[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 17:26:41 -0800

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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