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?

  • google nodemcu brownout – jsotola Jan 29 '20 at 20:51
  • does the cell have an integrated protection board? if so, then it will just cut-off. – dandavis Jan 30 '20 at 21:20
  • If you battery come with flat head with exact 6.5cm long, it likely is an unprotected type without the built-in over-votage, under-voltage shut down circuit. You will need to get your own protection circuit. Even with the protection circuit, it is likely that cut-off will be around 3v instead of 3.5v. ESP8266 operates at 3.0v to 3.6v, so when your regulator input fall below 3.3v, it won't cut off with or without protection circuit, your regulator will not function properly as it is supposed to be. – hcheung Oct 26 '20 at 6:41
  • If your regulator input voltage is lower than 3.3v, it will not shutdown, it simply unregulated. ESP8266 works between 3.0 to 3.6v, as it draw quite some current at power-up of the WiFi (170mA), it could cause the voltage drop to 3.0v, which will crash the ESP8266. To prevent this happen, make sure your MCP1700 output has a bigger capacitor than recommended 1uF on the datasheet. – hcheung Oct 26 '20 at 6:53

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).

  • I totally agree..i don't want in any case let esp work under 3v..so..from your answer i understand that the ldo i suggest does NOT provide you this safety.. Could you suggest a circuit that turn will power of the whole system if ldo output is less than 3V? – uzer123 Jan 30 '20 at 21:20
  • As far as I can see the MCP1700 does not have an undervoltage protection so I would expect the output voltage to gradually drop when the input falls below the LDO threshold. There are numerous Battery Protection boards available that could provide you with an automatic shutdown safety. You could also use the ADC of the ESP8266 to let it monitor its own supply voltage and take appropriate action. – StarCat Jan 31 '20 at 10:16
  • i guess something based to TP4056 that will shutdown at a specific voltage? about the ADC, so i connect LDO ground to esp ground, then batteries positive (LDO in) to a esp pin, and in this way "analogRead" my battery voltage? – uzer123 Feb 1 '20 at 20:58
  • TP4056 is a charging IC, it is not a voltage regulator. The TP4056 itself does not provide the protection circuitry, some of the TP4056 board has extra circuitry that come with a DW01A chip and 8205 dual-MOSFET to provide the battery over-charge (4.2v), over-discharge (below 3v) protection. You still need the MCP1700 regulator to get 3.3v. – hcheung Oct 26 '20 at 7:03

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