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Battery Life

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Gene

New Member
What is the formula for calculating battery life. In my specific application, a square 9v battery is used to back up a circuit that draws 10 ma. If the main power fails, the current is supplied to the circuit by the battery. So, the question is how many hours will the battery keep the circuit running?
 

Gene

New Member
Got it! battery Amp hr / running Amps = hours Now all I have to do is find a battery Amp hr table for common batteries (D, C, AA, AAA, 9v). I will try the manufacturers first. Thanks.
 

bogdanfirst

New Member
the best place to find the battery capacity is from the website of the maufactures. but you also have to take in your calculations that as the battery gets consumed the vltage drops, and the way it drops it depends ont the type of battery.
 

Dean Huster

Well-Known Member
In other words,

The amp-hour equation is not linear, nor does it handle extremes. For instance, a battery rated for 200 AH may supply 20 amps for 10 hours, but won't necessarily supply 2000 amps for any time at all (i.e., 200/2000 is 0.1 hours, but don't count on the battery being able to supply a current like that for even an instant because of its internal resistance) and it may not supply 2ma for 100,000 hours, because internal leakage may shorten that time considerably.

Dean
 

bogdanfirst

New Member
thats what i am trying to say, there are limits within you can aply the formula. i found something like that on the website of a battery manufacturer, dont remember wich, but it was saying the limits for wich you could aply the formula.
 

McGuinn

New Member
This is all good info. Another thing I noticed was that the battery voltage will deplete as the battery discharges or the current rises. What that can mean is that the voltage output from the battery will drop below the required Vcc supply voltage of the circuit.

This occured to me with a 9v battery, they are often rated as ~250mA and anytime I placed a load on the battery, the circuit would drop out.

Most bench power supplies compensate for this (It's called linearity I think), and it means that no mater what the current draw, the voltage remains the same...
 

Gene

New Member
Here's where I am on this. The oscillator that I want to keep running in the event of a power failure is a 555 and it can stand a drop in voltage. I chose a 9volt alkaline rated at 625 mah and the circuit draws 10 ma. So 625 mah / 10 ma = 62.5 hours (max). This should easily keep the circuit running for 2 days - that's fine. On to the next problem . . .
 
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