Tim Hutcheson wrote:

> Actually I was thinking the authors meant that the driver absorbs about 5/6
> of the signal at frequency, while the resistor "absorbs" the remaining 1/10,
> at any frequency. Is that incorrect?

That is correct.

> Keep in mind that only a resistor can "absorb".  A PCB trace cannot absorb
> anything.  If you really need a reflection absorbed at the source, you
> really need a resistor to absorb it.

Yes, but!

A resistor can be made many ways.  One way is to bias a transistor into its
linear region.  Another way is to put an amplifer in series with a sense
resistor,
and have a feedback loop adjust output voltage so that a specific voltage ratio
is seen across the sense resistor vs. the output pin.  This second method is
probably what your chip is doing.  The first method seems to be what some
other special drivers do.

The typical CMOS output driver has an output impedance around 1-2 Ohms
in the normal current range of the chip.  For short circuit protection, the
output impedance goes up with higher current.  This can actually be a
point to consider, that the pin is not ALWAYS a 50 Ohm load on the line,
only within a certain domain.

So, Ivan's comments about ONLY a resistor can absorb is correct when talking
about conventional "hard output" CMOS chips.  Due to these transmission line
problems, many newer chips are putting active resistors on critical output pins.

Any digital chip with a 50 Ohm output impedance is definitely specialized for
this sort
of problem.

Jon

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