Doug,

My colleague, an RF Engineer looked at your question at my request and
offered this explanation:

The factor W/(H-T) according to 5.38b in 317A must be less than 0.35 in
order for the equations to apply. He solved for that factor in your
equation and came up with 0.93. His conclusion is that your thinking is
correct, the formula is correct (within the limits stated in 5.38b) but
it doesn't appear to apply to your circuit arrangement. I hope this
helps.

Gary Shoemaker
Sr. PCB Designer
Allen Telecom
Site Products


Doug Brooks wrote:
>
> Ooops. I apologize that the first message had
> two W's and no H's! (No wonder it didn't work!)
> A corrected message follows:
> Sorry for the inconvenience
> Doug Brooks
>
> There is a question I have had for several years;
> I though I understood the answer but now recognize
> that I don't. Perhaps some of our resident "Designer
> Council" experts or some IPC gurus can help.
>
> The problem involves the formulas for intrinsic
> capacitance of a trace, as published in IPC 317 and
> 317A. For stripline, the  formula for Co is:
> Co=1.41Er/Ln[3.81H/(.8W+T)]  (5.37 in 317A)
> for Asymetric stripline, the formula is:
> Co=2.82Er/Ln[2(H-T)/(.268W+.335T)]  (5.38b in 317A)
>
> This later formula can be manipulated to
> Co=2.82Er/[Ln[6(H-T)/(.8W+T)]
> which can be compared with Eq. 5.37
>
> Now assume a stripline case with
> H = 10  (mils)
> Thickness = 1 oz
> W = 8
> Er = 4.3
>
> Co is calculated as 3.8 pf/in
>
> Now, simply move the lower plane very far away and use the
> asymetric stripline formula. You get 6.42 pf/in, roughly
> twice as much. But since you have only moved one plate
> AWAY from the trace, intuitively you'd think the capacitance
> would go DOWN by a factor of two, not up.
>
> Is there something wrong with my thinking, or with the
> formulas?????
>
> Thanks for the help.
>
> Doug Brooks
> .
> ****************************************************
> Doug Brooks, President              [log in to unmask]
> UltraCAD Design, Inc.           http://www.ultracad.com

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