# Signal reflections at the driver?

This question is related to How do I use directional couplers on a differential signal?

I understand that if I have an impedance mismatch at the end of a signal cable, E.G. from a terminator which doesn't match the cable impedance, then I can expect to get reflections back.

What I don't understand is exactly what happens when you have an impedance mismatch at the driver end of the cable.

If a very short part of the cable at the driver end has an unknown impedance, is this a problem? Will is cause noticeable reflections? My instinct would be 'no' because the length is so short.

I mean, this must happen in all situations anyway. The impedance of the driver IC's legs are not impedance matched, but they are so short that it's not a problem. Obviously, the length in that case would be very short indeed. If that's the case, then the question would be: "How long is too long?"

This cable will be carrying 100Mb/s LVDS (which stands for Low Voltage Differential Signal). This is my estimation of the spectrum of the signal.

Which is based on this wave shape:

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I don't think you would get significant reflections. If the load impedance is matched to the transmission line, there would be no reflections on that transmission line. In reality there's a tiny transmission line between the driver load and the long transmission line, so there's an opportunity for reflections there.

Think of it this way: "looking into" the transmission line, you will "see" a certain impedance (the wave impedance). Likewise, "looking into" the driver impedance from the driver itself, you will "see" another impedance.

Your driver transmission line is 15mm. The relative permittivity of most metals is 1, so we can assume the propagation speed is $c$. Your signal is 100 Mb/s, so let's assume a signal frequency of 100 MHz. The wavelength is thus:

$\lambda=c/f=(3 \times 10^{8} \mathrm{m/s})/(100 \times 10^{6} \mathrm{Hz})=3 \mathrm{m}$

What matters is the length of the tiny transmission line in multiples of the wavelength:

$l/\lambda=0.015/3=0.005\lambda$

This is too small to cause significant reflections. While you might get reflections due to the possible mismatch (as explained here, the reflection depends on the impedance values and not the transmission line length), the line is too small for the standing wave pattern to change.

The only thing you have to worry about is the voltage divider. Since the main transmission line is matched, you'll "see" it as a 100 ohm load: this will form a voltage divider with the driving load. For optimal power transfer to the load you'll want to minimize the driver impedance.

EDIT: If you want further reading, I'd recommend a textbook on the subject. I've been reading Fundamentals of Applied Electromagnetics by Ulaby et al., it goes over reflections and matching in detail in Chapter 2.

EDIT 2: I'm not in the field, I'm an EE student and we just happen to be studying this subject at the moment. I'm sure that someone with professional experience could give a more practical answer.

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Thank you. This is a very helpful answer. –  Rocketmagnet Apr 11 '12 at 21:55
@JeffE, approximating a digital signal by its clock rate is a very very poor decision. It is determined by the rise time of the signal, called the knee frequency, which in this case is not much higher, only 1/.3 times faster. A side note, the best book for learning these things is generally accepted to be High Speed Digital Design: A handbook of black magic. It is a bit older but unbelievably great and quite a tough read at times(makes you think). –  Kortuk Apr 11 '12 at 22:17
As not to be all negative, Thanks for taking the time to visit the site and become brave enough to write an answer. I hope you continue to visit us. –  Kortuk Apr 11 '12 at 22:18
Not negative at all, I appreciate the comments. –  Jeff E Apr 11 '12 at 22:27
@Kortuk I would love to know your thoughts on this question. –  Rocketmagnet Apr 12 '12 at 10:04