Low current

[rawrdino]

So yesterday i wrote this little skecth :

Code: Select all

const int LeftSensor = A1;
int SensorLeft;

void setup(){
  pinMode(LeftSensor, INPUT);
  digitalWrite(12, LOW);
  digitalWrite(A1, HIGH);
  Serial.begin(9600);
}
void loop(){
  SensorLeft = analogRead(LeftSensor);
  Serial.println("LeftSensor = ");
  Serial.println(SensorLeft);
  delay(100);
  if (SensorLeft = 30 || SensorLeft < 30){
    digitalWrite(12, HIGH);
  }
    else
    {
    digitalWrite(12, LOW);
    }
  }



And it seems that when i power LED from 12th pin it like receives 1mA only. (By the way the LED should light up from the laser but it always stays on.) Thanks for help

[adafruit_support_bill]

How is your led connected? Post a wiring diagram.

[JamesC4S]

Pin 12 is never set as an OUTPUT.

[philba]

Same with A1.

But there is also a newbee bug in the if statement. It will never execute the else code.

This leads to the question of what are the defaults on I/O pins. I think they default to input but, in true arduino form, this doesn't seem to be documented. If they do default to input, then the bug in the code causes a digitalwrite(high) to the input pin which turns on the internal pullup and, viola!, a small amount of current will flow.

[adafruit_support_bill]

I think they default to input but, ...

That is right. They do default to input.
http://arduino.cc/en/Tutorial/DigitalPins

Arduino (Atmega) pins default to inputs

[Stephanie]

The digitalWrite to set A1 high will set the internal pullup on that pin too, and ensure analogReads on A1 only return 1023.

[JamesC4S]

I think they default to input but, in true arduino form, this doesn't seem to be documented.

http://arduino.cc/en/Tutorial/DigitalPins
"Arduino (Atmega) pins default to inputs, so they don't need to be explicitly declared as inputs with pinMode()"