Franc Zabkar wrote:
On Wed, 29 Apr 2009 12:28:36 +1000, Sylvia Else
put finger to keyboard and composed:

Franc Zabkar wrote:
On Sun, 26 Apr 2009 22:48:56 -0700, UCLAN put
finger to keyboard and composed:

There is a big 400vdc capacitor (or two 200vdc caps in series) just after
input rectifiers on the AC input. Note C5 and C6 on:

http://www.pavouk.org/hw/en_atxps.html

They charge to the peak value of the input
AC voltage, or 1.414 times the RMS value. Since the cap(s) draw their maximum
current when at lowest charge (zero cross-over point), and draw their least
amount of current when charged to their highest point, the current waveform
*leads* the voltage waveform by 90 degrees. [Maximum current is at the same
time as minimum voltage; minimum current is at the same time as maximum voltage.]
As much as I dislike the man, he's right.

Look at Fig 6 on page 2 of the application note you linked to
elsewhere in this thread.
It does show that the current leads - look at how it's not positioned
symmetrically about the 90 degree point, for example. It's not a 90
degree lead, of course, but it's still a lead.

A bigger capacitor would result in less ripple, which means less lead.
If you were relying on intuition alone, you would expect that the PF
would move closer to unity.

In fact the PF actually becomes worse.

With a larger capacitor, you get less lead, but higher instantaneous
currents. My intuition would be that the net result is far from obvious.

However, the higher current for shorter periods means higher harmonic
currents, which don't contribute to the power, so it wouldn't surprise
me that the PF would go down.

Sylvia.