To doorknob or not to doorknob.

[abstruse1]

I think I’ve about reached the potential of the CW multiplier I’ve been working on.  It has nine stages, fully rectified, powered by 5x 12v. SLA batteries, an inductive heating driver, and a homebuilt flyback.  It uses 27, 2,200 pF, 50 kV blue plastic Chinese capacitors and various high voltage diodes (mainly 2CL106, 12kV, 450mA).  The FB and the CW are run in mineral oil.  I’ve fried numerous driver MOSFETs (260N) and diodes.  I’ve gotten 14” sparks between points, but I think that’s about its limit.  The sparks are blue, noisy and vigorous.







I’ve thought about building another one just like the current one and polarizing them differently, for ~28” arcs.  But it may be time to take what seems to be the next step, using doorknob caps and expensive diodes.  But I don’t want to do so unless there’s the possibility of much longer sparks.
This may be a no-brainer, but it could be a well-it-depends answer.

I’ve accumulated an inventory of 24 each doorknobs: 2200pF, 20kV; 2200pF, 30kV; 540pF, 40kV; 320pF, 40kV; and a few others in smaller numbers.  What would be the best choice?

I have a machine shop and a vacuum chamber, so I can fabricate pretty much whatever I need.

Your thoughtful wisdoms will be most appreciated!

[Weston]

Ideally there is not much of a difference between doorknob capacitors and ceramic disk capacitors. They are both disks of ceramic dielectric that are metalized on both sides. Doorknob capacitors are a bit nicer because the screw terminals are more mechanically robust. I think they get used a lot because they are more common as surplus and easier to reuse due to the screw terminals.

The biggest difference would probably be the dielectric material of the capacitors. The performance of the CW multiplier is dependent on the capacitance. A higher capacitance will lead to a lower voltage drop under load. Most ceramic capacitors have a dielectric that has a dielectric constant that decreases with applied voltage. This can lead to a significant reduction in the capacitance at voltages that are an appreciable portion of the rated voltage.

A lot of the low cost disk ceramic capacitors or doorknob capacitors (probably anything from china without a datasheet) use something like a Y5P dielectric that looses 75%+ of its capacitance at rated voltage and temperature. Doorknob capacitors may be a bit less cost optimized and have a better dielectric, but you cant really know without a datasheet. The gold standard here would be capacitors with a N4700 dielectric. You can find them on ebay sometimes. There are a few different product lines by Murata, here is an example datasheet from one.  https://www.vishay.com/docs/22210/715c-kt.pdf

As long as it has a sufficient voltage rating and enough capacitance any capacitor should work.


Whats your expected spark length based on the number of stages and input voltage? There is an upper bound and you might already be getting close to that.

Making a second CW of the opposite polarity would most likely be the biggest bang for your buck. It would reduce the problems with insulation for sure.

Also, if you are blowing up diodes do you have a discharge resistor on the output of the CW? I see something that looks like that in your picture but unless you are grossly exceeding the voltage or current rating of the diodes it should be pretty effective in protecting the diodes.

[abstruse1]

Thanx, Weston.  As to the last question, I have a 1 Gig resistor across the outout of the CW.  Is that about the right resistance?  I also have a 1G resistor across the output of the flyback. I dunno if that's needed.

My doorknobs are Russian, the green ones (from Ukraine, actually).  The blue ceramics i'm using are Chinese and they are 50kV rated.

As to spark length (spark as opposed to arc), I want as much as I can reasonably afford.  I don't mind spending a couple of thousand $ if I'm assured that I'll get proportionate results.  Another idea would be to add stages.  My current CW has nine, but I can go much higher than that as long as the incremental gain per stage gets trivial. Your thoughts on this?

Any other tricks, such as jacking up the CW input voltage until components often fail, parallel or series connecting 2x capacitor, 2x diodes, etc., or -- get this -- running the spargs in a high humidity box?

I doubt my values across the board are optimum, e.g. capacitance, input voltage & frequency.

[MRMILSTAR]

Are you certain that your 1G ohm resistor can withstand your output voltage? You probably have an output of 250 KV to 300 KV. I can't tell from your picture what the resistor consists of. Making a resistor that can withstand that voltage is not simple. Make sure that your output isn't just arcing across your resistor. It may be hard to see. If so, this could be the cause of your blown diodes.

[abstruse1]

Milstar,the 1G resistor is a couple of resistors in series, several inches apart, potted in paraffin.  I recently measured its resistance at 1G.  I think that if the voltage was arcing down the resistor string, I'd see damage thru the clear plastic tube it's in.

Another question: I have a large stash of diodes: 12kV @ 450mA, and others at much lower current rating (don't remember mA ratings and they're not where I am now) but 2x - 3x the voltage rating of the 12kV ones.  Any way to know if I ought to be using the higher voltage v. the higher current ones?

[MRMILSTAR]

Assuming that your VM is putting out 300 KV and you have a 1G ohm load, the current would only be 0.3 ma. With that, even a 5 ma diode should be plenty. Since you are blowing diodes that is why I am suspicious about your 1G ohm resistor. Are you using fast reverse recovery time diodes (100 ns or faster)? If you are using microwave oven diodes, they are much too slow and that could be your problem. A 100 ns 12 KV 450 ma diode would be quite expensive.

[abstruse1]

Very helpful info, Steve.  Some of us are slow to learn.

[MRMILSTAR]

I just noticed that you stated "a couple of resistors" for your load resistance. I take that to mean two resistors. What is the voltage rating of each resistor? If your output is 300 KV the rating for each resistor would have to be at least 150 KV and even more for margin. I haven't seen any single resistors that can handle close to 150 KV. The highest-rated resistors that I've seen are typically 30 KV to 50 KV.

Just to give you an idea of the difficulty of building a load resistor capable of very high voltage I have attached a picture of my load resistor. Its rated for 400 KV and 300 watts. The resistance is 6.7M ohms and uses 24 resistors in series-parallel. I implemented it in air for convenience. It could be made smaller if I had used oil immersion. This uses about $250 worth of high voltage high power resistors and that price was a good deal, believe it or not.

Another thought is your topology. I see that you are using a full-wave implementation. Unless you need an output with minimum voltage ripple, a half-wave topology uses 33% fewer parts with the same output voltage although it requires a higher AC power supply voltage. Its generally cheaper to boost the AC input voltage than adding 33% more capacitors and diodes.

[abstruse1]

Perhaps I misspoke. The 1G resistor I’m referring to is across the output terminals of the multiplier. I was (am) thinking that this may prevent damage to the multiplier.

What am I missing?

I also have a 300 M across the output of the flyback for same reason. Is this useless?  FB is homemade with 7 turns 12AWG as primary and 1,000 turns 28AWG on the sectored secondary. For quite a while now, it’s been faultless. It was vacuum evacuated after winding. Don’t know anything about the ferrite.

What I’m doing now is trying to decide whether doorknobs are better than the blue Chinese caps.

[MRMILSTAR]

Perhaps I misspoke. The 1G resistor I’m referring to is across the output terminals of the multiplier. I was (am) thinking that this may prevent damage to the multiplier.

What am I missing?

I also have a 300 M across the output of the flyback for same reason.

I am referring to the 1G ohm load resistor across the VM output. I see no purpose in using a resistor across the output of the flyback transformer as long as you have the multiplier stack connected to it as its load.