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Thevenin equivalent of a half-wave rectifier

  1. Oct 30, 2005 #1
    [SOLVED] Thevenin equivalent of a half-wave rectifier

    Hello folks,

    I'm trying to wrap my head around the process of reducing a half-wave rectifier (w/ filter) with an AC input into a simple Thevenin circuit with a DC equivalent voltage. My brain seems to be stuck in "but the input is not DC!" mode. What would the equivalent voltage be? I'm thinking it's the mean voltage across the load (assuming an ideal diode, with 0 resistance), but I don't see why it can't be RMS. The output impedance is not as tricky a concept, as it's simply the impedance of the capacitor, but the Thevenin voltage is giving me grief.

  2. jcsd
  3. Oct 31, 2005 #2
    I cant quite imagine what exactly your circuit looks like. If your caps are filtering does that mean you are choking the input? or is it just a resevoir cap at the end of the DC input?

    Either way I think I understand your confusion. A half wave rectifier crapily converts AC to DC as in it only allows positive voltages. A half wave rectifier doesnt produce a steady DC current, which is what you must be imagining. If the circuit is just an AC source, a diode, and a capacitor it might actually just be a primitive steady voltage source, with the capacitor discharging from peak to peak.
    Last edited: Oct 31, 2005
  4. Oct 31, 2005 #3
    elhinnaw, your guess was right on. My apologies for not being clear.
    Yes, the sole capacitor would be connected in parallel to any given resistive load, but even with insanely high capacitance (and resistance), the voltage still would not be steady enough to warrant an approximation with a single DC source. However, my TA has told me to take the mean voltage across the load over one cycle to be the equivalent Thevenin voltage[1]. I don't understand why, but I'll do it anyway if he says so. I mean, isn't the voltage ripple affected by the load resistance?

    Thanks a ton for your reply though.

    [1] That sure is one tedious series of integrations.
  5. Nov 2, 2005 #4
    A few things to remember from my experience with electronic texts:

    DC just means always positive voltage.

    Thevenin equivelent circuits are useless.

    In the field of electronic engineering, I could be so wrong its laughable, In my experience with electronics, a Thevenin equivelent never been needed. When I need to break down a circuit I did it through my own logical ideas.

    As far as your loading question goes, in theory every resistive load affects the voltage source. Its whats called "loading." But standard practice from what I learned is anything less than 1%-10% loading is considered acceptable.

    The fancy part of circuit design (or being an EE) is that you can design circuits that can have a huge fanout. The circuit in this case does not have a good fanout, which is why you will never see that circuit in use.

    I think its kind of a stupid question to assign, personally. Its much more important to know the theory behind whats going on.
  6. Nov 2, 2005 #5
    Thanks for the insight! It's much clearer now.
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