RE: [SI-LIST] : Source termination of transmission line

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From: Ingraham, Andrew (Andrew.Ingraham@compaq.com)
Date: Mon Apr 23 2001 - 15:58:52 PDT


> Source (series) termination of transmission lines to absorb
> reflections is well known. However, how long does the transmission line
> have to be to be recognised ,by the signal, as a transmission line?
 
There is no minimum length. A t-line of any length is a t-line. When it is
very short, we might not bother to model it as a t-line (purely for the sake
of mathematical simplicity); but at any length, it pretty much has to obey
the Telegrapher's equations, Maxwell's equations, etc.

But what I think you are asking is, can you somehow magically "fix" an
impedance mismatch problem over here by adding or lengthening a trace over
there until it behaves like a transmission line? If so, then I think you
misunderstand. (Unless your problem is narrowband, i.e., RF; as opposed to
logic signals.)

When the driver's impedance matches the t-line connected to it, all it does
is eliminate reflections that might otherwise happen RIGHT THERE at the
driver/t-line boundary. It does nothing for the impedance mismatches
elsewhere, i.e., between the connector and the t-line that comes before it,
or the one after it.

The reflection from the 40 ohm / 50 ohm interface will move towards the
source. Upon reaching the 40 ohm / 75 ohm interface, it will reflect again,
causing a forward-moving wave. It doesn't matter whether that "75 ohms" is
just the driver, or the driver plus any arbitrary length of transmission
line. (Well, it might, as a second- or third-order effect, partly because
the driver's actual impedance is probably nonlinear and/or time-varying,
etc. But for basic considerations, no difference.)

So don't bother lengthening the trace; you will just get more delay.

I wonder if you would have better luck with a 50 ohm driver, to match the
other trace, and accept the (usually short) impedance discontinuity through
the connector. If the driver is fixed at 75 ohms, you might try adding a
shunt resistor there to bring the combined impedance down. Or maybe not.
It might not actually help; but there are various things you can *try*
doing. Do you have the means to simulate any of this?

Regards,
Andy

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