From: Scott McMorrow (email@example.com)
Date: Tue Nov 14 2000 - 12:58:45 PST
Slew-rate is dependent upon amplitude. It therefore encapsulates
frequency domain information, as well as amplitude information.
There is one variable, slew-rate, and two unknowns (amplitude and
frequency response). To solve for frequency response, you need
to know the amplitude.
Rise-time is independent of amplitude. (i.e. it has been normalized
to the full swing amplitude.) As a result, the frequency response is
a known quantity.
At a signal swing of 1V, slew-rate is
slew-rate = 1V/(rise-time * 0.6) ** assuming 20-80% measurements
A 0.6 ns rise time has a 1V/ns slew-rate.
However, if the signal amplitude is 3V, then for the same rise-time
the slew-rate = 3V/ns
It is probably easier to work with a pulse that has rise-time shaping,
since it is amplitude independent. Here you are dealing with only
the harmonic content of the waveform. From this, the response
at a particular amplitude can easily be deduced through scaling.
-- Scott McMorrow Principal Engineer SiQual, Signal Quality Engineering 18735 SW Boones Ferry Road Tualatin, OR 97062-3090 (503) 885-1231 http://www.siqual.com
"Loyer, Jeff W" wrote:
> A question came up here that goes something like this: > If I'm going to buy a filter to slow my TDR pulse down to better mimic the > bus driver, should I slow it down to match the "slew-rate" or "rise-time"? > > For instance, a drive signal going from 1.0V to 1.8V in 0.4nS has a > slew-rate of 2V/nS, and a rise-time of 0.4nS. > > To mimic this driver's slew-rate, I would want my TDR (which has a 0.5V > voltage swing) to go from > -0.5V to 0V in 0.25nS. I would specify a 0.25nS rise-time filter. > > To mimic this driver's rise-time, I would specify a 0.4nS rise-time filter. > > Which is more important, the slew-rate or rise-time? > > The first intuitive answer is that slew-rate is the important characteristic > - that a signal with higher slew-rate has more high-frequency components > than one with a lower slew-rate. Unfortunately, this implies that our > "rules of thumb" which talk about rise-time only are assuming a specific > voltage swing (and therefore a specific slew-rate for that rise-time), and > valid only for specific voltage swing levels. > > The non-intuitive answer is that rise-time is the important characteristic, > regardless of slew-rate. This leaves the rules-of-thumb intact, but > violates our gut sense that a signal switching from 0 to 5V in 1nS is much > more "aggressive" (and sensitive to smaller discontinuities on a > transmission line, for instance) than one switching from 0 to 1V in 1nS. > > My conclusion at this point is that the correct thing to do is to mimic the > rise-time, not the slew-rate. > Some simple simulations show that doubling the risetime cuts the ringback in > half, while doubling the slewrate has no effect. > > Does anybody know of a good explanation (more than mere "gut feel" etc.) of > why the rise-time is the critical parameter, not slew-rate? > > Note: at this point, I believe the major flaw in thinking slew-rate is > important is in the sentence "...a signal with higher slew-rate has more > high-frequency components than one with a lower slew-rate." I'm pretty sure > that, when you do the Fourier Transform of a pulse, the amplitude of the > pulse is merely a constant and comes out of the integration. > > Jeff Loyer > (253) 371-8093 >
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