Cockroft-Walton question.

Are "Real Programmers" sure about that ?:-)

...Jim Thompson

--
|  James E.Thompson, P.E.                           |    mens     |
|  Analog Innovations, Inc.                         |     et      |
|  Analog/Mixed-Signal ASIC's and Discrete Systems  |    manus    |
|  Phoenix, Arizona            Voice:(480)460-2350  |             |
|  E-mail Address at Website     Fax:(480)460-2142  |  Brass Rat  |
|       http://www.analog-innovations.com           |    1962     |
             
I love to cook with wine.      Sometimes I even put it in the food.

The usual CW multipliers are *additive*:

Cockroft Walton Voltage Multipliers.

So after one doubler the voltage would go from V to 2V. If you add another stage accoding to the usual circuit, it becomes 4V. So after 10 stages it's only 20V. I'm asking why can't you change the wiring so that it's 1024V.

Bob Clark

So, did it go from V to 2V 'cause you added V, or multiplied by 2?

When you add stages, will you add V or multiply by 2?

Because the input to a c-w stage must be AC, and the output is DC.[1]

John

1. Except that DC doesn't exist.

The amplitude of the single working source is limited and the energy storage elements are all capacitive, in this topology. Disregarding possible resonant effects or intentional use of inductive elements, the source voltage amplitude is the only active voltage available to charge or discharge capacitances in any stage of the multiplier string.

If more sources or switches are present, different voltage amplifications are possible in each and any specific stage, but then it's not a Cockroft-Walton circuit any more.

RL

Hmmm You answered your own question. what do you think it does now ?

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

to add to that, if you look you can repeat the CW stages.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

It's certainly possible, but of course you can't just feed the output of one CW into another; you need an inverter that will turn the DC[1] output into AC (a square wave is best for the CW and easiest to make).

That's a problem, because it throws away the biggest advantage of the CW - the fact that no component has to withstand the entire output voltage.

Try designing one on paper paying attention to the voltage across each component and you will see that the stage1-stage2 inverter is expensive, the stage2-stage3 inverter is really, really, expensive, and you can't build the stage2-stage3 inverter at all.

The reason you don't even see a two stage system in practice is that there is a much better way to do it; use a step-up transformer to drive your first stage with a higher AC voltage.

Excellent thinking, though. It's always good to think "out of the box" and to look for something that everybody missed before.

----------------------------------------------------------------

(Those who are interested in a technical discussion only are encouraged to move on to the next post without reading the OT material below.)

[Note 1] Ignore the brain-dead flamers who think that they have discovered a weakness in their technical superiors when actually all they have done is to invent a nonexistent difference between DC and a pulse that starts when you turn on the DC power supply and ends when you turn off the DC power supply. True, realizing that the two are simply different ways of describing the same signal is a rudimentary mental skill that many of us "normal" people take for granted, but we sometimes forget that there are "challenged" persons in this world who find these things to be difficult to understand. The sad part is that they have the raw mental capability to arrive at a rational conclusion, but are so controlled by their emotions and fears that they grasp at any straw to attack the target of their schoolyard bullying. It's best to just killfile them and to let them flame away and shout into an empty hall.

--
Guy Macon

Try goole for "charge pump"--a variation of the normal doubler circuit that can be staged as you suggest. Output of Vinsin(wt) input would be the multiplier x Vin (DC voltage of course.) The limitation to the idea is poor voltage regulation and low-current capability.

Good luck

Tut

That's silly, and it's wrong. You often see them, and I often use them, for example. Frequently it's inconvenient to get a higher-voltage transformer. For example, check the meager HV offerings from Signal Transformer. Now that tubes are not so commonly used in industry, the available selection of ac-line transformers with outputs above say 250V is very meager indeed.

[ snip @)$*&^#$@&$ ]

--
 Thanks,
    - Win

What about neon sign transformers and oil burner ignition transformers?

Those that I see mostly have AC output of 6 to 15 KV, at least usually center tapped - 3 to 7.5 KV from either end to ground. That's a lot more than 250V.

Modern CRT flyback transformers put out pulses in one direction of a few KV, and "triplers" double this, "quintuplers" triple this, and so on. Even though two diode-capacitor "steps" add only one increment of peak input voltage (add to this increment whatever the peak voltage in the other direction is, presumably not zero), this may be practical in systems requiring low weight, small size of the initial high voltage generator, and/or battery power.

Get an older flyback transformer for color TV sets that did not use multipliers, and you have pulses of a good 25 KV or so, possibly 30 (I have achieved 30 KV from some of those).

How about use of an automotive ignition coil? Those generally produce

25 KV or more.

How about a Tesla coil? AC of voltage mostly from 40 KV to hundreds of KV!

Now, let's worry about selecting suitable diodes for high AC (or pulsating DC) voltages and AC (or pulsating DC) voltages for such diodes.

- Don Klipstein ( snipped-for-privacy@misty.com)

You, sir, are a Real Programmer!

Voltage doublers and multipliers work with an AC input from a transformer or other source of AC voltage. The AC charges a capacitor on one half of the cycle to the peak value of the AC voltage and adds that peak voltage to the peak voltage of the other half cycle to form a DC output. The DC voltage, ignoring losses, is the sum of the two half cycle voltages which is the peak to peak value of the AC. Once DC, there is no way to repeat the cycle again. But, if the AC voltage is capacitively coupled to a second stage, a second DC, peak to peak value will develop. If that stage is wired on top of the first stage, the DC values will add together forming 2 X the peak to peak value of the AC. Three stages will develop 3X the Pk-Pk value and so on. There is no way to "multiply" the voltages as you mentioned because the output of each stage is DC which cannot charge and add in a repetitive cycle because DC has no cycle. Bob

Instead of restricting it to CW, lets call it a voltage "exponenter" instead of "multiplier" that yields (mathmatically) X^n times whatever the original voltage was. The ugly real world steps in, current decreases, and each stage adds more and more loss. I don't know what the practical limit of such voltage multipliers is, but logically at stage ten, V will have increased by less than the mathmatical X^10 times due to accumulated losses.

Specifically to the CW and your question: Each stage in the CW rectifies and stores the input peak Vac by charging 2 capacitors. One is charged to the negative peak, the other is charged to the positive peak. The capacitors are charged individually, but are discharged in series, so each stage *adds* the voltage it has stored to the previous stage. The net effect is Total_V = (Original_V * 1.414) * N where N is the number of stages.

If you examine the schematic of a CW you will see that it can't multiply the voltage stored in the previous stage - it can only add to it. If you don't examine the schematic, then just recognize that the ac input is required by each stage. In essence, all stages are charged in parallel and discharged in series.

Ed

That's roughly 333 stages. The breakdown voltage in air is roughly 25kV to 75kV per inch (depending on electrode shape, humidity, etc) so lets give each stage an an inch to be really sure. That's a bit under 40 feet total, which will fit in your garage with sepentine construction.

As for the output, here is a fellow making 4-foot sparks in air: [

] So I can see why one would want to have an insulator better than air.

When you snip out part of an answer, it's traditional to leave a blank line. The above makes it look as if my reply starting with "The reason you don't" directly follows the "multiply the voltage by 2^10 = 1024." above it, when I actually wrote three paragraphs between the two.

|>> Why couldn't you have the output of a CW voltage doubler lead into the |>>input of a another doubler? It seems to me that instead of the voltages |>>being additive with additional stages as done now, with this method you |>>could double the voltage each time. |>>So with 10 repetitions you could multiply the voltage by 2^10 = 1024. |>

|>It's certainly possible, but of course you can't just feed the output |>of one CW into another; you need an inverter that will turn the DC[1] |>output into AC (a square wave is best for the CW and easiest to make). |>

|>That's a problem, because it throws away the biggest advantage of the |>CW - the fact that no component has to withstand the entire output |>voltage. |>

|>Try designing one on paper paying attention to the voltage across |>each component and you will see that the stage1-stage2 inverter is |>expensive, the stage2-stage3 inverter is really, really, expensive, |>and you can't build the stage2-stage3 inverter at all. |>

|>The reason you don't even see a two stage system in practice is |>that there is a much better way to do it; use a step-up transformer |>to drive your first stage with a higher AC voltage.

Than being said, I wasn't aware that anyone was using a dual-stage CW. Thanks for correcting my misconception. Are you limiting yourself to applications where the stages are few and the voltages fairly low, or do you have a method of doing the DC to AC inversion without subjecting any parts to the full output voltage of the first stage?

I find that neon sign transformers work well. Of course they don't come in the wide variety of sizes that you get from Signal, so you might be size-constrained.

--
Guy Macon

to be or not to be.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

count on your fingers. 2,4,8,16,32,64,128,256,512,1024 lets see, our RDI vaults produce a little over 1 M volts using a CW multiplier in an insulated pressured gas sourcing from a 3000 Kv OSC via pie RF transformers. i can tell you that we don't have a mile rectifier stacks in there.

--
Real Programmers Do things like this.
http://webpages.charter.net/jamie_5

Yes I agree, at least for some instances.

One finds the various multiplier techniques used with voltages from under 5V to the high voltages you mention. It's simply a matter of size and cost, convenience, and practicality. Often one requires something very small. I found Guy Macon's assertion silly and wrong. For myself, projects needing dc voltages in the region between 250 and 2000V lack many readily-available small 60Hz power transformers to choose from, and are thus excellent candidates for one of the specialized voltage-multiplier techniques.

I'd love to find an off-the-shelf supplier for small ac transformers in the 250 and 2000V region. I mentioned Signal Transformer. With their broad standard catalog, their 24-hour delivery, etc., they're often thought of as a good source for hobbyists and small production places like my lab. But they're also a popular transformer source in the higher-volume industrial-production world. I've been surprised by the number and range of instruments I've purchased on eBay that have a Signal Transformer inside when I open them up for inspection. But it's still true that despite the popularity of the company, their selection in the voltage range I mentioned is completely empty. It's a problem I've found easily solved with different types of voltage-multipliers.

--
 Thanks,
    - Win