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I am trying to build a capacitor bank for a project in order to feed an inductor with "high" voltage(500V) however I only have a dc generator rated at 100V at my disposal. Charging 5 identical capacitors in series all equipped with bleed resistors would not cut it as I would get 100/5=20V on each cap. My question is whether charging each capacitor separately with 100V and then connecting them in series would get the voltage at the two ends of the circuit at 500 Volts? I understand that the capacitance would be divided by 5. So far charging each capacitor separately while they are already connected in series gives me a total voltage of 270 volts.Any idea why that happens? Any answer will be appreciated !

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  • \$\begingroup\$ You can increase AC-input voltage using just capacitors and rectifiers in a voltage multiplier. See en.wikipedia.org/wiki/Voltage_multiplier and, if really ambitious, en.wikipedia.org/wiki/Cockcroft%E2%80%93Walton_generator \$\endgroup\$ – DrMoishe Pippik Jul 16 '18 at 19:39
  • \$\begingroup\$ Thanks a lot for the answer! So using a Cockcroft–Walton (CW) generator, I could get an ac generator up to 500 Volt if I understand correctly. However I also want to know if charging the caps separately and then connecting them in series would also give me 500 V. Thank you again! \$\endgroup\$ – Kostis Papadakis Jul 16 '18 at 19:58
  • \$\begingroup\$ Yes, it will. You could use a multi-contact switch to make it easier - charge them in parallel and then switch to series. \$\endgroup\$ – DrMoishe Pippik Jul 16 '18 at 20:17
  • \$\begingroup\$ Today i went along and built a 7 stages CW generator and fed it with 7VAC at 500Hz. The output was a promising DC at 100 V. So thanks a lot for the idea. Right after i connected a 630 V rated cap at the output of the CW and could only get 46 V on the cap(probably due to the resistance of it). Any idea of how the voltage drop can be calculated. Is it only the resistance of the cap that is to blame? \$\endgroup\$ – Kostis Papadakis Jul 17 '18 at 15:42
  • \$\begingroup\$ Also do you happen to know a good way to power the CW generator with DC pulses rather than AC \$\endgroup\$ – Kostis Papadakis Jul 17 '18 at 15:53
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You could use a Marx generator. This is a circuit that charges capacitors in parallel and then connects them in series to get a high voltage. In practice, you need to worry about the leakage current through the capacitors as that will limit how long they can store charge effectively. It may be leakage current that caused you to only get 270V in your experiment.

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  • \$\begingroup\$ Ι ll have a look at Marx generators. However i too was worried about the leakage adn after searching online i found out that polypropylene caps are known for very low leakage. So would it be possible to loose almost half of the voltage in a matter of less than 30 seconds. I also thought that the time it takes the voltmeter to measure the voltage it also "used up" some of the voltage! \$\endgroup\$ – Kostis Papadakis Jul 16 '18 at 20:54

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