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Re: calculating ripple
Original poster: "davep by way of Terry Fritz <twftesla-at-qwest-dot-net>" <davep-at-quik-dot-com>
> 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).
Just so. I'll stay with RMS. 8)>>
> I've always designed with series pass regulators, so the important thing
> for me is the "bottom" of the ripple.
Indeed, in that case. Gotta keep the regulator
in its active region. I was inferring that the
case under discussion involved an unregulated, filtered,
> Tesla list wrote:
>>Original poster: "davep by way of Terry Fritz <twftesla-at-qwest-dot-net>"
>>>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_
>> 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
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