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Re: Help in calculations


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>From mmdf-at-wwnet-dot-com Thu Apr 25 01:13 MDT 1996
>Received: from main.wwnet-dot-com (main.wwnet-dot-com [206.62.36.1]) by uucp-1.csn-dot-net (8.7.5/8.7.3) with SMTP id UAA21971 for <tesla-at-grendel.objinc-dot-com>; Wed, 24 Apr 1996 20:15:24 -0600 (MDT)
Date: Wed, 24 Apr 1996 20:04:06 -0700
From: Skip Greiner <sgreiner-at-mail.wwnet-dot-com>
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To: tesla-at-grendel.objinc-dot-com
Subject: Re: Help in calculations
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tesla-at-grendel.objinc-dot-com wrote:
> 
> >From jim.fosse-at-bdt-dot-com Wed Apr 24 01:16 MDT 1996
> >Received: from bdt.bdt-dot-com (root-at-bdt-dot-com [140.174.173.10]) by uucp-1.csn-dot-net (8.7.5/8.7.3) with SMTP id UAA29576 for 
<tesla-at-grendel.objinc-dot-com>; Tue, 23 Apr 1
> From: jim.fosse-at-bdt-dot-com (Jim Fosse)
> To: tesla-at-grendel.objinc-dot-com
> Subject: Re: Help in calculations
> Date: Wed, 24 Apr 1996 00:49:33 GMT
> 
> On Tue, 23 Apr 1996 08:30:09 +0700, tesla-at-grendel.objinc-dot-com, you
> wrote:
> 
> [snip]
> >My question(s) to the group are for help in trying to determine the
> >voltage across the primary capacitor at any time "t". My problem is how
> >to handle the current limiting / impedance of the neon and how to fit
> >this into the classical capacitor charging time constant.
> >
> Skip,
>         I'd suggest that you treat the neon as a current source. No
> hard info here, just WAG.

Jim,
I accept that the neon be treated as a current source, but how do you 
rewrite the classical voltage vs time charge equation to use a current 
source. I know its more complicated than just replacing the "e" (charging 
voltage) with 120ma and a "z" calculated from the neon's voltage and 
current capacity.

> 
> [snip]
> > Also if I know the the
> >maximum voltages to which I can get the cap charged, it becomes trivial
> >to determine if additional fires per half cycle will gain any increase in
> >power transfer.
> power available:        15kV * 0.120 =  1.8KVA
> power used:
>                 1/2 * c  *  V^2 * pps =
>                 .5 * 0.021*10^-6 * (0.707 * 15kV)^2 * 120 = 142 watts!

I like your math. This means that I am getting discharges at the rate of 
about 3 watts per inch. Tesla never did that well. 

But seriously...This can't be right. Something is missing and I can't put 
my finger on it.


Skip



> >
> >Why do all this? Basically I believe that bigger discharges can be wrung
> >out of neons and this may be one way of doing it.
>  You are only using 12.7% of the power available in your neon. You
> should be able to increase the firing rate by a factor of 10. However
> the LC time constant of the neon secondary may limit this. Combine
> this with the first and last 45 degrees of each half cycle being low
> voltage may limit the total power that you can extract.
> >
> >My present system using the above mentioned components and a .021 cap
> >delivers 54" point to point discharges. Since I have gone to the
> >synchronous gaps I have not blown up any more neons or capacitors either.
> >
> >I would appreciate any help that anyone in the group can offer.
> >
> cheers,
> 
> jim




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