Question: I was reading an [article](http://www.home-automation-community.com/arduino-low-power-how-to-run-atmega328p-for-a-year-on-coin-cell-battery/) someone modified an arduino and is running in low power mode.

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> An Arduino Uno runs less than one day on a 9 V battery because it uses about 45 mA current. Using an Arduino Pro Mini, with a simple modification, the power consumption goes down to 54 μA (0.054 mA) with the 3.3 V version or 23 μA (0.023 mA) with the 5 V version, in power-down sleep. That is 4 years on a 9 V battery with 1,200 mAh capacity or 2,000 times more efficient than the Arduino Uno. After removing the voltage regulator, the power consumption is only 4.5 μA for the 3.3 V version and 5.8 μA for the 5 V version, in power-down sleep. 
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He claims he can run the arduino (obviously it is doing nothing) for 4 years. Ignoring leakage and other things aside, how can I do the math to verify this claim?

Answer: 
The answers thus posted about using mAh = (hours of current draw) \* (current draw) are for ideal batteries and will not be good for a calculation like this. 

Do you know of any batteries that you can recover all of the energy that they hold? I don't. And at a certain voltage level the battery becomes unusable for you. A 9V battery will become useless after 3V if that is what your circuit needs. You have to 'derate' the equation for these cases, digikey uses a factor of 0.7 in their calculator. This especially holds for rechargeable batteries as there is a voltage point of no return on the discharge cycle that will kill the electrodes. The current draw will also make a difference because a battery is not an ideal voltage source, it has resistance and it is dependent on the [current draw](http://batteryuniversity.com/learn/article/what_is_the_c_rate). 

At the uA level, you also have to start to worry about leakage current, the battery has uA-nA's of leakage current. You'll have to factor that in also. If you need a ball park number, the ideal case is a great place to start but its the maxium ammount of time that your battery will last in the ideal case.
The author has done three things: 

1. Remove the power consuming Power Indicator LED
2. Avoid using an inefficient linear regulator
3. Keep the micro in sleep mode as much as possible through software design.

He states that the 9v battery has a capacity of 1200mAh. This means that it can supply 1200mA for one hour or say 120mA for ten hours. 

If the circuit is ising 0.054mA then it would be able to run for 22222 hours (1200 divided by 0.54) which is about 2.5 years.