Re: [SI-LIST] : Why can a resistor can reduce noise in output buffer

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From: Michael Nudelman ([email protected])
Date: Fri Dec 01 2000 - 12:45:55 PST

Because what you are doing is serilally terminating your output.

Explained on fingers it is like this:

Your signal (let's say - low-to-high transition), if a
low-output-impedance driver is directly connected to a transmission
line, will in its full amlitude travel to the end of that line, and
then, the line being unterminated at the end (infinite load resistance)
reflect back in its full amlitude; when it comes back to the source, it
adds itself to the existing signal and you will see the signal double
(that's your overshoot). Same for high-to-low transition - it will cause

Now, by plugging in a resistor in series with output, you divide the
signal between the resistor and transmission line. Suppose, the resistor
is equal to trans. line impedance. Than you divide it by half. So, half
of your signal travels down the line, then reflects and goes back,
adding to the existing signal (half added to half), making normal signal
amplitude (no overshoot).

In non-ideal situations you will see some overshoots/undershoots, but
significantly reduced.

Do not use this type of termination for clocks if you have more than one
clock receiver on line; series termination for clocks is only good for
point-to-point connection.

Wai-Ming Hung wrote:

> Hi
> I wonder why if I added a resistor into the output buffer can reduce
> overshoot and undershoot
> significantly ?
> Kenneth

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