# Re: [SI-LIST] : Critical Parameter: rise-time or slew-rate?

From: Scott McMorrow ([email protected])
Date: Tue Nov 14 2000 - 12:58:45 PST

Jeff,

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.

regards,

scott

```--
Scott McMorrow
Principal Engineer
SiQual, Signal Quality Engineering
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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