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# Re: [femm] Comparing inductance predictions: femm vs standard formula

- To: femm@yahoogroups.com
- Subject: Re: [femm] Comparing inductance predictions: femm vs standard formula
- From: David Meeker <dmeeker@ieee.org>
- Date: Sat, 18 Jan 2003 11:51:52 -0500
- References: <avn680+1sbr@eGroups.com> <3E1F3465.4050201@ieee.org> <001801c2bef2$0d89e860$cf745142@robertmacy>
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Robert Macy wrote:

> `Not having good results when trying to predict inductance of a multi layer
> solenoid coil.
> dimensions are
> inside radius 0.25 in
> length 1 in
> thickness of coil 0.44 in
> Did femm analysis using asymetric style in a 10 inch diameter sphere with
> over 14k nodes. Then assuming that inductance is 2 times the magnetic field
> energy divided by the total current squared resulted in an inductance
> prediction of 7.638 nH per turn squared.
> However using the formula to predict the inductance of a multilayer solenoid
> from ITT Reference Handbook for Radio Engineers 6th ed which is
> L = 0.8 *(r*N)*r*N)/(6*r+9*length+10*thick)
> produces a prediction of 3.356 nH per turn squared
> `

The equation that I have in my notes I got from http://home.earthlink.net/~jimlux/hv/wheeler.htm

> `L (uH) = 0.8 * a^2 * n^2 / (6*a + 9*b + 10*c )`

> where
> `a = average radius of windings
> b = length of the coil
> c = difference between the outer and inner radii of the coil.
> all dimensions in inches.`

> It states that it is accurate to 1% when the terms in the denominator are
> about equal. This is also an equation by Wheeler. It applies as long as the
> coil has a rectangular cross section.

This is the same thing as your equation, except that a is the average radius of the coil, not the inner radius. When I plug your numbers into this equation, I get 10.895 nH.
I made a quick FEMM model (see attached) with one amp evenly distributed over the cross-section of the coil. When I do the A.J integral and divide by 1 A^2 to get inductance, I get 10.59222 nH. If I integrate energy over the volume, I get 5.234074e-009 Joules, which would imply 10.4681 nH. The direct volume integration of energy provides a slighly lower inductance, because this doesn't include the extra 1% or so of the energy that is stored in the exterior region that is modeled via the asymptotic boundary condition. The A.J integral accounts for the contributions of the exterior region to the stored energy, giving a more correct value.
At any rate, things would appear to be pretty consistent with the empirical formula.
Dave.

    --
    David Meeker
    dmeeker@ieee.org
    http://femm.berlios.de/dmeeker

Attachment: `zip00046.zip`
Description: Zip compressed data

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