power nodemcu with battery and LDO, what if battery is empty? I've read a lot but im still confused..
i want to battery power my nodemcu while it will be mostly deep sleeping and waking up every 1-2 minutes.
i've seen the best way to power it with the least wasting power is though a lithium battery.
so i've bought a 18650 rechargable battery and a MCP1700 LDO which will give me constant 3.3V, that ill connect straight to 3.3V pin and bypass nodemcu's regulators as i understand from here
My question is: if all the previous are correct, what will happen when 18650 batter power fall down the 3.5V (3.3V + 0.2V (from MCP1700 dropout))
Will it stop automatically giving power at all? Will it provide less than 3.3V but it will harm nodemcu?
As with any electronics, once you get outside of normal operating parameters, behaviour becomes undefined. It's up to you to avoid that.

The smartest thing to do is to employ a battery protection circuit or some means to shut down the system (or prevent it from waking up from deep sleep) before the voltage drops too low. This will also prevent your battery from damage by overdischarging it.

The ESP's minimum operating voltage according to its datasheet is 3.0V. Below that, behaviour is undefined but you can expect unreliable operation, crashes or even permanent damage. If you're writing to Flash memory when below the minimum voltage, the Flash might become corrupted (including the program the ESP is running).