# High Voltage Module for Capacitor Charging

I just ordered a DC to HV DC converter (the F40 model, datasheet here: http://www.emcohighvoltage.com/Fseries.PDF [Broken]) from eBay for $50. My intent is to charge capacitors such at the 400V 12,000uF one located here http://www.amazing1.com/capacitors.htm. The converter has output up to 4,000V that is proportional to the input which is up to 12V. So if I wanted to charge the 400V capacitor, would it be best to input around 1.8V to the converter so that its output would be 400V, or would it be better (quicker maybe?) to go ahead and input a higher voltage to the converter, perhaps the full 12V, and simply open the charging circuit once the capacitor has reached 400V? Any help would be much appreciated! Thanks in advance for your time. Last edited by a moderator: ## Answers and Replies The 4,000 volt dc-dc converter you purchased is limited to 2.5 milliamps rated output current. This limits the rate at which you can charge your capacitor. Reviewing equations: Q = C V dQ/dt = I = C dV/dt So dV/dt = I/C = 0.025 amps/12,000 uF = 2 volts per second. So it will take 200 seconds to charge your capacitor. Also, be sure the dc-dc converter can withstand 400 volts reverse voltage if is turned off while it is connected to a charged capacitor. Thanks so much for doing that for me. I knew there must have been a way to calculate how long it would take to charge, I just didn't even know where to start. So, I'm assuming to even get the 2.5 mA output, I'd need to input the full 12V? berkeman Mentor I just ordered a DC to HV DC converter (the F40 model, datasheet here: http://www.emcohighvoltage.com/Fseries.PDF [Broken]) from eBay for$50. My intent is to charge capacitors such at the 400V 12,000uF one located here http://www.amazing1.com/capacitors.htm.

The converter has output up to 4,000V that is proportional to the input which is up to 12V. So if I wanted to charge the 400V capacitor, would it be best to input around 1.8V to the converter so that its output would be 400V, or would it be better (quicker maybe?) to go ahead and input a higher voltage to the converter, perhaps the full 12V, and simply open the charging circuit once the capacitor has reached 400V?

Any help would be much appreciated! Thanks in advance for your time.

What is the minimum input voltage specification of the DC-DC converter? I doubt it will run down to 1.8V input. It does not sound like a good match for making 400V.

What experience do you have working with high voltages? What are you trying to make?

Last edited by a moderator:
What is the minimum input voltage specification of the DC-DC converter? I doubt it will run down to 1.8V input. It does not sound like a good match for making 400V.

What experience do you have working with high voltages? What are you trying to make?

It's turn-on voltage is 0.7V.

I've built a few coilguns.

This particular inquiry is for the charging of (a) capacitor(s) for a railgun.

berkeman, do you still think this is a bad solution for charging my capacitor? and why?

berkeman
Mentor
From the datasheet, the F04 would seem to be a much better match for 400V, and has the benefit of higher output current to give you a faster charge. Just don't put your eye out, okay?

From the datasheet, the F04 would seem to be a much better match for 400V, and has the benefit of higher output current to give you a faster charge. Just don't put your eye out, okay?

Hahahaha

I actually called the company about the F04. They said it was around $120, and since that didn't meet their minimum dollar amount, it would cost another$25. So, I went ahead and got this brand new one on eBay for a mere \$50.

berkeman
Mentor
So you saved a hundred dollars, and increased the charge time x10 or so. Maybe a good tradeoff, but I can't make that call.

I'm still curious, though, if I input the full 12V into the converter, I risk over-charging the capacitor if I, perhaps, get distracted and forget to disconnect the charger. Right?

So, wouldn't it be as wise of an option to input the proper proportion of voltage into the converter so that the output voltage would better match the voltage to which the capacitor should be charged?

berkeman
Mentor
I'm still curious, though, if I input the full 12V into the converter, I risk over-charging the capacitor if I, perhaps, get distracted and forget to disconnect the charger. Right?

So, wouldn't it be as wise of an option to input the proper proportion of voltage into the converter so that the output voltage would better match the voltage to which the capacitor should be charged?

Yes, you absolutely have to be careful about that. Your charging DC-DC is not well matched to the application, and has the potential to overvoltage the caps (boom, or at least zzzzt).