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# Calculation of CR2032 battery life

Jakub17 4725 6
This content has been translated » The original version can be found here
• Jakub17
Level 6
Hello.

I need to get an estimate of the theoretical power consumption of the device and how long the battery will last before going on to experiment to confirm the estimates.

Current consumption for the systems included in the device:

Sleep state
HM-10 (400 uA)
Atmega8L (1.25 uA)
Active state
HM-10 (9 mA)
Atmega8L (1.6 mA)

I have two CR2032 batteries connected in parallel, so their total capacity is 240 + 240 mAh.

Battery discharge chart from documentation:

The battery voltage cannot drop below 2.7 V in the system, which, according to the diagram, gives approx. 1050 hours of operation under the tested conditions.

There are 2 batteries, so a total lifetime of 2100 hours. But the power consumption in the sleep state is about 400 uA and 190 uA for the graph. So I think it can be assumed (but I'm not sure) that the endurance is 2 times lower, i.e. 1050 hours for the sleep state.

The device works in the active state for about 0.5 seconds, say 10 times a day, which gives 5 seconds of the total duration of the active state per day, i.e. consumption of 10.6 mA.

Now I need to calculate somehow the average daily current, in order to estimate how long the battery will last for working both in active and sleep mode. How to count it? And are my simplifications in the presented analysis acceptable?
• TvWidget
Level 35
Temperature is also a key parameter. As it decreases, the voltage decreases and the internal resistance increases.
Check whether the given current for the HM-10 is average or peak current. It may turn out that the voltage drops below 2.7V at the first turn on.
• Jakub17
Level 6
The given HM-10 current is average. I don't have to be that precise to take the temperature into account.
Do you have any suggestion on how this should be calculated?
• TvWidget
Level 35
You need to know what the maximum current will be consumed by the system.
• Jakub17
Level 6
Well, I wrote above, a total of 10.6 mA of average current. I do not take into account impulse currents.
• TvWidget
Level 35
Jakub17 wrote:
I do not take into account impulse currents.

Without taking this parameter into account, the calculations do not make much sense. Estimated for two new batteries at room temperature, the voltage will drop to 2.7V when the device starts to consume 30mA for a while.
• Jakub17
Level 6
I am only interested in the calculation methodology. Let us assume that the impulse current is X. While it is easy to find such information in the microcontroller documentation, the hm-10 documentation is truncated.

I do not know how to calculate the average value of the daily current, but since there is time as a parameter, because the current 400 uA runs smoothly for (23h - 5s) and the current 10.6 mA for the remaining 5 seconds, then I can approach the matter from the energy side, from the power side which includes time by definition? I just don't know how to do it yet ...

Added after 12 [minutes]:

It can calculate the current work for two cases with the dependence W = UIt. Then add them together and average for the entire period, i.e. 24 hours. This way I have the average power calculated. Knowing that the voltage is constant (although it is tightly stretched because the battery voltage will drop gradually) from the average power, I can determine the average daily current. It will be fine?

Added after 16 [hours] 6 [minutes]:

When I transformed the formulas, it turned out that the average daily current does not depend on the voltage and everything turned out nicely.
The battery capacity that is given in the documentation, unfortunately, concerns the situation of a battery discharge to 2 V, and for me the limit is 2.7. There is only a problem here. If I knew the discharge capacity to 2.7 V at the calculated average current, I could easily calculate the battery life ...