I have a semi-serious hobby project with a couple of friends where we need to charge a pulse operated capacitor rated to around 4kV with 1500nF capacitance.

For this I need a high voltage supply and correct configuration to do the charging. Now I've done this before at around 300V but simulating this process with multisim is nontrivial because in this case I don't know the inner workings of kV-level sources the and now I have some questions:

If I were to use an off the shelf high voltage source of around 10kV maximum voltage and 20W power, how will the system react when the source is connected, possibly through appropriate resistor? Someone suggested me that the namely excessive 10kV would immediately break the capacitor as the voltage drop is all seen through the insulation inside the capacitor. I think this is incorrect and the voltage drop will mostly happen inside the HV supply, but I'm not entirely sure.

Let's use this cheap individual as our HV source example: https://www.ebay.com/itm/New-High-Voltage-Electrostatic-Precipitator-Power-Supply-With-Output-300W-30KV/163264230335?epid=27023884036&hash=item26034e73bf:g:lvwAAOSwQWxbIdWA:rk:1:pf:0

to summarize:

-I- Will an excess voltage (measured at the source) break the capacitor before actually charging it?

-II- If not, how does a typical HV source react to the situation? If connected straight to a capacitor (with effectively zero resistance), will the source just see the connection as about short circuit and promptly break or blow a fuse? Or will it just slowly start to ramp the voltage up, until the external capacitor is at equal voltage to source target (or until the capacitor breaks)

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    \$\begingroup\$ A 5 kV, 1500 uF capacitor is huge!. 0.5 m x 0.5 m x 0.5 m. Use a high voltage supply, their source impedance is high and you charge it up - do this behind a plastic shield. Have a bleed resistor. Watch out for dielectric absorption, soakage, or memory effect. 1.2x10^4 J. That's like having a 62 kg mass dangling 20 metres above your head. \$\endgroup\$ – D Duck Nov 23 '18 at 23:52
  • \$\begingroup\$ Reading your question indicates to me that your knowledge in this field is very limited. As pointed out above, the voltage and energy levels involved are well into the lethal category. My suggestion is to either find someone who is knowledgeable or find another project. \$\endgroup\$ – Barry Nov 24 '18 at 0:01
  • \$\begingroup\$ that example high voltage source can be turned down to 5kV , but you want 4kV \$\endgroup\$ – Jasen Nov 24 '18 at 0:04
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    \$\begingroup\$ Oh, Excuse me, I accidentally wrote 1500uF instead of 1500nF. Yeah, that would have been around 12kJ indeed. Edited the question now. \$\endgroup\$ – Elmore Nov 24 '18 at 0:28
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    \$\begingroup\$ 1500nF is merely 1.5uF. I think if you hack a bug zapper you will probably get a decent step up converter which will exactly match your need. \$\endgroup\$ – soosai steven Nov 24 '18 at 0:50

The example power supply you show has a cuurent limit and a voltage limit, it will limit is output to whichever limit is encountered first.

Into a, short-circuit, or a discharged capacitor it will supply 0 to 10 miliamperes as set by one of the knobs on the end panel.

If you get a supply like that, but can be set to 4000V instead of the 5000V lower limit of the example device it will do what you want.

I see your correction from 1500uF to 1.5uF that's now of the approximate scale of the capacitors present in microwave ovens, (slightly larger and slightly higher voltage)


If you really want to do this, first get a good life insurance policy.

Then it looks like you can have the output set to 5 kV. One simple solution would be to make a resistor divider, maybe this is the only application where resistor dividers are good for:

enter image description here

Power rating should be about 10-40 watts, and the resistors will cost you about $75-$120 each. The cap will be charged in about a minute or two.

However it is not clear what do you mean under "pulse operated" capacitor.

  • \$\begingroup\$ R1 not needed, just set the current limit to 0.05 mA \$\endgroup\$ – Jasen Nov 24 '18 at 4:27
  • \$\begingroup\$ @Jasen, no, without R1 the cap will be overcharged to 5kV if the source is not disconnected in time. \$\endgroup\$ – Ale..chenski Nov 24 '18 at 19:32
  • \$\begingroup\$ it won\t go past 4kv because with the limit set to 0.05mA, at 4kV all the supply current goes thorigh th 80M. \$\endgroup\$ – Jasen Nov 24 '18 at 19:43
  • \$\begingroup\$ @Jasen, I am not so sure. The "specification" says: "1. shortcut protection: when the positive output line and the negative output line are shorted ,the output voltage will be cut off". So, connecting uncharged cap to this PSU will likely cause it to shut down, so nothing will be charged. You need a current limiter, which will be provided by R1 (at 0.25mA). \$\endgroup\$ – Ale..chenski Nov 24 '18 at 20:18
  • \$\begingroup\$ that descriotion is not inconsistent with a current limit. In fact all power supplies work that way, it's pretty-much the definition of a short-circuit. \$\endgroup\$ – Jasen Nov 24 '18 at 20:25

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