[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*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 11:35:03 -0600*Resent-Date*: Mon, 30 Apr 2001 11:42:30 -0600*Resent-From*: tesla-at-pupman-dot-com*Resent-Message-ID*: <bUSiPB.A.-NF.CQa76-at-poodle>*Resent-Sender*: tesla-request-at-pupman-dot-com

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)

- Prev by Date:
**New web pages added** - Next by Date:
**Re: Basic TS curcuit conflict** - Prev by thread:
**Re: Balancing L/C Sizes** - Next by thread:
**Re: Balancing L/C Sizes** - Index(es):