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Re: Equdrive circuit, was Re: SGTC
Original poster: "M G" <gt4awd@xxxxxxxxx>
So to sum it up... If using the Equidrive capacitor setup, with a
spark gap or rotary coil, bleeder resistors would be a must. If you
don't have resistors across the caps you should just short out the
spark gap and/or rotary for a quick moment after use? Using Equidrive
capacitor setup with a VTTC should never hold a charge? I'm not sure
of the other coil designs, except for SSTC, which is a lot different
(no equidrive setup?). By the way, is it "Equdrive", or "Equidrive",
for some reason I felt the need to add an "i" to it.
Thanks,
Matt G.
---------[ Received Mail Content ]----------
Subject : Re: Equdrive circuit, was Re: SGTC
Date : Mon, 30 Apr 2! 007 23:17:28 -0600
From : "Tesla list" <tesla@xxxxxxxxxx>
To : tesla@xxxxxxxxxx
Original poster: "Barton B. Anderson"
Hi Phil,
>Why would one cap "conduct" (discharge?) without the current going
>through the other one as well?
It would have to. The caps are series connected through the primary.
If a single cap retains a charge, that charge will be felt across
both caps as soon as alternating current ended. The transformer will
provide R to and will discharge the caps. I was thinking along the
lines of a long discharge situation. But, if we think about DCR of a
transformer (any of the ones we use from NST's to Pigs), it's no
where high enough to cause a large discharge time. As a matter of
fact, it should be immediate. Your right Phil, the cap can't retain a
charge without an ope! n circuit. It's that simple.
I looked at the theory fro m a discharge rate, but didn't throw
numbers at it (my bad). 5RC is a fraction of a second even for a 4500
ohm 12/60 NST secondary. There's no way to retain a charge without an
open circuit condition (floating secondary or not).
I hate to waffle on a subject. That's what I get for trying to figure
out (in what possible way) could there be a retained charge. Then I
thought, it must be a long discharge time and just threw those
thoughts out. But the numbers don't lie and a remaining charge is
only possible in an open circuit. An open circuit can only be caused
by a failure or a series gap configuration.
Take care,
Bart
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