> maybe the impedance mismatch is a problem, the ethernet cards are
> designed to drive 25 ohm load when they're sending. (50 on each
> end of the run)

JT> As I understand it, it's still 50 ohm. The terminators make the
JT> ends look like an infinite length cable so there are no
JT> reflections.

 That's true, but from the card's point of view, it is driving a
constant current into the 50 ohm terminator in parallel with the
50-ohm line - 25 ohms. Normally, the *source* resistance of the card
is 50-ohms, in which case it only sees the 50-ohm cable and the
voltage is stepped down a half. In this case, the current is stepped
down a half.

JT> Reflections garble the data enough to make it useless for
JT> everyone let alone the transmitter trying to detect a collision
JT> or not.

 That's also true, but with lossy cable the reflections are lost,
anyway. I'm not talking about an ideal world (as taught to techs at
TAFE), I'm talking about the real world.

JT> In theory, cable length should not be an issue (aside perhaps
JT> from degraded signal strength by the time you get to the end of
JT> that 300 metre stretch), although I've seen some cards have a
JT> _minimum_ cable specification (about two metres if I recall).

 Minimum?

JT> Again in theory, if you had a long enough cable length on one
JT> (or both) ends, the system should still work, because there are
JT> no reflections, thus no corruption of signal.

 Thin ethernet cable loses 3dB over 10 metres, so any refection is
halved, which makes the cable think it is pretty-well terminated
anyway.

JT> What you would need is a cable length long enough at the ends
JT> that the inevitable reflections are degraded enough by the time
JT> they come back, that they offer little or no effect on the
JT> closer valid signals.

 No... that's the *ideal* situation. In reality, half is good
enough, as long as it's terminated on *one* end.

Regards,
Bob
  

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