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

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From: Tom Dagostino (tom_dagostino@mentorg.com)
Date: Mon Jan 10 2000 - 11:56:20 PST


Don't forget the old CMRR, common mode rejection ratio. The differential
reciever may see a common reflection but may not be able to handle it as you
would like. CMRR degrades with frequency.

To expand on the differential impedance

Zcm= Z11 + Z12
Zdiff = 2(Z11 - Z12)

Where:
Zcm is the common mode impedance
Zdiff is the differential impedance
Z11 is the self impedance of the line
Z12 is the mutual impedance between the two lines

On a typical circuit board Z12 will vary between 0 to 15% of Z11 depending
on the coupling (read spacing) between the differential lines.

Tom Dagostino
ICX Modeling Group
tom_dagostino@mentor.com
503-685-1613

-----Original Message-----
From: owner-si-list@silab.eng.sun.com
[mailto:owner-si-list@silab.eng.sun.com]On Behalf Of Joel Kolstad
Sent: Monday, January 10, 2000 10:52 AM
To: si-list@silab.eng.sun.com
Subject: RE: [SI-LIST] : Physcially-small far-end LVDS terminations?

> In general, doesn't one need to terminate both the
> differential and the
> common modes? The 100-ohm-only termination terminates the
> differential
> mode but not the common mode.

You'd argue that two 50 ohm resistors to ground would terminate both
common
mode and differential mode?

It seems to me there's little need to terminate common mode -- a
differential receiver will see the same common mode reflections on both
of
its inputs, thereby nullifying any common mode mis-termination
(reflections)
so long as the common mode input voltage maximum of the receiver isn't
exceeded.

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.

---Joel Kolstad

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