[femm] Re: Drawing circles & Another AC analysis question

[Dcm3c@xxxxxxx]

In a message dated 11/11/99 9:54:31 AM Eastern Standard Time, Marctt2@xxxxxxx 
writes:

> Ah, I see; you're specifying a voltage drive at the coil terminals which 
in 
> turn sets the coil current. This is why my calculation works fine at DC.

Yes, you could interpret it that way. That is, the J that you specify is 
equal to 
- sigma*grad(Phi), where Phi is voltage.

> I believe other solvers do this a bit differently.

Well, some other solvers have more flexibility in the eddy current 
formulation. Some nice additions to femm would be the ability to define 
regions in which eddy currents are either conserved or in which the total 
current is constant. These features would be useful to me for some work I've 
been doing, so I've been working on adding them. Hopefully these features 
will be available soon.

> So, what's the simplest way to calculate AC resistance of the one-turn 
wire ?
> I'm trying to generate plots of L vs. frequency and R vs. frequency 
> (basically to find the AC resistance and inductance of my coil).

Well, the way you should do it is to take the A.J integral over the volume of 
the coil. This integral includes in J both the induced and eddy current. 
Then, divide by the square absolute value of the total current in the coil 
(from the integral "total current", which combines both the source and eddy 
currents) to get inductance. 

However, I tried this on a quick example problem that I cooked up just now, 
and there seems to be a bug in this integral. It doesn't include the 
contribution of the eddy currents in the correct way for harmonic 
axisymmetric problems--bummer. I'll fix this and post a modified version 
soon. Looking further, there is also a similar bug in computing Lorentz 
forces on induced currents in the harmonic axisymmetric case. I'll fix this 
one, too.

In the mean time, you can just take the "stored energy" integral. This gives 
the right result, as long as you don't apply a "mixed" boundary condition to 
the outer boundary of your domain. Denote the result of the stored energy 
integral as E. To convert E to inductance, get the total current in the coil, 
i, by taking the "total current" integral. Now, inductance is:
L=4*E/abs(i)^2
The 4 gets in there (rather than a 2, like in the mstatic case) because we 
are working in terms of amplitude, rather than rms quantities. Likewise, 
resistance is:
R=2*(Total Losses)/abs(i)^2 

Dave.
--
http://members.aol.com/dcm3c