A full-wave rectifier is operating at a frequency of 60Hz, and the rms value of the transformer output voltage is 15 V.
(a) What is the value of the dc output voltage if the diode voltage drop is 1 V?
(b) What is the minimum value of C required to maintain the ripple voltage to less than .25 V if R = .5 ohm?
(c) What is the PIV rating of the diode in the circuit?
(d) What is the surge current when power is first applied?
(e)What is the amplitude of the repetitive current in the diode?
Vdc = Vp - Von
Vr = Vdc/R * T/2C
PIV = 2 * PIV
Vp = Vrms * 2^.5
The Attempt at a Solution
I've solved a-d.
(a) Vrms * .5^2 - Von (20.2 V)
(b) Rearrange Vr to solve for C (1.3475 F)
(c) PIV = 2 * Vp (42.4 V)
(d) Isc = 2 * pi * f * C * Vp (from deriving and such, 10777 A)
I know all of those are right.
(e) I have tried a bunch of number juggling, V = IR, I = C * dV/dt, a bunch of different variations, etc., and can't get the correct answer. The final answer should be about 1650 A.
Point me in the right direction?