Battery duration

[raff5184]

Hi all,

this battery http://www.powerstream.com/p/GMB300910.pdf has a capacity of  12mAh at an average voltage of 3.7V

which corresponds to: 3.7 * 0.012 * 3600 = 160 [J]   

Theoretically, if my load absorbs 6mA (at 3.7V) the battery should last 2h.

If instead, the load works intermittently for only 1 millisec every minute, the battery lasts about ... 1230 years (without considering non-ideal behaviors of the battery) Is this correct?

[IanB]

which corresponds to: 3.7 * 0.012 * 3600 = 160 [J]   

Why do you think this is surprising? The calculation seems correct to me.

Theoretically, if my load absorbs 6mA (at 3.7V) the battery should last 2h.

If instead, the load works intermittently for only 1 millisec every minute, the battery lasts about ... 1230 years (without considering non-ideal behaviors of the battery) Is this correct?

It would be easier to answer this question if you showed the working, rather than just quoting numbers. How can we tell if your calculation is correct if you do not show how you arrived at the answer?

[ataradov]

With intermittent mode your duty cycle will be 1 / 60000, so ideally you will get 2 * 60000 hours with the same load, which is 60000 * 2 / 24 / 365 = 13.7 years.

[raff5184]

I was surprised because it seems quite a lot of energy.

Sure here are the calculations:
instantaneous power: Pi = 3.7*0.006 = 0.022 W ( = 22mW)
Load energy: E = Pi * 0.001 = 22e-5 J

Duration: battery_capacity / E * 60 =
=160 / 22e-5 * 60 =
= 4.4e7 minutes    // "* 60" because the load works every minute
= 837 years
There was an error in the approximation therefore I had 1200.. but still it seems to be too long

[ataradov]

Your calculations are way too complicated. If you have 12mAh of capacity, then with 6 mA load, it will last 2 hours, ideally regardless of power consumption profile. With duty cycle lower than 1, you spread those 2 hours over time. A minute is 60000 ms, and you work for 1 ms out of that time, which means that battery will last proportionally longer, or 60000 * 2 hours.

I don't really follow your math, but where do you factor in 1 ms there?
No need to involve energy here for this basic scenario.

[raff5184]

thank you for your clarification, much appreciated.

the 1ms is here:
Load energy: E = Pi * 0.001 = 22e-5 J   (: absorbed energy = instantaneous power * 1ms).

I was using energy exactly because my calculations with energy didn't seem right. So I was trying to understand what I was doing wrong

Big thank you to everybody!

Mistake was here: Duration: battery_capacity / E * 60 =