How to calculate output impedance of a cockcroft walton voltage multiplier.  - Page 1
Forum section: electronics / projects

author: Fflint
author_rank: Frequent Contributor
timestamp: January 04, 2022, 12:20:47 pm
content: I've been searching everywhere including in old electronics books as well as online for equations how to calculate the output impedance of a cw voltage multiplier. I found only two pages (blaze labs and a blog written using blaze labs as one of its sources) online that describe the same method that doesn't work (it gives results not matching reality).I can't believe it is an unresolved scientific mystery. I think a method exists, but it is not documented anywhere. Hopefully someone who knows it will read this and answer.A Cockcroft-Walton voltage multiplier is a very simple circuit used a lot in old vacuum tube devices. Devices with a CRT etc. In practice it looks like this (on the right)In many devices it is used only when high voltage with very low currents are needed (CRT anode voltage etc), but in vacuum tube power amplifiers it is used with currents up to 1A (the one I show above can supply 2800W allegedly).I'm hoping there is a method other than build and test to design those. How can we even rely on simulation results when there is no straightforward mathematical way to verify the results?I'll mention briefly the best attempt to solve this I saw came from the blog I mentioned. The author came up with an equation like thisCimp - is the capacitor impedance at the working frequencyn - is the number of stagesCimp * (4n^3+3n^2-n)The result is close, but not quite. For single diode stages the error oscillates around +3kOhm (for 1 stage it is 2800ohm, 2 stages 3200 etc).No doubt it is because the equation doesn't take diodes into account at all.I hope someone can help and can shed some light on this mystery.Or if a straightforward equation doesn't exist, perhaps there are some "rules of thumb" to follow when designing such voltage multipliers.My purpose in asking this is not purely theoretical. I'm designing a high voltage transformer based lab power supply and I would like to be able to include an option to use a doubler-rectifier. For this PSU for this option to make any sense the output impedance of the doubler-rectifier has to be no higher than 1.5kOhm (at 50hz). I'm not sure if this is possible using commonly accessible components.

author: Capernicus
author_rank: Frequent Contributor
timestamp: January 04, 2022, 01:16:27 pm
content: they have 0 impedance on the output, isnt their output is a pure short?

author: Ian.M
author_rank: Super Contributor
timestamp: January 04, 2022, 01:43:30 pm
content: Consider a Cockcroft-Walton voltage multiplier as a charge pump. The charge that is transferred per cycle is limited by the voltage swing across the stage's coupling capacitor, which is limited by the drive voltage swing.However neither of the O.P's circuits is aCockcroft-Walton voltage multiplierwhich is characterized by the series chains of capacitors up both sides of the diode ladder, carrying the drive waveform up to the higher voltage stages, such that no component has to stand-off the full output voltage (or even more than one stage's voltage increase).

author: jonpaul
author_rank: Super Contributor
timestamp: January 04, 2022, 02:20:47 pm
content: schema were of doubler not a CW.CW mult Zo increases with the square of the # stages.Both the capacitance and frequency determine the Zo.As a reference, we designed Avionics CRT power supplies using 13 stage CW at 40 kHz, and 1000 pF caps. Finished PSU was 0.75" x 1.25" x 3"12 kV at ~ 100 uA.Jon

author: Ian.M
author_rank: Super Contributor
timestamp: January 04, 2022, 02:35:47 pm
content: Also, there is nothing 'magic' happening in a CW multiplier, so there's no reason to suppose SPICE simulation would be unreliable.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 04, 2022, 02:56:12 pm
content: Use the AC equivalents.If the input is sinusoidal, then within some factor, it must be true that the input current can't be more than Vin / Zc, where Zc is the impedance (mostly reactance) of the first series cap. Which thus sets a limit on output power. The same is true for each subsequent stage in the stack, where that capacitor drives the connected diodes, plus the current drawn by the stage above, and this loads the proceeding stage, etc.(Input current could be more in case of resonance, but we know that C has been selected much larger in comparison to stray inductances, so this is a correct conclusion.)(Hmm, does that do anything, anyway? Can a resonant CW stack be made? Do the tanks couple too tightly to each other, making the response all weird? If nothing else, the regulation will be poor, so there's that.)Anyways, this approach works, even though the circuit is nonlinear (full of diodes!). It can be made more rigorous by taking the cycle-averaged response from a given set of diodes, say, and using that voltage and current as an impedance which the capacitor acts against as an impedance divider. In general, the Vrms(Irms) characteristic of something nonlinear, is also nonlinear, so it's not that this helps us solve things very much -- you're still better off doing the transient simulation -- but as long as we can assume something close to proportional, it helps greatly just as a basic starting point, and then we can refine things from there.Or really not, because who bothers refining these things anyway, you're going to want to know simply what capacitor and diode to buy, get a bunch of them, and let that be that, right?So, for the same reason, it's very common that simply a straight (all elements equal) ladder is used, and that's just that. And that certainly doesn't work out so poorly in practice!Note that current increases with every stage, going down towards the source. This implies that most voltage loss should be in the bottom stages (for an equal-value stack), and so we might consider a tapered structure -- build a strong foundation to support the rest, right? But how shall we design it? If we increase the values geometrically, we probably keep losses distributed evenly; but this requiresexponentially many capacitors, so is rather far from economical! Well, maybe -- even the mighty exponential remains tractable at small values, and maybe this still pays off up to, quadruplers or so?The question is, is it better to distribute values somehow, or just use oversized values in the first place?I honestly don't know, offhand. I use them so infrequently, I haven't had much need to think about it.Going back to the impedances, it's a rule of thumb that the AC impedance factors in heavily, with respect to DC output. That is, say we have a 1 ohm source impedance; the DC output resistance might be ca. 4x for a half-wave rectifier, 2x for full-wave. And should be even worse at low power factor. That is, the output resistance increases at light load. Which makes sense enough, what with the conduction angle being smaller, and the diodes carrying less current -- diode internal resistance is roughly inverse with current flow.I don't know that you can produce more than a rule of thumb, or approximation, for the problem. For sure, a diode and resistance doesn't admit a closed-form solution (see: Lambert W function), so it seems very unlikely that this system would. There are few enough parameters that a useful amount of design space can simply be simulated, or empirically measured, and tabulated. (Covering variables like: capacitor value relative to supply frequency and impedance; diode internal resistance; and equivalent output resistance vs. load, number of stages, etc.)So, there you have it -- it's a complicated enough system that you're not going to get an exact result, though there isn't really anything deeply insightful to find here, either. It's also an example of a system that's just complicated enough to be easier to solve numerically, or just go to the lab and test. So, maybe not very satisfying, but at least the answers aren't super crazy (it's not like we're solving for polynomial roots), and it's very tractable as an engineering problem rather than a theoretical one.Tim

author: SiliconWizard
author_rank: Super Contributor
timestamp: January 04, 2022, 05:30:39 pm
content: Quote from: Ian.M on January 04, 2022, 02:35:47 pmOf course. Simulating this is absolutely no problem. If anything, make sure you add parasitics to your capacitors to make the simulation realistic.

author: jonpaul
author_rank: Super Contributor
timestamp: January 04, 2022, 06:42:15 pm
content: There are many fine papers on this exact topic.https://www.ijareeie.com/upload/2014/august/25_Analysis.pdfJust one example.Jon

author: Fflint
author_rank: Frequent Contributor
timestamp: January 05, 2022, 01:31:44 am
content: Thanks for replies.Tim, this is very interesting. I didn't consider the rule of thumb DC output impedance after a half wave rectifier can be 4* the AC output impedance(for sinusoidal ac) , but thinking about it now it makes sense. (So the important takeaway from this is to make sure to simulate correct AC source impedance if one has to simulate it.)Then, exactly, if one had a good design method one could optimise the circuit even for exponential growth in capacitance. Consider for example a 2 stage multiplier. The exponential growth in 2 stages can be quite practical and cost saving. If for example it turns out the last stage can use a capacitance measured in single digit uF while the previous stage needs hundreds of uF that can save lots of money on expensive caps (especially in high voltage applications where multiple expensive caps have to be tied in series to meet the voltage rating and one has to work with low frequency like 50Hz)Perhaps if no straightforward equation-like solution exists the best way would be to generate a large number of variations of component values for the desired circuit, by using a simple program in one's favourite scripting language. Then feed all variants to spice automatically and graph the results.For the above to make sense one would have to have a pretty solid way to simulate it including use of all relevant diode parameters, multiple caps in series etc.I still find it hard to believe it is such a difficult problem, but perhaps it is...Regarding other replies let me say that yes, the circuit I posted (on the right - on the left we have just a rectifier) is a voltage doubler, but after reading everything I could find I believe it is essentially the same thing if you consider it as a single stage half wave version of a cockcroft-walton multiplier. Both work by successively charging capacitors through diodes and summing resulting voltages for the next stage. A doubler has just one stage (half wave).If there was a simple solution I think it would apply to both.Ian.M and jonpaul, I realise how it works in general. Unfortunately I can't derive a method of calculating output impedance based on just the principle of operation. Also the article linked simply simulated the circuit and "analysed" it by talking about the results of their simulation. There is no attempt to find a mathematical solution of the problem we're discussing described there.Saying "simulating is no problem" misses the point. Yes, if indeed there is no other way, simulation will be the only option before ordering expensive parts, building the prototype then no doubt finding the simulation was way off due to some quirk (models used not quite exact in some crucial, but under appreciated parameter) redesigning, ordering more parts, etcWith voltages I'm working on now I need 20 400V caps per stage.The closest I got to any equation is this (all diodes and caps the same):Cz - capacitor impedance at working frequencyn - number of stages (Use 1 for a doubler)(Cz*(4n^3+3n^2-n))+(2800~3100)It seems to hold true for all designs I saw so far, but no doubt it will be wrong for a lot too.

author: Someone
author_rank: Super Contributor
timestamp: January 05, 2022, 02:26:19 am
content: Quote from: T3sl4co1l on January 04, 2022, 02:56:12 pmLike with many switching converters (capacitor charge pumps, or inductors), you hit the nail on the head, its going to vary (possibly considerably) with load.Quote from: Fflint on January 05, 2022, 01:31:44 amBeware magic numbers, 2800~3100 is just some addition to make thearbitrarymodel better fit your data. There is no underlying basis for that equation that I can see. Switching converters have notoriously complex output (and input) impedances which can't be neatly simplified to that sort of level. If you're so adverse to doing the engineering experiments (in simulation or otherwise) then just swamp the output impedance with a big capacitor and/or linear regulator until it hits your target impedance (both of which do have simple impedance relationships).

author: Ian.M
author_rank: Super Contributor
timestamp: January 05, 2022, 02:59:47 am
content: Here's a multi-stage Cockcroft-Walton test jig sim for LTspice. Its easy to change the number of stages, and everything else is either parameterized or is easy to configure.Unfortunately it doesn't support easily tapering the coupling capacitor and output capacitor values for each stage - if you need that I'll have to add some stuff to let a function get a voltage equal to the stage number so you can try different taper functions.

author: Fflint
author_rank: Frequent Contributor
timestamp: January 05, 2022, 07:48:46 am
content: Quote from: Ian.M on January 05, 2022, 02:59:47 amThank you for the ltspice file. I'm not adverse to simulating as I now know it is not an easy problem to calculate "by hand" due to diode non linearity (as it was said, impedance changes with current).

author: jonpaul
author_rank: Super Contributor
timestamp: January 05, 2022, 11:07:25 am
content: F Flint:a few moments please!1/ Simulation of diodes and non linear parts is long solved. Beware that most HV diodes have multiple dies , higher Vf 3..15V but still behave like a diode.I am not familiar with LTspice but suspect it is focusedas a tool for designs based on Linear Tech ICs.Our Power electronics colleagues use the excellent Spectrum Software MicroCap, 12, no support and free since creator has passed away. We found MicroCap to be excellent for all power and HV circuits.http://www.spectrum-soft.com/revisions_mc12.shtm2/ "simulation is the only option before ordering expensive parts, building the prototype....redesigning, ordering more parts,.....need 20 400V caps per stage".The construction of a model is not r complex or expensive and seldom needs iteration.Use scaling! Just divide the voltage by a factor of 10 or 50, scale all the cap's accordingly, use regular fast recovery diodes.For instance our 12 kV 13 stage used 600V/stage and 1000 V 100 pF caps. If we instead scale it to 600V out the caps are 50V. 100 nF.( caps will scale by V exp 2 factor )If you have some ordinary diodes and caps, you can build and test it in a scaled model in much less time than this very long thread takes to read!You can also scale for frequency.Bon Chance,Jon

author: Fflint
author_rank: Frequent Contributor
timestamp: January 05, 2022, 11:58:50 pm
content: Quote from: jonpaul on January 05, 2022, 11:07:25 amThanks. I'll check it out.Quote from: jonpaul on January 05, 2022, 11:07:25 amThats a good idea. I do have smaller low voltage parts I can use for a model.