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

- To: femm@yahoogroups.com
- Subject: Re: [femm] Re: Self and mutual inductances of coupled microstrip lines in RF frequency range
- From: David Meeker <dmeeker@ieee.org>
- Date: Fri, 22 Nov 2002 17:34:49 -0500
- References: <arlt1f+mmqc@eGroups.com>
- User-agent: Mozilla/5.0 (Windows; U; Windows NT 5.1; en-US; rv:1.1) Gecko/20020826

`Samuel wrote:`

> `Hello,`

>     Sorry, I used my friend's e-mail address in the previous mail, now I
>     have my own :-)
>     Thanks Dave for detailed explanation. Enforcing zero net circuit
>     current on the "passive"
>     line really changed everything. Here is what I'm getting now directly
>     from Dave's structures:
>     ------------------------------------------
>     Single Line

> `F L,nH/m`
> `0 559 1 MHz 559
> 100 MHz 557
> 1 GHz 539
> ---------------------------------------------------------------------
> Coupled lines`
> `F L11 L21,1st method L21, 2nd method`
> `0 559 267 352 1 MHz 559 269 352
> 100 MHz 549 270 344
> 1 GHz 513 281 315
> ----------------------------------------------------------------------
> In this table 1st method is integral of A21 over strip 2 with I2=0
> divided by cross-section; 2nd method is diference between
> cases I1=1A, I2=0 and I1=I2=1A (both cases use Int{A11*J1} over
> the area of the strip #1), done according to Dave's suggestion.`

`
I think that you've done the second method wrong--it's my fault for not describing it very well. What I meant is that you have to define a second circuit with 1 A in it and apply it to the other line. That is, you have two "circuit" properties, each with an amp, one applied to each line. Then, do the Int{A11*J1} integral. You will find that the result is exactly the same as the 1st method at zero Hz. It looks like what you did is applied the same circuit property to both lines, which splits the current so that there is 1/2 amp in one line and 1/2 amp in another line, so that the Int{A11*J1} wasn't the result that you thought that it was. See the attached two_lines_limit1.fem as an example of how you'd do the analysis for both lines on, and two_lines_limit0.fem for the case where there is only 1 line on.`

> ``
> `My favorite electrostatic solver is LINPAR ( A. Djordjevic et
> al,"Linpar for Windows", Software and Manual, Artech House, 1996),
> specialized software that analyzes coupled microstrip lines. I've been using it
> for years, its very robust. I trust it even up to 40 GHz. It calculates`
> `L-matrix from capacitance C-matrix as:`
> `[L] = 1/[C_0]/v_light^2`

>     where [C_0] capacitance matrix of conductors placed in air, v_light
>     is
>     the speed of light. [C] is found from charge distribution which, in
>     its turn is obtained from solution of integral equation by method of
>     moments.

>     To my knowledge, Linpar is very accurate in getting [L] at f~=0 for
>     our simple example. Note that these inductances are:

>     ----------------------------------------------
>     Single line L = 536 nH/m
>     coupled lines L11 = 503 nH/m, L21 = 295 nH/m
>     ----------------------------------------------

>     FEMM results at f=0 are 4% off for a single line (well, its
>     acceptable at least).

>     What kills me in FEMM results for coupled lines is that L11 at f=0 is
>     exactly the same as L for a single line! Linpar gives L11=503 against
>     single line L=536. It's normal because lines are only 10 microns
>     apart. So L11 from FEMM is 11% off LINPAR; L21 from FEMM is 9.5% off;
>     L21 is underestimated in FEMM.

``
`At DC, the self-inductance of one line /shouldn't/ be affected by the presence of another line that has a unit relative magnetic permeability. FEMM is giving the right result at DC.`
`Linpar is also probably giving the correct result--however, the assumptions used in doing the calculation are different. I think that your electrostatics program is assuming that the conductors are perfect conductors--which is perfectly reasonable for a capacitance calculation. (Can anyone tell me for sure if this is what Linpar is assuming?) If the conductor is a perfect conductor, all of the current is carried on the surfaces of the conductors and flux is excluded from entering the conductors. However, this isn't what's going on at DC. At DC, the current is evenly distributed over the cross-section of the conductors and flux can penetrate the conductors as readily as it does air. The results are different than Linpar because the problem that is being solved is different. At DC, the FEMM result is the "right" one because the conductors aren't perfect conductors. At high frequencies, things converge to the Linpar result as current is forced to the surface of the conductor by eddy current effects.`
`I've also attached a solution here where I've defined the permeability of the wire material to be just about zero, so that the flux is just about excluded from the lines. See two_lines_limit2.fem This ought to be analogous to the problem that Linpar is acutally solving.`
`Dave.`

    --
    David Meeker
    email: dmeeker@ieee.org
    www: http://femm.berlios.de/dmeeker

Attachment: `zip00040.zip`
Description: Zip compressed data

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