RE: [SI-LIST] : Physcially-small far-end LVDS terminations?

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From: Greim, Michael (mgreim@mc.com)
Date: Mon Jan 10 2000 - 11:58:50 PST


Joel Said:

        BTW, WRT the guy who said to route two 50 ohm lines instead of a 100
ohm
        differential pair... two 50 ohm lines as close together as
differential
        pairs usually are is going to produce something more along the lines
of
        90
        ohms of differential impedance, I believe.

Hi Joel,

I believe it was Lee Ritchey who brought up this topic. He
has also written a paper on this topic (differential signaling
doesn't require differential impedance). The quickest way
to describe this is as follows: Where the following comes from
is through simplifications of the matrix equations:

                Z diff = Z(11) Z(12)
                        Z(21) Z(22)

The first simplification occurs is that since the pair are routed
with the same routing rules in the same stack up, the equation
simplifies to the following.

                Z diff = 2 ( Z(11) - Z(12))

As you stated below, if the lines are close together, the mutual
coupling reduces the impedance of the line. However......

As the separation between the two traces becomes large in
comparison to the height above the plane. the mutual coupling
approaches zero. This is sometimes refered to weak coupling
model. Therefore the equation simplifies to:

                Z diff = 2*Z(11) or 2 * Zo

For traces routed with a 50 ohm impedance this effectively yields
a 100 ohm differential impedance.

Best Regards,

Michael

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