digitalWrite(1, LOW) == Ground?

I'm Reading here and for what I understand setting an output pin to low will make that pin to out put ground signal.
Is that right or am I wrong??

This is what it says: If the pin has been configured as an OUTPUT with pinMode() , its voltage will be set to the corresponding value: 5V (or 3.3V on 3.3V boards) for HIGH , 0V (ground) for LOW .

0V (ground)???

An LOW output does not "output ground". An output that is LOW is at ground potential will sink current. An output that is HIGH is at Vcc potential and will source current.

also, maybe you are using pin1 as an example, but pin 0 and pin 1 are reserved for the USB communications. Connecting anything to it will not allow programming of the board.

Thank you!! I understand now :sweat_smile:

That's a good point and yeah pin 1 was just an example.

Fig-1 shown below says at the right top corner -- it is guaranteed that the MCU will not assert more than 0.9V (VOL) at its output pin when the following code is executed:

digitalWrite(8, LOW);


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