Guys, I need to take a break for about 20 hours, sorry.

See you later.

On Mon, Oct 5, 2009 at 2:58 AM, Marechiare <marechiare@gmail.com> wrote:
>>> >>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.
>
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist