# Re: Electrum output impedance (fwd)

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---------- Forwarded message ----------
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.
--

-GL
www.lod-dot-org

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