Re: calculating ripple

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

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,
	supply...


> 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

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