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I've recently gotten into eletrochemistry, and I found out that if you break through \$\epsilon_{0}\$ with a high enough voltage you can ionize various gasses. I've already wound a 1:1250 EE core that I plan to operate at 350 kHz on a 25 V supply which should give me 31.25 kV in the arctube, but there's this problem where I think I need to rectify the output and uh, I have yet to see a 50 kV diode.

Is there a way to do it that I'm just not remembering? I thought there was a way to do that back in the vacuum tube days.

Peter Mortensen
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user14828
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    There was. Start with 1kV AC (or whatever you CAN find a diode for) and build a Cockroft-Walton multiplier. Each diode only sees the original AC voltage. –  Jul 28 '19 at 13:46
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    There are rectifier stacks that can handle (say) 100kV. Forward voltage might be 120V meaning that they're 100+ individual diodes in series. – Spehro Pefhany Jul 28 '19 at 13:51
  • @BrianDrummond Cockroft-Walton multipliers aren't that good for high multiplication factors, if I remember right. – Hearth Jul 28 '19 at 13:51
  • You know I thought about that. Just feed the transformer 3V, half-bridge and use some SiC Diodes @1.2KV I have. The problem is that for some ionizations you need real power and the multiplier wont cut it. You know those things aren't really meant for serious output. – user14828 Jul 28 '19 at 13:54
  • @Hearth that depends on the current requirement (unstated). You are correct that they have a rather high output impedance. (Which is either a disadvantage or a safety feature) –  Jul 28 '19 at 13:54
  • @BrianDrummond the electrodes will be enclosed in an arctube, set in a second layer of clear epoxy. No danger at all. The parameters are all MCU controlled. – user14828 Jul 28 '19 at 13:58
  • @SpehroPefhany You might wanna post that as an answer, you got it! I totally missed that. I should have realized that from LED strings and Zener strings – user14828 Jul 28 '19 at 14:00
  • . Any surface contaminant ions can reduce your ionization ( called Partial Discharge) leading to a breakdown voltage (BDV) This includes the diode, caps and wiring. Consider that a 100cm HV bushing rated for 200kV BIL arcs at around 60kVdc so your insulators unless pristinely packaged with polycarbonate layers may cause PD before the arc occurs. THis is solely due to invisible surface contaminants. So cascading HV diodes is trivial, but protecting the Diode cap surfaces with multipliers is critical using polycarbonate shields. PD breakdown may begin around 500V/mm unless extremely clean- – Tony Stewart EE75 Jul 28 '19 at 14:42
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    Good luck with that transformer not breaking down under 31kV... At 100kHZ normal insulation will not do well even if rated for 31kV. – MadHatter Jul 28 '19 at 19:14
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    Are you sure what you are getting into? I once had 30 kV punch through Teflon insulation that was not thick enough (supply for a [MALDI](https://en.wikipedia.org/wiki/Matrix-assisted_laser_desorption/ionization) [TOF mass spectrometer](https://en.wikipedia.org/wiki/Time-of-flight_mass_spectrometry)). – Peter Mortensen Jul 28 '19 at 23:42
  • There are some DIY designs for gas lasers that used mechanical rectifiers where a synchronous motor would connect the desired pole of a neon transformer to the output in time with the mains polarity. Noisy, large and a bit scary but probably cheaper than some other solutions. – KalleMP Jul 29 '19 at 12:47

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There are rectifier stacks that can handle (say) 100kV. Forward voltage might be 120V meaning that they're 100+ individual diodes in series.

Here is some data from a random Chinese supplier (no experience with those guys, but it should give you an idea):

enter image description here

Spehro Pefhany
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  • Thank you very much, I have about 200 FFSH20120A SiC Diodes I guess I can sacrifice. You know those things are expensive, but I'll only lose a volt each, so to handle 12kv it'll only cost me 10V. – user14828 Jul 28 '19 at 14:10
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    You answered my first question on this message board, how to drive a MOSFET with a push-pull (emitter follower). That was great and thanks again. I've actually improved on that circuit since then adding 2 additional shunted transistors to equalize the voltage at the output, you know, the diode drops. – user14828 Jul 28 '19 at 14:18
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    What's the motivation to use SiC rather than silicon diodes? – pericynthion Jul 29 '19 at 01:26
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    @pericynthion Performs better at higher voltages. You can get away with less diodes. – DKNguyen Jul 29 '19 at 05:11
  • Seems like the SF4007 or similar would work fine, not have significant forward conduction loss at these operating voltages, can handle the switching frequency and would be a lot cheaper. The SiC Schottkys would have lower switching loss though and I was wondering whether the OP expected that to be relevant. – pericynthion Jul 29 '19 at 15:02
  • @pericynthion I'm sure they anticipated the relevance in design, but it's probably just an added bonus of working with that particular semiconductor. I can't wait to see IC's being manufactured using Cuprates. They are high temp superconductors, while exhibiting semiconductor qualities. – user14828 Jul 30 '19 at 03:54