From: Dennis Tomlinson ([email protected])
Date: Fri Oct 13 2000 - 12:49:25 PDT
Bob, etc.;
Yes, this is a great discussion. It's a question my cohorts
and I have discussed on more than one occasion prior to this
thread. In fact, this thread lead me to attempt to formalize
the following...
Below is an analysis of the interface between a coupled
and an uncoupled differential pair. Both have the same
differential impedance (Zd), but different individual
Zo's. The analysis includes the path of both signal and
return currents at the interface. Kirchoff's Current Law
is used to enforce current conditions at the interface, and
to therefore infer a reflection coefficient.
"A" "B"
________ _______
---------()________()-----()_______()-------
\Z0a Z0b
\
\c
____\___ _______
---------()________()-----()_______()-------
Z0a Z0b
The sections with characteristic impedance "Z0a" have a
backward coupling coefficient "c", giving a differential
impedance of:
Zda = 2*Z0a(1-c)
If 2*Z0b = Zda, then we seemingly have a matched differential
interface between section A and section B.
Re-drawing with currents:
"A" "B"
Isa---> R*Isa<--- (1-R)*Isa--->
________ ___________
---------()________()-----------()___________()-------
Ira<--- R*Ira<--- (1-R)*Isa<---
Isa<--- R*Isa---> (1-R)*Isa<---
________ ___________
---------()________()-----------()___________()-------
Ira---> R*Ira---> (1-R)*Isa--->
In the above; R = reflection coefficient, Isa = forward
signal current, and Ira = return current (of forward).
In section "B", since there is no coupling, the return
currents are the same magnitude as the forward currents.
In the "A" section, the coupling gives rise to a disparity
in the magnitude of the return current flowing in the
reference plane, when compared to the forward current:
Ira = (1-c)*Isa
The total return current for either conductor in section
"A" is made up of the return current in the reference plane,
(Ira), with the remainder being in the coupled signal trace.
At the interface, Kirchoff's Current Law requires:
(1-R)*Isa = (1+R)*Ira
Substituting for Ira from above:
(1-R)*Isa = (1+R)*(1-c)*Isa
1-R = (1+R)*(1-c) = 1-c+R-Rc
Rc-2R = -c
R*(c - 2) = -c
c
R = -----
2 - c
I don't have hardware measurements to provide at the moment,
but we have simulated these effects using XTK. A 6" coupled line
section with Zo = 55.53 Ohms, Ze = 50.09, Zd = 2*Ze = 100.18 Ohms,
and a reverse coupling of Rvsx = 0.098 was used for the "A"
section, and a 6" length of uncoupled 50 Ohm lines were used for
the "B" section. To drive this, a differential TDR having 0.5V
pulse amplitude, 35 ps rise time, and 50 Ohm single ended
(100 differential) output was modeled. An ideal (Zin = infinite)
differential receiver was modeled for use at the "B" end.
The simulation showed a reflection at the TDR output at the interface
between the "A" and "B" sections, as expected. The coefficient of
reflection calculated from simulated date showed R = 4%. The expected
R calculated from the equation should have been 5.15%. The cursors
used to gather voltage data from the XTK simulation were limited
to one significant digit. The simulation thus showed the mismatch,
but was inconclusive at verifying the relationship between reflection
coefficient (R) and coupling (c).
Another curiosity results from looking at the reflection
coefficient of each single ended that signal without regard
to coupling:
Z0b - Z0a Z0b/Z0a - 1
R = ----------- = ---------------
Z0b + Z0a Z0b/Z0a + 1
By forcing a differential match between "A" and "B" sections we
get:
Zda = 2Z0a(1-c) = 2Z0b
from which:
Z0b/Z0a = 1-c
Substituting into the single ended R equation above gives:
1-c - 1 -c
R = --------- = -------
1-c + 1 2 - c
which has the same magnitude as R derived from Kirchoff's Current
Law, but opposite sign. The XTK simulation definitely showed a
positive voltage reflection. I'm still struggling with this one...
Perhaps it's something about common mode impedance???
Comments, suggestions, corrections,
Dennis (with help from Steve and Mike)
[email protected] wrote:
>
> Hi -
>
> This is a great discussion.
>
> Can someone post either theoretical or experimental evidence to
> show that the coupling type (loose, tight) along the entire path
> must be the same? In particular,
>
> 1) If I have loose coupling in the connectors and tight coupling on
> the boards, what's the consequence?
>
> 2) If I have broadside-coupled traces on the backplane and edge-
> coupled traces on the daughtercards (or vice versa), why would this
> matter?
>
> Granted, there might be a common-mode impedance discontinuity,
> but if we have a significant amount of common-mode voltage on
> these signals, aren't we in trouble already? After all, a lot of the
> 2.5Gbps links I've seen people propose have no meaningful
> common-mode termination in the first place.
>
> Thanks,
> Bob Perlman
>
> > I support Vinu's advice. If the connector is loosely coupled or not
> > coupled at all, then the traces on the backplane should be layed out
> > the same way.
> >
> > Chris
> >
> > -----Original Message-----
> > From: Vinu Arumugham [mailto:[email protected]]
> > Sent: Wednesday, October 11, 2000 12:41 PM
> > To: Michael Nudelman
> > Cc: Todd Derego; [email protected]
> > Subject: Re: [SI-LIST] : 2.5Gbps across a backplane?
> >
> >
> > Broadside coupling usually means significant coupling between the
> > signal conductors of the differential pair. However, many backplane
> > connectors may have little coupling between the signals of a
> > differential pair.
> >
> > For optimal performance, the type of coupling used on the board may
> > have to be compatible with the type of coupling in the connector.
> >
> > Thanks,
> > Vinu
> >
> > Michael Nudelman wrote:
> >
> > > Todd:
> > >
> > > First, do not use FR-4.
> > > Try something else. We do, and it works without equalization. And
> > > the
> > backplane
> > > is huge. The traces' runs are up to few feet. Also try to use
> > > broadside coupling; if your skin losses are high, this helps a bit.
> > >
> > > Mike.
> > >
> > > Todd Derego wrote:
> > >
> > > > I'm working to take a present 1.25Gbps backplane to
> > > > 2.5Gbps or
> > > > faster. Any thought on transceivers with preemphasis and/or
> > > > adaptive equalization to help get 2.5Gbps or better across a long
> > > > FR4 backplane?
> > The
> > > > eye looks pretty collapsed at 2.5Gbps but I have seen some
> > > > products the drive across a twisted pair and seem to recover a
> > > > signal from something
> > that
> > > > does look much like a signal.
> > > > Thanks!
> > > > Todd
> > > >
> > > > Todd DeRego
> > > > Senior Signal Integrity Engineer
> > > > Lucent Technologies INS
> > > > 200 Nickerson Road
> > > > Marlborough MA 01752
> > > > (508) 786-2168
> > > > [email protected]
> > > > >
> > > >
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