Marechiare wrote:
>>>> Exactly. But to get the same speed, meaning same signal
>>>> impedence, you would need 1.2Kohms since
>>>> 2Kohm // 3Kohm = 1.2Kohm.
>>> On rising edge I don't see how does mentioned in your calcs
>>> 3Kohm divider's bottom resistor help to speed the rising.
>> This is very basic electronics.  Look up something called
>> "Thevenin Equivalent".
> 
> Interesting, obviously, removing the bottom resistor from the divider
> would increase the speed of rising (assuming inputs would tolerate
> it). But, according to your logic/formula removing the bottom resistor
> would increase impedance and, thus, decrease the speed of rising.

Not exactly. With two resistors the impedance is 1.2k and with only one
it's 2k. But the equivalent voltages are 3.3 and 5 respectively. If you
are driving logic with TTL(1.2 volt) thresholds the rise times will be
very close.

Cheerful regards,

Bob

-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist