Marechiare 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;

Yes.

> 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.

I can make a much faster circuit too if it doesn't have to work right.  Your
hypothetical circuit is pointless since it's violating the maximum output
voltage spec.

Think of it this way:  Compare two circuits, one is a perfect 0 to 5V square
wave followed by the 2K,3.9K ohm divider.  The other is a perfect 3.3V
square wave followed just by 2K ohms in series.  Now assume each is driving
the same capacitive load.  Which one will have faster rise and fall times?

For extra credit show the Thevenin equivalent of each circuit.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.
-- 
http://www.piclist.com PIC/SX FAQ & list archive
View/change your membership options at
http://mailman.mit.edu/mailman/listinfo/piclist