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# Re: [femm] Re: inductance - parallel strips

- To: femm@yahoogroups.com
- Subject: Re: [femm] Re: inductance - parallel strips
- From: "Jim in_Dallas" <jim271@hotmail.com>
- Date: Thu, 15 Nov 2001 14:26:26 -0600
- Bcc:

``

>     From: Tomislav Borzic <tborzic@yahoo.com>
>     Reply-To: femm@yahoogroups.com
>     To: femm@yahoogroups.com
>     Subject: [femm] Re: inductance - parallel strips
>     Date: Thu, 15 Nov 2001 10:36:33 -0800 (PST)

>     Hello Jim, I solved this problem.
>     Problem was that part of the flux was closing through
>     conductor.So I changed boundary conditions to force
>     the
>     flux to stay BETWEEN conductors, not in them.
>     Femm file is attached.
>     Results:
>     L(femm)=1.1517 e-6 H/m

> `F1 L=lambda* N^2`

>     lambda=mio*b/h,
>     b=distance between conductors (not their radius)
>     h-height

> `L(F1)=1.1 e-6 H/m`

>     ***
>     i have another question, regarding complex inductance.
>     inductance doesn't depend of current, so often in
>     numerical
>     examples it is not given.so if I want to achieve same
>     result
>     as numerical (with current not given, but other
>     parameters are)
>     how can I solve similar problem in femm?
>     thanx
>     Tomislav

`
I think you have found the reason for the difference between FEMM and the simple formula for L. The formula works for the limiting case where the spacing becomes very small compared to the conductor width. In that case all the magnetic field energy is between the conductors. That is the condition you forced by your special boundary condition.`
`I set up another case where the conductors are much closer together and the result from FEMM is much closer to the formula results. For example, FEMM=44nH and the formula gives 57nH when the spacing is only 0.1cm and the conductors are 0.05cm thick.`
`Looking back at your original configuration, I found only about 1/2 the energy is between the conductors:
Energy between the conductors = 7.33e-5j/m
Energy inside the conductors = 4.50e-5
Energy outside the conductors = 2.55e-5`
`So we see that about only about 1/2 the energy is actually between the conductors, hence the difference of about a factor of 2 in inductance.`
`I think this formula is useful in actual practice because the spacing between conductors is often very small, corresponding to a thin insulating layer between the heavy conductors. I haven't seen the exact formula, but I imagine it's pretty complicated :-)
=========================================`
`If the current in the coil is specified, set J for the copper equal to I divided by the cross sectional area of the conductor.
Note that J is entered as MegAmps/m^2. You can check this by highlighting the copper in green and reading the total current. In a planar case where the current is both positive and negative, highlight only one of the two regions.`
`-- Jim`
``

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