From: Ingraham, Andrew (Andrew.Ingraham@compaq.com)
Date: Fri Feb 25 2000 - 14:19:54 PST
>Does series source termination defeat the purpose of selecting the fast
>slew rate option? Will it slow the edge transition time down, or will the
>signal still transition and travel as fast as if there was no termination,
>but simply be divided (thereby depending on the reflection at the
If you imagine driving ideal transmission lines, then there is no difference
on edge rates of the voltage at the far end of the line.
But with a small capacitive load, series termination would slow the edge
rate down a little.
If the line length is small, you have an RC network with some time constant.
Adding series termination increases R. Adding parallel termination reduces
For a long line, it's more complicated. But if you looked just at the load
end and small time intervals compared to the line delay, your time constant
at that node is Zo*C without parallel termination, Zo*C/2 with it.
> Could it be that it
>reduces EMI, not by slowing the edges down, but rather just simply because
>it terminates? And that, therefore, as far as EMI goes, parallel
>termination would have a similar EMI result?
Either termination can help, but there is more to it than just that.
Series termination reduces the size of both the voltage steps (delta-V) and
current steps (delta-I) of each pulse. The signal driven into the line is
approximately a half-sized replica of what it would have been even with
parallel termination. dv/dt and di/dt are about half. At the end of the
line, the voltage doubles, so the magnitude (and dv/dt) there are the same
as they would be for parallel termination, except for capacitive loading
effects noted above. di/dt at the load end is nearly zero for series
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