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Short introduction:

A Marx Generator is a capacitor arrangement that transforms a DC input voltage of let's say 1 to 20 kV to a multiple of that. Capacitors get charged in parallel and when they reach the break down voltage, they discharge themselves over spark gaps and essentially get connected in series leading to a much higher output voltage (theoretically input voltage times capacitor amount).

I built such an arrangement, but the sparks it produces are kind of small. The easiest thing would be to measure the output, but I don't have a multimeter for voltages that high. Additionally I also want to understand the theory behind it, which is why I wrote this question.

This image is a circuit diagram of a Marx generator. During my research I found several diagrams for Marx generators. They differed a bit, but all of them had the problem described further down. The image will help me describing my problem:

Marx Generator circuit diagram

Let us assume that the DC input is 8 kV and the resistors (Rb and the 10 resistors without text) have a value of 1 Mohm. According to my knowledge, that would mean that the first capacitor is charged with 8 kV and a resistance of 1 Mohm which would be perfectly alright. However as far as I know, the 2nd capacitor isn't charged with the same resistance; it is charged with 8 kV and a resistance of 3 Mohm (Rb + 2 times resistor without text). The third would get charged with 8 kV and 5 Mohms and so on.

My question is the following:

  1. Is my assumption about the charging behaviour correct?
  2. If yes, why would someone do that? Why would you connect the resistors in series instead of in parallel, just like the caps? Why would you want the caps to not get charged with the same current, so they achieve break down voltage at the same time? If no, what is my mistake?

Thank you for reading this far.

I am pretty new to the field of electrics, so please be patient with me.


Edit: As I don't know how to post an image in a comment, I will put in here. Some of you mentioned that the resistors would be exposed to the full voltage if placed in parallel too. Have a look at the picture. Wouldn't that fix the problem?

Edit again: New circuit diagram

new circuit diagram

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  • \$\begingroup\$ The capacitors are in series when the device fires. The resistors are too - if they were fed directly from a source via resistors when charging then the resistors would be across increasingly higher voltage points as you progress up the "cascade" and dissipate more power. You COULD use increasing resistor values but you get the same effect this way. \$\endgroup\$ – Russell McMahon Dec 13 '19 at 15:46
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You could use a parallel arrangement of resistors, but you would need very high voltage resistors, and clearances round them. The main point of a Marx is to get a very high output voltage with modestly rated components.

When the Marx fires, each resistor still only sees the charging voltage. If you used parallel resistors, the ones to the end stage would see the whole output voltage.

Eventually, all stages of the Marx will reach the same voltage. You can improve on the charging time by using inductors instead of resistors. This limits the length of the output pulse, but it's usually fairly short anyway.

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If you wait sufficiently long all capacitors will charge up to the input voltage.

When the "diagonal" spark gap goes off all capacitors are connected in series through their spark gaps and the total voltage is the input voltage * capacitor stages. The idea is that the spark gaps are at that time very low resistance.

To understand this, draw the diagram again, but with all capacitors charged and all "diagonal" spark gap replace with a short-circuit.

I built me one of these and learned not to let go off whilst your talking next to it on the phone.


In your second diagram you have a bottom resistor in series with the discharge circuit. That would severly limit the output current.


I worked with one of these at the high voltage lab of the university of Delft. It was 20 meters high and could produce several Mega volts. They did not use your adapted scheme so I would assume there are some disadvantages but I honestly can't see which. Apart from this one: For safety you want the ground of your feeder voltage and the end spark to be the same.

By the way: before you were allowed to work on it, it had to be discharged. You did that by turning a wheel which pulled a metal wire past the capacitors. Each time it hit a capacitor you would hear the 'chack' of the discharge.

I did not want to be pedantic but your initial diagram is wrong. The bottom capacitor does not add to the circuit. The 'left' arrow should come from the bottom which makes that it is at ground level. Which is important for my remark about safety and ground above.

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  • \$\begingroup\$ I edited my question. Please have a look at it. \$\endgroup\$ – hanslhansl Dec 13 '19 at 16:32
  • \$\begingroup\$ I edited it once again \$\endgroup\$ – hanslhansl Dec 13 '19 at 21:56
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The capacitors are in series when the device fires.
The resistors are too.

However, if they were fed directly from a source via resistors when charging then when the device fired the resistors would be across increasingly higher voltage points as you progress up the "cascade" and dissipate increasingly more power and increasingly load down the capacitors as the voltage points increased.

You COULD use parallel feed and increasing resistor values but you get the same effect this way.

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  • \$\begingroup\$ I edited my question. Please have a look at it. \$\endgroup\$ – hanslhansl Dec 13 '19 at 16:32
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to get the last capacitor more fully charged what you do is make all the other spark gaps slightly bigger than the last. In that way the last capacitor has time to get fully charged before the spark gaps fire.

When a single spark gap fires all the rest are triggered by the voltage pulses transmitted through the capacitors, so it does not matter if the have slightly different gap distances.

if you use resistors to a common input bus the resistors see a very high voltage when the gaps fire, with the series resistors the voltage on the resistors is only about the same as the voltage on the capacitors.

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  • \$\begingroup\$ It needs the combined voltage from all the capacitors being fully charged for a spark to jump any of the gaps, at that point a spark can jump all the gaps. Remember all gaps "see" the combined voltage. \$\endgroup\$ – Ian Ringrose Dec 14 '19 at 14:46
  • \$\begingroup\$ yes, but by triggering of the last gap it gives control over the charge level in the last capacitor, allowing time for it and all other capacitors to charge \$\endgroup\$ – Jasen Dec 14 '19 at 20:31
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All your observations are correct. However ...

1) If you wait long enough, all capacitors will charge to more or less the supply voltage

2) There are two common ways to get a Marx to erect. One is to have one spark gap set less than the others. When this breaks down, stray capacitance of the other nodes to ground will transfer some of the step to the other gaps, causing them to break down promptly. If this smaller gap is the last one in the chain, it will ensure all the caps are charged to at least that voltage. The other way is to use a triggered gap as the first gap. You can wait until the requisite charging time has elapsed before triggering the gap.

3) People build Marx generators to get verrrrrry high voltages. This means space and components are at a premium. The conventional layout of a Marx uses its length to good effect, adding the voltage withstanding of the space and the resistors in series. Your parallel layout forgoes that advantage. There is little point building a Marx for modest voltages.

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  • \$\begingroup\$ I don't understand why the resistors in my diagram need to withstand more energy. It makes sense to me that some of the resistors in my diagram are exposed to the full 32kV but why isn't that the case with the original one? After the the gaps fired the electricity can "choose" between 2 "ways" after leaving the last cap. Either it goes to the bigh spark gap or it takes the way back to the original source (Right?). In my schematic it would go through 1 and in the original through 4 resistors. Is that the difference? That the 4 resistors have to do the same work as 1 and therefore don't break? \$\endgroup\$ – hanslhansl Dec 17 '19 at 21:34

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