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yuhao
United Arab Emirates
4 Posts |
 Posted - Oct 14 2015 : 23:37:45
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I always expect the Z matrix to be symmetric, however I got one that is not. Does any know why would this happen?
Computed matrices (R+jL)
Row 0: n1_1 to n1_21
Row 1: n2_1 to n2_13
Freq = 205000
Row 0: 2.87586+2.45334e-005j -4.85723e-017+7.60142e-008j
Row 1: 6.93889e-018+8.015e-008j 0.211823+7.02882e-007j
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Enrico
545 Posts |
Posted - Oct 15 2015 : 00:25:33
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Can you share the input file?
Best Regards,
Enrico |
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yuhao
United Arab Emirates
4 Posts |
Posted - Oct 15 2015 : 12:27:29
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Hi, there is two loop of different size and turns
.units mm
.Default z=0 sigma=5.8e4
N1_1 x = 0 y = 0 z = 0
N1_2 x = 300 y = 0 z = 0
N1_3 x = 300 y = 300 z = 0
N1_4 x = 0 y = 300 z = 0
N1_5 x = 0 y = 2 z = 0
N1_6 x = 298 y = 2 z = 0
N1_7 x = 298 y = 298 z = 0
N1_8 x = 2 y = 298 z = 0
N1_9 x = 2 y = 4 z = 0
N1_10 x = 296 y = 4 z = 0
N1_11 x = 296 y = 296 z = 0
N1_12 x = 4 y = 296 z = 0
N1_13 x = 4 y = 6 z = 0
N1_14 x = 294 y = 6 z = 0
N1_15 x = 294 y = 294 z = 0
N1_16 x = 6 y = 294 z = 0
N1_17 x = 6 y = 8 z = 0
N1_18 x = 292 y = 8 z = 0
N1_19 x = 292 y = 292 z = 0
N1_20 x = 8 y = 292 z = 0
N1_21 x = 8 y = 10 z = 0
N2_1 x = 130 y = 130 z = 1
N2_2 x = 170 y = 130 z = 1
N2_3 x = 170 y = 170 z = 1
N2_4 x = 130 y = 170 z = 1
N2_5 x = 130 y = 132 z = 1
N2_6 x = 168 y = 132 z = 1
N2_7 x = 168 y = 168 z = 1
N2_8 x = 132 y = 168 z = 1
N2_9 x = 132 y = 134 z = 1
N2_10 x = 166 y = 134 z = 1
N2_11 x = 166 y = 166 z = 1
N2_12 x = 134 y = 166 z = 1
N2_13 x = 134 y = 136 z = 1
E1_1 N1_1 N1_2 w = 1 h = 3.500000e-02
E1_2 N1_2 N1_3 w = 1 h = 3.500000e-02
E1_3 N1_3 N1_4 w = 1 h = 3.500000e-02
E1_4 N1_4 N1_5 w = 1 h = 3.500000e-02
E1_5 N1_5 N1_6 w = 1 h = 3.500000e-02
E1_6 N1_6 N1_7 w = 1 h = 3.500000e-02
E1_7 N1_7 N1_8 w = 1 h = 3.500000e-02
E1_8 N1_8 N1_9 w = 1 h = 3.500000e-02
E1_9 N1_9 N1_10 w = 1 h = 3.500000e-02
E1_10 N1_10 N1_11 w = 1 h = 3.500000e-02
E1_11 N1_11 N1_12 w = 1 h = 3.500000e-02
E1_12 N1_12 N1_13 w = 1 h = 3.500000e-02
E1_13 N1_13 N1_14 w = 1 h = 3.500000e-02
E1_14 N1_14 N1_15 w = 1 h = 3.500000e-02
E1_15 N1_15 N1_16 w = 1 h = 3.500000e-02
E1_16 N1_16 N1_17 w = 1 h = 3.500000e-02
E1_17 N1_17 N1_18 w = 1 h = 3.500000e-02
E1_18 N1_18 N1_19 w = 1 h = 3.500000e-02
E1_19 N1_19 N1_20 w = 1 h = 3.500000e-02
E1_20 N1_20 N1_21 w = 1 h = 3.500000e-02
E2_1 N2_1 N2_2 w = 1 h = 3.500000e-02
E2_2 N2_2 N2_3 w = 1 h = 3.500000e-02
E2_3 N2_3 N2_4 w = 1 h = 3.500000e-02
E2_4 N2_4 N2_5 w = 1 h = 3.500000e-02
E2_5 N2_5 N2_6 w = 1 h = 3.500000e-02
E2_6 N2_6 N2_7 w = 1 h = 3.500000e-02
E2_7 N2_7 N2_8 w = 1 h = 3.500000e-02
E2_8 N2_8 N2_9 w = 1 h = 3.500000e-02
E2_9 N2_9 N2_10 w = 1 h = 3.500000e-02
E2_10 N2_10 N2_11 w = 1 h = 3.500000e-02
E2_11 N2_11 N2_12 w = 1 h = 3.500000e-02
E2_12 N2_12 N2_13 w = 1 h = 3.500000e-02
.external N1_1 N1_21
.external N2_1 N2_13
.freq fmin=205e3 fmax=205e3 ndec=1
.end
This is the output
Computed matrices (R+jL)
Row 0: n1_1 to n1_21
Row 1: n2_1 to n2_13
Freq = 205000
Row 0: 2.87586+2.45334e-005j -4.85723e-017+7.60142e-008j
Row 1: 6.93889e-018+8.015e-008j 0.211823+7.02882e-007j
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Enrico
545 Posts |
Posted - Oct 16 2015 : 18:30:44
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From physics point of view, you are right of course. However, in general, being the solution numerical, you can have some residual errors. This is particularly visible when you have input structures of very different dimensions and in the smaller terms (i.e. the smaller off-diagonals), as the underlying algorithm will converge as a whole (i.e. when the overall residual is small enough).
So the question is: is this error negligible or should it be considered? If we look at the real part of the L12 and L21 elements, well I would say this can safely be neglected. Your self-resistance is 2.8 ohm, while the mutual is 17 orders of magnitude lower in the best case. Now to the imaginary part. Here we see a discrepancy of 4 nH, resulting in an average error of 2 nH, when the smallest self-inductance is in the order of 703 nH.
So depending on your application, this error of 0.2% over the self-term and of 2 nH / 78 nH = 3.8% may be acceptable; note that if you decide to use the result as it is, you should use as mutual term the average of the two values, as this will minimize the error.
If instead for your application this error is too high, you can force FastHenry to a higher precision in the multipole expansion, at the price of a slower simulation (actually, not noticeable in your simple case) For instance, increasing the order of multipole expansion (-o) from 2 to 4, you get:
Freq = 205000
Row 0: 2.87586+2.45374e-005j 8.67362e-017+7.61195e-008j
Row 1: -3.46945e-018+7.61195e-008j 0.211823+7.02882e-007j
Now the mutual inductance terms are symmetric enough.
In general however be skeptical of any simulation software giving you perfectly symmetrical inductance or capacitance matrices. This only means that the off-diagonal terms have been mathematically symmetrized under the hood, i.e. the software already calculated the average and presented you that.
Best Regards,
Enrico
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Unequal mutual inductance
