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Homework Help: Full Wave Rectifier (2 diodes)

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    Vdc = Vp - Von
    Vr = Vdc/R * T/2C
    PIV = 2 * PIV
    Vp = Vrms * 2^.5

    3. 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?
  2. jcsd
  3. Mar 31, 2010 #2


    User Avatar

    Staff: Mentor

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

    10777 A
    1650 A
  4. Mar 31, 2010 #3
    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.
    = 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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