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

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- Subject: [femm] Re: Drawing circles & Another AC analysis question
- From: Dcm3c@aol.com
- Date: Thu, 11 Nov 1999 09:43:15 EST
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    In a message dated 11/11/99 12:27:51 AM Eastern Standard Time,
    marctt2@aol.com writes:

    > I'm wondering if there's a simple way to draw circles in FEMM.

    You have to build a circle out of two 180 degree arc segments. Although this
    might seem to be a weird way to do it, it turned out that it was a lot easier
    to write the code to work in this way.

    > I'm attempting to model a coil made of round wire, and to estimate losses
    > due to skin and proximity effects. Are there any tricks here in the
    > drawing and/or analysis of the losses in AC magnetic problems ? I
    > assume the simplest way to find inductance of the coil is to integrate
    > the stored energy over the problem area; resistive losses in the coil
    > can be estimated similarly.

    The best way to get inductance is to integrate A dot J over the volume of the
    conductor, and then divide the result of this integral by the square of the
    current in the coil. Note that when I say current, I mean the current in one
    turn of the coil, not the total amp*turns in the coil (unless you just have a
    1-turn coil, in which case these are the same thing). Basically, this is an
    energy method for getting the inductance. The energy stored in the magnetic
    field of a coil is:
    (1/2)*(Integral of A.J over the volume of the coil) = (1/2) L i^2

    The reason to calculate inductance in this way is that it works in the case
    where you have impedance boundary conditions applied. In this case, some of
    the energy is stored outside the solution domain, so if you just integrate
    B.H over the solution domain, you won't get all of the energy. An example is
    the case of a 1-turn coil with an asymptotic boundary condition applied to
    the outer boundary of the solution region to approximate an "open" domain.

    > Also, on another note, is there a way to have a coil made of a single
    > turn, but to have FEMM assume that there are no eddy and proximity
    > losses in the coil ? i.e. to have constant current density throughout ?
    > I think the lamination function may work here but I'm unclear on the
    > concept.

    If you don't want any eddy current effects, just assign the coil material to
    have a conductivity of zero.

    > Maybe I'm missing something simple:
    >
    > I'm attempting to model a single-turn coil. I would like to get the AC
    > inductance and resistance of this coil.
    >
    > I have made my coil out of copper wire. I set the necessary J in the
    > coil to get 1 Amp in the coil (to make calculation of the inductance
    > and resistance simpler).
    >
    > Now, when I use the solver to find stored energy for various operating
    > frequencies, I find that the stored energy in my workspace goes down as
    > frequency goes up. This makes sense, since the inductance of the loop
    > will decrease as frequency goes up (the current crowds to the inner
    > radius of the wire due to skin effect, and hence inductance goes down).
    >
    > However, the FEMM results show that power dissipated in the workspace
    > also goes down. This is counter-intuitive, as I expect the resistance
    > of the single turn to go up as frequency goes up (due to skin effect).
    > Am I missing something simple here ?

    The trick is that as frequency increases, the currents that are induced flow
    in the coil in a direction that tends to oppose the "source" current density
    that you impose by specifying J. The result is that less total current flows
    in the coil. Think about the equation for an inductor without skin effects
    driven by a voltage source:

    (s L + R) i = v

    what you are doing by assigning J is basically specifying the voltage. Say,
    v= R*idc, where idc is the current that would flow in the coil at dc. As you
    go up in frequency,

    i = R*idc/(s L + R)

    so the total current decreases as frequency increases. If you have skin
    effects, the idea is the same, but L becomes a function of frequency with
    real and imaginary parts. When L is complex, you interpret the real part to
    be the "apparent" inductance, and (imaginary part of L)* j*omega is the
    apparent increase in resistance.

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

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