I've just started with MCUs, so would like to at least to know whether I'm in the right direction...
Trying to power MCU STM32L100RC (just the MCU by itself) from a Panasonic Lithium Ion NCR18650B battery.
However, I'm trying to test it on the integrated circuit board first. See below.
The battery is constantly charging/discharging from PV solar array via DC-DC boost converter, mosfet driver, MCU and other components that make up - Maximum Power Point tracker.
I looked at the battery's data-sheet, and it seems that it can't ever output more than 5V, and over 200mA, battery's internal resistance is 29-39mohm.
So MCU is safe either way, since it said in MCU IC's documentation it can handle up to 1000mA.
MCU discovery board's manual also says that
LD1 (red) for 3.3 V power on
Does it mean MCU discovery board only requires exactly 3.3V?
What if power input varies? I actually tested with a DC power supply, and the MCU discovery board worked between 2.2V and 5V (I didn't risk higher voltages), but strangely enough I couldn't find the 2.2V minimum limit on the manual...
Btw, regarding that power input, it's just gonna be some voltage in, right? Like if I choose to do LDO regulator, output of LDO is just some voltage, correct? That voltage is gonna be connected through a wire to pin called "5V" on the top right of MCU discovery board?
Would this MCU discovery board still work if voltage in is below 3.3V? How about if it's above 3.3V? Say, like 6V? I guess it'd still work, but if it can work with 3.3V, then for efficiency it's better to stick with 3.3V, right?
So I guess I will use an LDO regulator circuit to get power in for MCU discovery board.
I was gonna use a voltage divider circuit first for its simplicity, but it seems that the battery's voltage will vary between 2V and 5V(?), so it's better to use LDO circuit.
So, using LDO regulator. Do I just attach a wire at this point and then the other end of the wire connect to pin "5V" (top right) on MCU discovery board?
And in the LDO circuit schematic, the 5V voltage source - that's gonna be the positive terminal of a single NCR18650B battery, correct?
There are actually 27 of the 18650B cells in parallel, and 23 of those modules in series.
But I only need to use one of those batteries to power up MCU discovery board, don't I?
Last dumb question:
What does 1.65 V to 3.6 V power supply in datasheet mean? Does it mean the power into MCU or the output of MCU (what MCU can provide?) I think it's the latter.
I will stop with questions here for now.