| 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
* The segments connecting the nodes of Inductor #1 [E Defines Segments]
E1 N1 N2 w=1 h=1
E2 N2 N3 w=1 h=1
E3 N3 N4 w=1 h=1
E4 N4 N5 w=1 h=1
E5 N5 N6 w=1 h=1
* The nodes of Inductor #2 (z=0 is the default) [N Defines Node]
N7 x=210 y=0
N8 x=210 y=100
N9 x=110 y=100
N10 x=110 y=10
N11 x=200 y=10
N12 x=200 y=0
* The segments connecting the nodes of Inductor #2 [E Defines Segments]
E6 N7 N8 w=1 h=1
E7 N8 N9 w=1 h=1
E8 N9 N10 w=1 h=1
E9 N10 N11 w=1 h=1
E10 N11 N12 w=1 h=1
* 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
|
| 1 L A T E S T R E P L I E S (Newest First) |
| 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.
kislay |
|
|
