a bridge rectifier fed from a single-phase 240-V (Au) supply with a 4:1 step down transformer. A 10-2 load is connected to the output. Assume a simplified model for the diodes with VoN 1V per diode and zero leakage current. (a) Draw voltage, current and p(t) waveforms for one diode. (b) Derive the diode voltage spec from the voltage waveforms, and add 50% as a safety 1. factor. (c) Derive the (average) diode current spec from the current waveforms (integrate). After obtaining the symbolic solution, put values in and add a 30% safety factor. (d) Derive the (average) diode power spec from the power waveforms (integrate). After obtaining the symbolic solution, put values in and add a 20% safety factor. (e) Calculate the output RMS voltage and output power. (f) Calculate the circuit efficiency (assuming losses only in diodes). (g) Now, from first principles (RC discharge curve), calculate the filter capacitor required to limit the output voltage Ripple Factor (peak-peak ripple : dc level) to 50%. Hint: watch lecture video for discussion on Ex. 3.12, p.137.

Answer 1

a4oV i(man) ?mS ottege D3 maximum input voltage Vi(ma iCmax) - 34.853 vots max- O2 Diod. → Curren. max) Poutr plt)

,D(max) u(max) = d[may) 84.853 uoHS Diode voltage speci-fication Vi(mox)5 84-853t o.5xsy-853 こ|127-291S volts Aueraqe diode Cument i(max) - yo) (- Dode onwar d g4.853 2x1)uoHaae =To@ug)'tGax: Jaig)) 2632-63) Diode cument spec' fication d) Auerage Dlode pewer ma X RL IT

Ou oLoer Spe cifcaton - avg) 2.63구 0, 2 x 二 2-637 3.1644 toatts output Rms oage. 84.85 3-ех Vo (Rms) Mo(Rms) 58-586 uolts OCRMS 3y3-23) watts Averaqe dc cunent I 2(84 853-2x)) datニ5, 2746

output Rms Crent - orm 58-586 6.5SA Pac To(imj) RL - 6.658)x10 - 343-16 w owe r circuit ffecte Pac 278.212 343.16 9) Ripple -factoY e fuency mot 9vein 0.5 =