[SI-LIST] Re: Inductance of wire section
- From: sdb@xxxxxxxxxx (Stuart Brorson)
- To: si-list@xxxxxxxxxxxxx
- Date: Fri, 5 Sep 2003 11:57:12 -0400 (EDT)
Hi Guys --
Thanks for all your answers. I used the following formula from Bart:
L = K * length * [ ln(4*length/Diam) - 0.75 ]
L = Inductance in nanohenries
D = Diameter of wire in cm
length = Length of wire in cm
K = 2
stuck it into Gnumeric (Linux-based Excel replacement), and got the
following answers:
Unit qnty qnty qnty
K 2 2 2
Length cm 1 1 1
Diam cm 0.08 0.06 0.04
Inductance nH 6.324 6.899 7.710
depending upon the wire diameter. Using the below figure of 20nH/in,
and knowing that 1in = 2.54cm, I get L = 7.87 nH.
Therefore, I think we have nailed it: L = 7nH/cm +/- 10% or so.
Thanks again for all the answers!
Stuart
>
> Hi Stuart: The ROT I use is 20nH/in.
> BTW, I verifed this using a 2D extractor
> and increasing the boundary distance far
> away from the wire to minimize mutual indutance.
>
>
> Hope this helps
>
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