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:
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:
- Is my assumption about the charging behaviour correct?
- 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