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Re: Electrum output impedance (fwd)
---------- Forwarded message ----------
Date: Tue, 14 Jul 1998 13:03:48 +1200
From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Electrum output impedance (fwd)
Greg, Terry,
Thanks for this interesting thread:
> Date: Sun, 12 Jul 1998 10:55:26 -0700
> From: lod-at-pacbell-dot-net
> To: Tesla List <tesla-at-pupman-dot-com>
> Subject: Re: Electrum output impedance (fwd)
>
> Terry Fritz wrote:
>
> > In another post (TC Output Impedance Matching) I propose that the
> > "Load Match Point" in a Tesla coil occurs when the load on the coil is equal
> > to the following equation:
> >
> > Rl = 2 * pi * SQRT ( Ls / Cs ) = 195k ohms
> >
> > For the Electrum:
> >
> > Ls = 0.130 H
> > Cs = 135 pF
> > Xls = 31K ohms (the reactance of the secondary at 38kHz)
> >
> > So the Load Match Point (or Tesla point) calculates out to 195k ohms.
> >
> > By knowing the current to the sphere, the frequency, and the inductance of
> > the secondary, the voltage at the top terminal can be calculated as 1.55
> > million volts in the 100, 20, and 10 uS pictures. By dividing the output
> > voltage by the sphere to arc current, the arc impedance can be estimated as
> > follows:
> >
> > In all cases the 50 amps in the secondary multiplied be the reactance of 31K
> > ohms gives an output voltage of about 1.55M volts.
> >
> > 100uS Io = 15A Zarc = 103k ohms
> >
> > 20uS Io = 5A Zarc = 310k ohms
> >
> > 10uS Io = 8A Zarc = 194K ohms
> >
> > You will notice that these numbers average out very close to the
> > calculated value of 195k ohms. I suspect that the arc may begin at the
> > 1.55M volt level but the voltage quickly reduces to ~750k volts and matches
> > the arc impedance. I don't know if the coil finds this point naturally or
> > if the Electrum was designed to operate this way. The close match of the
> > Electrum to these calculated numbers is obviously very interesting!
>
> This is an excellent analysis of the coils' output. I wonder
> though about the "2 * pi" in the top equation. I think that
> a rational constant, perhaps somewhere between 4 and 6, is a
> closer fit. Electrum's secondary was designed to be about a
> factor of three (not pi) below 100K ohms, which was my initial
> guess for the impedance of a burning arc that size.
> Given that, a value of 30 to 35K ohm for Xs would allow the
> sec to develop a high voltage initially, then act a reasonable
> current source as the burning arc drops in impedance.
>
> An arc will find the max power point naturally, provided that
> the coil characteristics are "in the right ballpark." The design
> effort on Electrum was to simply place the coil specs within
> this ballpark.
>
> I was not satisfied with the original design, which used 10AWG
> in the secondary, since it yielded only 19K ohm for Xs. Going
> to 12AWG on paper did not appreciably change the skin losses, and
> raised the (calculated) Xs to 36K ohm, so I went with the #12.
>
> I don't know if it's useful to try and define "Load Match Point",
> since the load has a negative resistance curve, and the coil will
> fall into it naturally. Perhaps a "Load Match Range" would be better.
It would be most interesting to see the effect of a relatively low
source impedance on the resulting arc. I suspect the arc would be very
bright. One route to this would be a large topload with a low
inductance secondary. A practical "garage sized" coil would then be
operating somewhere above the low frequencies encountered in the
largest machines. Must try. I made a suitable toroidal topload not
long ago and so far it's gathered dust. I have a space wound 6.5"
secondary with an h/d around 5 that could be useful in such a test.
Cs(total) would have to be restricted to allow sufficient voltage
production from a reasonable Ep. It could be most instructive to
compare a couple of secondaries, one with low inductance and one with
high but of the same physical size and with primary inductance changed
to compensate.
Malcolm