----- Original Message -----
Sent: Saturday, January 18, 2003 8:51
AM
Subject: Re: [femm] Comparing inductance
predictions: femm vs standard formula
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 ofcoil
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 isthe
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 modeledvia
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@xxxxxxxx
http://femm.berlios.de/dmeeker
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