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Re: Peak Primary Current

Original poster: "Jim Lux" <jimlux@xxxxxxxxxxxxx>

----- Original Message ----- From: "Tesla list" <tesla@xxxxxxxxxx> To: <tesla@xxxxxxxxxx> Sent: Wednesday, April 20, 2005 9:14 AM Subject: RE: Peak Primary Current

> Original poster: "Mark Dunn" <mdunn@xxxxxxxxxxxx> > > > 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.

So the duty cycle is quite low... 40E-6 seconds of current flow with 8666E-6
seconds between bangs.  Ball park there is about 1/200..  Even though the
power is the current squared, the small duty factor (0.005) means that
average power dissipation is quite low.  For instance, if your rms current
(during the bang) is 100 A, and the resistance of the wire is 0.01 ohm, the
peak power dissipation is 100*100*.01 = 100W.  The average power dissipation
would be only 1/2 watt in this case, though.

> > 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 is the peak current on the first half cycle.  If the pulse were a nice
rectangular envelope (which it is not), you'd still have to divide by 1.414
to get the rms current, which is what you should be using for power
dissipation calculations. There are handbook tables for damped sinusoid rms
values (or you can solve them from first principles... it's
I=exp(-k*t)*cos(omega*t) for a damped sinusoid in an RLC)... watch out
though, TC's don't necessarily have nice exponential decay. See below.
> This assumes voltage drop across spark gap is neglible which I guess is
> true once the current flow is established.

Actually, this is pretty far from the truth.  The voltage drop is probably a
hundred volts or so (perhaps more, particularly in a segmented gap).
There's an initial "cathode drop" of 20-30 Volts (depending on the materials
of the electrodes and what gas it is) plus some ohmic loss in the spark
column.  So, the gap's actually dissipating a fairly high power (at least
peak wise...)  Using your example of 865 Amps, a 100V drop will be a peak
dissipation in the gap of some 86 kW, albeit lasting only a few tens of
microseconds, so the total energy is fairly low.. Using a duragion 30 uS as
an example, and assuming constant current during that time (not a valid
assumption, but an overestimate!)  about 2.5 Joules per bang lost in the gap
(i.e. 300W, average)

This, in fact, is probably the single biggest limitation on efficiency in
spark gap coils.
There are some models of this in the pupman archives.
> Terry - Is this where you were going to take me or was there more?
> Thanks.
> Mark