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T O P I C R E V I E W
RonJ103
Posted - Oct 26 2011 : 22:54:42 I made a simple input file with 2 inductors. Each inductor has a port(.external command) associated with it. I was able to run FastHenry and I got a 2x2 impedance matrix. From this I am hoping to extract the Mutual Inductance (M) and coupling coefficient (k).
We know that for 2 coupled inductors: M=k*sqrt(L1*L2)
The output from FastHenry is a 2x2 impedance matrix (R+jL format). Are the following statements correct?
1) L1 = imag[Z11] 2) L2 = imag[Z22] 3) M = imag[Z12] or imag[Z21] (These values are close, but not exactly the same. I am not sure why.) 4) k = imag[Z21]/[sqrt(imag[Z11]*imag[Z22])]
Below is my input file: * 2 Inductors * 10/26/2011
* The following line names millimeters as the length units for the rest of the file .units um
* Make z=0 the default z coordinate and copper the default conductivity .Default z=0
.default nhinc=5 nwinc=7
* The nodes of Inductor #1 (z=0 is the default) [N Defines Node] N1 x=0 y=0 N2 x=0 y=100 N3 x=100 y=100 N4 x=100 y=10 N5 x=10 y=10 N6 x=10 y=0
* All of the above segments contain only 1 filament (No discretization. In order to properly model proximity and skin effects, finer discretization is needed) * Below is an example of E1 being broken up into 35 filaments: 5 along its height[nhinc=5] and 7 along its width[nwinc=7] * E1 N1 N2 w=0.2 h=0.1 nhinc=5 nwinc=7
* Define one 'port' of the network .external N1 N6 .external N7 N12
* Frequency range of interest. (ndec=Number of Points per Decade) .freq fmin=1e8 fmax=1e10 ndec=1
* All input files must end with: .end
Here is the output from FastHenry: Computed matrices (R+jL) Row 0: n1 to n6 Row 1: n7 to n12 Freq = 1e+008 Row 0: 6.72415+3.49553e-010j 3.31547e-009+1.82043e-011j Row 1: 3.31909e-009+1.82248e-011j 6.72415+3.49552e-010j Freq = 1e+009 Row 0: 6.72491+3.49552e-010j 3.31393e-007+1.82043e-011j Row 1: 3.31708e-007+1.82248e-011j 6.72491+3.49551e-010j Freq = 1e+010 Row 0: 6.80085+3.49448e-010j 3.0962e-005+1.82041e-011j Row 1: 3.09938e-005+1.82246e-011j 6.80085+3.49446e-010j
Any guidance would be greatly appreciated. Thanks, Jeremy
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kislay1947
Posted - Jan 23 2012 : 17:00:26 The code seems to be correct and the values will give u the mutual inductance between the coils.