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I'm doing a project with Arduino. I have a small Arduino board that every so often takes measurements from a sensor, but it's idle most of the time -- consuming 6mA. I want to use a small battery to power the board. I have seen batteries like the following one:

https://www.adafruit.com/products/1570

It's a 3.7V 100mAh battery. The board requires an input voltage in the range 3.7-4.2V. So, I suppose once the battery starts to drain, the voltage will decrease and the logic on the board may stop working correctly.

Is this correct? Or will the battery sustain 3.7V during most of its capacity?

Assuming the voltage will go lower 3.7V relatively quickly, I can imagine two possible solutions:

  1. To use a battery with a higher voltage and then use some form of voltage regulator.
  2. To use a battery with more capacity, hoping that the voltage loss will take longer to occur.

Which option is better? Or what are the tradeoffs from an electrical and efficiency point of view?

Is there any other option?

UPDATE

As some of the comments mention, the logic board indeed runs at 3.3V. It accepts a 3.7-4.2V input voltage and it uses a regulator to bring the voltage down to 3.3V. So I suppose a 3.7V battery will be able to power the board at least while it stays in the plateau section of its discharge curve.

With this new information, I guess my original question only makes sense if the board didn't have its own voltage regulator.

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  • \$\begingroup\$ Start with the battery datasheet and figure out what the discharge characteristics are. That and the Arduino requirements will help you specify your regulator, etc. \$\endgroup\$ – Transistor May 20 '16 at 23:07
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    \$\begingroup\$ I'm not the right person to advise you fully, but I think the end voltage (3.0 V) is the number you should work with. I suspect you'll end up using a small boost converter to 4 or 4.5 V to keep the micro happy. \$\endgroup\$ – Transistor May 20 '16 at 23:20
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    \$\begingroup\$ That is not a normal input range for a device. I'd look more closely at it to see if it really can't go down to about 3.0V under normal operation. \$\endgroup\$ – Ignacio Vazquez-Abrams May 21 '16 at 0:24
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    \$\begingroup\$ A guess would be the micro takes a nominal 3.3v (min maybe 3.0), and there is an LDO on the board - but still this doesn't really make much sense- then the max would be at least 6v. Rig the arduino as a voltage sensor, have it plot a battery discharge current to a terminal (and log it), sampling every 30s. This will give you an idea of what is possible - then add I guess 0.1v of margin. \$\endgroup\$ – Sean Houlihane May 21 '16 at 9:19
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    \$\begingroup\$ @SeanHoulihane, when the LDO drops out, the reference for the ADC will probably track the input voltage, and the ADC will return a constant value from there on out. Yes, I have done this (not with any arduino or relative, though). If you think about it and sketch it out, you will see why. It is still a useful excercise, though, because you can see where the constant value is. You know you can at least get there. \$\endgroup\$ – mkeith May 22 '16 at 4:22
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Each battery have own discharge behaviour:

enter image description here

The part of graph where it's more acclivous - it's the life cycle of battery. So real voltage of battery will drop down in short time after battery plugged into device and for good battery (ex.: E2 on the picture) it will save this voltage on the most part of life cycle.

So if you need to power device with voltage in range 3.7-4.2V - you need to select battery by higher threshold as minimum. Battery capacity about 100mAh is too small, even if you will use strongly power save (with consumption about 5uA) beetween two measurements. I use battery 3.3V 1000mAh for power board with STM8L151 series MCU and with all my efforts to power saving it will work only about 1 year.

If it's difficult or unreal for arduino to achieve good power save characteristics, so you need battery with big capacity and may be, it will be easy to use some linear regulator as LM317 series and 9V battery.

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