What is the Amplitude of the Repetitive Current in a Full Wave Rectifier?

AI Thread Summary
A full-wave rectifier operating at 60Hz with a transformer output voltage of 15V has been analyzed for various parameters. The calculated DC output voltage, considering a diode voltage drop of 1V, is 20.2V, and the minimum capacitance required to maintain ripple voltage below 0.25V is approximately 1.3475F. The peak inverse voltage (PIV) rating of the diode is determined to be 42.4V, and the surge current upon power application is calculated at 10,777A. The discussion focuses on finding the amplitude of the repetitive current in the diode, which is expected to be around 1,650A, but the user struggles to derive this value correctly. Further clarification on calculations and potential typos in previous results is sought to resolve the discrepancies.
shalzuth
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Homework Statement


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?


Homework Equations


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?
 
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shalzuth said:

Homework Statement


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?


Homework Equations


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?

There have got to be typos in your current numbers. Are they missing decimal points?

10777 A
1650 A
 
Positive there are no types :/, I know the values seems large, but the first one is a peak value of the surge current. It's 2*pi*f*C*V, which is derived from...

I = C * dV/dt
= C * d(Vcos(2*pi*f*t))
= C * -V*2*pi*f * sin(2*pi*f*t)
and since it's a peak value, the sin component can only max out at 1.
therefore
= C * 2 * pi * f * V
= 1.34 * 2 * pi * 60 * 20.2

And I can't figure out how to get the amplitude of the current through the diode.
And the solutions say 1650A. The other answers have been correct, so I'd assume e is correct as well.
 
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