I was reading an article someone modified an arduino and is running in low power mode.

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.

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?

  • \$\begingroup\$ Just an FYI, I removed the LED trace and used the RocketStream library and can confirm I'm drawing .054mA in sleep mode when calling LowPower.powerDown(SLEEP_FOREVER, ADC_OFF, BOD_OFF); \$\endgroup\$ Feb 19, 2016 at 9:05

3 Answers 3


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.

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.

  • \$\begingroup\$ Then for some circuits the "mAh=(time)*(current draw) doesn't apply?Can you explain the "derate" term? \$\endgroup\$ Feb 18, 2016 at 5:44
  • \$\begingroup\$ This kind of explains it: electronics.stackexchange.com/questions/207774/… \$\endgroup\$
    – Voltage Spike
    Feb 18, 2016 at 6:06
  • \$\begingroup\$ Its not that you can't use all the energy in a battery, the energy is hard to get. Lets say you have a camera, it uses two double AA's so 3V. It shuts off after the voltage drops to 2.5V. There is still energy in the battery, in the case of the AA maybe 10% or 5% but its energy you cant use because there isn't enough voltage in the battery. So you take your batteries out of the camera and throw it in a flashlight. The flashlight is dim but you can still somewhat use it, you use all of the voltage and suck the last 5-10% out and the light finally becomes so dim you can't see it. \$\endgroup\$
    – Voltage Spike
    Feb 18, 2016 at 6:12
  • \$\begingroup\$ I understand.So to after using the mAh formula I should take a look at the battery's voltage decrease graph to calculate run time more accurately. \$\endgroup\$ Feb 18, 2016 at 13:29
  • \$\begingroup\$ Yeah, and that is totally dependent on battery chemistry. An alkaline battery will be able to deplete most of its energy because there is no need to maintain a certain voltage level. With a lead acid, once you pass a certain point, the battery dies and you have to recondition it. With other battery chemistries there is no going back. \$\endgroup\$
    – Voltage Spike
    Feb 19, 2016 at 5:42

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.


He's giving uA draw rates in his claims, and gives you the mAh rating of the battery as 1200mAh.

For the math:

  • 1200mAh = 1200hrs run-time at 1mA draw
  • 54uA = 0.054mA || 1200mAh / 0.054mA = 22,222hrs run-time (almost 926 days)
  • 23uA = 0.023mA || 1200mAh / 0.023mA = 52,174hrs run-time (almost 2,174 days)
  • 4.5uA = 0.0045mA || ...266,667hrs (11,111days)
  • 5.8uA = 0.0058mA || ...206,897hrs (8,621 days)
  • \$\begingroup\$ 4.5uA is almost 30 years??! Wow. \$\endgroup\$ Feb 18, 2016 at 2:43
  • \$\begingroup\$ Is || an operator? What does it mean? \$\endgroup\$ Feb 18, 2016 at 2:49
  • \$\begingroup\$ @WhiskerBiscuit In some programming languages, || is an operator, but in the above answer I was simply using it as a separator between the 2 equations on each line. \$\endgroup\$ Feb 18, 2016 at 2:53

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