[SI-LIST] Re: Inductance of wire section

  • From: Kim Flint <si@xxxxxxxxxxxx>
  • To: si-list@xxxxxxxxxxxxx
  • Date: Fri, 05 Sep 2003 12:35:38 -0700

Hi-
I'm curious about assumptions made in some of these inductance formulas. I 
understand inductance as requiring a coil of wire, wherein magnetic flux 
would intersect the area of the coil. "Inductance of a wire segment" 
doesn't really mean anything without understanding where the rest of the 
current path forming our coil is located. So I'm guessing these formulas 
are approximations based on some set of assumed conditions, like height 
above ground or something like that? Does anybody know what those 
assumptions are in this case? I'd like to understand where I can expect 
this formula to be reasonably valid.

thanks!
kim

At 08:57 AM 9/5/2003, Stuart Brorson wrote:
>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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