>> >>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.

To evaluate your school knowledge of physics:

For the circuit:

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

We decrease divider's bottom resistor 3.9K to 3.85K. The impedance as
per Olin (R1*R2)/(R1+R2) will decrease.

How will change the time to rise the signal on the capacitor from 0V to 2.9V ?

Will it decrease as per Olin's statement that - the lower impedance -
the higher speed of signal
rising on that capacitor.
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