battery life calculation

Discussion in 'The Projects Forum' started by pazpaz, Nov 14, 2013.

1. pazpaz Thread Starter New Member

Nov 14, 2013
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Hello all,

Please guide me how to calculate battery life

My circuit takes current from 2 1.5V Duracell batteries of AA LR6 MN1500

total voltage = 3.0V from 2 batteries

My circuits is ON for 9Hours and it draws current of 25mA

and rest of the time i.e if say a day (24Hours - 9Hours) = 15Hours it draws current of 35uA i.e sleep current

can you tell me how much is battery life how much time it will last?

regards.

2. WBahn Moderator

Mar 31, 2012
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It can be a bit tricky, but the best place to start is the manufacturers data sheet for the battery (which is often not as data rich as would be nice).

http://ww2.duracell.com/en-US/Global-Technical-Content-Library/Product-Data-Sheets.jspx

You'll notice that on of the first things that you have to decide is what you mean by "how long will it last"? You need to decide on a cutoff voltage. Looking at the first couple of graphs, you can see that there is a noticeable change in the behavior at about 1.2V. So if your circuit can work down to 2.4V, that is probably a good choice. Let's go with that.

You can see that one of the curves they have is for 25mA which shows a life to 1.2V of a bit under 100 hrs. Zooming in on the figure until it fit a scale nicely on the screen, the actual value is about 85 hrs (though notice that it is right at 100 hrs for a 1.1V cutoff). Comparing the 25mA curve to the 50mA curve, it appears that the battery life is reasonably linear in this region, so we can figure that at 30mA we can expect a live of about (85 hrs)(25mA/30mA) or 71 hrs. Thus, ignoring the draw when the device is asleep, you are looking at about 8 days if used 9 hours a day.

The draw at 35μA is nearly a factor of 1000 less, so the amount that this draw will have on the battery life will be negligible in comparison since it is at this level for not even twice as long as it is at the higher draw.

Keep in mind that temperature plays a significant role in battery life (see the figure that shows that, at 0°C, the battery life is roughly half what it is at 21°C) and that no two batteries are exactly the same, and that it depends on how long the things have been on the shelf, and on and on.

So you can probably expect the batteries to last about a week.

3. pazpaz Thread Starter New Member

Nov 14, 2013
19
0

I found one calculation from net:
Please tell me if it is correct.

calculation 1:
battery rating = 2600mAh
sleep current = 0.035mA
sleep time = 15 hrs
on current = 25mA
on time = 9hrs

thus battery life:
= 2600 / ((0.035 * (15/24)) + (25 * (9/24))) //24 is 24 hrs i.e per day
= 276 hrs
= 11 days

calculation 2:
but I feel as 2 batteries are used then mAh = 2 X 2600 = 5200mAh
and with this calculation battery life:
= 5200 / ((0.035 * (15/24)) + (25 * (9/24))) //24 is 24 hrs i.e per day
= 552 hrs
= 23 days

regards.

4. WBahn Moderator

Mar 31, 2012
17,777
4,804
You can't add the battery capacities because they are in series.

Using a fixed battery amp-hour rating gets you in the very rough ballpark, but the actual amp-hour capacity of a battery is very dependent on both the current draw and the cutoff voltage.

Where did you get the 2600mAh number? Under what conditions does it apply?

If we use a current draw of 25mA, then that says the life is 104 hrs, which is a close match to the datasheet curve for a 1.1V cutoff voltage. At this current draw, a 1.2V cutoff voltage would more closely correspond to 2125mAh.

Depending on how critical it is that the battery last a minimum amount of time, make your best estimate and then reduce the life expectancy by some factor. If it's not a big deal if the battery doesn't always make it to the end, then may reduce it by 10%. If it's pretty important that they do, then maybe reduce it by 50%.

5. pazpaz Thread Starter New Member

Nov 14, 2013
19
0
Hi,
I found that from wikipedia.
but I didn't understand the last two paragraphs.

6. WBahn Moderator

Mar 31, 2012
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What last two paragraphs? From the wikipedia article?

7. GopherT AAC Fanatic!

Nov 23, 2012
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I think he meant from post #4 above...

8. WBahn Moderator

Mar 31, 2012
17,777
4,804
I don't really know how to say it much more simply. I need a better idea of where the confusion is coming in.

Q1) If the battery rating is 2600mAh and the current draw is 25mA, then how long will the battery last? Just do the calculations the same way that they were done on that internet page.

Q2) From the graph in the data sheet, what battery cutoff voltage has to be used in order to get a life expectancy that matches the above computation?

Q3) From that same graph, what is the life expectancy for a 25mA current draw using a cutoff voltage of 1.2V?

Q4) What battery capacity does that correspond to?

As for the second paragraph, that seems pretty self-explanatory. If you estimate a battery life expectancy of 30 days using your expected data and you then use this in a system that causes significant damage or loss of data or something else if the battery actually goes dead in 29 days, then you are pushing the reliability of the estimate too hard. On the other hand, if it is just a bit of an annoyance to have to change the battery at 29 days instead of 30 days, then no big deal beyond annoying your customer a bit. So in the latter case, lower the estimate time by 10%, calling it a 27 day battery life expectancy. In the former case, lower the estimate by 50% and call it a 15 day battery life expectancy. The idea is to match, at least roughly, the risk of the battery not living as long as estimated to the cost associated with that happening. And, before anyone says anything, the 10% and 50% numbers are just thrown out as rough suggestions.