[R-390] Exhumation of dead horse...

[twleiper@juno.com]

On Fri, 11 Jan 2002 13:38:50 -0500 "rbethman" <rbethman@home.com> writes:
> Tom,
> 
> I've read the threads of this "horse".  If the AC wave form 
> halves have no relationship to each other, you have rewritten
> every text on AC generation and theory.  IMHO that just
> does NOT ring true.  Their is a DEFINITE relationship between
> the halves.

That is true, but it is a comparative relationship, not an "affective"
relationship, meaning, if you remove one of the halves, it doesn't
somehow change the power delivered by the other. Now you might
want to argue about transformer saturation and load, for instance
a heavily loaded transformer would output a higher amplitude half
cycle if there was no load on the other cycle, etc. And you can
also calculate in the forward voltage drop of the diode, etc. But
for the purposes of this argument (this IS the correct room for an
argument, isn't it Mr. Barnhard?), I am assuming no such ancillary
factors. Only the power delivered by a half wave instead of the
full wave.


>  The addition of a diode in an AC line changes a lot of things.
>  The first being that you only get ONE pulse per second in lieu
> of TWO pulses per second.  The voltage or shape of that
> SINGLE pulse does not change from the voltage or shape of the
> TWO...
> If 26 VAC is passed through a diode, then 26 VDC comes out - BUT 
> only half the time that the 26 VAC did.  The waveform is still
sinusoidal 
> in shape, just missing the other half of the complete cycle.

My point, exactly. So half the power is delivered, not a third. And the
desired result is to deliver a third.

Let's agree on a couple things:

1. If you power the 6080's (in series) with 25V they will (briefly)
dissipate 120 watts.
2. If you power the 6080's with 12.5 volts they will dissipate 30 watts,
and
this would be the ideal voltage with which to power them in series..
3. The diode cleanly removes half of the AC waveform.
4. No power is transferred or somehow "stored" at zero crossover.
5. The filament resistance at temperature is constant, or insignificantly
affected by the 50% duty cycle change from 60 hz to 30 hz.

Therefore:

If you insert a diode, you dissipate 120 watts half the time and 0 watts
for the other half of the time (Refer to your own paragraph above).
The average dissipation is thus 60 watts, and the instantaneous current
during the "powered" half cycle is twice what it was designed to be.

The duplicate key exists, and it was only through the disloyalty of my
officers, that the idea of inserting a diode was proposed in the first
place.

> Whatever voltage drop occurs is caused only by the ohmic losses
> of the junction of  the diode. That is unless you put in a Zener diode.
> Then we have an entirely different animal.

Ancillary stuff... .7 volt forward drop for the silicone diode, big deal.

> 
>     Go for broke - beat the horse - he has been exhumed!

"And...they're off..."