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Half-Wave Rectifier Average Diode Current

  1. Jul 2, 2009 #1
    http://img12.imageshack.us/img12/7084/helpphg.jpg [Broken]

    In the image shown above, how is the average diode current calculated? This is an old homework assignment and I'm taking notes. I would appreciate any help.

    Thank you.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jul 2, 2009 #2


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    You should still show your work / thoughts, even if it's an old assignment / homework.

    Learn to break down the situation and look at the problem analytically. A brilliant / lucky engineer is sometimes able to spot something immediately and/or produce an answer from seemingly nowhere. A good engineer should be able to break down the situation logically, and work their way through a problem. Here are some guiding questions:

    1) What's 120VAC mean in V_RMS and V_Peak?
    2) What's V_peak on the output side?
    3) How much voltage does the (assume ideal) diode drop?
    Last edited: Jul 2, 2009
  4. Jul 2, 2009 #3
    MATHLABdude those are all questions that I already answered. The 120 volts AC is the AC voltage from the source (outlet). Primary Peak current is 120 volts RMS / .707.

    The peak load voltage is that across the secondary coil. 680 volts peak.

    In a forward biasing rectifier such as this ,the voltage drop across the diode is .7 volts.

    Assuming it is ideal it has no resistance, and therefore, 0v across its terminals.
  5. Jul 2, 2009 #4


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    Sorry, this wasn't made clear. I'm unclear of what has 0V across its terminals. The diode (even an ideal diode, with 0 resistance) has a forward voltage drop (the internal resistance just models the shift in the forward diode voltage one sees with changing current). Regardless, since 680 >> 0.5 V, the impact of that is little more than a rounding error.

    Good, now that you have your voltage across the secondary, you can calculate the peak current, which I assume that you've done. And you know that the current must be sinusoidal, except for when the diode blocks out the negative going portion of the waveform.

    So now what's the area of that waveform divided by its period? That's the average value. (Hint, you don't actually need to get an expression for the full waveform with frequency and what not, just use a generic sin function, and scale the results up by your peak current).
  6. Jul 2, 2009 #5
    Thanks for the help, but I figured it out on my own. The average Diode current is Vp / pi which is then / the load resistor.
  7. Jul 2, 2009 #6


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    You mean V_dc / R_load? Using the peak values won't give you the average values you're looking for.
  8. Jul 2, 2009 #7
    Both of them, yours and mine come up with the same result.
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