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Re: Bottle Caps, part III



From:	Alfred A. Skrocki [SMTP:alfred.skrocki-at-cybernetworking-dot-com]
Sent:	Thursday, November 13, 1997 10:15 AM
To:	Tesla List
Subject:	Re: Bottle Caps, part III

On Wednesday, November 12, 1997 2:28 PM Greg
[SMTP:ghunter-at-mail.enterprise-dot-net] wrote;

 
> So, if I'm using a 15kv/30ma neon xfmr at 50Hz and fixed series
> gaps, how much capacitance do I really need?

OK Greg, this is how I calculate the size capacitor needed;

First you need to determine your neon sign transformer 
(or other transformer's) Impedence:
                      
               E
        Z  =  ---
               I

  Z = Impedence
  E = volts
  I = current in Amps
  Note: divide milliamps by 1000 to get Amps.

Then in theory he Impedence of the primary capacitor should match the 
Impedence of the transformer at it's operating frequency. Which is given 
by:                                  
                      100000
        C =   ----------------------
              2 x pi x Z x frequency
  
  C = capacitance in microfarads needed for primary capacitor.
  Z = transformer impedence
  pi = 3.141592654

I then use this as a starting point and adjust impericaly by adding
capacitance until I no longer get any improvement in output.

> What would it hurt to "over cap" the system a little?  It seems like
> a little too much capacitance is better than not enough.  A slightly
> oversize cap will drop the maximum voltage a little.  So what?  This
> is easily corrected for simply by closing the spark gaps a little.
> The reduced operating voltage translates into less stress on both the
> neon secondary windings and the caps.  Sounds like a "win-win" deal.
> Is there a flaw in my reasoning here?
 
Well Greg, it works like this - the more capacitance you use the more
current it will pull from your transformer and the more it will drop
the transformers secondary voltage but neon sign transformers have
magnetic shunts that current limit them so what will happen is you will
get more and more output up a point and then it will drop with any more
capacitance, of course that is assuming that the primary was sized 
accordingly.

                               Sincerely

                                \\\|///
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                           Alfred A. Skrocki
                   Alfred.Skrocki-at-CyberNetworking-dot-com
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