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Re: Solving the DC coil mystery



Original poster: "Kevin Ottalini by way of Terry Fritz <twftesla-at-qwest-dot-net>" <ottalini-at-mindspring-dot-com>

Steve:
        Sorry for the minor typo there ... the current is ~0.292amps and I
left an
earlier number (.250 amps) in the calculation ... only a small difference
though,
it raised the loss to 426watts (from 425) and lowered the efficiency to
87.8%.

The scenario here is that the power supply is charging a purely capacitive
load
(the tank cap) so there is no resistive divider present to decrease gross
efficiency or limit the max voltage going into the capacitor.

Certainly, if I was only running 50% efficient my resistors would go out in
flames, but this is not the case (sorry Bert!).

Given long enough, the capacitor would charge to the full output voltage as
if
the resistor wasn't there at all.  The only losses are dynamic ones caused
by
the current passing through the resistor, and mostly only during the peak
charge current phase when the capacitor is starting up from ground.

In reality, the capacitor is never fully discharged while running so the
resistor
never usually sees the absolute max possible current, although I see
something a little worse during the discharge itself and during ringing.

You can see the schematic for my system here:
ftp.mindspring-dot-com/users/ottalini/highvoltage/DCCOIL/DCSCH2A.PDF

Since this is a working system and the resistors don't get hot, the caps
and diodes are cool,  the caps charge up to around 20KV if I don't spin
the ASRG and I do run at power ranges up to 4Kw, I can only say that
conventional wisdom (and reality) holds true.

Best,
        Kevin

----- Original Message -----
From: "Tesla list" <tesla-at-pupman-dot-com>
To: <tesla-at-pupman-dot-com>
Sent: Wednesday, July 25, 2001 6:19 AM
Subject: Re: Solving the DC coil mystery


> Original poster: "S & J Young by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<youngs-at-konnections-dot-net>
>
> Kevin,
>
> Very interesting.  But how do you get much better than 50% efficiency?
That
> does not agree with Bert Hickman's proof that half the power is dissipated
> in the series resistor.  Why do you use 0.250 amp in your resistor power
> loss equation?  If you meant 0.292 amps, that is the same mistake I made -
> one can't use the average current to calculate the resistor loss.  If Bert
> is correct, and I believe he most certainly is, then your resistors are
> dissipating more like 1,750 watts if you are charging your tank cap
through
> them.  Maybe your circuit setup is different than I am imagining.  Can you
> do an ASCII rendition of it or let us know where it is posted?
>
> --Steve