[SI-LIST] Re: Impedance calculation of asymmetric coupled lines?
- From: "john lipsius" <johnlipsius@xxxxxxxxx>
- To: <dgun@xxxxxxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Fri, 18 Oct 2002 16:37:49 -0700
All,
I think there's some lack of definition here, and so some bit errors :-).
See gigatest.com for simple, clarifying tutorial on this.
1. hfchen is, I think, correct. Zeven is a quantity reflecting even mode
driving a
pair, not defined for a single conductor or a property of a single one.
2. Perhaps what DG intends is that Z11 isn't equal to Z22? I suspect it
since he appears to mean one trace closer to ref. plane than the other,
but the two are still *uniform*. In that case, they are *still*
symmetrical
but the Zeven and Zodd (for the pair) are changed; there's now a mixture of
the two signal modes, odd and even. Therefore, terminations would have to
change
(over the Z11=Z22 case) to maintain SI. And, "differential" and "common"
impedance ideas still apply; they are a driving signal definition, not a
description of
physical symmetry. You have to terminate both impedances instead of just
the
differential.
3. If, instead, DG means there's a non-uniformity in the pair at some point
such that
any or all Zij change, then there's *is* a change of pair symmetry and a
change in
Zeven and Zodd at that place. You then have some Vdiff and Vcm reflections
(some conversions between even and odd modes locally)-- a lack of SI then
results
no matter what terminations are made.
Just my opinion. Other clarification welcome.
--
----- Original Message -----
From: "D G" <dgun@xxxxxxxxxx>
To: <si-list@xxxxxxxxxxxxx>
Sent: Thursday, October 17, 2002 4:23 PM
Subject: [SI-LIST] Re: Impedance calculation of asymmetric coupled lines?
>
> As I learn more about it, there are still two modes, but they are not
> strictly even and odd. I
guess I was referring to the fact that the capacitances from each line to
ground will be
different.
>
> I found some interesting, but math-intensive papers that describe SPICE
> modeling of assymetric
coupled lines. If anyone is interested, I can give them here.
>
> From: houfei chen <hfchen73@xxxxxxxxx>
> > Zeven and Zodd is defined for coupled line together,
> > one is for even signal input, the other is for odd
> > signal input.
> > Correct me if I am wrong, but I don't think there is
> > Zeven for line 1 and Zeven for line 2.
> >
> > --- D G <dgun@xxxxxxxxxx> wrote:
> > >
> > > Fair enough. What do I get out of an asymmetric
> > > coupled line when I put a differential signal into
> > > it? How do I simulate this with transmission line
> > > models?
> > >
> > >
> > > From: Fred Balistreri <fred@xxxxxxxxxxxxx>
> > > > If they are not electrically symmetrical then they
> > > cannot be called
> > > > differential so the definition does not apply.
> > > >
> > > > Best Regards,
> > > >
> > > > D G wrote:
> > > > >
> > > > > The calculation of differential and common-mode
> > > impedances is well known for symmetric coupled
> > > lines. How about asymmetric coupled lines, where
> > > Zeven for line1 is not equal to Zeven for line2?
> > > > >
> > > > > --
> > > > > Daniel
> > > > > ZZZ-dgun-ZZZ-@xxxxxxxxxxxxxxx
> > > > > (remove the Z-'s to reply--they're what I do
> > > when I read spam)
> > > > >
> > > > > --
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> --
> Daniel
> ZZZ-dgun-ZZZ-@xxxxxxxxxxxxxxx
> (remove the Z-'s to reply--they're what I do when I read spam)
>
> --
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