[SI-LIST] Doug Brooks's Question
- From: "Eric Bogatin" <eric@xxxxxxxxxxxx>
- To: "Si-List" <si-list@xxxxxxxxxxxxx>
- Date: Sun, 16 Nov 2003 08:36:44 -0600
Doug-
I'll take a stab at answering your question about understanding the
termination for a transmission line.
There are two different ways of thinking about it. First is the view
of matched boundary conditions. This will allow us to derive the
reflection coefficient as (Z2-Z1)/(Z2+Z1). I wrote a column on this
for the August 2003 issue of Printed Circuit Design and Manufacture
Magazine. You can download a pdf version of the paper from my personal
web site:
http://www.ericbogatin.com/SignalIntegrityColumns/pdfFiles/0308NMA.pdf
The second way is, as you are trying to do, approximating a uniform
transmission line as an LC approximation. Keep in mind, this is an
approximation. It is perfectly ok, and accurate to about 1% in
predicting the impedance of a line up to some bandwidth as long as the
number of segments > the bandwidth x 10 x 1/(time delay of the line).
(this derivation is in my book)
If you have an n-section LC network, and add a resistor at the far
end, the real question you want to ask is, what value of R across the
end will critically damp the LC network. This question applies to 1 LC
section, or 100 LC sections in the line. It is the last LC section
that will be damped.
Now, we have the question phrased as a circuit problem. If you have 1
LC section and add a resistor in series with the C-L-R, at the end of
the line, what is the Q of this circuit?
The Q is the ratio of the max energy stored in a cycle to the energy
dissipated in a cycle. It is also the ratio of the loss, the
resistance, to the reactance. Q = R/(2 x pi x f x L) = R x sqrt(C/L)
For any LC network, the last LC section will be a damped LC section.
The ideal R value will critically damp the circuit, with a Q of 1.
Note that for Q = 1, the value of R should be R = sqrt (L/C).
No matter how many sections we use to approximate our transmission
line, as long as the line is uniform, both L and C scale with the
length and we get the same ratio for sqrt (L/C). Of course, this is
also the characteristic impedance of the line.
The bottom line is, if you want to think about using an R to damp out
the oscillations of the last LC section in an n-section LC circuit,
you want to select an R that gives a Q of 1. This R is numerically the
same as the characteristic impedance of the line.
Of course, more details on approximating T lines as LC networks and on
termination can be found in my book, Signal Integrity Simplified,
published by Prentice Hall.
Hope this helps.
--eric
********************************************************************
Recently published by Prentice Hall, www.phptr.com
Signal Integrity-Simplified, by Eric Bogatin
Attend the GTL Signal Integrity University in Sunnyvale, CA - Nov
6-13
GTL 122 - Fundamental Principles of Signal Integrity
GTL 250 - High Speed Board Design
GTL 260 - Interconnect Models from Measurement
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---------------
Dr. Eric Bogatin
CTO, GigaTest Labs
26235 w 110th Terr
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www.GigaTest.com
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Date: Sat, 15 Nov 2003 14:21:25 -0800
From: steve weir <weirsp@xxxxxxxxxx>
Subject: [SI-LIST] Re: Doug Brooks's Question
Sainath, no, that is not the way that I see it. Unless I misread
Doug,
like you he noted that the LC at the end of the line, like all the
increments along the line has Zline. The problem is that a
transmission
line does not behave like a lumped LC. I don't personally see a way
to try
to replace the Tx line with a lumped LC, nor do I see a useful point.
If the idea is to think about the behavior intuitively, I personally
like
the parallels to conservation of momentum.
Jim Knighten's discussion of Kirchoff's Laws fits that nicely as well.
What doesn't seem to fit is a lumped LCR. The response doesn't
match. With a lumped LCR the ringing time constant is determined by
LC,
and L/R. With an underdamped transmission line, it is determined by
the
length of the line. All we get at the end of the line is L/C for
Zline. L
and C get scaled for convenience to solve the equations. The higher
the
frequency range we want to evaluate, the smaller both L and C become.
So,
the LCR time constant moves towards zero.
Regards,
Steve.
At 01:08 PM 11/15/2003 -0800, Sainath Nimmagadda wrote:
>Steve,
>
>I suppose Doug's track assumes that the line is matched with upstream
>networks and his concern is about the termination- specifically,
about
>interpreting time constants when it is not matched to the line. The
end LC
>of the line also represents characeristic impedance of the line.
>
>Do you see it differently?
>
>Sainath
>
>
>
>steve weir <weirsp@xxxxxxxxxx> wrote:
>Sainath, trying to combine just the end LC with the termination R
ignores
>the effects of the upstream networks and yields the wrong behavior.
>Regards,
>
>
>
>Steve.
>At 09:54 AM 11/15/2003 -0800, Sainath Nimmagadda wrote:
> >I understand the responses(using different tracks) to Doug's
> >question- but don't know why his track is wrong altogether.
Someone,
> >please explain why and where his approach doesn't work.
> >
> >Thanks,
> >Sainath
> >
> >[ from Archives]
> >==========================================================
> >From: Doug Brooks [mailto:doug@xxxxxxxxxx]
> >Sent: Wednesday, November 12, 2003 4:43 PM
> >To: si-list@xxxxxxxxxxxxx
> >Subject: [SI-LIST] Transmission line match
> >
> >I am trying to develop a *qualitative* (NOT quantitative)
explanation of
> >what happens at the terminating end of a transmission line. I think
I am on
> >the track with this model and discussion:
> >1. The transmission line model is a lumped model of a string of Ls
and Cs.
> >2. The termination value forms a time constant with these. That is,
if
> >R=Z=(L/C)^.5, then RC=L/R. That is, when correctly terminated, the
time
> >constants are matched.
> >3. If the terminating resistance is high, then the RC time constant
is
> >relatively longer. If it is low, then the L/R time constant is
relatively
> >longer.
> >
> >At this point I am having trouble getting over the "so what?" and
> >completing the argument. What is the next step in this
argument..... or am
> >I on the wrong track altogether?
> >
> >Thanks
> >Doug Brooks
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