Steve N. wrote:
wrote in message
oups.com...

You're probably right! I did actually try to rip the heating element
out of a 2kW electric fire to use as just such a test load, but the
manufacturers had used some really devious sort of screw heads to keep
the two casings together. :-(
Failing some serious resistance wire, the only other 'to hand' option
is to hook up the HF mobile and key-up; but I don't have a sufficiently
gutsy dummy load or antenna to dump 100W into, I'm afraid. :-(




Paul,

Welcome to the world of design...



No need to make a load to measure the ripple on the caps. You can
calculate it. I did this long ago for a 5V 28 amp supply I re-wound from a
28V 5A supply.



You have to know that for a capacitor, I = C * Dv / dt.

When the 60 HZ wave drops, the diodes drop out of conduction and the filter
cap is now supplying *all* the load current...and the voltage drops in the
usual capacitive nature following that formula.

I don't recall if you said the current, but...if you will be drawing, say 20
amps - you have the "I". If you have a "C" you plan to use from the old
design, you have "C". Dt is the discharge time between the (full-wave)
peaks from the bridge rectifier. This is 8.333 ms. minus the conduction
time. I forget the typical conduction time I have seen, but I'll assume 5
ms max-load (the diodes don't conduct very long, typically.and.and the peak
current is pretty high to boot!) -- leaving 3ms for the discharge time.
Then Dv is the ripple, or more accurately, the sag. Also, consider that the
peak voltage will be lower when loaded due to the transformer winding
resistance and diode drop and anything else in there.

Since this is a full wave rectifier if we had ideal diodes they would be
conducting all the time (IE: one SET of diodes if a bridge or one diode
if a "CT" type rectifier). Also we have a SINE wave output, but in a
FW the negative cycle is flipped up. So your formulas don't reflect the
varying duty cycle out of the rectifier (it's not ON 5 ms and off 3 ms, but
goes from 0 to full voltage following a sine function). I guess you can
use the RMS function to determine the "effective" on and off times, and
your guess is probably ballpark enough to be correct.





So, solving for Dv. Dv = (I* dt ) / C



If you wanted 20 amps and had 20,000 uF, You'd have (20*.005)/.02 or 3
volts of sag.



BUT! This assumes *normal* line voltage. If you DO measure as some of the
others suggest, remember that you'll be using whatever line voltage is at
the time. Consider when there is a brown out. Pick a low-line voltage, say
105 volts and use THAT voltage, Or the ratio 105/120 and adjust your output
voltage number accordingly when calculating things.



You have to make sure that *Everything* that you want to power from this
voltage has enough voltage at the low point; series pass & Driver & driver's
driver (perhaps the 723).



Finally, what I did to keep the required overhead to a minimum and, as a
result, the dissipation in the series pass transistors lower (and guarantee
poorer brown-out protection, unfortunately), I *added* a few extra, lighter
gauge wire turns to the *OUTPUT* winding to supply the 723 and driver stuff.
Add the same number of turns to each side of the secondary (put one and
measure the volts-per-turn to figure out how many) and put a diode from both
of them (pointing to) to another smaller filter cap. This requires, of
course two more diodes and filter cap, but they're smaller. You also have
to watch the ripple in the same way.



Hope this helps. Let me know @arrl.net



73, Steve, K9DCI