From: Ingraham, Andrew ([email protected])
Date: Fri May 11 2001 - 08:01:37 PDT
> My simulations using XTK for most of the daisy chain topology is showing
> non-monotonicity on both the edges of the data/signal. I tried different
> termination strategies to remove the kink. But couln't.
> What I saw was that the worst affected load is the first load from the
We have a concept of "incident wave switching" and "reflected wave
switching". It's mainly a matter of degree.
If the driver's source impedance is much smaller than the effective line
characteristic impedance, then the driven waveform swings cleanly through
and past the threshold levels, and you have incident wave switching. ECL is
an example of this; the trace is about 50-75 ohms, but the driver is
typically <10 ohms. Incident wave switching usually requires parallel
termination at the far end, unless you like a lot of overshoot and ringing.
When the driver's impedance is close to the line impedance, it is "source
terminated", you don't get much overshoot, and we say it uses reflected wave
switching. The initial signal is only about half amplitude. It doesn't
reach normal amplitude until it reflects off the (unterminated) far end and
the reflected wave comes back toward the source. The receiver closest to
the driver sees the widest glitch or kink.
These glitches (and therefore, reflected wave switching) may be fine for
most non-clock data signals, but NOT for clocks or anything that is edge
sensitive. For clocks, you should use either a fanout part and
point-to-point traces, or a "hot" (strong) driver to get incident wave
> Suggest me how to remove the non-monotonicity while using the daisy chain
You can: Use a stronger driver; Slow down the edge rates; Shorten the total
trace length; or Use a fanout part and point-to-point traces.
> and is it ok if I bring that kink within the threshold levels.
Depends on the type of signals. If a kink bothers you, definitely DO NOT
bring it within the threshold levels. It is OK outside, but not inside.
Obviously, you should not do that if the signal is a clock.
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