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

Question

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?

Answer 1

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: -

Image from Tiny Buck-Boost Converter for Low Current Applications.

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.

Answer 2

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

Answer 3

If you want the device to operate all the way down to 3.3V just ditch the regulator. Your battery at full charge is 4.7V and the NodeMCU is specified 3.3-5V so you've got a perfect match there. A battery will provide a steady DC voltage that would only fluctuate if your NodeMCU sporadically draws heavy current (which is highly unlikely). Plus, you made an excellent battery choice as the Li-Po performs well under heavy load. It has the lowest voltage drop under load out of all the lithium batteries. So all you need to worry about is the mAh rating of your battery. The NodeMCU shouldn't draw any more than 220mA so if your battery is rated at 220mAh that means it will power your NodeMCU for one hour. You'll have to look up the actual max current draw in the NodeMCU's datasheet. If you're concerned about voltage sag and want a bit of a cushion, you'd probably want the mAh rating of the battery to at least double the max current draw of the device. A 1000mAh battery can supply 1 whole amp of current for an hour so it will be yawning at 220mA. mAh rating is usually printed right on the battery. Refer to the NodeMCU's datasheet to find out it's max current draw then use this calculator to see how long your battery will power it for. No need for a regulator. Hope this helped.

https://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-battery-life