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*To*: tesla-at-pupman-dot-com*Subject*: Re: Balancing L/C Sizes*From*: "Tesla list" <tesla-at-pupman-dot-com>*Date*: Mon, 30 Apr 2001 19:14:25 -0600*Resent-Date*: Mon, 30 Apr 2001 19:18:56 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <AJthCC.A.VVE.87g76-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" <Mddeming-at-aol-dot-com> In a message dated 4/30/01 1:45:49 PM Eastern Daylight Time, tesla-at-pupman-dot-com writes: > > Original poster: "Scott Fulks by way of Terry Fritz <twftesla-at-uswest-dot-net>" < > darkthing-at-earthlink-dot-net> > > William wrote: > > wire gauge shouldn't make much of a difference, since > > you can get twice as many turns with wire that is half > > as thick. Since twice as many turns means four times > > the inductance, while half as thick wire makes four > > times the resistance, the inductance/resistance ratio > > stays the same no matter how small your wire is. > > This is not strictly true, due to the "skin effect" at high frequencies. As > the wire gets thicker, only the outer surface conducts the HF current, so > for thick wires the resistance increases in inverse proportion to the radius > of the wire, rather than the radius squared. Oddly enough, this means > thinner wires are more efficient carriers of current in TC secondaries. > > As an example, at 100 khz, only the outer .2 mm or so of wire conducts > effectively, so any wire gauge heavier than 26 (radius of .2 mm) will start > to lose efficiency. The thicker wire has a lower inductance to resistance > factor (Q) on a given coil form than the thin wire. > > However, most coilers seem to use the "1000 turns" rule, which makes this > argument less relevant. For a fixed number of turns, the thicker the wire > the lower the resistance at any frequency. > > Regards, > Scott Fulks (darkthing-at-earthlink-dot-net) Hi Scott, All! Is High Q really the defintion of operating efficiency in a Tesla coil? Matt D.

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