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I'm designing a circuit, which includes a voltage multiplier. The schematics of voltage multiplier is well known, but it usually appears without any specific values for capacitors and diodes:

2-stage voltage multiplier

My quesitions are:

  1. How do I calulate necessary capacitance of C1-C4 with respect to output current and switching frequency?

I assume, the minimum capacitance should be high enough to store the energy which I plan to draw from the output over 1 period of time (1/f). But I'm not sure this is also the optimal value of capacitance.

  1. What voltage should the capacitors and diodes be rated for?

My assumption here is, that the parts "see" a smaller voltage than the output, so they could be rated to this smaller voltage, plus some safety margin.

  1. What type of capacitors and diodes are best suitable for this application?

So far, I considered using ceramic capacitors and fast diodes (e.g. UF4007).


As a guideline, you can assume following parameters of the multiplier: Vi = 250 Vp, Vo <= 1000 Vdc, Iout = 20 mA, f = 50 kHz

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    \$\begingroup\$ If you can tolerate a balanced output, then two opposite 2x work more efficiently than a single 4x. If you can tolerate the full output voltage across your capacitors, then consider the Dickson configuration, which has a lower output impedance. \$\endgroup\$ – Neil_UK Sep 4 '19 at 4:54
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The parts "see" just the voltage of one stage. So, the peak to peak voltage of the AC and add a bit to be on the safe side.

As for actually figuring the values, that's... somewhat complicated.

You have a good start on the parameters: input voltage, output voltage, output current, frequency of the input AC.

Add in the allowable ripple voltage (or rather, voltage drop,) and the forward voltage of your diodes I think you have all the needed parameters.

This equation is at the heart of the whole thing:

$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{6fC} (4n^3 + 3n^2 - n) $$

I got it from this site. I recommend you read it through. I learned a lot of stuff from it that I had been trying to figure out myself.

You need to extend it to include the forward voltage of the diodes:

$$ E_{out} = 2nE_{pk} - \frac {I_{load}}{6fC} (4n^3 + 3n^2 - n) - 2nV_{f}$$

What that does is to compute your output voltage from several parameters:

  • \$E_{pk}\$ the peak voltage of the AC input (not the peak to peak voltage, but the simple peak voltage.)
  • \$I_{load}\$ - the load current.
  • \$f\$ - the frequency of the AC input
  • \$C\$ - the capacitance of your individual capacitors (in farads.)
  • \$n\$ - the number of stages in your multiplier.
  • \$V_{f}\$ - the forward voltage of your diodes.

Plug in your capacitance and other info, and see if it meets your needs. If the output is too low, first try to use larger capacitors. Then try more stages.

Notice how \$n\$ plays a very large part in the impedance of the multiplier. Each stage drastically increases the resistance of the multiplier.

If I assume 100nF capacitors, four stages, and diodes with \$V_{f}\$ of 1V, and take your \$250V_{pk}\$ input and goal of 1000VDC output, then I come up with the following:

$$ E_{out} = 2\times4\times250V - \frac {0.02A}{6\times50000Hz\times(1.0\times10^{-7}F)} (4\times4^3 + 3\times4^2 - 4) - 8\times1V$$

If I haven't fumbled it somewhere, that works out to 1792VDC with the 20mA load on it. That's a pretty good bit too high, but you see the idea.

You'll want to use diodes rated for at least 500VDC. The UF4007 you mentioned is rated for 1000V, so that's good. From the recovery time and the mention of the junction capacitance being measured at 1MHz, I'd say it looks fast enough for use at 50kHz.


You worked an example yourself, and came up with 2.8 microfarads. I find 2.2 microfarad capacitors rated for 1000V at $190 each.

I find 100 nanofarad capacitors rated for 1000V starting at $0.50. If you can live with a higher voltage with no load, it will probably be cheaper to go with more stages with the 100nF parts.

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  • \$\begingroup\$ Thank you for the answer. However, if I try to express C from given equation, I got division by zero, since expression Eout - 2*n*Epk is zero, for an ideal multiplier. \$\endgroup\$ – simonov Sep 3 '19 at 17:23
  • \$\begingroup\$ I guess there has to be a difference between the Eout and 2*n*Epk... and that's probably the voltage ripple. So, for my values and 1 V ripple I got 2.8 microfarad. \$\endgroup\$ – simonov Sep 3 '19 at 17:32
  • \$\begingroup\$ Yes. There will be a difference between real Eout and the theoretical 2*n*Epk. You can't avoid it if you use real capacitors. \$\endgroup\$ – JRE Sep 3 '19 at 19:31
  • \$\begingroup\$ Just realized: That formula ignores the forward voltage of the diodes. If you use good diodes, then it won't matter much. It amounts to 2*Vf*n. Modern diodes (even rated for 1000V reverse voltage) will have a forward voltage of around 1V, so you'd lose 8V on a four stage multiplier. \$\endgroup\$ – JRE Sep 3 '19 at 19:34
  • \$\begingroup\$ I assumed there are 2 stages. Each stage consisting of 2 diodes and 2 capacitors... Definitely the circuit should produce 4-times Vi, not 8-times. \$\endgroup\$ – simonov Sep 3 '19 at 21:31

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