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

[rbethman]

> 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 only point here that you and I disagree on is the relationship.  It is
"very" definitive.  I'm ignoring core saturations.  I never got into load.

>
> >  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.

Not so.  Power is not dependent on "time".  The power delivered during that
pulse is full, not half.  P=EI (Minus the long version of E-line I-line
cosine Theta [we are talking single phase]).

The only time that "time" becomes a factor in power is when you pay for it
based on watt-hours or kilowatt-hours if you prefer.

>
> 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..."
> _______________________________________________
>

We just have to adjust the figures for the full power delivered during this
pulse.

"They're making the first turn...."

Bob