Re: looking for inductance formulae that include resistance effects

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Original poster: "Jim Lux" <jimlux-at-earthlink-dot-net> 


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From: "Tesla list" <tesla-at-pupman-dot-com>
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Sent: Sunday, October 03, 2004 11:47 AM
Subject: Re: looking for inductance formulae that include resistance effects


 > Original poster: "Jan Wagner" <jwagner-at-cc.hut.fi>
 >
 >
 > On Sat, 2 Oct 2004, Tesla list wrote:
 > >Original poster: "Jim Lux" <jimlux-at-earthlink-dot-net> I'm looking for
 > >"handbook" type formulas (that I can put into a spreadsheet
 > >fairly easily) that can calculate not only the inductance of a flat or
 > >helical coil, but also the resistive loss (which, of course, will vary
with
 > >frequency).  Preferably not something that uses a big table of constants
 > >(like Medhurst for capacitance) and not that requires running a numerical
 > >integration (like INCA).
 >
 > If you know the tube material, dimensions, and freq, you could get a good
 > estimate of the AC resistance by using the formula for skin depth. As
 > derived in, for example:
 >
http://www.physics.uq.edu.au/people/ficek/ph348/sols/sol12/node2.html#SECTIO
N00011000000000000000
 > (and asserting that the assumptions made in the derivation are valid in
 > your case too, e.g. diameter of the conductor is large compared to the
skin
 > depth, etc)

Several classic assumptions in the standard formulas aren't met:

1) The diameter of the conductor is not << 1 skin depth, so the assumption
of an annular ring carrying the current isn't valid.

2) This doesn't take into account the proximity effect (current in adjacent
turns results in a current density that is not uniform around the conductor.

For instance, calculating the Q of an inductor that's, say, 10 cm in
diameter, 10cm long, with 20 turns wound with 1mm radius Copper wire, for
3.5 or  7MHz using the "tubular conductor" approximation, you get a Q of
several thousand.  Since "real" coils of this dimension have Q's in the
several hundred range, there's something that's not quite right (and, it
could, in fact, be my calculations..)


 >
 > Is this what you were looking for? Or a more "precise" formula?
 >
 > cheers,
 >   - Jan