# Re: Safety gap issues

```Original poster: Terry Fritz <vardin@xxxxxxxxxxxxxxxxxxxxxxx>

Hi Jim,

At 03:03 PM 11/26/2005, you wrote:
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```.......that's the max voltage for smooth sphere with nothing around it).
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I have noticed that the "breakout voltage" does tend to correspond to the radius of curvature and all. However, the top voltage can then go substantially beyond that. If there is a lot of power behind the arcs. Then the "breakout loading" is just not enough to hold the voltage down. So it does not act like say a hard Zener diode, but rather a Zener with a big resistor in series with it.
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Hmmm... That is.. the secondary inductor is still putting current/charge into the topload, and the charge isn't being lost fast enough through corona and streamers. I suppose there's also a "inductance of the topload" effect that would slow down the movement of charge off the topload.
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So. the voltage would rise rapidly to breakout, and then, keep rising, somewhat slower, after that?
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Yes. Once the streamer starts up and begins to drain power (current) from the top terminal, the rate of voltage rise slows down from what it would normally be. However, the coil can put power in faster than the streamer takes it out. I think looking at "Q" at that point has the most meaning.
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If you imagine a coil that coil easily hit a top voltage of say 250000 volts but had a 4 inch diameter top terminal, the terminal voltage would be 2 x 2.54 x 30kV/cm = 152400 volts If the terminal was say 30pF of capacitance we can find the energies:
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E = 1/2 C x V^2
E1 = 0.5 x 30e-12 x 250000^2 = 0.937J
E2 = 0.5 x 30e-12 x 153400^2 = 0.348J

If the voltage were held to 153400 volts, the losses would be very high.

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The first cycle loss would be E1 - E2 = 0.590J or a "Q" of 0.937 / 0.590 = 1.59
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If the Q losses were that great, all of our pretty measurements would show drastic losses very early in the firing cycle which they do not.
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Right now, the 220K + 1pF/foot streamer load model seems to give the loading sort of OK. Maybe those working on the streamer models can give us better numbers.... But the radius and breakout voltages are not super critical IMHO.
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Cheers,

Terry

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