I am studying the Cockcroft–Walton generator and one question comes to mind.
Let's say we have a load consuming every watts the generator can provide.
Would it be wrong to state that the generator would need to disconnect it's output from the load and wait n AC cycles to be able to generate the same voltage again , where n is the number of stage(s) in the generator ?

Ex: Voltage source AC : [-3V, +3V]
CW Module 1 Output : [6V]
CW Module 2 Output : [12V]
CW Module 3 Output : [24V]

We see here that it takes 3 AC cycles to go from [3V] to [24] volts.
If a load then consume all this power stored in the CWMs capacitors in an instant, for the generator to be able to "charge" up to 24 volts again, it would need to wait for 3 more AC cycles, correct ?

  • \$\begingroup\$ The generator isn't connected to the load; there's a CWM in between. Please explain what you mean. \$\endgroup\$
    – Andy aka
    Mar 24, 2023 at 13:04
  • \$\begingroup\$ I edited the question, i hope this helps \$\endgroup\$ Mar 24, 2023 at 13:21
  • \$\begingroup\$ I think you need to draw a picture. The load is the thing fed by the CWM output and, the generator feeds the CWM's input i.e. the generator is the AC power source and isn't connected to the load (as far as I know). \$\endgroup\$
    – Andy aka
    Mar 24, 2023 at 13:28
  • \$\begingroup\$ haaa i think i see what you mean, it's just a matter of sementics. Wikipedia uses the term "generator" for the whole process, including the CWMs. What i actually mean by "generator connected to the load" is actually the last module, the one with the highest voltage. \$\endgroup\$ Mar 24, 2023 at 13:38

1 Answer 1


If the load is rapidly being removed and applied, then yes, the output voltage would vary. As the rate of output voltage growth is finite, it would take several cycles to regain its peak voltage, to replenish the energy, the charge (not the power) stored in its capacitors.

However, that's not how we would usually use a CW multiplier. The load would tend to be steady. Each input cycle, some charge would be transferred from stage to stage to meet the output current demand.

It's not clear where you get your 'per stage' figures from, but if the first two stages are 6 V and 12 V, the third stage would be 18 V, each stage only adds the input peak-peak voltage (less of course the diode drops).

Don't confuse a Cockcroft Walton with a Marx generator, which is intended for transient loads, and can look similar in photographs with 'two vertical towers cross-connected by diagonal stuff'. The CWM capacitors will tend to be 'small', little energy stored in them compared to the power output expected of the multiplier into its load. Marx capacitors OTOH are 'big', they store all the energy required for the output pulse.

  • \$\begingroup\$ (Duh ! my mistake, of course, 18 volts. ) Thanks , that's much more clearer ! \$\endgroup\$ Mar 24, 2023 at 13:39

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