Which UPS?
James Knott
james.knott-bJEeYj9oJeDQT0dZR+AlfA at public.gmane.org
Wed Jun 13 14:19:05 UTC 2012
Scott Allen wrote:
> Who said anything about two stages being more efficient than one? Go
> back and reread the post you are questioning. I'm saying that two
> stages is more efficient than three.
>
> Note that I was talking about the specific case where the user is
> limited to a 120VAC source, which is on line, and either a 120VAC
> output UPS is used, or a 48VDC output UPS is used.
>
> My statement, that you quoted and questioned, was for the 48VDC output UPS:
> -----
> With a 48VDC output UPS powered by mains AC:
> - While on mains (most of the time):
> UPS needs to convert 120VAC to 48VDC at, let's say, 85% efficiency.
> The 48VDC PC power supplies above are 70% efficient but let's say we
> find one that's 85%. UPS 85% and PC 85% is 72.25% overall.
As I mentioned in another note, the high voltage DC in power supplies
approaches 170V. Why should a power supply that runs on AC and be more
efficient than one that runs on a similar DC voltage? What is it about
that referenced DC supply that makes it so inefficient? Why should the
rectifiers in an AC supply be significantly more efficient than a
separate rectifier? Either way, you have to convert AC to DC and then
back to high frequency AC. Something about that DC power supply that
was linked to doesn't make sense. In this day & age, why is it so
inefficient?
> UPS needs to convert 120VAC to 48VDC at, let's say, 85% efficiency.
What is the conversion efficiency of the rectification part of the AC
supply? Why should a separate rectifier be significantly different
> Note that I was talking about the specific case where the user is
> limited to a 120VAC source, which is on line, and either a 120VAC
> output UPS is used, or a 48VDC output UPS is used.
An AC output UPS will have an additional stage to convert DC to AC. The
computer will then have an additional stage to convert that AC back to
DC. Even if UPS is not used, why should producing DC in a separate box
be less efficient than when in the same power supply.
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