Swend
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You can't have only a diode in series with a capacitor.
Think about it.
Once the capacitor charges through the diode, there's no path for the discharge current.
Well, your first circuit had diodes in series with some of the capacitors with no discharge path shown, and that won't work.
you're circuit does not look like the Marx Generators I found on the internet
1) https://en.wikipedia.org/wiki/Marx_generator
2) https://www.instructables.com/id/Build-a-simple-Marx-Generator/
3) https://www.penguinslab.com/marx.htm
your circuit appears to be the charging model, Figure 3, from this site:
https://www.ijert.org/research/soli...ies-a-comparative-study-IJERTCONV3IS01033.pdf
i expect it will work if you build it per items 2 and 3 above.
Hi Friends
The question is: How does C2 get charged? As you can see C2's charging path is from plus to minus (+) Ddc C2 Df1 (-).
View attachment 153794
C2 gets charged when switches Tdc, Ta1 and Te1 (in the schematic) are closed
I saw no inductors in the SSMG
but i think you're not understanding what LTSpice is telling you
Now I see the similarity with your erroneous Ltspice simulation. Voltage across C6 simply can't be measured ground referenced, it must be measured differentially.Voltage across C6 with ref. to 0 (zero), now why can't this be like C4?
Now I see the similarity with your erroneous Ltspice simulation. Voltage across C6 simply can't be measured ground referenced, it must be measured differentially.
I measured (post #1) voltage across D3 (DS3 in the simplified schematic in post #15) differentially, and it read 410mV and the datasheet says 1.7V - so D3 never turns on?
If D3 never turns on, how is C6 other terminal jumping to 600V? There's no other current path provided.
1.7V specification is for 1A diode current, you also have junction capacitance dynamically reducing the forward voltage. To see if C6 is charged, you'll measure its voltage with a differential probe.
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