Question

Mechatronics

void setup () Write the following code and upload it to the { Arduino, observe and comment on the pinMode (8, OUTPUT) function of the circuit and explain how the pinMode (2, INPUT); code is working. In addition, explain how to digitalWrite (8, HIGH) turn ON the LED in case it is OFF. void loop () { digitalRead (2) int A if (A 0) { digitalWrite (8, LOW)

Answer 1

To turn on an LED, the Arduino needs to send a HIGH signal to one of it’s pins. To turn off the LED, it needs to send a LOW signal to the pin. The Arduino has an on-board surface mount LED that’s hard wired to digital pin 13. Also, an external LED or any other powered module can be controlled in a similar way.

For example,in the code, we want to use pin 8 instead of pin 13:

pinMode(8, OUTPUT);

Next you’ll need to change the code that tells the Arduino which pins will get the HIGH and LOW output signals. That’s done everywhere there’s a digitalWrite() function. In the program above:

digitalWrite(8, HIGH);   

Now, pinMode(2,INPUT): The Second Pin of the Arduino is set up as an input pin.

In the loop part of the code:

int A=digitalRead(2) //The value of Pin 2 is being read which is either 1(High)or 0(Low).

digitalWrite(8,LOW)//

If the value of A, ie. if the value at Pin 2 is Low(0) then Pin 8 output will be low, hence turning off the LED at Pin 8.

Working:

Initially the LED at Pin 8 is ON. Now the input is given to Pin 2 which can be done by adding a pullup resistor (to +5V)(HIGH INPUT), or a pulldown resistor (resistor to ground)(LOW INPUT) on the input. A 10K resistor is a good value for a pullup or pulldown resistor.

Now the value of Pin2 is constantlt being checked.If the value is Low, the led at Pin 8 will be turned off and if the value of Pin 2 is high, then the LED will remain On.