Dividing by 1023 or 1024? The final verdict on analogRead

adwsystems:
What is the lowest voltage (in microvolts) that will produce an output of 1 count from the ADC?

The width of each "box" will be 5v / 1023 (or 5v / 1024 - take your pick)

So if the bottom point is 0v the next point will be 4.44876 millivolts. And the change-over point should be half of that. Give or take.

...R