Re: [SI-LIST] : 2.5Gbps across a backplane?

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From: J. Eric Bracken (bracken@ansoft.com)
Date: Fri Oct 13 2000 - 13:46:57 PDT


It seems to me that the "tight" versus "loose" coupling that
we're talking about here can affect both the common-mode
and the differential-mode impedances of the two coupled lines.
But if you somehow manage to match up the differential
impedances and not the common-mode impedances, then
you will observe that:

   * a pure differential signal (+1 on one line, -1 on the other)
     propagates across the discontinuity without any reflection

   * a pure common mode signal (+1 on both lines) exhibits some
     reflection when it hits the discontinuity.

   * A "mixed" common-mode/differential signal (such as a
     single-ended input signal) would exhibit some
     reflection, but only of the common-mode component. A
     differential receiver with sufficiently good common-mode
     rejection ratio won't notice the reflections/oscillations in the
     common-mode component.

The above applies if the two lines are symmetric--if they're
not symmetric, then they don't really support a true differential/
common-mode pair of signals, and you'll see some "crosstalk"
from the common-mode signal into the differential signal when
it hits the discontinuity.

The bottom line is: if you can keep the common-mode component
of your input signals small, and your differential impedances
closely matched, then there shouldn't be a problem.

--Eric

Dennis Tomlinson wrote:

> 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)
>
> bobperl@best.com 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:vinu@cisco.com]
> > > Sent: Wednesday, October 11, 2000 12:41 PM
> > > To: Michael Nudelman
> > > Cc: Todd Derego; si-list@silab.eng.sun.com
> > > 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
> > > > > tderego@lucent.com
> > > > > >
> > > > >
>
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