[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

RE: Peak Primary Current

Original poster: Terry Fritz <teslalist@xxxxxxxxxxxxxxxxxxxxxxx>

Hi Mark,

At 08:12 AM 4/20/2005, you wrote:

Antonio, Terry:

Thanks for response.  My testing confirmed Gary's observations(Gary's
tests very interesting).  This contradicts old common "coilers" belief
that heavy gauge wire is required for primary connections.

I see I forgot Pi in the formula(Thanks Antonio for pointing out).
So Imax = 2*Pi*Fres*V*C
This makes Imax Pi times the values I had calculated so current is even
higher than I thought.  I still can't believe I can push that much
current through #14 ga wires with no significant heating!!  I realize
each cycle is only 2uS and ring-up is only 10 uS with 1st notch at 40uS.

Terry asked for Primary Inductance.  My equipment is not accurate at
that low an inductance and I find primary inductance calculations to be
pretty rough, but for my small coil(which I have much data for because
of extensive testing):

Tank Cap is 18 nF and Fres = 383kHz(checked with scope)
So Xc = 23.1 Ohms thus Xprim = 23.1 Ohms
So Lprim = 9.6 uH.
Primary resistance is 0.1 Ohm(meter may not be accurate, but value
negligle compared to Xprim)
So Zprim =  Xprim = 23.1 Ohms

I = V/Z so 20kV/23.1 = 865 amps.
Really turns out to be same formula as above just obtained though X as
opposed to dv/dt.
This assumes voltage drop across spark gap is neglible which I guess is
true once the current flow is established.

Terry - Is this where you were going to take me or was there more?

Thanks.
Mark

Some systems are far less sensitive to primary resistance than others. Coil's with big high inductance primaries can have a lot of resistance. Big coils that have say 4 turns on the primary with a 200nF primary cap are far more sensitive and need very low resistance wiring. In general, most of use don't have to worry too much with primary resistance as long as we don't go too far.

Spark gaps eat up 30 to 50% of a coil's power! When doing power calculations, I just assume 40% is lost in the gap and then the voltages and currents work out pretty well. Sounds like you have a good understanding of everything!!

Cheers,

        Terry