## Skin Effect/Material Constants

The value of the skin depth represents the depth at which the
current density is 1/e of the surface density. Of course the
current density never goes to zero, even at the center of the
conductor, but it becomes extremely small at over 5 x the skin depth,
etc. Standard exponential stuff here.

For purposes of computing the effective resistance of a
circular conductor, you assume that all the current is flowing (at
a uniform current density) in the volume within one skin depth of
the outer surface.

The formula for the skin depth is:

sigma = 2.6 * K1 / sqrt(f)

where:
sigma is the skin depth, in inches
f is in Hertz
K1 is a function of the material:
K1 = [ (1/u_sub_r) * (rho / rho_sub_copper) ] ^(1/2)

Where u_sub_r is the relative magnetic permeability of the metal
(==1 for copper, aluminum, and non iron/nickel/cobalt materials)

The ratio ( rho /rho_sub_copper ) is the relative bulk resistivity of
the material, as referenced to that of copper.

Note: K1 ==1 for copper.

```Material   rho     rho_rel   K1
-----------------------------------
aluminum    2.69   1.51      1.23
copper      1.77   1.000     1.000
silver      1.59    0.90     0.94
tin        11.5     6.49     2.55
Tungsten    5.5     3.11     1.76
Brass      ~7.0     3.95     1.99 (depends on alloy and temp.)
Phos-bronze 7.8     4.4      2.1
Bronze     17.      9.6      3.1

Units of rho are *1e-8 ohm*meter
For any material, rho_rel = rho / 1.77e-8 ohm*meter
rho_rel is unitless.
Values extracted from "Industrial Electronics Reference Book", 1948.

```

Example: At 200KHz, the skin depth for
copper is 0.0058 inches. So, for a circular conductor, you'd
want to use something with a diameter greater than twice that,

If the conductor is thicker than that, it can be shown that the innermost
portions of the conductor carry very little current.

Most Tesla people are not making capacitors using anything that
thick! I think I'll point this fact out to the list and see what
happens...

The one thing I don't know is how this applies to a non-circular
conductor, like a thin plate! I'm not sure if the current would all
follow the outer perimeter, leaving none in the center of the
plate! This would be detectable by measuring the capacitance
of the plate at different frequencies. If the measured capacitance
decreases with increasing frequencies, then this scenario seems

Most capacitor failures occur near the outer edges of the plates.
Perhaps this supports the possibility that most current is flowing
along the perimeter of the plates!

Here is a rambling on the skin effect which I found on the tesla@pupman list:
```

------------------------------

Date: Sun, 29 Mar 1998 14:20:06 -0700 (MST)
From: Tesla List
To: tesla@pupman.com
Subject: Re: Primary and copper (fwd)
Message-ID:
Content-Type: TEXT/PLAIN; charset=US-ASCII

---------- Forwarded message ----------
Date: Sun, 29 Mar 1998 09:16:29 -0800
From: Jim Lux
To: Tesla List
Subject: Re: Primary and copper (fwd)

> Or:
> skin depth=sqrt(resistivity/(pi*frequency)/u0) meters
> where u0=pi*4e-7.
> For copper (resistivity=1.724e-8 Ohm-m):
> skin depth=0.06608/sqrt(frequency) meters, or
> skin depth=2602/sqrt(frequency) mils.
>
> This value is valid for flat sheets. For round wires the usual approach
> is to consider a sheet wrapped around the wire, what results in:
> resistance=(length/diameter)*8.31e-8*sqrt(frequency) (any length unit).
> This works if the skin depth is much smaller than diameter/2, but fails
> totally for thinner wire or lower frequency.
> Do someone know the expression for wire resistance as function of
frequency
> when the wire radius is not much greater than the skin depth?
>
> Antonio Carlos M. de Queiroz
> http://www.coe.ufrj.br/~acmq
>

This would be a pretty tricky problem, because the current in a particular
point is affected not only by the layer on the side of the circle closest
to it, but also across the diameter. The usual caveat: "valid for conductor
radius >> skin depth" is a manifestation of this. The whole skin effect
phenomena is due to the propagation of the EM radiation in the conductor
(i.e. it takes time to propagate in), so some sort of solution of Maxwell's
equations (but of course...) with the right boundary conditions should give
it. I suspect that there is a fairly simple empirical equation around. At
some frequency, the wire starts to act more like a transmission line than a
conductor with a skin. The question would be at what frequency.

FWIW here are some things from "Reference data for engineers"

Rac equation is valid for conductors spaced at least 10 diameters from
3%, were depth of penetration is small.

it then gives an equation:

factor =  D * sqrt(f) * sqrt(Permeability * conductivity of Copper /
conductivty)

if factor is >40 then

Rac/Rdc = .0960 * factor + 0.26

For an isolated wire or tubular conductor for T < D/8 or T1 < D/8 then

Rac = FactorA * sqrt(f) / D * (permeability * conductivity of Copper/
conductivity) * 1E-6

A value of T (thickness of tubular conductor in inches) that just makes A =
1 indicates penetration of current below the surface. Therefore,

T1 (depth of penetration)  = 3.5 / sqrt(f) * sqrt( perm * condCu / cond)

where T1 in inches.

There is a table in the book of factor A, ranging from 1 to 2 for solid
conductors. It would appear that for D*sqrt(f) * sqrt(perm * condCu/cond) <
3.0, the ac resistance is approximately equal to the dc resistance.

permeability is relative to free space.
```

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