Turn off the second builtin Nodemcu 0.9 led

Question

I'm making a low energy consumption device so I'd like to turn off all the led when I'm idle.

Unfortunately the nodemcu turn on a blue led while I'm idle (during delay(ms))

Here is my code:

#define LED    16

void setup() {
  initialSetup();
  //Turn off the built-in led
  digitalWrite(LED, HIGH);
}

void loop() {
  //Here I do everything
  doStuff();
  //GO TO SLEEP
  delay(3600000);
}

The led that turn on during delay is the one near to D0 pin.

The led that I can control and so turn off (pin #16 -> D0) is near the 5v pin.

It's possible to completely turn off all led to consume less?

Answer 1

See this link which shows your second LED exactly (the image is for v1.0, I have v0.9 and the code below works just the same in this situation):

'Onboard LEDs? NodeMCU's got two!':

https://lowvoltage.github.io/2017/07/09/Onboard-LEDs-NodeMCU-Got-Two

Quote:

ESP-12 LED Blink:

 void setup() {
        pinMode(2, OUTPUT);     //Initialize GPIO2 pin as an output
 }

 void loop() {
   digitalWrite(2, LOW);   // Turn the LED on by making the voltage LOW   
   delay(1000);            // Wait for a second
   digitalWrite(2, HIGH);  // Turn the LED off by making the voltage HIGH 
   delay(2000);            // Wait for two seconds
 }

Or write it like this (ie. using 'D4' rather than '2'):

ESP-12 LED Blink:

 void setup() {
        pinMode(**D4**, OUTPUT);     //Initialize GPIO2 pin as an output
 }

 void loop() {
   digitalWrite(**D4**, LOW);   // Turn the LED on by making the voltage LOW   
   delay(1000);            // Wait for a second
   digitalWrite(**D4**, HIGH);  // Turn the LED off by making the voltage HIGH 
   delay(2000);            // Wait for two seconds
 }

Regards, Geoff