# Re: HV power supply

```Original poster: "by way of Terry Fritz <twftesla-at-qwest-dot-net>" <Tesla729-at-cs-dot-com>

In a message dated 5/12/02 11:01:24 AM Pacific Daylight Time,
tesla-at-pupman-dot-com writes:

<< I may be wrong, but isn't overvolting any type of capicitor

Coiling In Pittsburgh
Ben McMillen
>>

Hi Ben, all

Yes, overvolting a capacitor is bad, but a large capacitor imparts
a big load on the power supply so until the capacitor begins to get
fully charged, the power supply will see the capacitor's low impe-
dance as a near short which will make the actual voltage across
the capacitor much lower than the rated output of the power supply.
As the capacitor gradually charges, the voltage across the capa-
citor will gradually rise. I used to charge large electrolytic capaci-
tors (400 volt, 2500 uFD) with the rectified output of an MOT. MOTs
usually put out ~2000 volts, but a discharged cap of this size is ba-
sically a dead short until it begins to take on a charge. I had the
input of the MOT controlled thru a variac but would turn the variac
up to at least 30% (probably 700 volts or so). I of course had a
volt meter at the terminals of the capacitor(s) and when the volt-
age measurement at the cap's terminals reached the rated 400
volts, i would simply shut off the power from the MOT. That's basic-
ally what I'm trying to do with my 10 kV,10 kJ quarter shrinker/ can
crusher capacitor bank now. If the supply voltage is only say 10
kV, then since it takes 5 RC constants for the cap to completely
charge to the supply voltage, it takes a looong time for the capa-
citor's charge to to completely equal the voltage of the power sup-
ply. The caps charge voltage can reach 80% of the power supply
voltage pretty quickly, but it takes almost forever for it to reach
> 99%. If the supply voltage is say only 20% higher than the cap
voltage rating, the cap could be charged relatively quickly to its
rated voltage and then removed from the power once its rated
voltage has been reached. Of course, the impedance and current
capacity of the power supply would also have a bearing on how
quickly the cap could be charged, too (a 50 mA supply will charge
faster than a 20 mA supply, assuming all other factors equal)

Does any of this make sense :-)??

David Rieben

```