C
Chris Carlen
Guest
Greetings:
The skin depth is given by:
d = sqrt[ rho / ( pi * mu * f ) ]
where:
mu = 1.25664e-6 N/A^2
f = frequency in Hz
and for copper:
rho = 1.723e-8 ohm*m @ 20C
The current density of course does penetrate below d, so what d actually
means is the distance from the outer surface such that if the current
density were uniform (DC) but with the same net current as is present in
the AC case, then the uniform value would be equal to that found at
depth d in the AC case. Ie., d satisfies the mean value theorem for
integrals as applied to the radial current density function.
Now what happens as the wire diameter approaches the skin depth?
I was tempted to define the effective AC cross sectional area like this:
Let D = the wire diameter
The AC area should then be:
Aac = (pi/4)*D^2 - (pi/4)*(D-d)^2
Aac = (pi/4)(2*D*d - d^2)
However, when the diameter is less than twice the skin depth, then it's
basically all skin depth. One might hope that it becomes simply the
piecewise function:
Aac = (pi/4)(2*D*d - d^2) IF D > 2*d
Aac = (pi/4)*D^2 IF D <= 2*d
When plotting this function, one notices a discontinuity. For D<2*d,
the area increases parabolically with D, as expected. Then when the
D>2*d crossover kicks in, the curve jumps downward, to an area less than
the value of the DC equivalent at that diameter, and begins increasing
at an almost linear rate. This doesn't work.
A hopefully more correct way:
If one extrapolates the AC area function downward from the point of the
discontinuity, it eventually reaches a tangent point with the DC area
curve. Perhaps the correct function describing the AC equivalent area
of the conductor would then be:
Aac = (pi/4)(2*D*d - d^2) IF D > tangent point
Aac = (pi/4)*D^2 IF D <= tangent point
Now what is this tangent point? It's the solution to:
D^2 = 2*D*d - d^2
Which turns out to be d!
Interesting.
So it's not until the diameter is reduced to the skin depth, that the
current density is uniform, and the conductor may be treated simply as
it's geometric cross sectional area.
Perhaps I have no question after all.
Good day!
--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov -- NOTE: Remove "BOGUS" from email address to reply.
The skin depth is given by:
d = sqrt[ rho / ( pi * mu * f ) ]
where:
mu = 1.25664e-6 N/A^2
f = frequency in Hz
and for copper:
rho = 1.723e-8 ohm*m @ 20C
The current density of course does penetrate below d, so what d actually
means is the distance from the outer surface such that if the current
density were uniform (DC) but with the same net current as is present in
the AC case, then the uniform value would be equal to that found at
depth d in the AC case. Ie., d satisfies the mean value theorem for
integrals as applied to the radial current density function.
Now what happens as the wire diameter approaches the skin depth?
I was tempted to define the effective AC cross sectional area like this:
Let D = the wire diameter
The AC area should then be:
Aac = (pi/4)*D^2 - (pi/4)*(D-d)^2
Aac = (pi/4)(2*D*d - d^2)
However, when the diameter is less than twice the skin depth, then it's
basically all skin depth. One might hope that it becomes simply the
piecewise function:
Aac = (pi/4)(2*D*d - d^2) IF D > 2*d
Aac = (pi/4)*D^2 IF D <= 2*d
When plotting this function, one notices a discontinuity. For D<2*d,
the area increases parabolically with D, as expected. Then when the
D>2*d crossover kicks in, the curve jumps downward, to an area less than
the value of the DC equivalent at that diameter, and begins increasing
at an almost linear rate. This doesn't work.
A hopefully more correct way:
If one extrapolates the AC area function downward from the point of the
discontinuity, it eventually reaches a tangent point with the DC area
curve. Perhaps the correct function describing the AC equivalent area
of the conductor would then be:
Aac = (pi/4)(2*D*d - d^2) IF D > tangent point
Aac = (pi/4)*D^2 IF D <= tangent point
Now what is this tangent point? It's the solution to:
D^2 = 2*D*d - d^2
Which turns out to be d!
Interesting.
So it's not until the diameter is reduced to the skin depth, that the
current density is uniform, and the conductor may be treated simply as
it's geometric cross sectional area.
Perhaps I have no question after all.
Good day!
--
_______________________________________________________________________
Christopher R. Carlen
Principal Laser/Optical Technologist
Sandia National Laboratories CA USA
crcarle@sandia.gov -- NOTE: Remove "BOGUS" from email address to reply.