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I am going to use a MCU + bluetooth BLE module. Its VCC pins are going to connect directly to the terminals of a 3V lithium CR2032 battery, for example a standard 170mAh one. 99% of time the module will stay into deep sleep mode on which the current consumption is <= 1uA, and it will only wake from deep sleep through interrups due level change of the GPIOs. The module work as low as 1.8. The datasheet actually show 0.4uA @ deep sleep.

But there is another SOIC-8 chip on the same board, and it only works with a minimum of 2.5V. My main question is: When the 3V lithium cell reach 2.5V and the second IC start to fail, will I have used a high percentage of the load charge of the CR2032? Maybe used 80% or more of its charge capacity?

Regards.

EDIT: specs of the module. enter image description here

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    \$\begingroup\$ Looking at the discharge curves in the datasheet in the answer, there's really not much battery life left below 2.5V. \$\endgroup\$ Apr 18, 2020 at 19:40
  • \$\begingroup\$ Hi. Thanks! And if there was a 2.7V minimum supply IC on the same board, would I do use the greater amount of the CR2032 also? \$\endgroup\$
    – abomin3v3l
    Apr 18, 2020 at 19:52
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    \$\begingroup\$ hint: if this will be installed in a non temperature controlled environment, consider battery performance degradation when iit gets cold. \$\endgroup\$
    – bobflux
    Apr 18, 2020 at 23:59

2 Answers 2

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Taking a look at a normal datasheet, at idle there will be no significant loss of capacity with 2.5V or the later edited 2.7V, (the graph below is actually for a much higher load than your sleep current)

However when you wake up and start drawing more current that is a different story, that is closer to the pulsed line for most BLE modules I have looked at, the pulses that it draws could pull down the battery voltage below where those IC's are specified for, and could cause lockups and other weird behaviour

In past projects, I've had to add some capacitance in parallel to the battery to reduce how much the battery voltage fell during my transmissions, In those cases I just worked out the current, and the length of time of those pulses, and simulated it to find the right values.

enter image description here

Energizer CR2032 Datasheet

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  • \$\begingroup\$ I will place a pair of 22uF 10V 0805 X5R caps very close to vcc of the module. The minimum voltage IC is really 2.5V, and not 2.7 \$\endgroup\$
    – abomin3v3l
    Apr 19, 2020 at 0:38
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    \$\begingroup\$ The other thing to check is the leakage current of the capacitors, the best 47uF capacitor I could find for my last project ended up being a 2uA load based on its leakage. \$\endgroup\$
    – Reroute
    Apr 19, 2020 at 0:43
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    \$\begingroup\$ Would need to simulate, what is your expected current pulse and for how long? mine was 12mA for 0.8ms, If you know that, its not too hard to work out how much of the capacity is usable, and how much capacitance you would need as a trade off, \$\endgroup\$
    – Reroute
    Apr 19, 2020 at 0:51
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    \$\begingroup\$ I never found a strong relation, seemed like higher voltage capacitors had lower leakage, but I was just working off the datasheets, AVX brand datasheets on average specified to half the leakage of other brands. \$\endgroup\$
    – Reroute
    Apr 19, 2020 at 1:03
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    \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$
    – Reroute
    Apr 19, 2020 at 1:11
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Your battery actually starts from 3.4V, its minimum voltage is 2V, so if we calculate linearly its about 30% left at 2.5V. There should be a circuit to stop the battery from overdischarging, or you could just replace it. 1 IC working and 1 IC not working can lead to undefined behaviour.

Datasheet.

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