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cw multiplier cap questions.

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ramses

Wed Feb 27 2008, 05:37PM

Registered Member #1208 Joined: Thu Jan 03 2008, 05:30PM
Location: Chesterland, OH
Posts: 154

is there a perfect value for the capacitors in a cw multiplier, based on frequency and voltage and current?

ex. 2kv 1a in at 2000hz, so 2000w /2000hz = 1 watt second (thus 1 joule) per cycle.

(1 joule/(2000v)^2)(2) =0.0000005 F or 500nf. per cap, neglecting loss.

would this prevent too small of a cap being "held" at max voltage, and too large of a cap not being charged all the way and holding back the Vout?

my made up formula:

((Vin*I)/frequency)/(Vin^2)(2000000000)= capacitance in nF

thanks,
ramses

Proud Mary

Wed Feb 27 2008, 05:56PM

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

For a series half-wave multiplier, the ripple voltage is given by

Vrip = [ I * (Nsq.+N/2) ]/8*f*C

where:

N = number of stages
f = AC input frequency (Hz)
C= capacitance per stage (farads)
I= DC output current (amps)

ramses

Wed Feb 27 2008, 07:02PM

Registered Member #1208 Joined: Thu Jan 03 2008, 05:30PM
Location: Chesterland, OH
Posts: 154

I meant the above as a formula to optomize the capacitance per stage so that the caps aren't

1: made to discharge before being fully charged and
2: held unnecessarily at full charge.

The main purpose is to determine the maximum capacitance per stage without the cap's being incompletely charged.

ramses

HV Enthusiast

Wed Feb 27 2008, 07:11PM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

The following information should help you:

(This taken from the following website:

Regulation and ripple calculations
The voltage drop under load can be calculated as:

Edrop = I1/ (f*C) * (2 /3*n^3 + n^2/2- n/6)

where:
Iload is the load current
C is the stage capacitance
f is the AC frequency
n is the number of stages.

The ripple voltage, in the case where all stage capacitances (C1 through C(2*n)) may be calculated from:
Eripple = Iload/(f * C)*n*(n+1)/2

As you can see from this equation, the ripple grows quite rapidly as the number of stages increases (as n squared, in fact). A common modification to the design is to make the stage capacitances larger at the bottom, with C1 & C2 = nC, C3 & C4= (n-1)C, and so forth. In this case, the ripple is:
Eripple = Iload/(f*C)

For large values of n (>= 5), the n2/2 and n/6 terms in the voltage drop equation become small compared to the 2/3n3. Differentiating the drop equation with respect to the number of stages gives an equation for the optimum number of stages (for the equal valued capacitor design:

Noptimum = SQRT( Vmax * f * C/Iload)

Increasing the frequency can dramatically reduce the ripple, and the voltage drop under load, which accounts for the popularity driving a multipler stack with a switching power supply.

ramses

Wed Feb 27 2008, 07:17PM

Registered Member #1208 Joined: Thu Jan 03 2008, 05:30PM
Location: Chesterland, OH
Posts: 154

no offence, but i was trying to calculate the best capacitor based on the desire to fully charge the cap's durring one full stage and not letting the input current ever drop to zero. this would let each cap be as large as possible and still be able to be completely charged in 1 cycle. By the way, those values were completely fake, just some easy numbers to work with.

This has nothing to do with calculating ripple.

Proud Mary

Wed Feb 27 2008, 09:01PM

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

ramses wrote ...

no offence, but i was trying to calculate the best capacitor based on the desire to fully charge the cap's durring one full stage and not letting the input current ever drop to zero. this would let each cap be as large as possible and still be able to be completely charged in 1 cycle. By the way, those values were completely fake, just some easy numbers to work with.

This has nothing to do with calculating ripple.

If you first set the ripple acceptable for your application, the number of stages to be used, and the input frequency, the formula I have given you will tell you what size the capacitors should be.

ramses

Wed Feb 27 2008, 09:06PM

Registered Member #1208 Joined: Thu Jan 03 2008, 05:30PM
Location: Chesterland, OH
Posts: 154

fair enough, but if the cap is big enough to not charge fully in one cycle, then you loose voltage. but it's not like I can afford multi nF caps for a multiplier anyway. at least at the voltages I had in mind.

CT2

Wed Feb 27 2008, 10:07PM

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

The capacitors will never charge in just once cycle, if you look at the circuit what happens is one cap charges on say the positive half cycle, then when current switches that capacitor (as well as the supply) are used to charge the next capacitor. But now the first capacitor isn't fully charged because it was used to charge the second one, and therefor you'll need a few more cycles to bring both capacitors up to fully voltage. I've also wonderd about what capacitance value would be optimal, then I whent to college, and I now know that what Harry said is true, the value will have to do with all those aspects.

ramses

Wed Feb 27 2008, 11:34PM

Registered Member #1208 Joined: Thu Jan 03 2008, 05:30PM
Location: Chesterland, OH
Posts: 154

well that makes sense, thanks.

Tesladownunder

Thu Feb 28 2008, 04:28AM

Registered Member #10 Joined: Thu Feb 02 2006, 09:45AM
Location: Bunbury, Australia
Posts: 1424

For my laser supply from a MOT, I have the first 2 stages use large caps (to drive the laser) and the next 5 stages to provide the high voltage at zero or low current to ignite it.
Works well for sparks like a Jacobs ladder, provided you recognise the need for a resistor unless you are operating into a fixed resistive load (and not a short circuit or capacitor)

TDU

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