# 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.

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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