Re: peak current when spark gap fires

["Tesla list" <tesla@xxxxxxxxxx>]

Original poster: John <guipenguin@xxxxxxxxx>

Thanks Terry,

although I'm still a little confused......I guess I have a /lot/ more 
homework to do before I even power on my coil.

If you have 172.8 watts at 16,970V,    and current = power / 
volts    then wouldn't current in amps = 172.8 / 16970 ?





On 10/22/06, Tesla list <<mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx> wrote:
Original poster: Vardan < 
<mailto:vardan01@xxxxxxxxxxxxxxxxxxxxxxx>vardan01@xxxxxxxxxxxxxxxxxxxxxxx>

Hi,

The coil power is the energy in the cap multiplied by the BPS:

Power = 1/2 x C x V^2 x BPS = 1/2 x 10e-9 x 16970^2 x 120  ==   172.8
watts.  If you have a 12,000V 30mA NST that is about right.

We draw more power than one might think since the cap values we use
force more current than normal out of the NST.  The actual AC line
current is still higher since the power factor is typically not real good.

The current does not indicate power.  If we remove the coil and just
dead short the cap at 120BPS, the current might be like 10,000 amps!!!

Cheers,

         Terry


At 06:03 PM 10/22/2006, you wrote:
>I don't understand how it would be 245 amps every time the spark gap
>fires.... it would make sense in an energy storage bank because the
>capacitors would have to take time to charge, but at 120 bps? Thats
>more energy then your putting in.
>
>
>
>On 10/21/06, Tesla list 
<<mailto:tesla@xxxxxxxxxx ><mailto:tesla@xxxxxxxxxx>tesla@xxxxxxxxxx > wrote:
>Original poster: Vardan
><<mailto:vardan01@xxxxxxxxxxxxxxxxxxxxxxx ><mailto:vardan01@twfpower 
electronics.com>vardan01@xxxxxxxxxxxxxxxxxxxxxxx >
>
>Hi,
>
>This page explains the SQRT(2) factor and how it is found:
>
>< http://www.ee.unb.ca/tervo/ee2791/vrms.htm> 
<http://www.ee.unb.ca/tervo/ee2791/vrms.htm>http://www.ee.unb.ca/tervo/ee2791/vrms.htm
>
>Cheers,
>
>          Terry
>
>
>At 10:27 PM 10/20/2006, you wrote:
> >Say was does the SQR of 2 come from. It easy to imagine .707 * peak or
> >rms/.707 hmm Root Means square...to the Inet. Never mind the math gets out
> >there for a "collection of n".
> >
> >Elementary but I don't remember covering it that way.
> >
> >-----Original Message-----
> >From: Tesla list [mailto: tesla@xxxxxxxxxx]
> >Sent: Friday, October 20, 2006 6:10 PM
> >To: <mailto: tesla@xxxxxxxxxx><mailto:tesla@xxxxxxxxxx> tesla@xxxxxxxxxx
> >Subject: Re: peak current when spark gap fires
> >
> >Original poster: Vardan <
> 
<mailto:vardan01@xxxxxxxxxxxxxxxxxxxxxxx><mailto:vardan01@xxxxxxxxxxxxxxxxxxxxxxx> 
vardan01@xxxxxxxxxxxxxxxxxxxxxxx>
> >
> >Hi,
> >
> >There is a list of such formulas here:
> >
> > 
<http://hot-streamer.com/temp/FormulasForTeslaCoils.pdf>http://hot-streamer.com/temp/FormulasForTeslaCoils.pdf
> >
> >The formula you want is at the top of page 6.
> >
> >However, you already figured it out almost correctly ;-)  The thing I
> >would change is the actual voltage the gap fires at.  If your NST is
> >rated at 12000V RMS then the peak voltage will be 12000 x SQRT(2) =
> >16970 V.  So the primary peak current is probably 245 amps.
> >
> >Cheers,
> >
> >          Terry
> >
> >
> >At 09:23 PM 10/19/2006, you wrote:
> > >Hello. I am new to a lot of this, so I want to make sure I got this
> > >correct. Did I do this right to find the peak current when the spark
> > >gap fires? Assuming the only inductance in the calculation is the
> > >primary coil....
> > >
> > >Surge impedance = sqrt(Lp / Cp)      my primary is around 48.029uH
> > >according to calculations, and capacitor is
> > >0.01uF    so  sqrt(0.000048 / 0.00000001F ) = 69.282     and then
> > >current peak = Vp / Surge impedance  so  12000 / 69.282   = 173.205
> > >amps?   This is most likely way wrong :P
>
>