"Ken Scharf" wrote in message
.. .

That Steve N. wrote:

Paul,
Welcome to the world of design...
No need to make a load to measure the ripple on the caps. You can
calculate it. ...
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....


Saying:
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).


Ken,
This is true only without the capacitor filter. When you add a capacitor
input filter, the diodes (ideal or not) only conduct for a rather short part
of the input cycle. Line current is in the form of rather short current
pulses as a result. This is a concern for the power company, by the way.
It also keeps the power company wave form rather flattened on the top,
because of all the supplies on the grid drawing current only there.

You can see and measure this on a supply if you have the capability. The
cap will charge, pretty much to the peak. When the sine wave then drops
rapidly toward zero, the caps "hold" that voltage and are now are supplying
the load current. The voltage on the cap drops according to the formula.
(the action of the regulator keeps the current at this point constant, so
*THAT* formula applies)


Also we have a SINE wave output,

Not once you add the capacitor input filter. You need the cap
filter to (help) keep the output at the constant DC we desire.


but in a FW the negative cycle is flipped up.

*** Yes and therefore you have 120, half cycles per second (in the US).
That is 8.33 ms period (one peak every 8.33 ms). Draw a full wave
rectified waveform, then plot some capacitor descharge on it and you will
see that for a peak-to-peak ripple of, say, 10% the time the cap is charging
is rather short. This is the diode conduction time.


So your formulas don't reflect the varying duty cycle out of the
rectifier (it's not ON 5 ms and off 3 ms,

I did mention that the diode conduction-angle duty-cycle does vary, but
this is due to load current varrying - and it is predictable We only have
8.33 ms per pulse and the conduction angle is only 1-5 ms. depending on
load. This *IS* the way linear supplies work. In theory and practice.
That's why *that* formula works. If there was current in the diodes all the
time, it would imply thta your capacitor filter ripple would be the full
supply voltage.


but goes from 0 to full voltage following a sine function).

Again, this is only without a capacitor input filter, if you refer
to the output voltage.


I guess you can
use the RMS function to determine the "effective" on and off times, and

No, this does not apply to this situation. RMS refers to the
effective voltage, not diode conduction times.
If you look at the *sine wave* waveform itself (full-wave rectified, in
this case) then you can determine the charge and discharge waveforms and
that is what I have described.
If you can get a scope and do some real measurements for yourself, you'll
see I am correct. For light loads on the power supply, the diode current
pulses are quite short ~~ 1 ms and go only to around 5 with a heavy load.



your guess is probably ballpark enough to be correct.

No guess here. What I described *is* the way it works. We do it
when designing linear supplies all the time. You can see it with a scope.
There are drawings on the web to show you what I tried to describe in words.

I didn't read all of the following web pages, so one may be easier to
understand than the others.

The first shows *JUST* the rect waveform of which you speak:
http://en.wikipedia.org/wiki/Rectifiers

However, the rest of these show a capacitor input filter and the resulting
waveforms and conduction characteristics I described.

The following has a movie showing very well and slowly, how the "Smoothing"
capacitor works: http://www.greenandwhite.net/~chbut/rectification.htm

Ignore all the formulas in the following and about 2/3 of the way down are
waveforms:
http://www.visionics.ee/curriculum/E...ectifier1.html

Perhaps a simpler explanation:
http://www.play-hookey.com/ac_theory/ps_filters.html

I applaud your desire to help Paul's understanding, but you missed here.
73, Steve, K9DCI



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