>> 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. =A0The 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?

-- =

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