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Homework Help: Into capacitor, half wave rectifier diode conducts for?

  1. Jan 18, 2017 #1
    1. The problem statement, all variables and given/known data
    The time for which the diode conducts is

    2. Relevant equations
    Using integration and differentiation Vc = 1 / C integral (current)(dt)

    3. The attempt at a solution
    Vs = -Vc
    Vm sin(wt) = -Vc = - (1/C) integral (Idt)
    now differentiate both sides to get expression for I that is current.
    Vm (coswt) * w = -(1/C) I
    So I = Vm(costwt) * C * w (-1)

    So now i draw current wave, and then how to proceed? How do you get an answer for this? Am i to find out when current goes negative?

    There was a similar question for RL circuit that was

    Vs = L di/dt
    so integrate to get expression for current.
    Vs sin(wt) = L di/dt
    Vs sin (wt) dt = L di
    Vs cos (wt) (-1) / w + k = L * I
    I = Vs cos (wt) (-1) /(wL) + k' where k' = k/L equation 1

    to find value of K' take initial condition. At t = 0, I = 0
    so 0 = Vs (1) (-1) / (wL) + k' from equation 1

    K' = Vs/(wL) equation 2

    so expression for current becomes from equation 1 and 2

    I = Vs cos(wt) (-1) / (wL) + Vs / (wL) = Vs/ (wL) * (1 - cos (wt))

    Drawing this we get I is always positive, so it conducts for 360 degrees when input goes 1 entire cycle.
    360 degrees was right for RL circuit but for RC answer is not given as the book is old and the page is torn.

    i'm not sure what to do in case of given question of diode + C with charged as shown.
    Expression is
    I = Vm(costwt) * C * w (-1)

    So what to now?

    Attached Files:

  2. jcsd
  3. Jan 18, 2017 #2


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    Staff: Mentor

    For the circuit shown, if the capacitor is initially uncharged then the diode will conduct for the entire first ¼-cycle of sinusoid V.sin ωt. After that, the diode will never conduct again.

    BUT your thread title says "RC" so shouldn't you have included a resistance in parallel to the capacitor?
  4. Jan 18, 2017 #3
    Thanks, Quarter means 45 degrees right?
    I was unable to find how to change title but found now. Learnt 2 new things today.:smile:
  5. Jan 18, 2017 #4


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    Staff: Mentor

    A full cycle is 360°, a half-cycle is 180°, so a quarter-cycle is ⬜❓
  6. Jan 18, 2017 #5


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    Staff: Mentor

    I don't like the question now. Are you sure it's how you want it? Almost no mathematics required.
  7. Jan 18, 2017 #6
    Quarter cycle is 90 degrees. Darn i am so stupid.

    We do need maths to find equation of current by integrating... Vc = 1/C (integration) (Idt)
    I = Vm(costwt) * C * w (-1)
    This is current waveform. So it'll flow from 0 to 90 degrees.

  8. Jan 18, 2017 #7


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    Science Advisor
    Homework Helper
    Gold Member


    Usually there is a load in parallel with the capacitor that causes it to discharge. This in turn causes the diode to conduct for part of each cycle (after the initial 1/4 cycle). In fact problems like this usually ignore the initial 1/4 cycle and only want you to calculate the time per cycle after that.
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