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Re: Re: Maxwell]
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- Subject: Re: Re: Maxwell]
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- Date: Thu, 10 Nov 2005 19:51:34 -0700
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Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>
At 07:58 AM 11/10/2005, Tesla list wrote:
Original poster: "D.C. Cox" <resonance@xxxxxxxxxx>
Can you really assume your current pulse is 50 uS long? Wouldn't it
depend on the effective quenching time, ie, knowing when the arc
strikes as the rotary electrode approaches the stationary electrode
(depends on sharpness of edge of electrode and electrode operating
temp) and when the spark extinguishes?
No... I was just coming up with some plausible numbers to illustrate
why a 1 amp supply can have an RF RMS current substantially
higher. I picked 50 uS because I figured it would take 5 cycles of
a 100kHz damped sinusoid to ring down, and then made the simplifying
assumption that the amplitude is constant during the 5 cycles.
5 cycles? A round number that seems to be plausible.
In reality, it's different, but probably not by huge amounts, and I
didn't want to bother trying to do the exact integral.
Original poster: Terry Fritz <terry-cpu4@xxxxxxxxxxxxxxxxxxxxxxx>
In a message dated 11/6/05 7:54:37 PM Eastern Standard Time,
Indeed... and that's why you can get a whole lot more than 25 A rms
through a cap from a power supply that puts out only an amp at HV
DC. The RMS counts the *square* of the current, so, to just use a
simple example with plausible (but wrong) numbers:
Assume you've got a pig type supply at 10kVA, putting out 0.5 amps at
20kV. RMS current is 0.5A.
Let's say you're firing at 100 breaks per second (i.e. every 10 ms),
and the RF pulse lasts 50 microseconds. To a first order, then, the
current through the capacitor is going to be 10000/50 * 0.5A during
the pulse and zero at other times. That is, you'll have a series of
100 Amp pulses with a 0.5% duty cycle.
Jim, I've been trying to figure this one out over the past few
days. Please forgive my little brain, but: How did you arrive at
the "10000/50 * 0.5A" for the cap pulse current?