So I am attempting to determine if the battery I choose can power the devices I need. This is my first day doing anything with ohms and current and such. So go easy.

I have:

2000mAh battery

45mA Bluetooth card

3.9mA GPS card

2.0mA IR Thermometer

Now the arduino mini only says that it has a max output of 150mA so I don't know how much it consumes. For this example I am disregarding it because if I am right then I have a ton to spare.

So based on my thought process I think I get 2000mAh/51mA = 39.2 hours?

enter image description here

  • \$\begingroup\$ You didn't account for the draw of the Arduino Mini. \$\endgroup\$ – Jon Watte Aug 20 '13 at 17:03
  • \$\begingroup\$ @JonWatte yeah as I stated in my question, there wasn't any data in the documentation I linked that clearly said its draw rate. The answer below found it though. \$\endgroup\$ – user27643 Aug 20 '13 at 17:08

The way of your calculation is basically correct.

Of course, the devil is always in the details. Normally, one calculates a "typical" outcome, and a "worst-case" outcome. For those calculations, you'll need the "typical" and the "maximum/minimum" figures from the data sheets.

A few factors to consider:

  • the battery might have a capacity of 2000mAh when new, but this will most probably go down as it ages,
  • the battery capacity highly depends on the temperature; if you want to use it only at room temperature, it might behave much better, but outside at -25C degrees, the capacity might be only a fraction of the normal,
  • the consumption of the components might as well change depending on the temperature,
  • the consumption of communication peripherals usually highly varies according to the state of the communication (active, waiting, standby, etc).
  • \$\begingroup\$ Okay awesome. Yeah I grabbed the max values to calc a worst case scenario. Most of these devices almost go completely off when they aren't in use but for my application they will be in constant use. Thanks again \$\endgroup\$ – user27643 Aug 18 '13 at 23:26
  • 3
    \$\begingroup\$ Battery capacity also depends on discharge current, oops! \$\endgroup\$ – Kaz Aug 19 '13 at 2:11

I recommend to refer the discharge characterstics. Discharge characterstics vary with tempearture, amount of charge.

I am not sure what type of battery you are referring. a example datahseet can be. http://industrial.panasonic.com/eu/i/21291/Handbook2011/Handbook2011.pdf

Sample discharge charactersics will be of this form. enter image description here

As battery discharges it's voltage will be reduced, i am not sure what is the minimum voltage required to proper operation of your circuit. All these factors needs to be looked, when deciding how many hours a battery can power the circuit.

This post has more information Does mAh measure how long a battery would last?

  • \$\begingroup\$ Great info! I love learning about this. I don't think that I will have an issue with discharge rate but this is very valuable to know. Thank you \$\endgroup\$ – user27643 Aug 19 '13 at 12:10

There's another 9.5 mA to run the Arduino, bringing you nearer to 60mA total, plus the draw of any peripherals you use on the Arduino board.
From the AtMega328P data sheet:

enter image description here

  • \$\begingroup\$ Oh man I wish I could up vote this more than once. I looked all over that thing and still missed it. THANKS! \$\endgroup\$ – user27643 Aug 19 '13 at 14:28
  • \$\begingroup\$ Don't feel bad - I had to hunt for it myself. (As you can tell from the section number, they stuck these things way at the end.) Glad to help. \$\endgroup\$ – JRobert Aug 19 '13 at 21:24
  • \$\begingroup\$ you would think since these devices are intended to be hacked together and since they put so much time into the documentation, they would make the power consumption easy to find. \$\endgroup\$ – user27643 Aug 19 '13 at 22:11

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