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I need to generate up to 2 kV by pulsing a DC 1kV source. The circuit I used is the following:

enter image description here

Clearly, this circuit is intended for an AC input that has both positive and negative halves where as in my case I have fully positive AC input. I obtained 2 Us with this circuit but this has pushed me to better understand what's happening in my case. I know to double my voltage I need for two capacitors in series to somehow be charged each to Us and for the diodes to block current flowing back to the voltage source. In AC, I understand that we alternate charging two capacitors alternatively thanks to branches d1 d3 and d2 d4. In my case, in this diagram the only two capacitors that matter would correspond to C2 and C4 correct? Is my understanding correct that D1 and D3 will not function in this case? How can I simplify this circuit for my usecase?

Do you have any other suggestion perhaps easier to step up my 1 kV to 2kV in a cheap way?

JRE
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ShiroGeek
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    I'm afraid you won't be able to do it with a DC pulsating voltage; AC voltage is required. The very principle of this type of multipliers is such that after charging the capacitor, the input source must reverse its voltage to add it to the capacitor's voltage. – Circuit fantasist Apr 19 '23 at 14:04
  • Do you already have a 1 kV source, that you are switching to make the pulses? If so, you only need 2 diodes and 2 capacitors to generate an extra 1 kV to 'stand on top of' your existing 1 kV supply. As the output impedance goes as the square of the number of stages, this makes for a much more efficient multiplier. – Neil_UK Apr 19 '23 at 14:36

4 Answers4

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The circuit is not dependent on the bias of the input. Assuming perfect diodes, the cathode of D2 will have a DC voltage equal to the peak-to-peak voltage of your input, and the cathode of D4 will have a voltage twice that of the cathode of D2. All 4 capacitors "matter" because the C1 capacitvely couples your drive signal, effectively eliminating any bias effects. This type of circuit is used for applications requiring little current such as photomultiplier tubes. If you try to run any power through it, its peak currents will become quite high. This circuit would really be a good candidate for SPICE.

If you need power, you will do better with a transformer.

John Birckhead
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The circuit you have will be perfectly happy to accept pulsed DC.

This is an LTspice simulation of your voltage multiplier example circuit:

enter image description here

The green trace is a 50Hz pulse - it ranges from 0 to 20 V. The blue trace is the output. It starts at 0 V and builds up to just under 40 V - it is a bit less than double the input pulse voltage because the diodes drop a little voltage.

All of the capacitors and diodes matter - all of them have a part to play at all times.

Don't look at the diodes as recitifiers. View them as pressure valves. When one side is at higher pressure, the diode conducts and allows current to flow through.

The circuit uses C2 and C4 to add the pulses together. C1 and C3 allow the pulses to pass from one stage to the next without shorting the DC stored in C2 and C4.

Normally, folks expect a voltage doubling on a single stage of a voltage doubler. That's with two capacitors and two diodes. With AC, what happens is that you get the peak to peak voltage as DC. If you used a simple rectifier, you'd get the AC peak voltage as DC. The "double" part refers to getting the peak to peak voltage instead of the peak voltage.

The peak to peak voltage from the pulsed DC is the same as the peak voltage. A single voltage multiplier stage with a pulsed DC input will give you simply the peak voltage (which is the same as the peak to peak voltage.) Nothing doubles.

You can stack two voltage multiplier stages together to get twice the output voltage - that's what your circuit does.

You need to keep in mind that the impedances of the capacitors mounts up rather quickly. That limits the current you can get through the multiplier.

You need to use either a high frequency for the pulse rate or large value capacitors.

Large value capacitors rated for high voltage can be expensive. You might have to increase the frequency so that you can use smaller value capacitors.

JRE
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Be careful about your 'pulsed DC' source. There is a difference between a DC source that is connected and disconnected, vs one that is connected, but returns to 0 V.

It is the 0 V that is required to emulate the original AC requirement. A DC that is pulsed by opening and closing a switch won't work for you.

jp314
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Clearly, this circuit is intended for an AC input that has both positive and negative halves ...

No, it does not.

enter image description here

Antonio51
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