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CW output with large AC component!

Discussion in 'High Voltage' started by mn1247, Aug 2, 2014.

  1. mn1247

    mn1247 Member

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    I have a 1-stage Cockroft-Walton being driven by a step-up transformer. The input to the transformer is a square wave of 200V peak-to-peak at 20 kHz. In TINA simulation, everything works as expected (DC output with a bit of ripple). However, in real life as measured by my scope, Vout contains the expected DC output PLUS a large AC component. Curiously, the RMS of the AC component is virtually equal in magnitude to the DC component! Coincidence? I think not, but I still can't seem to figure out what's going wrong. Is the extra AC signal purely reactive? If so, why is it there at all? Should I just ignore it? I've replaced all the parts - no improvement. Note: the diodes are 100 mA, 20KV devices, not the ones shown. The caps are rated to 30kV.

    I'm hoping more experienced minds can help solve this one.
    Thanks,
    Eric

    cw.jpg
     
  2. throbscottle

    throbscottle Well-Known Member

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    Are you sure this is AC you are seeing and not ripple?

    Also, more stages would mean you could use lower rated (= cheaper) components. Also, according to wikipedia, CW ripple rises sharply with load, so you should check what current is being drawn by your 101M. Sorry I can't help with the maths, but I know that the basic formula for CW is a horribly complex quadratic. (used to be into HV years ago)
     
    Last edited: Aug 2, 2014
  3. ronsimpson

    ronsimpson Well-Known Member Most Helpful Member

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    Please pull out your diodes and test them. I am most worried about D2.
     
  4. dave

    Dave New Member

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  5. mn1247

    mn1247 Member

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    Thanks for the replies. I don't think it's ripple - too large in amplitude, and wrong waveform (what I'm getting is a pure sine wave). As for D2, I wondered about that too, but I've replaced it twice now without change. I tried altering the spice model by putting some shunt resistance in parallel with D2 and D1, but this didn't reproduce the actual waveform. Same for shunting the caps.

    Of note, on the transformer, there are built-in physical gaps between the ferrite core halves, to limit output current (i.e., to provide leakage inductance). I have the two core halves shorted electrically with a piece of copper wire, so that they can be tied to ground. Do you think that would cause the problem?
     
  6. throbscottle

    throbscottle Well-Known Member

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    I doubt it. CW doesn't care what it's source is. What happens if you connect a cap across the output?
     
  7. mn1247

    mn1247 Member

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    Yep, that did it. It disappears with a cap, and I don't think there's much current flowing. The signal was reactive. Not sure quite how though.
     
  8. throbscottle

    throbscottle Well-Known Member

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    Someone else will have to go over the "how" with you, my knowledge is too limited! Glad it's resolved though.

    But I'm curious, what voltage do you have coming out of the transformer secondary?
     
  9. mn1247

    mn1247 Member

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    In theory, 9500V rms. (Ignore the diode part numbers... they're 20KV, 100mA)
     
  10. throbscottle

    throbscottle Well-Known Member

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    I read recently that for a sqaare wave you should be using the average voltage, not rms, as far as a transformer is concerned. I suspect it's more complicated than that though, especially if your wave is not actually square. Again, plenty of others know far more than I do.
     
  11. mn1247

    mn1247 Member

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    The input to the transformer is square, but the output is sinusoidal.
     
  12. throbscottle

    throbscottle Well-Known Member

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    Then you are losing the high frequency components of your input. To be honest, this might be a good thing for your application - I doubt it's a good idea to drive a CW multiplier with a square wave.
     

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