shaneshane1 said:
really?
if i have a 9V battery (1000mAH) then when it gets down to 4.5 (half voltage) then wouldn't i still have around 500mAH at 4.5V?, i thought this is how it works, i never really thought of it any other way
Not the way I understand it. One thing to clear up: it's not like we're draining the voltage, which is kind of what it sounds like you're saying with the "gets down to 4.5 (half voltage)" thing. We're draining the charge at a certain rate, which creates a current of a certain amperage. The closer to full capacity the battery is, the better able it is to generate the expected voltage across its terminals.
For instance, check out this datasheet for an Energizer 9V NiMH (really 7.2V but it's called a 9V) rechargeable:
https://www.electro-tech-online.com/custompdfs/2007/12/nh22-175.pdf
Look at the discharge graphs to the right of the datasheet page and you'll see what kind of voltage it can produce across its terminals for certain amounts of remaining charges and different discharge rates. At a 175mA rate of discharge you get about 30 mins out of the thing before it drops below 5.5V; at 17.5mA you get just about 10 hours. And the voltage plots aren't straight lines, as you might expect.
The rated capacity is 175mAH. So at a 17.5mA discharge rate, the math would say you get 10 hours out of it. So far so good; that's what the graph shows. But at ten times that discharge rate, the math would say you get 1 hour out of the battery--as the '175mAH' would suggest. But look at the lower graph: you get 30 minutes, and the whole time, the voltage is dropping like a stone.
A battery which comes closer to approaching the ideal (at least, for lowish currents) is the LR55 Lithium 9V:
https://www.electro-tech-online.com/custompdfs/2007/12/l522.pdf : With this, the load can draw up to 120mA and the voltage will stay above 5.5V for over seven hours. The same load on the other 9V would drop it to 5.5V in less than an hour. According to the datasheet, trusting the math and figuring that one could pull, say, 1.2A for 42 minutes will result in the perpetrator needing a new battery.
These numbers of course only deal with a drop to 5.5V across the terminals, but that's because we don't speak in terms of absolutes like the battery being "empty" or "full"--you determine what good values for "full" and "empty" are and use those. Sure, at 175mA you might still have a voltage at one hour, but it would be so low as to be useless to a circuit expecting a 9V source, even if that circuit were designed to work down to, say, 5.5V. We have to decide where we draw the line when designing a circuit: i.e. "this should operate OK with a supply between X and Y volts".
Torben