Ted Lemon writes:
> All this might make sense if the AC power was not being run through a
> step-down transformer before being regulated and smoothed to produce
> the nice 3.3/5/12V DC that your digital electronics are expecting to
> get.   When you run a square wave through a transformer, this is what
> it looks like on input:		        

This is a good explanation of what happens in a step-down transformer.

The dynamics of feeding a switching supply are a bit different.  The
AC input is immidiately rectified and filtered with a bridge
rectifier (*).  The 120v AC is converted to a dc voltage of:

	(* 1.414 120)    ; 1.414 is the ratio of RMS volts to peak volts
	169.68  volts

When the line input voltage to the switching power supply in the
computer is below the voltage that the capacitor is charge up to the
current flow is off.  When the input voltage finally exeeds the
voltage in the capacitor, then current flows and the capacitor is
(re)charged.  For the above numbers with a 120v AC line the capacitor
would be charge to about 170v DC.  Given a large enough capacitor, the
voltage on the capacitor wouldn't sag by more than a volt or two in
the 8ms or so between half cycles.  This means that the input
capacitor would be charging except when the input to the power supply
was within a volt or so of its peak.

If the powersupply was delivering 340watt (at say, 100% efficiency) it
would take an average of 2 amps at 170v.  If the capacitor's charging
only happened for 10% of the time, at best we'd see 20Amp spikes.
Whatever was delivering the power to this power supply would need to
be able to put out these 20+ amps.  This means you need much beefier
circuits than you could get away with if you could spread the load
over more of the cycle.

-wolfgang

(*) or in the case of the 120/240 volt supplies where it feeds into two
    half-wave recifier (aka a left-right) to make 340v DC.
-- 
Wolfgang Rupprecht    <wolfgang@wsrcc.com>     http://www.wsrcc.com/wolfgang/
	  Never trust a program you don't have sources for.