woelen
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Charging time needed for 10-stage Cockcroft-Walton voltage multiplier
This is not a chemistry-related question, but I think there are some people over here, who could help me with this. It definitely is a sciencemadness
question  .
I plan to build a 10-stage dual-branch Cockcroft-Walton voltage multiplier. The circuit looks as follows, but has more stages:
My circuit will use 30 capacitors and 40 diodes.
I want to connect it to a center-tapped transformer, each half giving 120 V AC. So, per capacitor, I can obtain appr. 350 V DC and the total output
voltage can be 3500 V. I want to use 220 uF/400 V capacitors. This allows for an energy storage of 400 J in the total circuit!
I have one problem. If I connect the device to my output transformer, then I expect it to blow out my fuse, due to the enormous initial charge
currents. I already built a 3-stage device with 100 uF/400 capacitors and when this is powered up, then sometimes the fuse is blown out, so with the
more than 3 times as large circuit and double capacitance I expect major power up surges.
I do not need high power output, I only want to charge the device and then use the fully charged device. So, I have the idea to place a series
resistor of e.g. 1 kOhm in series with the left branch and right branch, immediately in series with the trasnsformer output. This limits the charging
current, but also the time, needed to fully charge the device increases. THen I wait, until the voltage of 3500 V is reached and then start the
experiments. My question, however, is how long it will take before the circuit is charged to e.g. 95% or so.
I've no idea how long it will take to charge the device with 1K resistors, when a 10-stage circuit is used. My input is 50 Hz AC. If someone did
simulations of this circuit then that would be nice. Any suggestions? I did some experiments with my current 3-stage multiplier circuit and 2.2 kOhm
resistors in series with the transformer AC-leads (GND was connected directly). That already gives very long charge times.
Are there other safe ways of limiting the initial charge currents?
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Mr. Wizard
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Place incandescent light bulbs in series with each the AC legs. Make the bulb wattage and voltage as large as the fused circuit can safely handle.
Light bulb filaments are good current limiters and as the current drops after the initial in-surge they will cool and drop in resistance, allowing a
more steady charging rate, and allow your circuit to reach full voltage. Obviously they work quickly enough to keep fuses from blowing ;-) Another
possible benefit is you will be able to spot a shorted diode or cap as the bulb will stay lit longer on one side.
Edit. I realize you have 220 volt mains and may not have access to 110 volt light bulbs, which would be the ideal for that set up, as you could put
one on each leg. The setup should still safely work even with 220 volt bulbs, it just won't charge as fast.
[Edited on 12-6-2006 by Mr. Wizard]
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IrC
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A starting point for you would be to calculate the time constant for a single C and R and then multiply this by the number of capacitors. This is due
to the fact that as in a MARX device this circuit achieves higher voltage by "erecting" all the capacitors. In effect you are charging all the C's in
parallel and then discharging them in series which is how the multiplier reaches a higher voltage. If I had to guess I would say add the caps and find
out the forward impedance of a diode and add them to give you the R, and use a standard formula for time constant with one C value (Ct), and one R
value (Rt). At least this would get you close to what you are wanting to know, i.e. give you a starting point in your calculations. I would also think
about adding any series resistance since at high voltage (considering all C's near zero ohms initially) the dissipation would be immense. A relay on
the primary side of the transformer with a power resistor being switched out after charge builds would be one way to go. Many high power RF amplifiers
use this idea to limit initial turn on surge.
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unionised
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Lamps, which have large switch-on surges, are not good surge preventers.
I wonder if you have thought about mocking this up with a low voltage to see what happens? That way you could do it relatively cheaply and safely-
even check it with a 'scope if you ahve one handy.
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woelen
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So, if I understand correctly, just sum up all capacitors and multiply with the series R for the time constant? That is nice and simple. I however
have some questions. The capacitors only charge for part of the sine wave (only when the voltage is larger than the capacitor's voltage), but can this
be taken into account by using an additional factor?
I may not draw more than 0.25A from the transformer, at full load. With 220 V AC across both AC's, this means I have to use two 470 Ohm resistors in
series (one in each leg). With 30 caps of 220 uF, my time constant will be 30*220*470*10^(-6) = 3.1 second. Taking into account that charging is only
done at part of the sine wave this would mean a time constant of maybe 10 seconds or so?
I did some experiments with my 3 stage multiplier circuit and this shows that all capacitors are charged at the same time. So, if the first one is at
200 volts, then the last one also is at 200 volts. I think this is quite remarkable. I expected the lower stage to be charged first and the last stage
the latest.
I find it very hard to analyse this circuit easily in a numeric way. The basic idea is not hard to understand, but performing actual computations on
it is quite hard.
I also have the option to use 15 stages, but then only with the central capacitor array and one side array. Which would be better to your opinion? Do
you think the 15 stage device still is useful, or is it so feeble with so many stages that it is hardly useful anymore? I do not have much experience
with this type of circuits, these are my first steps in high-voltage electronics .
Up to now I did a lot of audio and digital electronics, but this is a whole different kind of thing.
[Edited on 12-6-06 by woelen]
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Mr. Wizard
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| Quote: | Originally posted by unionised
Lamps, which have large switch-on surges, are not good surge preventers.
I wonder if you have thought about mocking this up with a low voltage to see what happens? That way you could do it relatively cheaply and safely-
even check it with a 'scope if you ahve one handy. |
All the AC lamps I have seen will take a full load right off the mains without blowing the fuses. Are they different where you are? If he assumes the
capacitor bank is a dead short, the most current the circuit would draw is whatever the lamp would draw. It's not a FAST surge preventer, but it will
certainly keep a fuse from blowing. I used to use a 1500 watt light bulb in series as a ballast for a carbon arc furnace, and never blew a 20 amp
fuse. The best part is, as the caps charge up the current drops and the lamp allows more voltage to the circuit. It's cheap, low tech and,
automatic. Maybe I just don't understand the problem? 
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Twospoons
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I think you will find that your 10 stage array will take ten mains cycles to charge - its not quite the same as a three stage one.
Setting it up as 15 stages means you will have a higher output voltage, the same energy storage, but higher impedance and longer charge time.
Light bulbs as current limiters are cheap and robust, but remember that the resistance when cold is about 1/10th of that when hot. The do make good
fault indicators though.
A ballast inductor from a fluorescent light would also work.
I'm sure I don't need to tell you that getting stung by just one of those caps is potentially lethal. As a precaution you should put a 1 Meg bleed
resistor across each capacitor, so you don't get a nasty surprise when you come back to your 'toy' after a week. I also like to put a Neon with 1Meg
series resistor across each cap in such devices. That way you can tell at a glance which caps are likely to bite.
Put the device behind a blast shield, when in use. Polycarbonate sheet is good. 400J will fire diode shrapnel a long way.
This advice comes from my professional experience in electric fencing, and high power electronics (300kW inverters) - its not just scaremongering.
I have been hit by a 25J, 900V capacitor bank and it was a deeply unpleasant experience I have no wish to repeat.
Play safe, and have fun! High voltage is sooo cool!
Helicopter: "helico" -> spiral, "pter" -> with wings
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12AX7
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| Quote: | Originally posted by woelen
So, if I understand correctly, just sum up all capacitors and multiply with the series R for the time constant? That is nice and simple. I however
have some questions. The capacitors only charge for part of the sine wave (only when the voltage is larger than the capacitor's voltage), but can this
be taken into account by using an additional factor? |
Very good. Yeah, it's about a factor of 4 for half wave stuff. Maybe more for something as chunky as this.
| Quote: | | I may not draw more than 0.25A from the transformer, at full load. With 220 V AC across both AC's, this means I have to use two 470 Ohm resistors in
series (one in each leg). |
Yes, for short-circuit current. It can handle short overloads, and also, you aren't going to have it under short-circuit conditions for long, so 100
to 200 ohms would be fine with me. BTW, check the DC resistance of the windings, I bet you'll get half of it right there (50 ohms), if not all of it
(10% regulation, a typical value for a transformer, would drop 10% of the voltage at full load, or 240V would drop to 226V; the loss is 24V, which
divided by the load of 0.25A is about 100 ohms).
| Quote: |
I find it very hard to analyse this circuit easily in a numeric way. The basic idea is not hard to understand, but performing actual computations on
it is quite hard. |
That's because it has nasty phase angle conduction crap embedded in it. So you need to make assumptions and such, which is ugly.
For the jist of it: with the capacitors discharged, the first cycle only whacks the first capacitor, through the first diode. On the following half
cycle, the first diode is turned off, and the charged capacitor is thrust up into the second cap, delivering perhaps half its charge (since charge
distributes according to capacitance). On the next half cycle, the first cap is recharged again, while the third is charged to 1/4 peak. The fourth
half cycle delivers 1/8 peak into the fourth cap, while the second cap, which had a 1/4 charge on it, is charged to 5/8ths (by the 1st cap, which was
charged fully again). And so on it continues. It's clear to see that current quickly decays along such a chain, as it must since you can't get power
out of nowhere, after all. After a long period, the voltages level out and reach equilibrium with the load current. As you can tell from the loss,
if loading is high, only so many stages will remain useful. With relatively little load, of course, all the capacitors will eventually become fully
charged and you'll get n * 1.4 * Vrms output (or 10 * 1.4 * 240 ~= 3.4kV in your case).
Incidentially, have you measured your line voltage? Around here it always seems to be high, circa 125V. "110" and "220" are merely outdated
expressions referring to the line voltages of that magnitude. I know some countries still use that lower voltage standard, however.
| Quote: | | I also have the option to use 15 stages, but then only with the central capacitor array and one side array. Which would be better to your opinion? Do
you think the 15 stage device still is useful, or is it so feeble with so many stages that it is hardly useful anymore? |
If you just want low static, that's fine.
FYI, I'd use smaller caps, maybe 1-10uF, and leave it plugged in. You have it behind a transformer, you say, so ground isolation isn't an issue
(which, incidentially, you'll have to mind grounding anyway since that transformer isn't going to hold off several kilovolts if you
accidentially ground the live end of that thing!). A charged bank is just as deadly as a live one, so there's no difference there. It'll charge
faster, and store less energy. Some resistance, perhaps 100 ohms, can limit input current, while a beefy high voltage resistor can hold off output
current depending on the load you need from it.
Tim
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woelen
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Thanks for all your responses and thoughts. I now have more confidence in that this is really going to work. I'll investigate the idea of light bulbs
in series with the circuit. I like the idea very much, but indeed, over here, I have no access to 115V bulbs, that may be a drawback.
As far as safety is concerned, I'll build the entire thing on a PCB and mount it in a plastic case, and I'll bring out a few stages (e.g. stages 1, 2,
5, 10) by means of sturdy connectors, accepting banana plugs. In this way, I'll have a very versatile power supply with a nice set of output voltages,
and no spaghetti of wires and bare components all over the place, and the connectors have the metal pieces deep inside, so accidental touching of any
charged capacitor is not done easily. Also, the center-tapped transformer is not grounded at its output stage. In that way, the device is "floating"
and only when two different places in the circuit are touched at the same time, there is a risk of severe accidents.
The reason that I'll use those 220 uF capacitors is that I can obtain these at very low prices (just 20% of what normally needs to be paid for
capacitors of this size). This particular value is very commonly used in switched power supplies for PC's and as such, they are available cheaply
sometimes at eBay or surplus electronics parts sellers. Now, I could obtain 50 pieces of these for a bargain.
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Mr. Wizard
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You could use a 55 watt or less 220 volt light bulb on one AC leg of your charger. At a 'nominal' 220 circuit this will draw .25 amp or less. Figure
your own bulb size by voltage x current=watts. Having two bulbs only gives you symmetry and by a difference in brightness, an indicator if one side
of the circuit is drawing more current due to a bad cap or diode. It is not a necessity. You could put two 220 volt bulbs in the circuit and adjust
their size to limit the current to your .25 amp limit. Small wattage light bulbs will slow the charging rate a bit,and provide better current
regulation due to the smaller thermal mass of their filaments. I would try the small candelabra type bulbs or the small holiday decorative bulbs and
just parallel them until you get the desired draw.
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Twospoons
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Do put the bleed resistors in too - you don't want your outputs sitting there "hot" long after the thing is switched off - how are you going to know
when its safe to change your connections?
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woelen
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Twospoons, thanks for your concern. I indeed had forgotten this  . Right now, I
made a nice safety indication plan:
I plan to make the following terminals on the outside of the case, with resistors and an NE-2 in between, the NE-2 nicely mounted in the case between
the two connectors, for which it shows the presence of charge:
stage 0, GND (but not earthed)
1M + NE-2
stage 1, appr. 350 V
1M + NE-2
stage 2, appr. 700 V
1M + NE-2
stage 3, appr. 1000 V
2 x 1M + NE-2
stage 5, appr. 1700 V
5 x 1M + NE-2
stage 10, appr. 3400 V
With this setup I can see for each connector, whether there still is significant charge present or not. Even with only 100 V across, the NE-2 still
gives a clearly visible orange glow. In parallel to the NE-2 I'll add a 22 M resistor, in order to get rid of the last traces of charge on longer
standing, otherwise I still can have 70V or so across the connectors after a long time.
With this setup I think I can work fairly safely (of course, HV always has some risks). If you still have suggestions for improvements, then I would
be pleased to read about them.
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jpsmith123
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Hello Woelen,
You should probably d/l "CircuitMaker" or some other circuit simulator from emule and play around with values a little bit.
Attached is a "CircuitMaker" schematic and output vs. time waveform for a 10 stage full-wave multiplier (the other two curves show the voltage at the
first and an intermediate capacitor).
Attachment: 10 Stage Multiplier.zip (77kB)
This file has been downloaded 542 times
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unionised
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I have seen, somewhere on the 'net, an expression for the output impedance of a multiplier like this. If you use that to calculate the current
available and work out how long that would take to charge the cap bank You will probably get pretty close.
"All the AC lamps I have seen will take a full load right off the mains without blowing the fuses. Are they different where you are?"
Just for the record, yes, it seems they are.
Whereas the biggest lights I run from the lighting circuit are 150W, the fuse is rated for 6A, equivalent to about 1500W (strictly speaking it's an
MCB rather than a fuse).
If you switch on a kilowatt of tungsten lamps with a fuse rated just above the calculated current it will blow. The turn on surge is typically 10
times the normal running current. This is a real pain in the neck if you are trying to use triacs or thyristors for lighting control.
The point I was making is that they have exactly the wrong current vs time characteristic for surge prevention.
They do have the advantage of being cheap and easy to get.
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Twospoons
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The only other suggestion I have applies to everything we ever do here : aways ask the question "What could go wrong?"
Oh, one other thing - watch the voltage ratings on your resistors. An ordinary 1/4W resistor is only rated for about 250 - 300V. Philips make VR37
type resistors which are good for about 1500V, and VR68 type which can handle around 3500V. You can use several resistors in series to get your
voltage rating. Also film type resistors (and this includes the high voltage ones I've mentioned) do not like arc discharges. There's some kind of
funny resonance effect that blows holes in the side of the resistor. Its quite something to see! This can be avoided by putting a small value
capacitor across the resistor : 100pF seems to be enough.
[Edited on 14-6-2006 by Twospoons]
Helicopter: "helico" -> spiral, "pter" -> with wings
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woelen
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@jpsmith123: I tried to download your zip-file, but it doesn't work for me. I get the message that the website is non-existent or cannot be accessed
.
@Twospoons: I know of the voltage limitations of many standard resistors. I use the somewhat larger type (rated for 2W, safe to at least 500 V).
Larger resistances I make by series connection, e.g. between the 1700 V and 3400 V terminals I connect 5 of these resistors, each one of them 1 MOhm.
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jpsmith123
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Hmmm...I just tried it and I had no problems. Give it another try.
(BTW, if you get CircuitMaker 2000 or whatever, let me know and I'll give you the file and save you some time).
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woelen
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Jpsmith123, thank you very much for this simulation you did. Now I could download the file. Apparently, yesterday there was some long lasting hick-up
in the path between that server and my internet connection.
Now I get a feeling of how this circuit will behave. It surprises me that with only 10 Ohm resistors it takes such a long time to obtain the total
voltage at the output. From the blue curve "A" I estimate the effective RC time to be approximately equal to 1.1 second. After 1.1 second it is at
3400*(1-exp(-1)) (appr. 2150) volts.
I'm certainly going to look after that product (CircuitMaker 2000), with which you did the simulations. If I have found and installed it, I'll come
back on this. Thanks for the effort you already have put in this.
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