From: Zabinski, Patrick J. ([email protected])
Date: Fri Jun 23 2000 - 13:17:36 PDT
Haris,
I think the cross section is a bit different than
what you show:
________________________ PLANE
^
h2
| ------
| | | (note: trace has thickness)
v ------ trace
^
|
h1
|
____v___________________ PLANE
Where both h1 and h2 are referenced not to
the middle of the trace but to one of its surfaces.
Generally, h1 is the separation from the bottom
plane to the bottom of the trace, and h2 is the
from the top plane to the bottom of the trace.
For example, say you have a total plane-to-plane
separation of 10 mils, and you have 1 mil thick
copper (using easy numbers). If you set h1 to
8 mils, h2 will be 2 mils, but the spacing between
the top of the trace and the upper plane will be 1
mil (not the 2 mils as in the case of an infinitely
thin trace as shown in your original drawing).
If, however, if you swap h1 and h2 parameters, then
the separation between the lower plane on the trace
will be 2 mils. Here, the trace is twice as far
away from the nearest plane, which will result
in a higher impedance.
One quick way to confirm this is the case is to see if
your simulator asks for the trace thickness. If it
does, then this should be the problem/issue.
If your tool assumes infinitely thin traces
(some tools still do), then I have no idea.
Pat
> Gentleman,
>
> I have question regarding offset (asymmetric)
> stripline.
>
> ________________________ PLANE
>
> h2
> ------ trace
> H
>
> h1
>
> ________________________ PLANE
>
>
>
> Equation for the characteristic impedance, that I
> found and used, is:
>
> Z0~= [80/sqrt(Er)]ln(4.75(h1/w)*(1-h1/4*h2)
>
> MY QUESTION:
> How come that characteristic impedance of the trace
> relates differently to thickness h1 and thickness h2?
> What if we flip this geometry by 180 degrees? Does
> that mean that characteristic impedance of the trace
> is going to change?
> I used polar field solver to perform the experiment
> and I got two different characteristic impedances as
> the result. Why?
>
> Thanks in advance for your input regarding my
> question.
>
>
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