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Help understanding a semi-controlled full rectifier

  1. Nov 12, 2016 #1
    Hi, I'm analysing the the circuit here


    And I'm having some trouble understand the Vout waveform and why it goes to 0 when when pi<wt<pi+a

    If there is always an output current wouldn't there always be a voltage out? The only thing I could think would be the voltage across the inductor cancelling out the voltage across the resistor?
  2. jcsd
  3. Nov 12, 2016 #2


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    I swear this looks exactly like an old professor of mine hand writing (uddin?)

    To answer your question - the Vout waveform goes to zero because both controlled rectifiers (T1 and T2) are not conducting (reverse biased). Are you aware that inductors oppose changes in current by creating their own current flow by using their own stored energy? that is what is happening here.

    The path that io is flowing is through the resistor and inductor and through diodes D1 and D2. And you are correct, the inductor is creating the voltage that is dropped across the resistor. think of the path through D1 and D2 as a wire in a conventional circuit and Vo is basically putting your meter across it.

    FYI - diodes D1 and D2 are acting like "flyback" or "freewheeling" diodes during this part of the cycle.

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  4. Nov 12, 2016 #3
    It could be :)I'm working through a similar problem and found the notes here


    Thanks for clearing up the question, it's clearer now I'm a bit rusty with inductors.

    I'm still a bit confused with the voltage across the thyristors, we analysed a similar circuit in the lab and I got a different waveform. I added it to the link, it is the pink line on the oscilliscope? Any idea why there is a discrepancy?
  5. Nov 12, 2016 #4


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    Not sure where the waveforms are
  6. Nov 12, 2016 #5
    Hi, I added them here


    They bottom (pink) signal in each one is supposed supposed to be Voltage of T1, voltage of D2 and voltage of the inductor with a firing angle of 90 degrees - I think they may have been mislabeled

    Yellow -Vin
    Green -Vout
    Blue -Iout

    But I'm having trouble what was being measure on the pink signal in each
  7. Nov 12, 2016 #6


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    It would make sense to me if the pink was the voltage across T1 and the voltage across D2, since they conduct during the positive half cycle of Vs, and do not conduct during the negative half cycle of Vs.
  8. Nov 13, 2016 #7
    Thanks, that sounds right to me

    Do you know what the waveform would look like for the ac source current? I thought it would be the same as Iout but 0 from


    pi+a - 2pi

    Or would it be negative during the period pi+a to 2pi?

  9. Nov 13, 2016 #8


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    Only time the source current isn't flowing (is 0) is from $$0 - \alpha$$ $$\pi - (\pi+\alpha)$$ $$2\pi - (2\pi + \alpha)$$ $$\vdots$$ Unlike the output current, the source current flows in two directions depending on which rectifiers are on
  10. Nov 14, 2016 #9
    Thanks for the help, all much clearer now
  11. Nov 14, 2016 #10


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    Not a problem
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