Machine installed with arduino suddenly turns on when usb is unplugged

Question

Why is my machine suddenly turns ON when I unplugged the USB on my computer? Does it have connection on the auto reset feature of my arduino board?

String input = "";

void setup(){

  Serial.begin(9600);   // Sets up communication with the serial port

  pinMode(6,OUTPUT);
  pinMode(7,OUTPUT);
  pinMode(8,OUTPUT);
  pinMode(9,OUTPUT);
  pinMode(10,OUTPUT);
  pinMode(11,OUTPUT);   // for start of program operation


  digitalWrite(6,HIGH);
  digitalWrite(7,HIGH);
  digitalWrite(8,HIGH);
  digitalWrite(9,HIGH);
  digitalWrite(10,HIGH);

  digitalWrite(11,LOW);

}

void loop(){

  int recipe;
  //Serial.print("Enter recipe number: ");

  while(Serial.available()==0){    // Waits for data from serial port
  }

  while(Serial.available() > 0){
    int data = Serial.read();
      // convert the incoming byte to a char
      // and add it to the string
    input += (char)data;

    recipe = input.toInt()+'0';    //recipe=0 if not a digit
    if(recipe == '0')
      recipe = data;

    if(data == '\n'){
      //Serial.print("Recipe: ");
      //Serial.println(recipe);
      //Serial.print("String: ");
      //Serial.println(input);
      input = "";
    }
  }

  if(recipe == 's'){     
    digitalWrite(11,HIGH);
    delay(1000);
    digitalWrite(11,LOW);

    digitalWrite(6,LOW);
    digitalWrite(7,LOW);
    digitalWrite(8,LOW);
    digitalWrite(9,LOW);
    digitalWrite(10,LOW);    
  }

  else if(recipe>'0' && recipe<'32'){    

    recipe = recipe-'0';
    char binary[7] = {0};   // This is where the binary representation will be stored
    recipe += 32; // Adding 32 so that there will always be 6 digits in the string
    itoa(recipe,binary,2);  // Convert recipe to a string of base 2 and save it in array 'binary'
    char* signals = binary + 1; // Get rid of the most significant digit to get the 5 bits

  //Serial.print(signals);  // Print out the signals in binary

    int i;
    for(i=0; i<5; i++){
      if(signals[i] == '0')
        signals[i] = '1';
      else
        signals[i] = '0';
      digitalWrite(i+6, signals[i]-'0');    // write to pin; converts the bit of the string to HIGH or LOW
      //Serial.print(signals[i]);
    }

  }
  else
    loop();

}

Nick Gammon · Answer 1 · 2016-03-10T04:52:33.263

Do not call loop from loop. You will quickly run out of RAM and the sketch will crash:

void loop(){

  int recipe;
  //Serial.print("Enter recipe number: ");

  while(Serial.available()==0){    // Waits for data from serial port
  }

  while(Serial.available() > 0){
...

  }
  else
    loop();

}

For the pins that are supposed to be high, I suggest pull-up resistors:

  pinMode(6,OUTPUT);
  pinMode(7,OUTPUT);
  pinMode(8,OUTPUT);
  pinMode(9,OUTPUT);
  pinMode(10,OUTPUT);
  pinMode(11,OUTPUT);   // for start of program operation


  digitalWrite(6,HIGH);
  digitalWrite(7,HIGH);
  digitalWrite(8,HIGH);
  digitalWrite(9,HIGH);
  digitalWrite(10,HIGH);

When the Arduino powers up it enters the bootloader for a second or so. Those pins won't be set high yet. A pull-up resistor would make sure they are high until such time as you want to drive them low.

If you are saying that the machine powers up when you disconnect the power to the Arduino, that is still a good reason for using pull-up resistors, to make sure that it does not operated until commanded to do so.

where will i connect the pull up resistors and how?

Typically pull-up resistors will be around 10k, and are connected from the pin to +5V. Of course, if the Arduino is not powered, then you need another source of the +5V.

please to clarify the "do not call from loop to loop"

Remove these two lines:

else
    loop();

When loop exits, is it immediately re-called. You don't need to do it that way, in fact it is a Bad Idea.

Dave X · Answer 2 · 2016-03-10T05:05:45.277

It looks like the this line in your setup:

  digitalWrite(11,LOW);

... would trigger contact input 211, and do "Start of Program Operation when D11 switches from off to on" every time the sketch starts.

If the arduino resets, it will pull this line low. Autoreset would initiate this.

By changing the line to the following, the code will wait until it gets an 's' before pulling the line low.

  digitalWrite(11,HIGH);

I don't have an Arduino, but perhaps http://playground.arduino.cc/Main/DisablingAutoResetOnSerialConnection may help in managing the Auto-Reset issue.

Machine installed with arduino suddenly turns on when usb is unplugged

2 Answers2