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fed
Italy
1 Posts |
 Posted - Jul 17 2009 : 17:33:45
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Hello, I'm a new user.
I'm trying to replicate the results I have found in a paper published on an IEEE journal for the coil in Fig.1
[url]ht*p://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=04387341
The conductor is copper(conductivity 5.8108×107 Siemens/meter) with width 6.35mm and thickness 0.1778mm
Discretisation is : thickness divided into 10 filaments and width
into 50 filaments.
Frequencies are 63.86 MHz and 127.75 MHz
The calculated inductances should be L63.86MHz = 66.72 mH and L127.75MHz = 79.75 mH.
This is my input file but gives different results
.units mm
.Default sigma=5.8e4 nhinc=10 nwinc=50
N1 x=0 y=48 z=0
N2 x=41.5 y=24 z=0
N3 x=41.5 y=-24 z=0
N4 x=0 y=-48 z=0
N5 x=-41.5 y=-24 z=0
N6 x=-41.5 y=24 z=0
N7 x=0 y=47.99 z=0
E1 N1 N2 w=6.35 h=0.1778
E2 N2 N3 w=6.35 h=0.1778
E3 N3 N4 w=6.35 h=0.1778
E4 N4 N5 w=6.35 h=0.1778
E5 N5 N6 w=6.35 h=0.1778
E6 N6 N7 w=6.35 h=0.1778
* Define ports
.external N1 N7
* Frequency range of interest
.freq fmin=63.86e6 fmax=63.86e6 ndec=1
* All input files must end ....
.end
Can someone help please?
Best regards |
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chromatik
France
29 Posts |
Posted - Apr 27 2011 : 11:57:40
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It's an old thread but I've checked the simulation and obtained the following result:
Row 0: n1 to n7
Freq = 6.386e+007
Row 0: 1.25329+0.000185967j
So L=0.18mH
At the moment, no explanation but I'll check this out.
Regards |
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yuchenxi
China
7 Posts |
Posted - Aug 01 2014 : 11:27:17
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hello#65292;friends#65292;I'm a new user. I have encountered the same problem that you have ever encountered when I try to replicate the result of this paper. Have you solve this problem#65311;if you did,please help me out.
thanks in advance
Best Regards
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Enrico
550 Posts |
Posted - Aug 04 2014 : 19:30:20
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There are two issues here.
The first one is that the FastHenry file first line must NOT be the .units statement. The first line is ignored, so your units defaults to meters, giving you wrong results. The first line can safely be a comment (or a blank line, but definitely a title is better)
The second one, I'm afraid, is in the paper you cite. There is no way for an inductor so small to have an inductance of 66 mH. Running the simulation you get about 186 uH (micro Henry) that is a much more reasonable value. You can compare this value with a simple closed formula for a circular coil, as a sanity check, to get the feeling of the order of magnitude; or even with the flat spiral coil formulas in "Simple Accurate Expressions for Planar Spiral Inductances" by S. S. Mohan, M. M. Hershenson, S. P. Boyd, and T. H. Lee, IEEE JOURNAL OF SOLID-STATE CIRCUITS, VOL. 34, NO. 10, OCTOBER 1999 (you can even find online calculators for these formulas). I got about 220 uH with a crude approximation of a single-turn coil.
I suspect this apparently wrong result was not noticed since the paper deals in the end on how to minimize the mutual inductance, not on how to calculate or use the self inductance. I don't know if the error is the result of wrong coil dimensions, or of a wrong use of the Zc.mat results.
One last remark, you should use odd numbers for nhinc and nwinc, and definitely such an enormous discretization is not required. You should be happy also with much lower numbers, like nhinc=5 nwinc=11
(for precise numbers you should calculate the skin depth here, as indicated in FastHenry's user's manual)
Best Regards,
Enrico
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Hexagon self inductance
