D * 4.80 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 12 V on which a maximum of ±1-V ripple is allowed. The rectifier feeds a load of 200. The rectifier is fed from the line voltage (120 V rms, 60 Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: (a) Specify the rms voltage that must appear across the transformer secondary. (b) Find the required value of the filter capacitor. (c) Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. (d) Calculate the average current through the diode during conduction. (e) Calculate the peak diode current.

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Engineering

Kamren Stokes

13 May, 17:22

D * 4.80 It is required to use a peak rectifier to design a dc power supply that provides an average dc output voltage of 12 V on which a maximum of ±1-V ripple is allowed. The rectifier feeds a load of 200. The rectifier is fed from the line voltage (120 V rms, 60 Hz) through a transformer. The diodes available have 0.7-V drop when conducting. If the designer opts for the half-wave circuit: (a) Specify the rms voltage that must appear across the transformer secondary. (b) Find the required value of the filter capacitor. (c) Find the maximum reverse voltage that will appear across the diode, and specify the PIV rating of the diode. (d) Calculate the average current through the diode during conduction. (e) Calculate the peak diode current.

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Margaret Gibbs · Answer 1

Explanation: a) Resistance, R = 200 ohms; Voltage = 12 volts

Current, I = V/R = 12/200 = 0.06 A

rms current, i = 0.06/√2 = 0.06 X 0.707 = 0.042 A

rms voltage, v = i*R = 0.042 X 200 = 8.48volts

b) Capacitive reactance, Xc = 120/0.042 = 2.86 kilo ohms

Xc = 1/2πfc

Capacitor, c = 1/2πfXc = 1 / (2 x 3.142 x 60 x 2.86 10∧3) = 1/1078334.4 = 0.93 micro Farad

c) Maximum reverse voltage, Vr = (12 - 0.7) volt = 11.3volts

d) Average current, Ia = 0.7/200 = 0.0035 A

e) Peak diode current, Io = (0.06 + 0.0035) A = 0.064 A