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Re: [femm] inductance - axisymmetric model






From: Tomislav Borzic <tborzic@xxxxxxxxx>
Reply-To: femm@xxxxxxxxxxxxxxx
To: femm@xxxxxxxxxxxxxxx
Subject: [femm] inductance
Date: Tue, 13 Nov 2001 07:12:23 -0800 (PST)

Hello everybody, I am (again) coming with small
problem.
It is about inductance. I am callculating inductance
of a coil.
1. one turn
2. four turns.

In case (1) I use formula L= (mi0/pi)*(1/4 + ln(D/r0)
where is: mi0=4*pi*e-7
D- diameter of coil ( like this: 0-----D----0, where 0
is
conductor and D is space between them)
r0= radius of conductor

I know this is formula for two paralell conductors and
inductance
between them but with some assumptions it can work.
so in this case, r0=0.05m, D=0.8m,I+ = 1A, I-=-1A,
my callculations give 1.2 e-6 H/m, FEMM 1.16 e-6 H/m,
so it is ok.
I used AJ block integral (A j dV/2 = energy)
BUT, if I take four turns of wire, inductance should
be 16 times
higher because inductance of coil rises with n^2.
(n = number of turns)
I can not achieve acceptable solution.

*.fem file is attached.
plase help
thanks
tomislav.

If we use the axisymmetric model for the coil, the FEMM results is very close to the value calculated using the formula for a current loop.

Example:

current loop: L=N^2*R*mu0*Ln(R/r)
where R=radius of loop
r=radius of wire

Using your dimensions:
R=4cm
r=0.0005cm
4 turns
then L = 167.2nH

An alternative formula found in some references is: L=N^2*r*mu0*[Ln(8*R/r) -2]

Since the Ln(8) approx = 2, these formulae give similar results.

Using FEMM:
Let J=1.0 then
integral(A.J)=6.236e-9
I = 0.1954 Amps/turn

L=163.4nH which differs by less than 3 percent. I've attached my fem file.

HTH

Regards,
Jim



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Attachment: smallcoil2.FEM
Description: Binary data