Re: calculating ripple

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

I'll leap in here..

If you've got ripple, of say 2kV, on your 20 kV (no load) supply, the
average output voltage is really on the order of 19 kV.

There's also usually a fair amount of resistive loss in the transformer and
diodes for HV supplies.  The current being drawn when the cap input filter
charges is much higher than the average output current (particularly in a
high ripple situation), so the IR losses wind up being more than you would
think.

And, if you are using any sort of half wave or multiplier set up, you get
bitten, hard, by the limited charging time, that gets even more limited as
you move up the stack.  The effective duty cycle gets pretty low.


----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Thursday, August 22, 2002 10:37 PM
Subject: Re: calculating ripple


> Original poster: "Crow Leader by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<tesla-at-lists.symmetric-dot-net>
>
> hmm, where exactly will these losses take place?
>
> KEN
>
> > Original poster: "S & J Young by way of Terry Fritz
<twftesla-at-qwest-dot-net>"
> <youngs-at-konnections-dot-net>
> >
> > Ken,
> >
> > The reality of DC supplies is that the 20 KV DC is true only for no
load.
> > As you start drawing serious current, your average output voltage will
> drop
> > downward toward 15 KV.  It's probably more realistic to use 15 or 16 KV
in
> > your calculations.
> >
> > --Steve Young
>
>
>