From: Mike Jenkins (email@example.com)
Date: Thu Mar 22 2001 - 18:02:09 PST
I would argue that, by symmetry, there would be ZERO coupling
between the single-ended line (5-6) and a differential signal
on lines 1 and 2. Common mode coupling would occur, of course.
Also, differential coupling would occur to the extent that
there were asymmetries.
ps: Nice character graphics!
Hassan Ali wrote:
> Below is a poor sketch of a differential pair routed around a via in one
> layer of a PCB, call it Layer2. The via routes a single-ended signal from
> Layer1 (above Layer2) to Layer3 (below Layer2). I need to compute the
> crosstalk between the diff-pair and the single-ended signal. To do that, I
> have to obtain the equivalent circuit that takes into account self
> parasitics of the via and the coupling between the via and the diff-pair.
> Does anybody know how this equivalent circuit should be? I'd also appreciate
> it if someone can show me how to accurately compute all the parameters of
> the equivalent circuit.
> single-ended signal (from
> node 5 to node 6 thru the via)
> 3 4 5 ____________
> | | ||
> | | ---------- ||------------ gnd
> /via \ diff-pair ||via
> 5 ---|--0---|----- 6 normal to screen 1.||.2
> \ / ___________ ||____________ gnd
> | | ||
> | | diff pair ||
> 1 2 (line1 from node 1 to 3) ---------- 6
> (line2 from node 2 to 4)
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