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Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Steve McConner wrote ...
Er, by googling "Cockroft Walton design equations"?
The best equation I've managed to find relating to current passed by capacitors is Xc=1/(2 Pi f C), but this doesn't seem to produce any meaningful results.
Registered Member #30
Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706
Sure, but this circuit isn't just capacitors.
Here's what I would do:
wrote ... Noptimum = SQRT( Vmax * f * C/Iload)
Assume that the designer used the optimum number of stages, so you know Noptimum and C. Then guess two of the remaining variables and you can calculate the third. The transformer may have an obvious self-resonant frequency that you could measure to give you a clue to "f". And, you could guess Vout from the diode ratings.
Or use the "eripple" equation assuming some sensible ripple value like 5%, which is the closest thing I can see to your concept of "current passed by capacitors".
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Well, if Vmax is the output voltage (Vmax wasn't defined above), a frequency of 20kHz gives Noptimum as 22 0r 23, which is pretty close to 24, so 20kHz seems like a good starting point. Also I see what you mean by using the ripple equation now, but it was useless without some idea of 'reasonable ripple'. I'll try it with 5% and see what I get.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
I've entered figures into the calculator recommended by PM and obtained the following screenshot. While the capacitor values seem to be reasonable, the values for Vreg (voltage drop) seem to be a bit puzzling. Also, I can't find an explanation for the cap voltage/P-P Vin term.
EDIT: It's the half wave column that is relevant. Also the value I input for the capacitors is for the three caps at the lower end of the multiplier.
Registered Member #543
Joined: Tue Feb 20 2007, 04:26PM
Location: UK
Posts: 4992
I must say, Ash, that a half-wave multiplier with 24 stages, each sagging more than the last, would make me think of 2 or 3mA output, far less than the other guesstimates in your thread. .
If you re-enter 3mA and 30kHz you may get a different picture.
What is the total working voltage of the chain of polystyrene capacitors connected across the output? The intended DC output voltage is probably somewhere around a half of that figure at most.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Proud Mary wrote ...
I must say, Ash, that a half-wave multiplier with 24 stages, each sagging more than the last, would make me think of 2 or 3mA output, far less than the other guesstimates in your thread. .
I started with Steve's suggestion, PM. I don't know enough about these things to guesstimate myself, however, I didn't consider it to be unreasonable as I've seen circuit diagrams of Glassman 20mA units with capacitors considerably smaller than the ones here.
Proud Mary wrote ...
If you re-enter 3mA and 30kHz you may get a different picture.
I'll try this in the morning and post the results
Proud Mary wrote ...
What is the total working voltage of the chain of polystyrene capacitors connected across the output? The intended DC output voltage is probably somewhere around a half of that figure at most.
If you mean the cylindrical silver ones, there are no markings on any of them, but each one is in parallel with a 200M ohm resistor, 16,000M ohms total, if this gives any indication.
I'd previously been working on the assumption that if the capacitors in the multiplier are rated for 8kV, then the maximum theoretical multiplication would be 24X4kV=96kV. What would be a reasonable figure to assume after sag and safety factor? Steve suggested 50kV which seemed reasonable to me.
I realise that reverse engineering without even a circuit diagram is 'a bit of a challenge', but I'm hoping to come up with some figures for input voltage and frequency (to the multiplier) that won't damage it (and hopefully work out the theoretical output after sag etc) then I can (hopefully) work out what to feed the transformer.
Thanks for your advice so far, I'll post results after I feed some more figures into the calculator spreadsheet.
Registered Member #3414
Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245
Proud Mary wrote ...
The 200M resistors are there to equalize the voltage across the capacitor chain.
Now I'd assumed that the resistors, and capacitors, formed a resistive divider, tracing the circuit back I assumed this is what was used to measure the output voltage. I don't think I'm the only one here to make this assumption. Is this actually a capacitative divider network? I've read recent threads on both. Is there a certain 'overlap' between the two?
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