# Cutoff Voltage of Battery and Minimum input voltage needed by Voltage Regulator

I am working on a project for school and I'm having trouble identifying some variables. I am fairly new to electronics and need some guidance. I am trying to power a nodemcu via a Li-po battery rated at 3.7v. The battery has a maximum voltage of 4.2v when full and a discharge cutoff voltage of 2.75v. The nodemcu operates between 3.3v and 5v. I am using a voltage regulator(LD1117v33) to output a stable 3.3v to the nodemcu. The dropout voltage of the voltage regulator is 1v. The problem I'm facing is that I want the nodemcu to keep operating until the battery dies and needs to be recharged again. So my question is, will the battery keep powering the nodemcu even when its voltage falls below 3.3v? how does this work? and if it does stop working when the battery reaches below 3.3v what changes do I need to make to make certain that the nodemcu keeps working until its battery is fully drained?

• Why do you need a voltage regulator if the nodemcu works up to 5 volts? Commented Dec 14, 2023 at 10:36
• i just want to make sure that the nodemcu receives a steady 3.3v without any fluctuations
– Solo
Commented Dec 14, 2023 at 10:42

The problem I'm facing is that I want the nodemcu to keep operating until the battery dies and needs to be recharged again.

If you want to supply 3.3 volts to the nodemcu with a battery voltage range of 2.75 volts to 4.2 volts then you need a buck-boost converter. Maybe this (or a similar one) might work: -

However, if you were happy to let battery voltages greater than 3.3 volts through unregulated, you can probably find a boost converter that would do the job.

So my question is, will the battery keep powering the nodemcu even when its voltage falls below 3.3v?

You should read the nodemcu data sheet to find this out but, based on what you also said: -

The nodemcu operates between 3.3v and 5v.

I suspect it won't.

When using a LDO (Low Drop Out) regulator, the output voltage will always be lower than the input,

from the datasheet of LD1117v33 we can find that drop out is 1V typ for 100mA unfortunatly, there is no curve in the datasheet, if you know the current consumption of your nodemcu you can find the drop out of your regulator, Let say it is 1V,

The battery has a maximum voltage of 4.2v

this means that the maximum output voltage of the LDO will be 3.2V what is less than the minimum supply voltage of your nodemcu

The nodemcu operates between 3.3v and 5v.

You have two choices, 1- supply directly you node MCU with the battery (I won't recommand this, because you won't have a constant voltage depending on battery life

2- use a buck boost converter, in this case you can use the battery until it is empty

here you can find a description about buck boost

• okay thanks i will check it out. It seems like a better option than using a voltage regulator
– Solo
Commented Dec 14, 2023 at 11:18