The circuit that has been created is shown in the following link:
**broken link removed**
My question is that I would like to replace the 9V battery with the following 3.7 V rechargeable LiPo one and I don’t know how. Are the characteristics of this battery suitable for my project? The reason I am changing the battery is to minimize the circuit to a portable device.
The Pro Micro and Bluetooth Shield are both designed for 3.3V and won't be happy connected directly to a 9V battery. Is there some other circuitry involved - perhaps a voltage regulator to drop the 9V down to 3.3V?
It has a diode feeding a 3.3V regulator. I did not look up the dropout voltage of the regulator and I did not lookup if the processor works from a 3.2V to 4.2V Li-Po battery voltage.
No. A capacitor is not a battery.
A battery voltage stays close to a high voltage for most of the time of a discharge. Then when it is almost dead its voltage drops quickly.
A capacitor voltage IMMEDIATELY begins to drop quickly.
The graph is in Google Images.
It doesn't matter. Supercapacitors store less energy than LiPo batteries, for the same weight or size. They can deliver greater power than batteries of the same size or weight, but that's not what's needed here.
1) Lookup the maximum dropout voltage of the 3.3V regulator.
2) Lookup the minimum supply voltage for the processor.
3) Lookup the minimum allowed voltage of the battery or its low voltage cutoff from its protection circuit (if it has one).
4) Add the maximum forward voltage drop of the diode.
Note that if you allow a Li-Po battery cell to discharge to less than about 3.2V then it will be damaged.