Help modeling a high voltage rectifier diode?

Hey guys!

 I'm new to CircuitLab, and I'm using it for student purposes. If this question is a little on the obvious side, I apologize in advance for my confusion. I am trying to use CircuitLab to model a Cockroft-Walton multiplier driven with a Flyback, and I noticed that the diodes offered by this software are no where near the reverse voltage protection value that I need. (20kV).

 I opened up the option to modify the parameters of the diode manually,(With the intent to recreate THIS diode located here: "") but there are about 8 different modification options available to me including "saturation current", "emission coefficient", etc... anyway, none of which are Reverse Voltage. The schematic of what I am trying to model is attached to this message. It will help conceptualize this problem for you guys. Again, I am sorry that this is so basic. Please, answer at your leisure. Thanks in advance! (Perhaps the offered specifications can be related mathematically to the Reverse Voltage...?)
by killik
February 28, 2015

Wow, my apologies. The formatting system here is not what I expected, it made my upper message look terrible.. Anyways, here is the CW multiplier including the diodes that I am trying to create. I forgot to include it above.

by killik
February 28, 2015

Dear Signality,

Thank you so very much. All of those links were incredibly helpful. I really appreciate your taking the time to answer for me :)

If you don't mind, I have one last question for you.

I have noticed that on some schematics for Cockroft-Walton multipliers, a series resistor is recommended to be placed on the output, right where demonstrative spark discharging would occur, on the end leads. (HV ground side for diode protection?) My question is, is the primary purpose of this high ohmic output resistors to limit the current that flows through the HV arcing that takes place on the outputs, for the purpose of protecting the rectifier diodes from avalanche breakdown produced via the high-stress experienced with higher currents being present in the arcs? If this is the case, what happens to the overall output VOLTAGE rating?

Since I=V/R, it makes sense to me that the output CURRENT would be limited by the resistor. BUT, would the desirable high voltage output also be subtracted upon due to the presence of the resistor? Resistors are used to drop voltage (obviously) in simplistic LED circuits, so they don't touch just the current... In summary, what purpose is served by a high ohmic series (or parallel) resistance placed on the output leads of these high voltage (100-300kV, 1-10uA) Cockroft-Walton multipliers?

by killik
March 01, 2015


An idea what value the limiting resistance is?

I'm not an expert in EHT circuits but I suspect the resistor is to limit the output current: (i) damage to the multiplier circuits due to the very high initial current that could otherwise flow at the instant the spark breakdown is established, (ii) so that the same current limiting prevents the spark turning into a high current arc which would then also damage the electrodes, (iii) most importantly as a safety feature to limit the current that could be discharged into you!

Getting zapped by static as you climb out of a car on a cold dry day is different from getting a belt off the ignition coil HT lead. Even though the voltages may be similar, the latter is capable of delivering a higher current for longer.

Play safely.


by signality
March 02, 2015

Thanks again! That's why van de graffe generators are safe enough to touch in comparison to lower voltage sources... it's the amperage that kills, the way I understand it.

Anyway, that's totally OK that you don't know too much about the inner workings of these types of circuits. The resistance I described above is of arbitrary value. Let's say 20k.

So anyway, just in case you do know, in a purely conceptual sense, not quantitative at all but more qualitative, if you have a steady DC source outputting a very small current (around a milliamp or less) at a very high voltage of about 300kV, and you are compare the output with a 20k resistor in series, and then without the resistor, I'm just wondering what the difference would be in the current and voltage output ratings with and without the series resistance. My hope is that the voltage drop across the resistor would be negligible at this high of voltage, but by ohm's law, the current flow will reduce significantly, thus confirming my assumption that the purpose of the resistor is to limit current while preserving the majority of the high tension output.

I'm really brand new to this field, just trying to understand the conceptual relationships between these elements. Thanks again for the responses, they have really helped me out. Since I've pretty much digressed EVERYWHERE on this thread, (Apologies) I'll go ahead and readdress any follow-up questions I have on this obscure topic to the appropriate help forums, and cease to waste your time :)

Really though, your responses are of great value to me and they are immensely appreciated.

Warm regards

by killik
March 02, 2015

Oh, any closing thoughts on this final question though? :)

by killik
March 02, 2015

When the spark is established, the ionised air has a very low resistance in the order of hundreds of Ohms or less.

So initially you have the charged up high voltage out of the CW multiplier, Vini, applied effectively across the series resistor, Rser, plus the spark gap effective resistance, Rsprk, so applying Ohms law:


In practice I think you'll find that Rser is in the order of tens of MegOhms so

Rser+Rsprk ~ Rser

where ~ means approximately equal to.

A half wave CW multiplier (CWM) is effectively made from N caps, Ccwm, in series so the effective capacitance, Ccwm_eff, is Ccwm/N. This capacitance determines the charge that the CWM can deliver.

For a low Ccwm_eff, you can probably ignore the mains charging pulses since the spark duration may be much less than a mains half cycle and it takes many cycles to charge the CWM to Vini.

So, once the spark has established, depending on Ccwm_eff, the voltage out of the CW will tend to droop. More Ccwm_eff = less droop.

In effect you could model this as a cap = Ccwm_eff, charged to an initial voltage = Vini, connected to Rser in series with a spark gap.

This can all be described mathematically and gets complicated for high Ccwm_eff where the spark can exist for many mains cycles.

It is further complicated for full wave CWMs.

Soooooo, you are modelling this in a simulator.

So simulate it!

In effect you could model the CWM as a cap of value = Ccwm_eff charged to an initial voltage of Vini, connected to Rser in series with a spark gap.

A more advanced sim would be to build the CWM and for a simple spark gap model, use a voltage controlled switch driven by a PWS() source to model the spark on resistance and spark duration.

See CL Documentation about PWS() sources.

For a more complex but more realistic simulation, you can replace the switch by a model of a spark gap:

Just copy and paste it into your schematic.

Then you can measure the voltage out of the CWM and the current through Rser and see what is likely to happen in a real circuit.

After all, that's what simulation is for!

However, it is essential that you read and understand the following:

:About setting an initial voltage across a capacitor

About capacitors in CL


Capacitors in CL are not the mathematically ideal components of zero effective series resistance (ESR) and infinite parallel resistance (i.e. zero leakage current) that your problem assumes. CL places a very large but finite resistance across the capacitors.

If you replace each capacitor in your problem with this parallel resistance then you will get the same DC solution given by CL.

Which of course is of no help to you because it does not give you the answer you are expecting. However, if you replace the DC voltage source with a

PWS(0,0, 1n,90)

source then if you run a Time Domain sim for say:

Start Time = 0

Stop Time = 1s

Time Step = 1m

then you will get the answer for your problem as originally defined.

This is because the voltages across the caps all start from zero and are then stepped from an ideal zero source resistance voltage source and the very large parallel resistances have such a long time constant with the caps that you don't see the voltages drooping or climbing as they very slowly discharge or charge to where the DC solution ends up.

If you then run the sim with:

Start Time = 0

Stop Time = 1e15

Time Step = 1e12

then you will see the time constant. So in one sim you can see the initial solution (the one you want) and the steady state (for the conditions imposed by CL) solutions.

See this thread for more information:

DC voltage report is 33V/66V at a/b I calculate 54V across C1, 27V across C2, 18V across C3.

If CL applies the same sort of techniques as SPICE then CL probably places a very high value resistor in parallel with a capacitor as part of the model. This resistance will be somewhere between 1G (1e9) and 100T (1e14) Ohms.

Call it Rpar.

Rpar will be the same across each capacitor irrespective of value.

Hence the DC solution is correct.

Here's a proof:

The original circuit has been deleted and without it this thread is hard to understand.

Here's something that makes sense of it all again:

Note that this is a poorly conditioned circuit and even if the parasitic resistance across the caps were to be infinite, in practice you would not realistically expect to find the voltages as calculated unless the initial voltage across them was zero and the supply was ramped up from zero.

Any practical attempt to measure the voltages of course would set up a leakage current which would alter the charge on the caps which would effectively give a zero voltage reading after more than a few time constants.

The time constant here would be due to the effective series capacitance seen across the measurement instrument terminals to ground and the resistance of the measurement instrument.

About switches in CL

Note that it is not clearly documented but switches in CL do not have an infinite OFF state resistance. This can lead to unexpected results.

For example, a DC plot of this circuit:

with R1 removed from the circuit would show

V(output) = 1V.

Here's why.

The resistance to ground at the Output node is infinite because it is an open circuit.

However, the switches in CL are not ideal. They have a very large but finite OFF resistance.

Therefore when operating into an open circuit load there would be no difference between the ON and OFF state voltages at the Output node.

The example demonstrates this using a DC Sweep probing V(output) with a Decade Sweep of the value of R1 (R1.R) from 1e12 to 1e18 at 10 points/Decade.

It can be seen that V(Output) = 0.5V when R1.R is equal to around 1e15 Ohms.

This is the undocumented OFF state resistance of the CL voltage controlled switch.

Note that if a CL voltmeter were to be connected in place of R1 then there would be a clear ON/OFF state difference because the CL Voltmeter has a user definable finite resistance with a default value of 1G.


by signality
March 03, 2015

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