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Cockroft Walton Voltage Multiplier Problems

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LAN_403

Fronzbot

Mon Sept 20 2010, 07:47PM

Registered Member #2710 Joined: Tue Mar 02 2010, 06:04PM
Location:
Posts: 7

Hey everyone.

So I like to have electronics projects going on to keep me on my toes during the academic year and my project I chose for fall ended up being a Cockroft Walton Generator. Basically, I was just looking to make a simple negative ion generator from it. I finally got the parts in today but am having SEVERE problems getting it to work properly or even figuring out what the hell is going on. This is my first HV project, so I'm pretty unfamiliar with this territory. I'm hoping someone out there can help me figure this stuff out.

First, the calculations. I used this site as a reference.

I wanted an output of -40kV at a current of 0.845mA. I chose 20nF as my capacitance and, as per the suggestion on the site, I made C1 & C2 = nC, C3 & C4 = (n-1)C etc... To save on cost (since I'm a poor college student) I chose 5 stages and was just going to use the input from my Function Generator which has a peak of about 5V and the frequency I settled on was 100Hz.

Vout = 2n*Vpk - [Iload/(6fC)] * (4n^3 + 3n^2 - n)
Vout = 2(5)*5 - [0.000845/(6*f*20e-9)] * (4(5^3) + 3(5^2) - 5)
Vout = 50 - 70.416667*570
Vout = -40087.5V

Vripple = Iload/fC = 0.000845/(100*20e-9) = 422.5V

So those values were perfectly fine for me. When I attached it up, though, it was only outputting -47VDC. Problem. I tried increasing the frequency and the max output I could get was -101VDC at an 8.3kHz Square Wave. I then added 5 more stages in (Stages 1 and 2 with 100nF, 3 and 4 with 80nF, 5 and 6 with 60nF, 7 and 8 with 40nF and 9 and 10 with 20nF) This time I got it up to -205VDC with an 83.2kHz Square wave.

I'm confused.

I've concluded that my output problem likely lies with the fact that I have such a small input voltage and if the circuit was attached to a more beefy supply, such as mains, the output would be much greater. I've also concluded that these equations don't make any sense. I see no flaw with my arithmetic and so there are, as I see it, only two possibilities:
1) The equations are wrong
2) I have assumed something I shouldn't (such as Load Current, capacitance, etc)

Number 2 is the more likely case.

So anybody out there that can help me out?

Proud Mary

Mon Sept 20 2010, 08:23PM

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

Is the output of your function generator not perhaps 5V into 50R?

And what is the input impedance - Zin - of your C&W?

Fronzbot

Mon Sept 20 2010, 08:48PM

Registered Member #2710 Joined: Tue Mar 02 2010, 06:04PM
Location:
Posts: 7

Well it's the the 50Ohm output, but I measured this voltage on the circuit itself so there is definitely a wave with a 5V peak going in. As for Zin, I'm not entirely sure how to calculate it for a CW generator, nor why it matters for much beyond calculating the output current? The whole cascade of diodes really puts a strain on my ability to calculate the total impedance of the circuit.

Patrick

Mon Sept 20 2010, 09:47PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

you need a real switching oscillator at power, not a wimpy function generator, square wave is best for your app.

your problem is this .... your 50ohm impedence into a bunch of caps, is an RC circuit with a fundamental freq below your desired charge freq. im almost sure thats it.

ScotchTapeLord

Mon Sept 20 2010, 10:16PM

Registered Member #1875 Joined: Sun Dec 21 2008, 06:36PM
Location:
Posts: 635

I'm surprised nobody pointed this out... but one of the main problems is that you used the calculation for the voltage drop that would be occurring IN the multiplier itself as your output voltage. The basic formula is 2n*Vpk - (some losses). The example on the site shows that out of the 60kV it would ideally output, 8kV is lost. The losses cannot exceed the output... You used the current figure .000845mA, which as you suspected, may have been a faulty assumption.

It's surprising you got to 205V to be perfectly honest, and it implies that your function generator definitely has a higher voltage with high impedance loads. Just charge a capacitor through a diode with it and measure that to see its maximum peak voltage.

If you want 40kV, or -40kV, you'll need a lot of stages or a higher input voltage (just don't exceed the voltage of your capacitors!).

At such a low current you could consider an electrostatic device if you don't want to design a HFHV SMPS, which is the usual approach to *most things* in this particular forum. :)

radiotech

Mon Sept 20 2010, 10:20PM

Registered Member #2463 Joined: Wed Nov 11 2009, 03:49AM
Location:
Posts: 1546

Vout = 2(5)*5 - [0.000845/(6*f*20e-9)] * (4(5^3) + 3(5^2) - 5)

f=100 khz

Vout = 10 *5 - 0.000845/ 6e5 * 20e-9 * 125 + 70

Vout = 50 - 0.0704 = 49.93 which is what you got?

unless my arithmetic is wrong

Fronzbot

Mon Sept 20 2010, 10:38PM

Registered Member #2710 Joined: Tue Mar 02 2010, 06:04PM
Location:
Posts: 7

ScotchTapeLord wrote ...

I'm surprised nobody pointed this out... but one of the main problems is that you used the calculation for the voltage drop that would be occurring IN the multiplier itself as your output voltage. The basic formula is 2n*Vpk - (some losses). The example on the site shows that out of the 60kV it would ideally output, 8kV is lost. The losses cannot exceed the output... You used the current figure .000845mA, which as you suspected, may have been a faulty assumption.

It's surprising you got to 205V to be perfectly honest, and it implies that your function generator definitely has a higher voltage with high impedance loads. Just charge a capacitor through a diode with it and measure that to see its maximum peak voltage.

If you want 40kV, or -40kV, you'll need a lot of stages or a higher input voltage (just don't exceed the voltage of your capacitors!).

At such a low current you could consider an electrostatic device if you don't want to design a HFHV SMPS, which is the usual approach to *most things* in this particular forum. :)

Yeah, 11.84V peak on a square wave, 7.58 on a sine.

So, I figured my conclusion that I need a higher input voltage would be the case. Now, is there an easy way for me to step up the voltage out of my function generator without a need for heavy transformers (I wanted to make this as lightweight as possible. No real reason, it just seems more impressive to have something that can shoot sparks and is fairly lightweight)? If not, I'm pretty crap at making transformers, but where would be a good place to buy a core for the windings? Or, even better, are there any effective cores I can find lying around that would suffice?

Also, as a note, my caps are rated to 1kV so I have quite a bit of wiggle room to work with haha!

@radiotech- you're using 100kHz not 100Hz (which is what my original calcs used), and you forgot to multiply 5^3 by 4.

Antonio

Mon Sept 20 2010, 11:11PM

Registered Member #834 Joined: Tue Jun 12 2007, 10:57PM
Location: Brazil
Posts: 644

Don't use equations without checking their meaning first. And verify simple things as power. A 5 V generator with 50 Ohms output resistance can output at most 125 mW of power (over a matched 50 Ohms load), never the 33.8 W required.

Proud Mary

Mon Sept 20 2010, 11:15PM

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

Lookey here - 100Hz is just not a real world C&W frequency.

At 100Hz, your 20nF capacitors will have a reactance of 79K577.

This is why your contraption perked up a bit when you fed it with your "83.2kHz Square wave." where Xc for 20nF had fallen to 95R646.

Moreover, the more stages you add, the less efficient do they become, and the greater the sag - the wilting away of the voltage when
current is drawn.

With only 5V in, and a volt of that lost across the first diode, it's no big surprise that your results were less than electrifying, now is it?

Fronzbot

Mon Sept 20 2010, 11:41PM

Registered Member #2710 Joined: Tue Mar 02 2010, 06:04PM
Location:
Posts: 7

Proud Mary wrote ...

Lookey here - 100Hz is just not a real world C&W frequency.

[/quote]
Oh, i just chose that because I have seen a lot of CWs running off of mains which is at 50-60Hz. Do they step up the frequency then? I must not have realized that.

wrote ...

At 100Hz, your 20nF capacitors will have a reactance of 79K577.

This is why your contraption perked up a bit when you fed it with your "83.2kHz Square wave." where Xc for 20nF had fallen to 95R646.

Moreover, the more stages you add, the less efficient do they become, and the greater the sag - the wilting away of the voltage when
current is drawn.

With only 5V in, and a volt of that lost across the first diode, it's no big surprise that your results were less than electrifying, now is it?

Mhm, that all makes sense - it's just got really confused as to why it was outputting the voltage it was. Like I said, I realized that I probably didn't have a large enough input voltage and now, after all the discussion in this thread, see that my frequency is a bit on the low side.

@Antonio - yeah, you know I didn't take the power aspect into consideration at all. Makes sense seeing as how my current wasn't even measurable so it all balanced that way. As for the equations, it was very difficult for me to find decent documentation on them. I understand what they're doing it's just that not all of the aspect I was sure on. Though if I took your suggestion and didn't use the equations because I didn't understand them fully, then I wouldn't be learning from it at all, would I? :)

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