>> >> 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?
>
> This is getting silly. =A0Report back when you've looked up
> Thevenin Equivalent.

So, may I conclude that you insist that for your circuit

"perfect 0 to 5V square wave followed by the 2K,3.9K ohm
divider, driving some capacitive load connected to ground"

when we decrease divider's bottom resistor 3.9K to 3.85K, the time to
rise the signal on the capacitor from 0V to 4.9V will decrease as per
your statement that - the lower impedance - the higher speed of signal
rising on that capacitor.

-- =

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