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 ?

• jsotola: try pinMode(7, INPUT); and pinMode(7, INPUT_PULLUP);
• Mah Neh: not sure what i'm supposed to understand @jsotola it seems pull up is default but what is it?

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.