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BACKGROUND:

I'm working on a project where we use the nRF51 Bluetooth low energy chip, and our first prototype boards just arrived.

So I've made some power measurements trying to calculate the battery life: 

When idle: 3,1uA
When bluetooth connected: 4,5mA

The bluetooth won't be connected that often, so this I can actually live with the resulting battery life for the prototype (gives me around 17-18 weeks battery life).

We're using a Panasonic CR1216 battery: https://industrial.panasonic.com/cdbs/www-data/pdf2/AAA4000/AAA4000C277.pdf

And just now I realised that the battery voltage takes a good drop when the bluetooth is connected for a longer time. The max connected time is 10sec and there I got: 

Start: 2,99V drops exponentially to 2,7V over 10sec.

When the bluetooth disconnects it slowly climbs towards 2,99V again. I've waited 10min now and it is around 2,96V.

QUESTION:

In the battery datasheet I can't find anything about max discharge current? But clearly the 4,5mA is too much? What will happen to the battery's 25mAh? (I know that it is hard to give a specific number, I'm just interested in knowing if my battery will die within 2 days instead of my 17weeks).

EDIT :---- I've found the similar battery that says max cont. discharge current = 1mA http://www.rossmannweb.de/files/File/Kataloge/CR1216_Renata_Rossmann.pdf

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  • \$\begingroup\$ This question is very similar: Pulse-powering heavy loads with a coin cell. The app notes linked in that topic also provide useful information. \$\endgroup\$
    – SamGibson ♦
    Commented Jun 8, 2016 at 18:12
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    \$\begingroup\$ Have you considered a larger battery? If you go from 12mm to 20mm diameter, you can go to a CR2032 and have 10x the capacity. \$\endgroup\$
    – DoxyLover
    Commented Jun 8, 2016 at 21:16
  • \$\begingroup\$ The last chart "Capacity vs load resistance" is probably the most relevant. But you've gone off the edge of the chart for the current draw you're asking. So it's hard to say. \$\endgroup\$
    – Simon B
    Commented Jun 8, 2016 at 22:17

1 Answer 1

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The battery you want to use is not designed for a current draw that large. The manufacturer only tested to a maximum of 2mA.

The total capacity of the battery will be decreased with these larger current draws but it will most likely be more than 2/3 of the total capacity.

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