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Thread: 3-Phase rectification & Ripple

  1. #1
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    3-Phase rectification & Ripple

    I need a formula to calculate the AC ripple for an unfiltered 3-Phase Bridge ( 6 pulse ). The only formulas I can find are for circuits with a smoothing capacitor.

    Maybe I could just use an ungodly small capacitor for the calculation...?? :confused:

    The final intent of this exercise is to be able to measure the ripple to determine the health of the diode bridge. The supply is variable 0 - 300 VDC.
    Advise is a dangerous gift, even from the wise to the wise.

  2. #2
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    Quote Originally Posted by SG-1 View Post
    I need a formula to calculate the AC ripple for an unfiltered 3-Phase Bridge ( 6 pulse ). The only formulas I can find are for circuits with a smoothing capacitor.

    Maybe I could just use an ungodly small capacitor for the calculation...?? :confused:

    The final intent of this exercise is to be able to measure the ripple to determine the health of the diode bridge. The supply is variable 0 - 300 VDC.
    So what is/are the formula/s with the smoothing capacitor?
    I'll never get there. No matter where I go, I'm always here.

  3. #3
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    Quote Originally Posted by SG-1 View Post
    I need a formula to calculate the AC ripple for an unfiltered 3-Phase Bridge ( 6 pulse ). The only formulas I can find are for circuits with a smoothing capacitor.

    Maybe I could just use an ungodly small capacitor for the calculation...?? :confused:

    The final intent of this exercise is to be able to measure the ripple to determine the health of the diode bridge. The supply is variable 0 - 300 VDC.
    If it's unsmoothed, perhaps an easier way to check if all the diodes are conducting would be to measure the DC output voltage of the bridge.
    Assuming it to be a full-wave six-pulse bridge, the output DC voltage should be 1.35 times the AC input voltage. Any non-conducting diodes will reduce the output voltage significantly.
    You can measure the AC ripple content but bear in mind that it predominantly at six times the fundamental frequency (probably 360Hz in your case) and is a fairly low voltage compared to the DC. The meter may not respond well to that. Anyway, for what it's worth, the RMS ripple for a six pulse bridge operating correctly is a bit over 4% of the DC voltage.

    One other thought. I've been in the power electronics game since before the Dead Sea reported sick. I can't ever remember a power diode to fail by going open circuit. The usual failed condition in my experience is short circuit. And that you can easily check with most standard multimeters.

  4. #4
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    Quote Originally Posted by Besoeker View Post
    I've been in the power electronics game since before the Dead Sea reported sick.
    That's cute!
    Code references based on 2005 NEC
    Larry B. Fine
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  5. #5
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    Quote Originally Posted by SG-1 View Post
    I need a formula to calculate the AC ripple for an unfiltered 3-Phase Bridge ( 6 pulse ). ...
    Quote Originally Posted by Besoeker View Post
    .... Anyway, for what it's worth, the RMS ripple for a six pulse bridge operating correctly is a bit over 4% of the DC voltage.

    ...
    Don't know how accurate the formula is, I derived the formula myself... I did not reference any publications.



    FWIW, it yelds a waveform of about 6.7% p-p, which I imagine is pretty close to a bit over 4% rms.
    I'll never get there. No matter where I go, I'm always here.

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    Quote Originally Posted by Smart $ View Post
    Don't know how accurate the formula is, I derived the formula myself... I did not reference any publications.



    FWIW, it yelds a waveform of about 6.7% p-p, which I imagine is pretty close to a bit over 4% rms.
    Altogether unnecessarily complicated and not right anyway.

    Peak voltage, Vp, as always is Vp = root(2) * Vrms
    Minimum is Vp sin (60) or Vp*root(3/2)
    So peak to peak is about 14% of maximum voltage.

    I suppose I have posted this before. It gives the waveform, the ripple, and the 6th harmonic content.

  7. #7
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    Quote Originally Posted by Smart $ View Post
    So what is/are the formula/s with the smoothing capacitor?
    I was using this website:

    http://www.electronics-tutorials.ws/diode/diode_6.html

    V (ripple) = ( I(load) / ( f(60hz) x #P (6) x C )) x V
    Advise is a dangerous gift, even from the wise to the wise.

  8. #8
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    Quote Originally Posted by Besoeker View Post
    One other thought. I've been in the power electronics game since before the Dead Sea reported sick. I can't ever remember a power diode to fail by going open circuit. The usual failed condition in my experience is short circuit. And that you can easily check with most standard multimeters.
    This one diode ejected it's anode. ( I have a picture, but my uploads fail since the forum changed. I am probably doing something wrong.) Others were shorted. This resulted in arcing on a variable auto-transformer.

    10 bridges ( 9 now) with 12 diodes per bridge & I would rather not open the cabinet, but I can.

    I am digesting the rest of your posts, good stuff. ( Yum )
    Advise is a dangerous gift, even from the wise to the wise.

  9. #9
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    Quote Originally Posted by Besoeker View Post
    Altogether unnecessarily complicated and not right anyway.

    Peak voltage, Vp, as always is Vp = root(2) * Vrms
    Minimum is Vp sin (60) or Vp*root(3/2)
    So peak to peak is about 14% of maximum voltage.

    I suppose I have posted this before. It gives the waveform, the ripple, and the 6th harmonic content.
    When I stated 6.7%, my reference value was AC voltage peak to peak. Referenced to one bus of rectified voltage it would be double at 13.4%.

    Additionally, your formulas do not yield an instantaneous waveform. SG-1 did not specify what type of formula he was wanting. Plotted out, my formula's waveform is identical to yours.
    I'll never get there. No matter where I go, I'm always here.

  10. #10
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    Quote Originally Posted by Smart $ View Post
    When I stated 6.7%, my reference value was AC voltage peak to peak. Referenced to one bus of rectified voltage it would be double at 13.4%.
    Actually, not quite right. It is 13.4% referred to the peak of the AC voltage. Not the DC.

    Quote Originally Posted by Smart $ View Post
    Additionally, your formulas do not yield an instantaneous waveform.
    That sounds like an oxymoron.
    The formula yields an instantaneous value. The waveform is a time varying function, not an instantaneous value.

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