[SI-LIST] Re: Even and odd impedances
- From: "john lipsius" <johnlipsius@xxxxxxxxx>
- To: <sunil-chandra.kasanyal@xxxxxx>, <si-list@xxxxxxxxxxxxx>
- Date: Fri, 14 Feb 2003 20:34:53 -0800
All,
Oops. I mis-defined the Z matrix (not shown in my email) so my
results are garbage. The new, new, much better than new version
is below. As usual, the derivation is left as an exercise for the
reader :-> .
definitions:
I1 = I + Ic , I2 = -I + Ic
V1 = V + Vc , V2 = -V + Vc for mixed diff./common
then V = ZI
results in :
Z_d = 2(Z11 - Z12)/(1 + [Ic/I]) diff impedance
Z_odd = (1/2)Z_d = (Z11 - Z12)/(1 + [Ic/I]) ea. conductor
and
Z_c = (1/2)*(Z11+Z12)/(1 + [I/Ic]) comm. mode impedance
Z_even = (Z11+Z12)/(1 + [I/Ic]) ea. conductor
--------
The above also indicates the impact on the Z's for different values
of differential and common mode current. This will give a 1st order
estimate of reflections.
--------
Still, as in the original email, Z_odd decreases with Ic and Z_even increases
from 0. This is logical.
When Ic = I, as an example,
Z_odd = (1/2)*(Z11-Z12) , or half the pure differential value
Z_even = (1/2)*(Z11+Z12) , or half the pure comm. mode value
The better answer to the original question is:
If you've terminated for pure differential, for example, and some
common mode noise exists, then the termination will not match Z_odd,
which has decreased, so you'll generate differential noise as well, and
common mode reflections will increase since no Z_even termination was
provided.
-John
----- Original Message -----
From: john lipsius
To: sunil-chandra.kasanyal@xxxxxx ; si-list@xxxxxxxxxxxxx
Sent: Friday, February 14, 2003 5:26 PM
Subject: Re: [SI-LIST] Even and odd impedances
Sunil,
You are asking something that alot of the papers and presentations haven't
analyzed (ignoring the complete literature, of course):
What *are* the impedance changes for mixed mode signaling (or crosstalk)?
Likewise, the right answers will indicate what Vdiff, Vcomm signals will be
injected to affect the signal integrity. I haven't seen this analytically on
this list yet.
By item...
1) see the links provided by other replies already, esp. the one at
http://www.ewh.ieee.org/r5/denver/rockymountainemc/archive/2000/diffimp.pdf
2) Z_odd decreases since V_odd decreases as -(Ic/2)(Z11-Z12), where Ic is
the common mode current part of the total. See below.
In the same mixed case (Icm and Iodd are present), Z_even increases as
(Ic/I1)/2.
Z_odd and Z_even are the two impedances seen on a *single* trace by
the odd and even signals.
3) See the doc at
http://groups.yahoo.com/group/si-list/files/Technical%20Documents/
by Loyer.
------------------------------------------------------------------
Details of item 2:
It doesn't matter whether we drive a Vdiff+Vcomm (mixed signal) or are
a victim of crosstalk. The same results occur...
Using the standard matrix formulation V = ZI and using I2 = -I1 + Ic and
V2 = -V1 + Vc to express the mixture of even and odd modes, one comes
up with:
definitions:
Z_odd = (1/2)*(V1-V2)/I1
Z_even = (1/2)*(V1+V2)/I1
so
Z_odd = (1-[Ic/2*I1]) * (Z11 - Z12)
Z_even = Ic/(2*I1) * (Z11 + Z12)
By inspection, Z_odd decreases as Ic > 0.
For terminations:
Z_d = 2 * Z_odd = (2-[Ic/I1]) * (Z11 - Z12) differential
impedance
Z_c = (1/2) * Z_even = (Ic/[4*I1]) * (Z11 + Z12) common mode impedance
Also, when Ic = I1 we have half the Z_even of pure even mode and half the
Z_odd of pure differential mode, so
Z_even = (Z11 + Z12)/2 and
Z_odd = (Z11-Z12)/2 for Ic = I1.
In such a case the termination would have to satisfy both since the
signal is using both equally.
-------------
Reverse topic: What will Ic or Id (comm. or diff. current) changes
generate for V diff and comm?
From Z_diff and Z_comm above, just interpret differently...
Starting with
V_diff = V1 - V2 = 2*V1 - Vc = (Z11 - Z12)*(2*I1 - Ic)
V_comm = (V1 + V2)/2 = Vc/2 = (Z11 + Z12)*(Ic/2)
where Vc is common to both traces,
Clearly, V_comm goes as Ic/2 or Vc/2. It makes sense that diff. I
won't affect V_comm. However, V_diff *is* affected by common I:
V_diff changes as -Ic or -Vc, which is twice as strong a sensitivity
to common mode currents.
---------------------
Finally, putting this stuff into Excel and playing with Z's, I's, etc.
allows one to get 1st order sensitivities for mode conversion, a good
antidote to handwaving.
----------------------------------------------------------------
----- Original Message -----
From: sunil-chandra.kasanyal@xxxxxx
To: si-list@xxxxxxxxxxxxx
Sent: Friday, February 14, 2003 5:35 AM
Subject: [SI-LIST] Even and odd impedances
Hello all,
(1). How the odd and even mode impedances of a differential pair are
defined?
(2). In differential pair when noise is equal on both the traces,
impedance seen by it is even mode and this noise slightly increases the
impedance of both traces. How?
(3). How can one terminate odd and even impedances?
Thanks and Regards,
Sunil C Kasanyal
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