From: Shayle Hirschman (firstname.lastname@example.org)
Date: Fri Feb 25 2000 - 08:49:25 PST
That's interesting, Andy. Can I infer, then, that EMI or perhaps coupling
is reduced more by series source termination since the voltage and current
are halved near the source?
At 05:19 PM 2/25/00 -0500, you wrote:
>>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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