Hi Mike,

Say that two alcaline penlight cells have a capacity of 3000ma/h.
You don't want that 3A over one hour but only 0,5ma during X hours.
3000/0,5 = 6000 (x 1 hour). Which is ca. 250 days, or ca 10 month,
before the cells are empty. And empty is about  1.3v each.

Good luck, At vW.


>For example:  if I have a circuit which will consume 450 uA
>until the input voltage drops to 2.7V, how long can that sit
>until it reaches 2.7V (assuming two 1.5V AA alkaline cells)?
>Failing that, does anyone know where I can find equations
>to solve?
>Mike H.

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