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I want to build voltage multiplier using Cockcroft–Walton scheme.

I've got bunch of 3kV capacitors (2200pF though), and even bigger bunch of 1N5399 1kV diodes.

So, obviously I want 3 diodes in series to match capacitors rating.

I know that one cannot simply do that, and one need balancing resistors to match reverse current, but I see that for reverse current in datasheet (5uA) I need 200MOhm resistors, which seems to be something I can ommit. I know that at elevated temperatures diode reverse current would increase, but I won't work above 30 C.

Getting Higher-voltage diodes is probably not an option - they are ether too expensive or have too high forward drop voltage (like 20V for TV 14kV generators), and tiny current.

Suggestions, can I go without resistors?

PS. I know that I might want bigger caps, but they are not really common stuff. In this case I will have DC stage providing 3000V single polarity pulses @~30kHz or a bit more, so output current should be sufficient even with 2200pF. Target voltage is 15-20kV, so I will have ~ 7 stages.

BarsMonster
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    I've never used resistors, but admittedly I've only built CW circuits up to a few kV. Do you have a source that shows/recommends these resistors? – mng Oct 30 '11 at 21:09
  • @mng Hmm.... Probably you right, I've seen this on traning video which probably was outdated. – BarsMonster Oct 31 '11 at 09:42

2 Answers2

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I think the idea is that the voltage divider formed by the parallel resistors dominates the 'divider' formed by the reverse biased diodes. So in other words, you want considerably more current through the resistors than the diode's typical reverse bias current. Put another way, smaller resistors will do a better job of balancing the reverse biased diode drops. Of course you don't want too much leakage current because this will bring down the average voltage you can attain, so as usual there's a trade off to be made.

Also, you probably already know this, but don't try to get 1kV of drop across a single 1/8 watt sized resistor - it's better to series up a few smaller ohm values to get the resistance you want, so as to physically spread the drop out and reduce the E field. Might not matter too much until you get into 10's of kV but steep E fields cause corona.

JustJeff
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No one that I found will agree that, in 'real life', you can do without balancing resistors for GP diodes.

One of the alternatives discussed was to use avalanche diodes, as their breakdown characteristics are 'better controlled' vs. general purpose diodes like the 1N5399. This allows you to 'share' the voltage to some degree, but no one was comfortable with the idea of sharing at 100% of each diode's rating.

A datasheet I found for the 1N5399 (Vishay) shows variation in the leakage current from \$ 5 \mu A\$ to \$ 300 \mu A\$ as a function of diode temperature. Be careful with your lower assumption.

Ultimately, you'll have to source some alternate parts and try it yourself, or be prepared to wire lots of series resistors. If you want 100% certainty, either get appropriately-rated diodes or reduce the input AC to 1kV and add more stages.

Adam Lawrence
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