Note: I am aware of this question. It is 6 years old and I didn't find an answer there and I didn't want to necro the question.

As part of a school physics project I need a very large voltage. I will be generating sparks in a low-pressure environment which will require at least 10kV (we still need to test the jars to see how low a pressure we can realistically maintain).

In order to generate the high voltages required I intend to use a Cockroft-Walton Voltage multiplier


simulate this circuit – Schematic created using CircuitLab

I've seen several values for Capacitors dotted around the internet and no explanation for them. According to this video its normal to use a square wave, again, no explanation. Most diagrams I have seen require a load resistor with a high resistance, I assume this is because of the inability for the circuit to produce high currents without discharging the capacitors.

My questions are these:

  • Is the resistor necessary if I'm using this to drive a spark gap? I would actually quite like the output voltage to drop when there is a load for safety reasons. (My classmates are not as respectful of electricity as I am.)

  • Are there any good resources with equations describing the relationship between the capacitances and the behaviour of the circuit?

  • What effect does the shape/frequency of the waveform have on the behaviour of the circuit?

  • 1
    \$\begingroup\$ Didn't know "necro" was a verb, but I'll take it. OP didn't want to go necrophile on a six year old question. That seems understandable. But seriously, it's OK to improve old questions and thus raise them in the "active" queue of this site. \$\endgroup\$ – Marcus Müller Feb 7 '18 at 20:20
  • \$\begingroup\$ Is it not a common verb on forum-like sites? \$\endgroup\$ – Disgusting Feb 7 '18 at 20:21
  • \$\begingroup\$ I will bear that in mind for the future though. I also didn't want my first question on this stack to be editing somebody else's. It would feel rude for some reason. Its probably just being from the UK. \$\endgroup\$ – Disgusting Feb 7 '18 at 20:22
  • \$\begingroup\$ Well, not that I'd be aware of! Anyway, your questions are a bit confusing; what resistor are you referring to? add that to your schematic (this site has a built-in schematic editor, by the way). Regarding a good resource: Hm, I can only recommend imagining what happens if you first assume all capacitors are discharged, then get the first AC voltage peak, then the inverse peak, then the next, and so on, until all capacitors are charged, and then "count" the voltage between the terminals... it's very much a circuit that needs "temporal" imagination :) \$\endgroup\$ – Marcus Müller Feb 7 '18 at 20:25
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    \$\begingroup\$ Go here DIY Physics and also buy their book. It's worth it! \$\endgroup\$ – jonk Feb 7 '18 at 23:12

Each stage of such a voltage multiplier stacks a voltage the size of the input voltage on top of the circuit input voltage. But also, each stage has substantial losses. So the key to success is starting with an input voltage as high as possible.

Don't have it 1V but at least 200..300V. You can create that input voltage for the voltage doubler ladder using a flyback converter, which also has the nice feature of already having a "square" wave on output. The diodes and capacitors each have to withstand the double input voltage.

The diodes have to be fast recovery types. 50Hz diodes as the 1N4007 won't work with high frequencies. This also limits the frequency of your input.

Important for your life (and that of others)

The input square wave gains its "relative safety" even at 300V from its high frequency —your muscles cannot follow, so you won't stick to it– but this doesn't apply to the voltage doubler ladder. Each of the capacitors is charged to 300V, which can reasonably harm you.

So choose those caps as small as possible. 50nF is a good value for small sparks.

Parallel each individual cap with a 10MΩ resistor to discharge it as soon the input voltage is off. Discharge may take a few seconds.

Same for the output. It can substantially harm you. Use a HV resistor of at least 10MΩ, if you can't get your hands on one, use several 2.2MΩ resistors in series.

To address your questions

  1. As soon there is a spark, output voltage will drop quickly. That resistor limits the output current and that way, increases the time there is a spark. This is actually useful for studying it.
  2. The charge stored in the capacitors depends heavily on their ESR and ESL values, the AC parameters of the diodes, and the frequency of the input. Their capacitance only puts an upper limit on the stored charge.
  3. No practical ones. The waveform of the input current will be spiky in any case, because the capacitors can only be charged when the outside voltage is higher than their own voltage. As with any rectified power supply.
  • \$\begingroup\$ Thanks for the detail. A few follow up questions. Firstly, we are not allowed for the input voltage to be higher than 40V due to the arbitrary constraints that health and safety has placed on us, why do you say 300V? Secondly, as far as I was aware, they didn't geometrically double the voltage each step but arithmetically add twice the input at each stage? Just want to double check that I understand your words. Finally, am I to understand that in this specific case higher resistor and capacitor values simply increase the "spark time"? \$\endgroup\$ – Disgusting Feb 8 '18 at 9:48
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    \$\begingroup\$ When there is no spark, the output resistor is small in relation to the gas resistance, so it doesn't matter for experiments about electric field resistance of gases at various pressures. As soon there is a spark, the capacitors will discharge through it. If you have no resistor, the discharge time is governed by the ESR and ESL of the caps and as you want these very low for better charging and higher voltage, it's very short. Putting an output resistor in the circuit limits the discharge current and increases the time there is a spark, so you can see/film it without darkening the room. \$\endgroup\$ – Janka Feb 8 '18 at 10:45
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    \$\begingroup\$ Increasing the capacitance makes the circuit more harmful, plus, such caps usually have higher ESR and ESL values (or they are expensive), so you end up with a harmful amount of charge at 1000V instead of a smaller, less harmful amount of charge at 20000V. \$\endgroup\$ – Janka Feb 8 '18 at 10:47
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    \$\begingroup\$ Whatever these constraints on safety are, these are irrelevant for the circuit you are planning. Its HV output is a deadly risk in any case because it serves a high-voltage-high-energy pulse as soon someone touches it. It also makes no sense to have more than about 10 stages because losses and series inductance of the caps add up, so upper stages won't have any more voltage gain. \$\endgroup\$ – Janka Feb 8 '18 at 10:50
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    \$\begingroup\$ I thought again and yes, you are right, the output voltage doesn't double at each stage but it's only stacked with each cap. Caps also have to withstand the double input voltage. \$\endgroup\$ – Janka Feb 8 '18 at 10:52

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