CW Multiplier

Discussion in 'General Electronics Chat' started by jester1, Feb 12, 2011.

  1. jester1

    Thread Starter Member

    Feb 10, 2011
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    http://www.blazelabs.com/cw-brm-java.asp Based on this what should I rate the diodes at. The same as the capacitors or peak input voltage and would that be applied to the rms or peak voltage of the diodes themselves.Also if I use multiple diodes rated at 700v rms and 1000v peak in series would it go up at say two diodes = 1400v rms and 2000v peak and so on.
     
    Last edited: Feb 12, 2011
  2. Jaguarjoe

    Active Member

    Apr 7, 2010
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  3. jester1

    Thread Starter Member

    Feb 10, 2011
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    That document states that the reverse voltage is 2* input. But does that apply to the peak or rms value given for the diode. Also if the diodes were in series would the rms and peak values be added together per diode or is there some equation to calculate there rating together.
     
  4. Adjuster

    Well-Known Member

    Dec 26, 2010
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    When calculating reverse voltage ratings, always use peak values. The point is that you need to consider the worst stress. Diodes may be given RMS ratings for rectified current, but the voltage rating is PIV: Peak Inverse Voltage.

    At least under light loads or no load, the capacitors in the multiplier chain charge up to (almost) the peak value of the supply. On every second half-cycle, each diode feeding into a capacitor will face an opposite peak voltage at its other end. The maximum reverse voltage is therefore about twice the peak voltage of the AC input.
     
  5. jester1

    Thread Starter Member

    Feb 10, 2011
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    Last edited: Feb 13, 2011
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    You need to be very careful with this. Putting diodes in series to increase the PIV rating might work, but not necessarily. Differences in the diode characteristics could lead to failure.

    Using capacitors in series is a very bad idea, unless you use parallel resistors across each capacitor to swamp out the effects of leakage currents. This would waste power, why not just use more stages - is your drive voltage already decided?
     
  7. jester1

    Thread Starter Member

    Feb 10, 2011
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    We'll im hand winding a transformer so the output isn't decided. What would the optimal output for the transformer be if i were only using single 2.5kv capacitors and single 1n4007 diodes (so 2 diodes and caps per stage).thanks
     
  8. Adjuster

    Well-Known Member

    Dec 26, 2010
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    I am not sure about advising you any further. You are contemplating hand-winding a transformer for a dangerous voltage, so as to generate even higher voltages using a multiplier. This is not a good project for a novice.

    Given the questions you have been asking, it seems doubtful whether you are in a position to tackle this sort of project safely. It's better to be safe than to be sorry.
     
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