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>

RMS?  More like the average, I suspect... One could make a case on an
"energy balance across the filter" basis for RMS (constant energy flow out,
energy proportional to square of voltage going in). 

I've always designed with series pass regulators, so the important thing
for me is the "bottom" of the ripple.

Tesla list wrote:
> 
> Original poster: "davep by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<davep-at-quik-dot-com>
> 
> > hmm, where exactly will these losses take place?
> 
>         Depending on the supply, i would not use the
>         word 'loss'.  Specifically, assuming a filtered,
>         unregulated supply:
>                 The no load voltage will be set by the _peak_
>                 voltage from the transformer.
>                 The loaded voltage will be set by the _RMS_
>                 voltage.
>         Thus the loaded voltage will be roughly 0.707 times
>         the unloaded/no load voltage.
> 
> >>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.
> 
>         best
>         dwp
> 
> ...the net of a million lies...
>         Vernor Vinge
> There are Many Web Sites which Say Many Things.
>         -me