From: Martyn Gaudion ([email protected])
Date: Fri Jun 23 2000 - 15:18:55 PDT
Harris,
You need to remember that the trace has finite thickness..
If you use the Polar CITS25 Field solver and dimensions
are for example:
H (between Planes ) = 10
H1 plane to stripline = 5
T (line thickness) = 1
If you invert the stucture
H1 does not equal 5
H1 = 10 - 5 - 1 = 4
Try this in the CITS25 and your answers
will be identical...
Kind regards
Martyn Gaudion
Marketing Manager
Polar Instruments
Tel: + 44 1481 253081
Fax: + 44 1481 252476
www.polar.co.uk
email: <[email protected]>
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At 11:41 23/06/00 -0700, you wrote:
>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.
>
>Haris Japic
>
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