At 07:08 PM 10/4/2009, you wrote:
> >> >> He propopsed an idea that an impedance calculated
> >> >> as (R1*R2)/(R1+R2) would reflect the "speed" of the signal -
> >> >> the lower impedance - the higher speed.
> >> >
> >> > Yes.
> >> >
> >> >> I objected it with the example to remove the bottom
> >> >> resistor, - the impedance will be higher, and the speed
> >> >> will be higher. That's it, nothing more, no rules changed.
> >> >
> >> > Yes they have.  The output will now go to 5V instead of 3.3V,
> >> > which is a violation of the rule that the output must drive
> >> > a 3.3V input within its spec.
> >>
> >>Well, the idea is scalable as one might expect. Decrease divider's
> >>bottom resistor R2 by 0.1% (to stay within specs). You'll get lower
> >>impedance according to your (R1*R2)/(R1+R2).
> >>Is not it obvious that despite the fact that impedance gets lower, the
> >>speed of rising will be lower, not greater as you stated above?
> >
> > Does the number 1-1/e mean anything to you?
>
>Why are you asking that?

To evaluate your background knowledge.

>Best regards,

Spehro Pefhany --"it's the network..."            "The Journey is the reward"
speff@interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com



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