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Re: [femm] Self and mutual inductances of coupled microstrip lines in RF frequency range

Sergey Somov wrote:

Hello,

Here goes the question that perhaps isn't very common among this group: application of this great software to microwave problem. Actually, what I'd like to evaluate is an [L] and [R] per-unit-length (p.u.l.) matrices for a system of rectangular-cross-section lossy metal strips located within an arbitrarily stratified infinite lossy dielectric substrate that potentially may be conductive. Usually mu is 1 for all participating materials. This case covers PCB, LTCC, semiconducting chips etc. Operating frequencies may be from 0 Hz to dozens of GHz and eddy currents in the substrate aggravate the stuation making [L] frequency dependent (actually,lossy strip metall also slightly contributes to the frequency dependence of L ).

If substrate is purely dielectric electrostatic solvers may do the job and give pretty accurate results that match those obtained from full EM solvers. Cross-sectional dimensions are small relative to wavelength (say, width 500 microns and highest frequency 500 MHz for PCB or width 50 microns and highest frequency 10 GHz for RF chip) so application of static and quasi-static solvers is fairly justified. So I hoped to check the accuracy of FEMM comparing results with my favorite electrostatic solver.

It generally makes me nervous to apply a FEMM at frequencies like this, but so long as you've done the comparison to wavelength to make sure that the quasi-static approximation is justified....


Unfortunately, results for coupled lines are out of whack :-(( could be that I'm doing something wrong...

I attached two files. First file describes single line, second file describes two edge coupled lines.

In both cases perfectly conducting infinite ground plane backs the single layer substrate. Substrate material is named Silicon but actually it is air (sigma=0, mu=1). For single line I applied circuit current (I = 1A) to the conducting strip and enclosed the structure in a circle that models open space. Boundary condition A=0 is applied to the bottom of substrate. So structure is very simple (W=50 microns, T=5 microns where W stands for strip width, T is the strip thickness).

For the Single_Line structure my electrostatic solver gives p.u.l. L=536 nH/m.

Remember this solver implies that L is frequency independent.

Now go FEMM results (post-processed L=\int{A*J}/I^2 over the strip cross-section and rounded to 1 nH/m)

F L, nH/m

0 Hz 560

1 MHz 560

100 MHz 557

1 GHz 540


Not bad, at least in the ballpark.

I don't really think about this type of problem that often, but the way that you've set it up seems reasonable. If the ground plane is perfectly conducting, the vector potential on the ground plane is constant so that no flux passes normal to it. Normally, the inductance of a single wire isn't well-posed. However, in this case, the perfectly conducting ground plane provides a current return path so that the inductance calculation is well-posed. The only difference that I'd do as opposed to how you drew it is maybe to draw it using the Kelvin Transformation approach to model a perfectly conducting half-space. I've attached a modified version that does it this way.

The single-line inductance results are just about the same as the ones that you listed above. I don't know how you got the comparison result, but the change in the FEMM result versus frequency is because the field is excluded from the line, leaving only the portion from the flux that flows outside the wire. If you assumed that the wire was perfectly conducting for your comparison result, it would discount the energy stored inside the wire at low frequencies, giving a high-frequency asymptote result for inductance--I think this could be what's going on.



For the Two_Lines structure my electrostatic solver gives L11 = L22 = 503 nH/m; L21 = L12 = 295 nH/m.

L11 L12
[L] =
L21 L22

With FEMM I evaluated [L] as following:
First, I appied circuit current = 1A only to the strip #1, no current applied to the strip #2.
So
L11 = \int{A_11*J_1}/I_1^2
L22 = \int{A_21}/(I_1^2*W*T)

Here A_11 is potential over strip #1; J_1 is current density over strip #1; A_21 is potential over strip #2; W*T give strip cross-section area.

Here go FEMM results:


F L11, nH/m L21, nH/m

0 Hz 560 272 almost nice :-)

1 MHz 561 272 :-)

100 MHz 427 23 :-(

1 GHz 359 0.63 :-((((


Field is noticeably distorted for F>100MHz.

So my question to all people who used FEMM for PCB evaluations (and of course to Dave): what I'm going wrong and are there any ways to get reasonable results? My experience with FEMM is only week long and I never worked with magnetic solvers. May another choice of boundary conditions help? Actually, I tried rectangular box but this circular enclosure seems to work better.

Any help and ideas are greatly appreciated,

Samuel

Again, you might try the Kelvin Transformation approach to exactly obtain the perfectly conducting half-space ground plane. I modified the "Silicon" definition so that the conductivity was exactly zero--just makes things easier to debug so that there are no induced currents in the "Silicon"..

I think that when you want to consider one line by itself, you'd want to define a "circuit" with a zero current and apply it to the line that is supposed to be off. Otherwise, you'll be inducing some net current in the other line (rather than zero), which is probably not what you intended. I've attached an example of this, too.

With the zero curent constraint defined, the self-inductance of just one line will be slightly lower than in the total absence of the other line, because at high frequencies, the flux can't penetrate the other line. The flux path is just more reluctant.

I'd guess that we'll have to be sort of careful in getting the mutual inductance contributions. In getting to eq. (24) in the manual (i.e. the L22 = \int{A_21}/(I_1 *W*T) result), I'd assumed a uniform current in the second coil, which isn't the case here. You might do better just doing two cases: (I_1=1, I_2=0) and (I_1=1, I_2=1) and evaluating:
L = \int{A_11*J_1}/I_1^2
You'd then look at the difference between the results in the two cases to get the mutual inductance.

Dave.

--
David Meeker
dmeeker@xxxxxxxx
http://femm.berlios.de/dmeeker


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