# Am I calculating correctly?

If I have measured for 5 minutes (between 2 measurements) as attached picture.
For how long will a 2700mAh battery last?
I calculate (20,1uA) x 12 (5 minutes x 12 to get 1 hour)
= 241,2uA
Then 241,2uA x 24 (hours/day) = 5788,8uA=5,788mA/day
Battery 2700mAh / 5,788mA = 466 days

Is that correct?

Hi,

your average power consumption 20.1uA = 0.0201mA seems to be a valid assumption based on the graph; you are capturing the average for a full period of what looks like a repeating pattern.

Theoretically you should then get 2700*mAh / 0.0201*mA = 134328*h, or about 15 years.

(In your calculations you mixed current and energy; if you do it with energy you get 6.03uWh per 5 min * 12 = 72.3*uWh per hour. Assume the battery is 3V then it contains 3V * 2700*mAh = 8100*mWh.
Then you get 8100/0.0723=112033\h, or about 13 year, which is close to the 15 years above)

In practice you need to consider the following;

• Is your current consumption stable ? Or does it vary as battery voltage, environment etc varies ?
• What is the battery leakage ? Is it fresh and new ? Some battery chemistrys have significant leakage.
• Can you actually use the full 2700mAh ? In some cases when the battery has discharged and the voltage goes down then peak current need cannot be provided by the battery at the right voltage, and the device does not perform correctly.
• Also remember that you when you mass produce/deploy you get statistical variations in components, batteries, and environment; some devices will fail earlier, some later.