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How does a cockroft walton multiplier work?

  1. Nov 24, 2012 #1
    Could someone help me out? I've looked at the two following diagrams:
    Two stage multiplier
    Full wave multiplier
    I'm not understanding how it works in terms of current flow and such. COuld someone explain this? Also, what's the difference between these two setups insofar as how well they work? I understand that it has something to do with using diodes to send the current in a specific way, and then charging capacitors in each stage, but beyond that, I'm not sure.
  2. jcsd
  3. Nov 24, 2012 #2


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    Last edited: Nov 24, 2012
  4. Nov 24, 2012 #3
    Been there done that :). I stared at this image for about 10 minutes before coming here.http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltage_multiplier.png [Broken]
    To be more specific, here in this image. THe power source charges the top left capacitor, but how does this effect stack up to add to the voltage? There's a path from the voltage source to every capacitor, so every capacitor is charged, but do the capacitors then charge each other? Is that why the "stacking" effect works? I feel like I'm missing something as far as the overall swing of things is.
    Also, does the current source have to be AC current? This diagram shows it like so, but could you use a rectifier and an AC source to convert, say, 120 volts of AC wall current to 168 volts DC, then use a Cockroft-Walton multiplier to stack the 168 volts?
    Last edited by a moderator: May 6, 2017
  5. Nov 24, 2012 #4


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    Been where, done what? Staring at a schematic diagram seems an unlikely method to learn the theory of operation.

    At the Wiki site mentioned above, the words "Operation of the CW multiplier, or any voltage doubler, is quite simple. Considering the simple two-stage version diagramed within, which is attached to an AC power source on the left side of the diagram. At the time when the AC input reaches its negative pole the leftmost diode is allowing current to flow from the ground into the first capacitor, filling it up. When the same AC signal reverses polarity, current flows through the second diode filling up the second capacitor with both the positive end from AC source and the first capacitor, charging the second capacitor to twice the charge held in the first. With each change in polarity of the input, the capacitors add to the upstream charge and boost the voltage level of the capacitors downstream, towards the output on the right. The output voltage, assuming perfect conditions, is twice the peak input voltage multiplied by the number of stages in the multiplier." seem to explain the workings of the CW multiplier fairly well.

    Have you studied the first section in the above "all about circuits" site? It provides a simplified theory of operation.

  6. Nov 24, 2012 #5
    I honestly didn't see that section in the link you gave me. At any rate that site seems really good for this, so I'll give it another go. Thanks!
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