OT question for electronics wizards

From: Eric Smith <eric_at_brouhaha.com>
Date: Mon Dec 21 23:14:27 1998

"Daniel T. Burrows" <dburrows_at_netpath.net> wrote:
> It would have to be an awfully inefficient switched to draw 7 amps at full
> load.(300w).

The 7A figure (for the AC input) printed on the supply is the peak current;
the power supply will not ever draw 7A continuous.

Keep in mind that for AC, it is NOT necessarily the case that
watts = volts * amps. That relation is only true for a perfectly resistive
load. For such a load, the voltage vs. time and current vs. time waveforms
and sinusoidal and exactly in phase.

Most computers and other devices with switching power supplies present a
highly non-linear load and draw a disproportionate fraction of their power
consumption at the peaks of the cycle.

If the computer consumed 300W on 115V mains and presented a perfectly
resistive load, you would expect the average current to be about 2.6A, and
the maximum to be about 3.7A (at the peak of the cycle when the line voltage
is around 170V). But because it is not a resistive load, it actually will
draw more than 3.7A peak.

This is why the maximum load of a UPS is rated in VA or KVA, rather than in
watts.

Inductive loads are even less well-behaved than switching power supplies.
Their current waveform is sinusoidal, but out of phase with the voltage.
The electric utilities don't like this very much, which is why Power Factor
Correction has become such a big thing recently.

Eric
Received on Mon Dec 21 1998 - 23:14:27 GMT