>> For your 3.3V it will be >=2.6V.
>> Obviously the presence of the bottom resistor of the
>> divider would slow the signal rising to that level, the
>> times with/without the resistor should not be "very close".
>>
>
> Using the word "obviously" doesn't make it so. Show me the math.

Well, perhaps I'm missing something, let's look:

Olin wrote:
> My standard answer is 2K ohms in series
> followed by 3.9K ohms to ground for 5.0V to 3.3V

Do I understand it correctly:
- The output connects to 2K resistor;
- The other leg of that 2K resistor is connected to 3.9K resistor;
- The other leg of that 3.9K resistor is connected to ground;
- The input is connected in between 2K and 3.9K resistors;

If so, do you realize that the signal on the input would approximate
5V * 3.9 / (2 + 3.9) = 3.305 V forever?
In fact, due to the output is not exactly 5V, it's sort of 4.7V, the
signal on the input capacitance would approximate 3.1V forever.

The threshold voltage 2.6V is quite close to that 3.1V and the
approximation curve is quite smooth, so the input signal would reach
2.6V later than it would reach when there is no that 3.9K resistor. In
the latter case it would approximate to 4.7 V and would pass 2.6V at a
good speed thus much earlier.

You don't need math to get it. What you need is just very basic
understanding how voltage on a capacitance is changing when the
capacitance is connected to DC source through a resistor.

Kinda can't get rid of a stupid feeling that I participated in a
similar discussion in my previous reincarnation many years ago.
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