# Calculating Battery Life with different working voltages

I have device that draws 30 uA in current continuously, and it is powered by a Panasonic CR2032 device (240 mAh - continuous standard load current: 0.2 mA ).

There are two components on this device, once is a sensor module and the other is a MCU:

MCU draws: 2 uA (working voltage range from 3v - 1.8v) Sensor unit draws: 27 uA (working voltage range from 3v - 3.5v)

I calculated my battery life to be 8000 hours (~11 months) - (240 mAh/0.03 mA = 8000 hours) - however I doubt I would get full 11 months if sensor's voltage range is 2.5 v which is discharging ...

How should one think about this, when determining battery life with components of different voltage range? I don't believe I would get 11 months, it must be lower

Would this be an appropriate way to calculate battery life with components of different voltage ranges?

Any reference or thoughts would be helpful.

• Nominal voltage of CR2032 is 3V, so any drop puts you below the range of the sensor. It looks like you can ignore MCU for this purpose May 1, 2020 at 21:17

>> How should one think about this, when determining battery life with components of different voltage range? I don't believe I would get 11 months, it must be lower. Would this be an appropriate way to calculate battery life with components of different voltage ranges?

The battery is "dead" when the output voltage exits the operating range of the highest 'lower voltage rating'. The MCU works from 1.8-3V. The sensor is good from 3V-5V

There's only 1 voltage they can both work at ... 3V. The fact your MCU can go down to 1.8V is irrelevant. Once the battery drops below 3V, the sensor (supposedly) will stop functioning.

Here's the datasheet from an Eveready 2032 cell: https://data.energizer.com/pdfs/cr2032.pdf

Look at the first chart. Notice almost immediately, the output voltage (with a very small load) drops below 3V. That 3V specification is "nominal", not "real". Only good for a brand new battery right out of the package. Use it, and you're not at 3V anymore.

That means your system will work briefly, or not at all.

However, all is not lost. I recommend a simple solution. Use TWO lithium coin cells, stacked on top of each other (in series). That will give you 6V nominal to start. Then, use a simple linear voltage regulator to drop it down to 3V. There are plenty to choose from that will have quiescent currents in the same range as your other devices. Use tantalum or ceramic capacitors, not electrolytic (EL's bleed current, ceramic do not).

If you can not accept the extra space of a second battery, note that a 2016 cell is exactly 1/2 the thickness of a 2032 (1.6mm vs 3.2mm). You can "definitely" find room because you've got it there already. (However, that means 1/2 the lifetime of course...)

As to how long the batteries last... That's a much stickier question. The 'mA-Hr' thing really only applies to more significant loads... a few microamps is right at the level of self-discharge for most battery chemistries (lithium coin cells are better in this regard, which is why they're used for computer memory backups, but they still 'self discharge') That means basically, you can not accurately calculate the battery lifetime in your application. The battery life curves are different depending on how much current you draw. The only thing you can be sure of is that it probably will be "a little less" than your calculation.

Here's a great application note from Eveready: https://data.energizer.com/pdfs/lithiumcoin_appman.pdf

If you really wanna maximize lifetime, consider a different battery. Even just bumping up to a CR2450 (2 pieces in series of course) will triple your lifetime. There are also cylindrical lithium cells that would last years to decades if you can afford the space.

• How much of the energy stored in the two batteries is lost in the linear regulator? Apr 9, 2022 at 9:48