What really matters is the energy the battery can supply, and the efficiency
in its use. As you said, V x mAh is more interesting that mAh alone. Volts x
Amperes x time is a measure of the energy the battery can give you:
 . 12V,100mAh   ->  12V x 0.1A x 3600s/h=4320J
 . 4.5V, 200mAh -> 4.5V x 0.2A x 3600s/h=3240J

So, theoretically, you could get 4320 Joules off the first battery, and 3240
Joules from the second.

This numbers are made supposing the voltage and the current remains constant
during the discharge, but this is not usually the case, so this numbers must
be taken as aproximations only.

If you use an 80% efficient converter to drive the circuit from the 12V
battery, you get 0.8*4320=3456 useful Joules out of the battery, and waste
0.2*4320=864J, which are finally transformed in heat (enthropy) in one way
or another.

In this case, 3456J is slightly better than 3240J.

> Other posts have also 
> been talking about getting a lower
> voltage battery with a higher mAh 
> value (in this case "twice"
> the capacity).
> 
> Now, Let's say we have a 
> 12V/100mAh battery and one
> 4.5/200mAh battery. Now, the 4.5V battery 
> *seems* to have
> double capacity (which most equals with "lifetime" 
> under
> similar conditions). But, if using a well designed step-down
> converter, will that not draw *less* current from the higher
> voltage 
> source, then the target is using ? Or in other words,
> if one could 
> expect a 100% efficient converter, is it not realy
> the V x mAh value 
> that is interesting, not ? Even if the
> converter runs at something like 
> 80-85%, I think that
> one can not just compare the mAh values 
> directly...
> 
> I'm not claiming to be an expert in this, just liked to 
> know...
> 
> Jan-Erik.


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