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Re: Unexplained arcing current
Gary Lau wrote:
>
> I just saw something I can't explain. For as long as I've been coiling,
> the resistors in my NST protection circuit have been getting hotter than
> expected. The power defined by I**2 R, I being the NST current, never
> came close to explaining how hot they got, so I assumed that the
> additional wattage was due to a reverse transformer action. After the
> gap quenches, the energy in the secondary could couple back to the
> primary, although it won't be tuned any longer now that the gap is open...
> Not a very good theory, I know.
>
> Tonight I shorted my gap, essentially shorting the NST, but having the
> current pass through my RC network. My R's are each 1.6K, 113W, non-
> inductive. C's are 450pF. Metering the current, I got 73mA (mine is a
> 15KV/60mA NST), and the resistors barely got warm.
>
> Next I removed the short across the gap but broke the connection between
> the tank L and C, so that the gap arcs, but no cap charging/discharging.
> Now the resistors get hot as a pistol in no time. I tried measuring the
> current to the gap, but my digital meter just wigged out, I'm probably
> lucky it still works.
>
> So, what's different that causes the R's to see so much more power going
> to an arcing gap? It's just down to 60 Hz stuff now.
A very possible reason is DC in the resistor due to repetitive firing
at low voltage:
Assuming that the series reactance of the transformer is much higher
than the resistance or the resistors, with the gap shorted you have
only the current limited by the reactance:
Ioutrms=Vrms/(w*L)
Where Vrms is the open circuit rms voltage of the transformer,
w=2*pi*60, and L is the transformer output equivalent series
inductance. Vrms=Vpeak/sqrt(2).
But, if the gap fires at low voltage, instead of the steady-state
current, what appears is a sequence of transient currents.
Assuming a sinusoidal input, and the gap firing at 0 volts
at every cycle (simplification that results in the worst case),
we have:
Vin=Vpeak*sin(w*t)
Iout=Vpeak/(w*L)*(1-cos(w*t)) (cosinusoid + DC)
Ioutrms=Vrms/(w*L)*sqrt(3) (rms value of the above waveform)
The dissipation R*Ioutrms^2 at the resistors is 3 times higher!
Make a test: If the gap fires at high voltage, the current
waveform shall be close to the steady state waveform, and the
resistors would heat normally. With the gap firing at low
voltage they shall heat more, up to 3 times more.
Antonio Carlos M. de Queiroz