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Effective wire AC cross sectional area

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

Reply to
Chris Carlen
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Ah yes, I have blundered. The 1/e point definitely doesn't satisfy the mean value theorem. :-(

I was hoping to avoid a more thorough analysis. I don't fully understand the discussion in my EM text, which only deals with finding the AC resistance of a conductor where D>>d.

Ok, I'll do some more calculating. Thanks for the heads up.

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.
Reply to
Chris Carlen

Interesting. The skin effect is an exponential and the skin depth means the current reduced to 1/e. Again...

Rene

--
Ing.Buero R.Tschaggelar - http://www.ibrtses.com
& commercial newsgroups - http://www.talkto.net
Reply to
Rene Tschaggelar

"Chris Carlen" wrote

You wouldn't happen to be looking for 'Litz Wire'?

-- Nicholas O. Lindan, Cleveland, Ohio Consulting Engineer: Electronics; Informatics; Photonics. Remove spaces etc. to reply: n o lindan at net com dot com psst.. want to buy an f-stop timer? nolindan.com/da/fstop/

Reply to
Nicholas O. Lindan

Just to help you along..

The exponential decay of curent versus vertical depth applies only to large FLAT conductor surfaces. The surface of a round wire is curved.

However, at a sufficiently high frequency, when skin depth is less than about 1/8th of wire diameter, the radius of curvature relative to skin depth is sufficiently large to make the current think it is flowing on a flat surface and the decay is sensibly exponential.

It turns out that the HF resistance of a round conductor when skin effect is fully operative ( ie., when depth < Diameter/8 ) is equal to the DC resistance of a copper tube of Diameter D and wall thickness d. Which is easy to calculate.

=====================

You may find the following formula useful. It applies to round copper wires, of diameter D mm, at 20 degrees Celsius. F in MHz. It is within about 1 % accurate at any frequency from DC upwards.

Let K = 7.566*D*Sqrt( MHz )

When K4, U = K^2 / ( 2*K - 1 + 0.6 / K^2 )

====================================

Other useful quantities for copper wire at 20 degrees C.

DC resistance = 0.02195 / D^2 ohms per metre.

Skin depth in copper = 0.0661 / Sqrt( MHz ) millimetres.

Approx HF wire resistance, d < D/8

R = Sqrt( MHz ) / 12 / D ohms ohms per metre.

--
Reg.
Reply to
Reg Edwards

Yes. The discussion in my text starts with an infinitely deep rectangular slab.

Right.

Yes.

I will be interested in comparing the results of your formula below to my formulations when I am finished.

Thanks!

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.
Reply to
Chris Carlen

Do we need another expression using "U" to compute Rac/Rdc, or is "U" already equal to Rac/Rdc in this case?

Reply to
The Phantom

When K>4, U = K^2 / ( 2*K - 1 + 0.6 / K^2 )

U is a 'typing error' when copying from my old note book to screen.

So U = Rac / Rdc

There are two curves, one for K4. They smoothly meet at K = 4.

At least, I hope they do. ;o)

---- Reg.

================================

Do we need another expression using "U" to compute Rac/Rdc, or is "U" already equal to

Reply to
Reg Edwards

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