FreeLists / si-list / [SI-LIST] Re: GND is perfect conductor?

["Jack" <mediwheel_js@xxxxxxxxxxxxx>]

All,

Forgive me, but I don't understand how your models are constructed???
My assumptions would be that your ground plane is part of the structure
in the model and not connected to "0" an ideal ground???  Is it commonplace
to analyze point to point or multidrop in this manner???

Any input would be appreciated.

jack ~


----- Original Message -----
From: "Ingraham, Andrew" <a.ingraham@xxxxxxxx>
To: "Signal Integrity Mailing List" <si-list@xxxxxxxxxxxxx>
Sent: Saturday, September 06, 2003 9:59 AM
Subject: [SI-LIST] Re: GND is perfect conductor?


> The in-house field solver we had at Compaq/HP did not use perfect
conductors
> for ground.  It rolls both the ground and signal conductor losses together
> into the model you create.  I would guess that many other field solvers do
> the same.
>
> SPICE t-line models (and W-line models are probably in the same boat)
might
> seem to have a perfect ground because you normally connect the reference
> nodes at both ends together to the same ideal GND (node 0).  But that's
not
> the same thing as treating the ground as ideal.
>
> T-line models can be thought of as being transformer coupled; what comes
out
> on the far end is isolated from ground and you get the same differential
> output signal whether you connect the far end's reference node to GND or
to
> any other arbitrary node.  The SPICE t-line model represents only the
> differential propagating mode, but it can correctly include ground losses
> with that mode.  The Berkeley SPICE documentation talked about using two
> T-line models to represent both even and odd modes of a single
two-conductor
> transmission line.
>
> Regards,
> Andy
>
>
>
>
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