Using skin effect

To: tesla-at-poodle.pupman-dot-com
Subject: Using skin effect
From: Tesla List <tesla-at-stic-dot-net>
Date: Sun, 27 Apr 1997 23:38:29 -0500
Approved: tesla-at-stic-dot-net
Subject:  Re: Using skin effect
  Date:   Sun, 27 Apr 1997 15:48:15 +0500
  From:   "Alfred A. Skrocki" <alfred.skrocki-at-cybernetworking-dot-com>
    To:   Tesla List <tesla-at-pupman-dot-com>


On Tue, 22 Apr 1997 08:03:26 +1200 Malcolm Watts

> HI Alfred, all,
>                 One skin depth in copper is 66/SQRT(f) mm.

I had checked my notes and found the equation 2.6/SQRT{f} inches.
Which when multiplied by 2.54 to convert to mm. yields 66.04/SQRT{f} 
mm. which compares favorable with your equation.

> The definition of a skin depth is that distance inwards to the point 
> where current has dropped to 1/e of its total value. The equation 
> above has resistivity factored in so will be different for silver.

Hmmm, in my notes the equation is listed as being independent of 
materials or resistance.

> In fact, the skin depth on silver is not as deep due to the higher 
> conductivity. If a silver film is not thick enough I have heard
> things could actually be worse than bare copper.

This again compares favorably with my notes. I have an entry that 
showed that starting with bare copper and plating a progressively 
thicker layer of silver yielded a bell shape curve in regards to 
resistance above 100 Khz. I think that in plating silver onto copper 
that a boundary layer is formed at the point where the silver is in 
contact with the copper and it is this layer that has the higher 
resistance.

> And it still requires enough conducting cross-sectional area to carry
> the current without significant heating (resistance).

I would think that the {I^2}*R losses due to heat would be the same 
regardless of frequency {taking the skin depth into consideration}
so we should be able to use this to calculate total cross sectional 
area starting with the Skin Effect depth and then calculate the 
surface area which divided by pi would yield the needed diameter of 
the conductor.

                               Sincerely

                                \\\|///
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                           Alfred A. Skrocki
                   alfred.skrocki-at-cybernetworking-dot-com
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