Question

到目前为止，我已经编写了代码，如果1然后输入按下或发送点击，然后再次输入1并输入按下或发送点击导致LED 1打开，如果'1''0'以类似的方式输入然后LED 1关闭，依此类推LED 2和3，即：'2''1'打开LED 2，'3''0'关闭LED 3。

 int incomingVal;
    int ledPin = 16; 
    int ledPin2 = 15; 
    int ledPin3 = 14; 

    void setup()
    {

      Serial.begin(9600);     // opens serial port, sets data rate to 9600 bps
      Serial.println("starting");
      pinMode(ledPin,OUTPUT);
      pinMode(ledPin2,OUTPUT);
      pinMode(ledPin3,OUTPUT);
    }




    void loop()
    {
      if (Serial.available() > 0 ) //then chars are in the serial buffer
      {
        incomingVal = Serial.parseInt();
        Serial.print("You entered: ");
        Serial.println(incomingVal);

        if (incomingVal == 10)//turns off led 1 
        {
          digitalWrite(ledPin, LOW);  

        }

         if (incomingVal == 11)//turns on led 1 
        {
          digitalWrite(ledPin, HIGH);  

        }

         if (incomingVal == 20)//turns off led 2
        {
          digitalWrite(ledPin2, LOW);  

        }

         if (incomingVal == 21)//turns on led 2
        {
          digitalWrite(ledPin2, HIGH);  

        }

         if (incomingVal == 30)//turns off led 3
        {
          digitalWrite(ledPin3, LOW);  

        }

         if (incomingVal == 31)//turns on led 3
        {
          digitalWrite(ledPin3, HIGH);  

        }

      }

    }

我怎么能改变代码所以我可以输入第三个值来将亮度从0改为250？例如，输入“2,1,125”会使LED 2以50％的亮度点亮。

Answer 1

要使LED亮起一半亮度，您需要使用pulse width modulation。确保您使用的引脚位于它们旁边（Uno上的引脚3,5,6,9,10,11）。

我会略微改变你的输入法，因为0亮度基本上与关掉LED一样，你应该只需要2个数字。第一个数字对应LED，第二个数字对应亮度。

int led_pins[3] = {9,10,11};

int incomingVal;
int brightness;
int parsed_value;
int x = 0;
void setup()
{
  Serial.begin(9600);     // opens serial port, sets data rate to 9600 bps
  Serial.println("starting");
  pinMode(led_pins[0],OUTPUT);  // You could also use a loop here
  pinMode(led_pins[1],OUTPUT);
  pinMode(led_pins[2],OUTPUT);
}




void loop()
{
  if (Serial.available() > 0 ) //then chars are in the serial buffer
  {
    incomingVal = Serial.parseInt();
    Serial.print("You entered: ");
    Serial.println(incomingVal);

    //figure out which LED we selected
    parsed_value = incomingVal;
    while (parsed_value > 9){
      /*
      because of integer division, this line will remove the last
       number from the integer. ie 11 / 10 = 1 (the result is rounded down)
       The loop will continue until only 1 digit remains
       */
      x++;
      parsed_value = parsed_value / 10; 
    }
    // x represents the number of times parsed_value was divided by 10
    // this line removes the first digit of the incoming value, leaving
    // the brightness
    brightness = incomingVal - pow(10,x)*parsed_value;
    /*parsed_value will be 1 greater than the array key value because
     the array is 0 based.*/
    //set the LED to the right brightness
    analogWrite(led_pins[parsed_value-1],brightness);
  }

}

使用3个LED的串行通信改变亮度

1 个答案: