Here is a diagram of the circuit described in the OP.
I am currently using 100 ohms for the resistor near the LED and 470 ohms for the other resistor. The total resistance will affect the voltage at which the coin cell settles and the charge speed. The top resistor might affect the very low modes of the light. The bottom resistor will affect the parasitic drain of the coin cell when the flashlight is on. As described in the above posts.
If you are referring to the system Lexel described: as Lexel wrote, I think there must be a capacitor or something in parallel with the MCU in order for the MCU to be powered while the light is on. In his diagram the MCU essentially receives the voltage across the FET/driver, which would be mostly zero when the FET is 100% on, and at some non-zero value when the FET is not completely on.
There would be some nice things about using a capacitor instead of a coin cell, like not having to worry about over discharging and ruining the cell while storing the light without a main cell, for instance. But I suspect that capacitors in this size range just don’t store enough charge to reliably perform the function of powering the MCU, at least without making severe power restrictions on the flashlight.
Yes, I was referring to Lexel’s drawing. I still think that the only way to avoid a coin battery or a capacitor is to turn the battery by 180° as I tried to illustrate.
I did a bit more research in the tailcap driver
actually its not using a capacitor it seems to use a inductor to power the MCU while the LED is on
Interesting. It must have a lot less drain than the ~2mA of the attiny13A while on.
It is not using an attiny 13A
it is using a much smaller smd MCU
I think the Inductor with some caps and diodes are the power supply for the MCU,
even PWM the FET a bit on highest level will result in a negative voltage in the inductor powering the MCU