[R-390] 6080 in place of 6802 - RMS ???

[twleiper@juno.com]

On Thu, 10 Jan 2002 10:24:00 -0800 "Roger L Ruszkowski"
<rlruszkowski@west.raytheon.com> writes:
> Try the following ideas on for size.
> 
> Bench test #1
> Set your RS 12 volt 1 amp transformer (with a center tape secondary) 
> up on
> the lab bench.
> Clip your single power diode to one 12 volt secondary lead
> Clip your filter cap to the diode output and the other 12 volt 
> secondary lead.

Don't you mean the center tap? If not, then you are not in
agreement with your conclusions:

> With the 12 watt 12 volt 1 amp transformer,
> a half wave 1 diode rectifier will yield 3 watts (6 volts .5 amps)

Across the full secondary, that voltage above would be double.

> a full wave  2 diode rectifier will yield 6 watts (6 volts 1 amp)
> a bridge 4 diode rectifier will yield 12 watts (12 volts 1 amp)

In addition, what is the purpose of the filter cap? With a resistive
load does it really matter? Would the integration of the half wave
output really change anything, except to throw the voltage and
current out of phase? Or is the idea to tune out the transformer
inductanceI thought the power factor only affected the efficiency
of inductive loads.

In any event, center-taps and filter caps aside, I still believe that
in the 6080 discussion as proposed, insertion of a diode will only
reduce the power consumed by half instead of two thirds. I just
don't understand how you can restore half the AC waveform
and get three times the power, which is the corollary to the diode
insertion argument. I fail to (again, in a resistive load) see how
the current flow in one direction has any relationship at all to the
current flowing in the other direction, and what the elimination
of one could possibly have to do with the other.

If you agree that no power is being delivered at the moment of
zero crossover in the waveform, any other conclusion would have
to be flawed. How about THAT one?

Tom :-)