Same LED but different brightness on same PIN

Question

I connected a led to pin 7 (with a small resistance to avoid burning it) and tried the code below, with and without pinMode as output

void setup() {
  pinMode(7, OUTPUT);
  Serial.begin(9600);
}

void loop() {

  digitalWrite(7, HIGH);
  delay(5000);
  digitalWrite(7, LOW);
  delay(200);

  Serial.println("hello world");
}

Why does commenting the line pinMode(7, OUTPUT); make the led dimmer ?

Answer 1

All the pins are configured as INPUT by default on startup. So if you leave out the pinMode(7, OUTPUT);, pin 7 will still be configured as input. An Input doesn't provide any current (or only a very very low leak current), so the LED will be off in this time.

When you write HIGH to a pin, while it is configured as INPUT, you are activating the internal pullup resistor. This is a resistor (typically about 10kOhm) from the pin to Vcc (HIGH level) inside the microcontroller itself. Now a small current - set by the total resistance (your resistor + internal pullup) - can flow through the diode, making it light up very dim.

Configuring a pin as INPUT with the internal pullup resistor activated can be done in one command: pinMode(7, INPUT_PULLUP);, which makes the intent more clear, but does exactly the same under the hood as setting to INPUT and writing HIGH to the pin.