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200KV PLUS voltage multiplier

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LAN_403

CT2

Wed Aug 05 2009, 07:53PM

Registered Member #180 Joined: Thu Feb 16 2006, 02:12AM
Location: Ontario, Canada
Posts: 187

I believe you're correct in your drawing. What you are trying to do is basically get rid of any "sharp" points which can bleed off charge, ie. corona. The spheres you added are exactly what you need (although a little too small), however the toroidal rings would provide better field control. The connections from you diodes and caps would be hidden in the middle of the ring (the ring and the caps and diodes would all be electrically connected, such that the ring will have the potential that that stage of the stack has), since they are in the middle of the ring though, the field strength would be low and there wouldn't be any corona or charge leakage. That is what you are trying to accomplish. There is no risk of ground arcing unless you brought something near the stack, but you would get a ground arc with or with out the ring if you did that.

Hope this helps, keep asking question, and great build so far!

Sparks

Wed Aug 05 2009, 08:35PM

Registered Member #2263 Joined: Mon Aug 03 2009, 04:57PM
Location:
Posts: 20

Thanks Chris,

If my drawing is correct I intend to put the rings on over the weekend, I'm also thinking of adding five more stages to make a million volts at an input of 40kv. I have no need for a million volts but what the hell, I have a slight tendency for overkill LOL

Firefox

Wed Aug 05 2009, 08:43PM

Registered Member #1389 Joined: Thu Mar 13 2008, 12:50AM
Location: Pittsburgh, PA
Posts: 346

If you're thinking about putting 40kVAC into this multiplier, don't. You're already running your 40kV caps at their absolute maximum with 20kV peak in, because each component in a cockcroft walton multiplier will see double the peak voltage in, 40kV in your case. This is where the name 'doubler' comes in. Each stage adds 2x the voltage, as I said in my first post in this thread.

Sparks wrote ...
Assuming the output is 20kv then the multiplier produces something like 200kv.

Your above statement is incorrect for the same reason. 20kVpeakin * 2 * 10 stages = 400kV no load output.

Sparks

Fri Aug 07 2009, 08:42AM

Registered Member #2263 Joined: Mon Aug 03 2009, 04:57PM
Location:
Posts: 20

The data sheet for the diodes I used is attached, each diode was constructed from 20 of these soldered together and potted in paraffin wax as a insulator. As you can see these diodes can withstand a surge of 20 Amps. Most CW use the smaller high voltage diodes which cant handle many amps. Taking this into consideration any I ideas of what size current limiting resistor would be appropriate, I don't want to unnecessarily reduce the current but keep it in the operating range of the diodes I have used.

The driver is 20kv 15-30KHz at 150 Watts

Sparks

Fri Aug 07 2009, 09:03AM

Registered Member #2263 Joined: Mon Aug 03 2009, 04:57PM
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Posts: 20

Firefox wrote ...

If you're thinking about putting 40kVAC into this multiplier, don't. You're already running your 40kV caps at their absolute maximum with 20kV peak in, because each component in a cockcroft walton multiplier will see double the peak voltage in, 40kV in your case. This is where the name 'doubler' comes in. Each stage adds 2x the voltage, as I said in my first post in this thread.

Sparks wrote ...
Assuming the output is 20kv then the multiplier produces something like 200kv.

Your above statement is incorrect for the same reason. 20kVpeakin * 2 * 10 stages = 400kV no load output.

Hi Firefox,

The reason why I originally thought the output would be 200KV is because I was following the format presented in a book, as you can see the for mat shows a driver with an output of 15kv and the sucessive stages of voltage obtained ie 15,30,45 etc

Using the same format but with a driver of 20kv I thought it the result would be 20,40,60,80,100,120,140,160,180, 200KV

Sparks

Fri Aug 07 2009, 09:06AM

Registered Member #2263 Joined: Mon Aug 03 2009, 04:57PM
Location:
Posts: 20

Sorry forgot attachment

The other discrepancy is that in this book the driver produces 15kv and the capacitors used in the stage are also rated at 15kv...the reason why I thought I could power my cw stage with 40 kv as I am using 40 kv caps

Steve Conner

Fri Aug 07 2009, 09:36AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

If the output waveform from the driver is asymmetrical, as it is from a flyback driven by a TV circuit, then the "double the peak voltage" rating for components doesn't apply. (a cookie for anyone who can explain why this is, and how to calculate the actual rating)

Also, go easy on the double posting Sparks, it's against the forum rules.

Dr. Dark Current

Fri Aug 07 2009, 09:50AM

Registered Member #152 Joined: Sun Feb 12 2006, 03:36PM
Location: Czech Rep.
Posts: 3384

Steve McConner wrote ...

If the output waveform from the driver is asymmetrical, as it is from a flyback driven by a TV circuit, then the "double the peak voltage" rating for components doesn't apply. (a cookie for anyone who can explain why this is, and how to calculate the actual rating)

Because the caps charge to the peak-to-peak voltage? The actual voltage on the caps is the forward+flyback pulse multiplied by the turns ratio, but I don't know how to calculate the flyback pulse other than to measure it directly. (well it should be possible to calculate but that would need much more data on the input

such as frequency/primaryL/duty cycle/resonant cap value etc.)

LutzH

Fri Aug 07 2009, 09:55PM

Registered Member #1721 Joined: Sat Sept 27 2008, 08:44PM
Location:
Posts: 136

Hello:

Nice Construction!!!

I see one major flaw however which WILL cause you to blow your diodes eventualy, no matter how big they are:

You have to put resistors in the circuit to limit any possible discharges to less than the single cycle surge rating of the diodes. This can be done in two ways, I prefer the second.

1. Using a HV resistor network on the output end to limit the current. Or a water resistor ect.

2. Placing the resitors between each capacitor in both of the stacks, you do not need fancy HV resistors for this, because they are effectively all in series. If for example each stage produces 10KV, and the single cycle surge rating of the diode is 10 amps: Then you would want 2000 ohm's of resistance between each capacitor for a 5 amp surge limit (50% Safety margin) What the big boys at the major manufacturers do is put 2-4 cheap 2-5 watt ceramic resistors in series, to make each resistor. They arrange them in a zig-zag fashion. So each 2K ohm one would be (2) 5 watt 1K ohm ones in series, or (4) 5 watt 500 ohm ones in series.

Last you normaly double up on the diodes in the first and last stage, although for the huge diodes that you have this may not be needed. I would however still add an additional HV input resistor at the bottom to eat any inductance related surge from your power supply. Another form of cheap insurance is a spark gap at the input end for the same protection.

I hope you find this usefull, I really like the look of your VM, also these are nice capacitors, and very sweet diodes :)

Aloha....Lutz :)

Proud Mary

Sat Aug 08 2009, 04:20AM

Registered Member #543 Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992

Hi Sparks, welcome to our online community!

You can reduce the output ripple voltage on a half-wave C&W considerably by increasing the capacitance at the base of the stack.

Thus in a four stage device you would have 4C in the first stage, 3C in the second, 2C in the third, and finally C at the output.

I cannot agree too much on the need for damping resistors - preferably disseminated throughout the whole circuit rather than having a single resistor at the end - if you want to save your diodes. The risk with having a single diode at the output is that it does not take account of the unruly nature of highly energized electrons, which will often insist on going their own way (what they see as the laziest way) despite our best efforts to impose order and discipline on them. A loud crack somewhere will tell you your diodes have probably gone! And when one goes, it's helter skelter, failure in cascade, so the whole lot are fit only to be recycled.

A C&W isn't intended as a low Z source like a Marx, so it does no harm to put a fair amount of resistance in its path.

You can wind your own non-inductive resistors with nichrome or manganin resistance wire on a mica card or teflon sheet if you are keen.

I'll be interested to see how you get on. It's good to have you with us.

Lastly, have you noticed the wonderful advertisement-free fresh air in here? That's because the forum is funded wholly by its members, the good folks whose names appear at lower left.

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