# Battery Life Calculation, Sleeping and Wake modes

I have a circuit that consumes 1.6μΑ during sleep mode and 40mA during awake mode. The awake period lasts for 3 seconds and the circuit is on awake mode every 30 minutes. The circuit is powered by this battery (the link leads to the datasheet).

I used this battery life calculator because i do not have the knowledge to do the maths on my own.

I know that as the circuit working the capacity of battery is decreasing as well as the voltage. But as the battery voltage is decreasing there are a few components that cant work under 3.0 volts.

I am trying to say that there is point where the battery will be capable of feeding power the circuit but some components are not going to work because the voltage will be lower than their voltage threshold.

For example, if i have a battery at 8500mAh 3.6V and at 2.9V are still left 1000mAh i dont care because my circuit doesn't work at 2.9V.

Is the rest capacity "wasted"?

Is that thought wrong?

Does the attached calculator take care about that too?

Because i didn't see in the equations nothing about voltage.

Many thanks

• "if i have a battery at 8500mAh 3.6V and at 2.9V are still left 1000mAh" - How did you arrive to this conclusion? According to datasheet discharge curves, at 3V the battery will be COMPLETELY drained. – Ale..chenski Jul 12 '18 at 14:18
• I just said that. Like an example, i didn't read that. Sorry for the misunderstanding. – alexisicte Jul 12 '18 at 14:22

This isn't that hard that you need an on line calculator.

First determine the average current consumption:

In sleep mode, the current (1.6 uA) is so much smaller than the wake mode current that we can assume that this 1.6 uA is used 100% of the time.

In wake mode we have 40 mA for 3 seconds every 30 minutes = 30 * 60 = 1800 seconds. That is a duty cycle of: 3 / 1800 = 1/600 = 0.00167 So those 40 mA peaks average out to: 0.00167 * 40 mA = 66.7 uA

Total average current consumption: 1.6 uA + 66.7 uA = 68.3 uA

Now we look in the datasheet to see what battery capacity that gives us given a discharge to 3.0 V.

The nominal capacity of this battery is 8.5 Ah. Let's look in the graphs if that is the value we can use. Battery capacity decreases with increasing load current but the load current for that 8.5 Ah is 4 mA, a lot less than the 68.3 uA we need. So yes, we can use the 4 mA value, our 68.3 uA is so small that the battery's capacity is not deteriorated by it.

From graph 1 we can determine to which voltage the battery is discharged in the capacity test. All curves are quite flat above 3.0 V so when the voltage is below 3.0 V then the battery is quite empty already. So that 8.5 Ah is valid for discharging to 3.0 V.

So let's continue with that 8.5 Ah. 8.5 Ah means the product of current and hours is 8.5 Ah. So: 8.5 Ah / 68.3 uA (from above) = 124451 hours = 5185 days = 14 years !

In practice batteries for long life applications are often only guaranteed for a 10 years lifetime. The 14 years exceeds that. So when used in a product you should instruct the user to replace the battery every 10 years for optimal performance.

• Many thanks for your comment. Your explanation is very clear and i think i can understand now. – alexisicte Jul 12 '18 at 13:49

Yes, the rest of the battery is "wasted," but it's not so bad as it seems. If you look at the discharge curves on page 2, the 60mA curve is most similar to what you'll see in your 30mA "active" mode. That curve is rated at 7.2A-H, but that's based on 60mA average current...you're only drawing about 68uA average, so time is more your enemy than power drain as far as battery life goes. (If not for self-discharge, you'd get a good 15 years out of the battery. However, it will probably last only 2-4 years anyway, depending on environmental conditions.)

The 60mA curve shows voltage under load, so it can be used to determine how much energy is "wasted." The remaining battery capacity at any point on the curve is the area under the curve from that point to the end. At 60mA, you can see that when the battery voltage reaches 3V, the curve is nearly vertical and there's only a tiny sliver under the curve between 3V and 2V. Compared to the total area under the curve, it's miniscule and not worth worrying about.

You in particular won't have to worry about it because the battery's life will be determined largely by self-discharge. The second chart on page 6 here indicates that at 0.4mA, or 1/47500C, the battery loses 4X as much capacity due to self-discharge at 60C as it does at 25C. Keeping in mind that the loss is over a shorter time period, run the numbers and it indicates that the self-discharge rate at 60C is almost 6X the self-discharge at 25C.

• Thanks for the comment. I didn't understand why you said about environmental conditions. I mean at ideal condition we can expect about 14 years, about 10% of self-discharge let's say 12 years. Can the environmental conditions be so critical? The device is placed outdoors and is exposed to the sun up to 60 degrees Celcius. – alexisicte Jul 12 '18 at 13:48
• Updated answer with more detailed self-discharge information. – Cristobol Polychronopolis Jul 12 '18 at 15:49

one important specification is the cut off voltage of your system, the voltage at what your system stops working, 3V in your case. So as you say the rest of the capacity will remain in the battery but your circuit won't be able to use it, so yes a lower cut off voltage will help to use more capacity of the battery and enlarge the life of your device. I don't see that that is taken into account in the calculator, it uses a generic 0.85 to reduce the nominal capacity. Also doesn't take into account the self discharge, on that technology depending of the manufacturer can be around 1%.

Another point to have in mind, is the temperature, it can highly affect the battery and in your case taking a look to the datasheet will stop working below -10ºC.

But of course depending on your application that can be important or not.