[Ian Iveson]

Looking at the content of a rectified sine wave, this is
what I get, where F1 is twice mains frequency, and values
are harmonics of F1 as proportions of rms source V:

Mean voltage Vdc: 0.9 - diode drops
Ripple Vrms:
F1: 0.424
2H: 0.085
4H: 0.032
6H: 0.016
8H: 0.0086
10H: 0.0045

Leaving around 0.014 unaccounted for, widely and
comparatively evenly spread over the next 30 or so
even harmonics.

Note that AFAICS those first few harmonics are not
significantly changed by diode drop as long as it's less
than about 15% of transformer secondary voltage. The higher
harmonics are reduced.

All frequencies lead the mains sine by 90 degrees, such that
when the rectified sine is at zero, the fundamental and all
harmonics are at their peak negative values, which should
sum
with the mean voltage to give zero.

Currents and voltages at any point in a power supply circuit
can be calculated by analysing for each frequency and
combining the results. The algebraic formula for a
generalised circuit is a bit long
but once done, a spreadsheet can be constructed so different
component values can be tried easily.

Here's a reminder of how to deal with vectors:

http://www.kwantlen.ca/science/physi...s/Phasors.html

Don't forget the effective series resistance of the
transformer, which can be estimated from its rated current
and regulation.

Ian