[SI-LIST] Re: Diff.Pairs - Return current distribution

  • From: "Knighten, Jim L" <JK100005@xxxxxxxxxxxxxxxx>
  • To: doug@xxxxxxxxxx, si-list@xxxxxxxxxxxxx
  • Date: Wed, 15 Oct 2003 18:28:49 -0400

Doug,

I a subscribe more to the view of your second paragraph than the first.

First of all, currents probably don't know whether or not they are "return"
currents.  Currents have but one mission and that is to flow back to their
source along the path of least impedance. When you have three conductors in
proximity, two traces and a ground plane, the only magic that makes them
coupled is if electric field lines begin on one conductor and end on another
and if magnetic lines of force encircle more than one conductor.  The terms
"loosely coupled" and "tightly coupled" are just descriptions of how strong
these interactions are, with respect to the two traces, not the plane.

Each trace is coupled to the other trace by E and H fields and each trace is
also coupled to the plane.  A signal current in one trace will induce
current in the ground plane, some of which is trying to head back to the
source along the length of the trace and some current heads off to the
region underneath the second trace since it has a path (non-conductive) to
the second trace by the EM coupling from plane to trace.  In addition, you
also have current (not conduction current) flowing from one trace to the
second trace.  

In the perfectly balanced case, the currents that one trace induces in the
ground plane are equal and opposite to the currents induced by the second
trace.  So, even though you have currents flowing in the plane, there is no
net current.  Therefore, it stands to reason that if the ground plane had a
gap in it, there would probably not be a large effect on two traces.  (A gap
would disturb these currents that are induced in the plane, but since these
currents all net out to zero, then if the gap is symmetrical, you will find
that no imbalance is introduced.)

In the case of tightly coupled lines (perfectly balanced), such as closely
spaced lines, or ground plane move farther away, the currents in the ground
plane are reduced in amplitude from what they would have been if the traces
were farther apart or if the ground plane were moved closer to the traces.
But still, the currents balance out to zero on the plane.

If you induce imbalance in any aspect of the environment of these two traces
next to a ground plane, then you wind up with a non-zero net current in the
plane (usually called common-mode) and then crossing a gap in the plane is a
bad thing from an EMI perspective - but probably not from the signal's
perspective.  The differential signal waveform is fairly insensitive to
imbalance.

Jim

________________________
James L. Knighten, Ph.D.
Teradata, a division of NCR                 http://www.ncr.com
17095 Via del Campo
San Diego, CA 92127
tel: 858-485-2537
fax: 858-485-3788


-----Original Message-----
From: Doug Brooks [mailto:doug@xxxxxxxxxx] 
Sent: Wednesday, October 15, 2003 8:45 AM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Diff.Pairs - Return current distribution

I would like the group's opinion on something.

One view is that the currents on the plane are "return" currents, as they 
would be if we treated the traces simply as a pair of microstrip traces. 
But what if we don't treat the differential pair as simply two microstrip 
traces? After all, microstrip traces have their return currents leaving 
(for example) the ground pin of the device and then traveling on the plane. 
With a differential pair, there is no current that flows through the ground 
pin.

What if we took the view that the currents on the plane are NOT return 
currents. What if we believed that the *return* current is on the other (-) 
trace and the currents on the plane are circulating, coupled, eddy currents 
(formed by something analogous to transformer action). The further apart 
the traces are, the further apart the current loops are. And if we move the 
plane very far away, the main signal current loop doesn't change, but the 
coupled signal on the plane gets smaller (because the coupling gets weaker 
with distance.) This view also explains why differential circuits can 
perform without planes and with planar discontinuities.

The distinction between these two views is *very* significant, because the 
*current loop* definition (relevant for EMI) turns out to be very different 
depending on which view you consider to be correct.

The situation is actually even more complicated than this, because there is 
a combination of effects at work here. If we had a *perfect,* very high 
frequency square wave, the signal during the transitions would couple to 
the plane through the "distributed caps" along the transmission line. But 
during the "flat" parts of the waveform, the return signal would not couple 
to the plane (the "distributed caps" would be fully charged) but would be 
on the other trace.

Doug Brooks




At 10:55 PM 10/14/03 -0700, Charles Grasso wrote:
>Referring to Eric Bogatins book (SI simplified)
>there are nice pictures that show the current distribution
>in a pair of coupled transmission lines over a ground plane.
>Fundemetally, when the coupling between the signal line
>and the return plane is much larger than the the coupling
>to the adjacent signal, there are seperate and distinct
>return currents with very little overlap in the plane.
>
>(to paraphrase from the book)


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