>> How will be the time to rise the signal on the capacitor from
>> 0V to 80% of 3.0V (that is 2.4V) in the case of 2.95K
>> bottom resistor compared to the rise time when the bottom
>> resistor is 3.0K?
>
> Once again you are asking for the rise time to a fixed voltage,
> not a fixed fraction of the steady state voltage.
>
> Your problem statement was a 5V source driving a 2.000Kohm,
> 3.000Kohm divider with a capacitive load. =A0That means the
> steady state output is 3.000V. =A0If we are measuring the rise
> time to 80%, then the threshold in that case is
> 2.400V.
>
> If you change the bottom resistor to 2.950Kohms, then the
> steady state voltage becomes 2.980V and the 80% threshold
> level therefore becomes 2.384V. And yes, in the second case
> the result of a step input will reach the 80% threshold (2.384V)
> a little quicker than it will reach the 80% threshold (2.400V) in
> the first case.

Still beyond my comprehension. This thread is about converting output
of 5V part to input of 3V part. Thus the 3V Vdd is fixed. The  part
CC1101 is fixed too. Its logic "1" input voltage is from VDD-0.7 to
VDD according to the datasheet (see 4.8 DC Characteristics, Table 17:
DC Characteristics). So the bottom logic "1" input voltage is fixed
too and is equal to 3V-0.7V=3D2.3V

What we can do is just slightly vary the divider's resistors, and see
how this would affect the time the rising voltage would reach that
2.3V. (That's it, we don't need discussing those 80% and introducing
other new rules like the divider ratio must be constant) .

That variations are exactly what I did in real life. For larger
variations of the divider's ratio I used Schottky diodes on inputs to
limit the input voltage (was within the specs for that parts). Within
these absolutely valid to OP's specs variations you can easily find a
pattern when the positive variation of impedAnce of the divider would
result in increase of rising speed, not in the decrease which
contradicts your statement: "lower impedance - higher speed".

-- =

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