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

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From: Ingraham, Andrew (Andrew.Ingraham@compaq.com)
Date: Tue Apr 24 2001 - 05:20:07 PDT


Dave,

Not enough time for a decent response, so this may be kind of rough....

> > 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.
>
> I have a hard time visualising that in practice. When I TDR lines I
> know that any discontinuities which last less than the TDR risetime do
> not appear with their full value. So yes, to a zero risetime signal an
> infinitely short line would appear as a line, but for signals with real
> risetimes I think the line has to have a minimum length depending on the
> actual risetime.
 
OK, a t-line has to be some minimum length for you to *notice* it as a
t-line. But a shorter t-line is a t-line nonetheless.

As for reflections at impedance mismatches, or lack thereof when you match
the impedances, the principle holds no matter how short the line is. It's
just that you might not notice a MISmatch if the t-line is short compared to
a risetime.

When a t-line is very short compared to the risetime, then its
characteristic impedance doesn't much matter either, and you can mismatch
impedances to your heart's delight.

---

> True but while some of the energy is reflected back towards the load the > greater part carries on towards the driver and if the driver is source > terminated it ends there and doesn't get re reflected so I get just one > ring and not a whole sequence. Now my thinking is - if the driver is > hard on the connector it's just 75 ohms trying to terminate 40 ohms, > doesn't work and in any case putting the driver hard on the connector is > generally not possible. If the driver is some way from the connector > then it needs a t-line to the connector and if that t-line matches the > driver impedance then anything that gets back to the driver stops > there. Saves the shunt resistor and it's imperfections Um, I think there's some flaws in this reasoning.

If you start with a backward-moving wave in the 40 ohm medium, then what you want to do is match the *40* ohms, not the 75 ohms. The fact that you are connecting 40 to 75 ohms, means that tiere is a mismatch right there, with consequent reflections.

Whether you have just the 75 ohm driver, or put it behind a 75 ohm t-line of any length, it looks like 75 ohms. Remember, a 75 ohm t-line of any length with a matched (75 ohm) load on the far end, looks just like 75 ohms. So your connector sees 75 ohms, looking back toward the source, NO MATTER HOW LONG THE LINE IS. (Well, ideally.)

On the other hand, if you used a 40 ohm driver, and optionally connected to it with a 40 ohm trace, then there is (ideally) no mismatch between it and the connector. The only reflections you get are at the 40 / 50 ohm interface. (Ideally.)

If you consider just the driver, connector, and the trace between them (and ignore what comes after the connector), then you have a time-dependent reflection problem. Think of it like a TDR, looking back toward the driver. At first, you have a reflection due to the mismatch between the 40 ohm connector and 75 ohm line. This part of the reflection lasts roughly twice the delay of the t-line. After that, and from then on until infinity, the reflection continues at the same magnitude (so that you don't notice any difference), but is now due to the mismatch between the 40 ohm connector ... considered in this case as your reference impedance for the TDR ... and the 75 ohm driver. If you somehow changed it to a 40 ohm driver, you would see only the t-line reflection, which would last only a finite time. If the 75 ohm trace were much shorter than the risetime, you wouldn't even notice its reflection.

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