Continue to Site

# Battery Life

Status
Not open for further replies.

#### 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?

Read this application note from MAXIM. Its related to lithium-coin batteries but same fundamentals can be applied to other battery types

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.

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.

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

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.

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...

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 . . .

Just a quick one.. I see that there is a 7555 chip available which this pin-for-pin to the 555 timer. It's benefits are low power consumption, in the uA range:

Status
Not open for further replies.

Replies
8
Views
1K
Replies
8
Views
4K
Replies
4
Views
432
Replies
35
Views
4K
Replies
9
Views
1K