[SI-LIST] Re: Circle bus topology; Circular Firing Squad?
- From: olaney@xxxxxxxx
- To: si-list@xxxxxxxxxxxxx
- Date: Sat, 28 Jul 2007 10:57:39 -0700
Again, no. I described it exactly as wave theory predicts. For a
circular trace driven at one point, the two waves launch in each
direction, meet at the electrical halfway point, continue through each
other, and finally meet again at the origin, where the termination
impedance determines what happens next. At the halfway point, the linear
superposition of the two waves causes the voltage to double as they pass
through each other, i.e. reinforcement, and current to droop to zero, but
that is only a virtual open circuit, not a physical one.
Wave theory teaches that a step launched into a single ended trace will
propagate to the end, and meet a virtual wave headed the other direction
(i.e. reflect), because the open circuit demands that the end point
current must be zero. The incident wave and its own reflection add by
linear superposition, causing the instantaneous voltage to double at the
trace end. Assuming a step waveform in, the doubling propagates back to
the origin, yielding the classic two step waveform we see on
oscilloscopes. For a CW sine wave, we see nodes and antinodes along the
trace that are used to calculate VSWR. But we are talking about a looped
trace, where the two branches have minimal coupling between themselves.
It is almost the same except that the return path for each direction is
the entire loop.
Here's an experiment that you can do with a TDR and a looped trace. At
the drive point, make the waves launched in each direction slightly
different from each other. This can be done by placing a small piece of
ferrite on one of the branches as close to the origin as possible, or
some other means of slowing the step edge (or making it ring, whatever)
for that one direction as an ID marker. This allows us to distinguish
the waveform launched into one direction from that launched in the other
direction.
Now, take your oscope probe and look at the waveform at each branch. Let
us call them A and B. You will see the wave launched into B arrive at A,
and the one launched into A returning to B. In a sense, the halfway
point could be called an open circuit per your interpretation, based on
the observation that where the waves have passed through each other, the
local current drops to zero. But this is due to the superposition of
equal but opposite currents, not any open circuit per se. To the extent
that the waveforms in each direction differ (and we arranged for that)
the instantaneous sum is not zero, and the difference is observable.
Take your scope probe and move it around the trace to view this effect.
Let's not be fooled by viewing a looped trace as a single ended one with
a slot down the middle. That viewpoint puts the round trip on the same
piece of copper, caused by an open circuit at the far end. In a looped
configuration, the waves go all the way around the loop, and there is no
far end as such. Consider that if you cut one of the branches next to
the origin, that is where the far end of the just created single ended
trace will be. Cutting the trace creates the ultimate directional ID,
all in one direction and none in the other, which allows you to view the
single propagating wave without superposition. As before, once the
signal traverses the entire loop, the (open circuit for the severed
branch) terminating impedance determines what happens next.
You might wish to attend the three day course (bessercourse.com course
193) I will be teaching in San Jose, starting August 13. The discussion
includes TDR theory, and I can make this example part of the demo.
Orin Laney
On Fri, 27 Jul 2007 23:22:20 -0700 ronald miller <ron@xxxxxxxxxxx>
writes:
olaney
The two waves will not reinforce each other. They see an apparent open..
Ron
olaney@xxxxxxxx wrote:
No, the waves in each direction have the same phase at the source. They
would meet and *reinforce* each other at the electrical halfway point,
continue around the loop in both directions, and meet again at the
source. Unless the source was back terminated, they would reflect from
the low source impedance, this time in inverted phase from the original
sense, and repeat the exercise until losses cause the waveform to die
out. This is just a wierd example of exactly what happens on a regular
single ended trace with an open end, except that you are driving two
traces of identical length in parallel that have their ends shorted
together. Also note that to back terminate the source, its impedance
would have to be half that of the transmission path, because it sees both
ends of the line in parallel. Example: 50 ohm source to drive a 100 ohm
looped trace. Or, you could terminate the halfway roundtrip point the
same way. Or both, if you double the drive level (great for video but
not good practice for logic signals). Basically, ring layouts in copper
foil are strange birds, useful for narrowband microwave purposes, but not
of much use for wideband, time domain waveforms for all the reasons
mentioned.
Orin Laney
On Thu, 26 Jul 2007 14:21:57 -0700 "Salkow, Steven"
<steven.salkow@xxxxxxxx> writes:
Mr Townsend et All made good points.
DFM software would flag this as an error.
Electrically, I think the idea is that the propagating standing
magnetic
wave of each arc path would exactly cancel each other when the waves
met
and no reflections would reflect back to the source. If one side of
a
path was not exactly equal, the point at which the waves cancel
would
not be at an physical midpoint rather at an electrical midpoint
which
may fall on a driven node. The point made by others already is
better
schemes exist already such as star distribution where end point
termination may be used to easily terminate a line with quite
predictable results.
ss
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]
On Behalf Of Townsend, Fred
Sent: Thursday, July 26, 2007 9:05 AM
To: David.Carney@xxxxxxxxxx; si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Re: Circle bus topology; Circular Firing Squad?
David:
I have to ask what would be gained from such a topology? Ron and
Scott
both make some very good points. In the case of the Mux bus
(1553)the
ring encompasses the airplane and the double ring structure gives
increased reliability. Microwave can make use of some structures
like
directional couplers to help reduce reflections. If rings are good
why
aren't we still using the token ring? Again the token ring was over
an
area much bigger than a PCB. A ring in a PCB would have all of the
problems with no apparent gain. Think about your PCB router. Rings
would
drive the router nuts?
Fred Townsend
-----Original Message-----
From: si-list-bounce@xxxxxxxxxxxxx
[mailto:si-list-bounce@xxxxxxxxxxxxx]
On Behalf Of David Carney (Neenah)
Sent: Thursday, July 26, 2007 6:47 AM
To: si-list@xxxxxxxxxxxxx
Subject: [SI-LIST] Circle bus topology
Has anyone ever experimented with a circle bus topology. The basic
concept would be a bus with several devices attached. They would be
routed in a daisy chain topology, and then the two ends of the daisy
chain would be connected together. The PCB routing would look like
a
circle or a loop for each net on the bus. Pointers to references
such
as papers or application notes would be greatly appreciated. I'm
particularly interested in signal integrity and EMC implications of
this
topology.
=20
Thanks.
=20
=20
----------
David Carney
Senior Hardware Engineer
Plexus Corp.
Phone - 920-751-5646
=20
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--
Ronald Miller
Ghz Data, Signal Integrity Consulting
7721 Sunset Ave.
Newark CA 94560
tel 510-793-4744
cell 510-377-9380
fax 510-742-6686
www.ghzdata.com
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