Question

我试图同时运行两个＆＃34; Blink＆＃34; -esque功能。

我在其他地方找到了一些代码，可以让我这样做并修改它！大。

然而，我要做的是按照惯例每隔1000毫秒打开和关闭一个LED，但另一个LED以奇数模式闪烁，例如ON为3000ms，OFF为100ms，ON为100ms，OFF为200ms然后循环回来。

为了实现这一点，我尝试在我找到的代码中添加一个随机函数，但我不喜欢它的外观。它看起来像一个闪烁的LED干扰或其他东西，如无线电发射器或其他什么。我试图复制一个旧的闪烁灯泡，这意味着它总是必须打开的时间长于它关闭的时间。基本上它需要为ON / HIGH以便更长时间＆＃34;一段时间然后被几次短暂的ON和OFF闪烁中断

所以我正在寻找帮助，帮助我如何在更具体的一系列闪烁中协调第二个LED的开启和关闭。

这是我目前使用的代码：

/* Blink Multiple LEDs without Delay
*
* Turns on and off several light emitting diode(LED) connected to a digital
* pin, without using the delay() function.  This means that other code
* can run at the same time without being interrupted by the LED code.
*/
int led1 = 13;                // LED connected to digital pin 13
int led2 = 12;
int value1 = LOW;                // previous value of the LED
int value2 = HIGH;                // previous value of the LED
long time1 = millis();
long time2 = millis();

long interval1 = 1000;           // interval at which to blink (milliseconds)

void setup()
{
 Serial.begin(9600);
 randomSeed(analogRead(0));
 pinMode(led1, OUTPUT);      // sets the digital pin as output
 pinMode(led2, OUTPUT);
}

void loop()
{
 unsigned long m = millis();

 if (m - time1 > interval1){
   time1 = m;

   if (value1 == LOW)
     value1 = HIGH;
   else
     value1 = LOW;

   digitalWrite(led1, value1);
 }

long interval2 = random(100,1500);

 if (m - time2 > interval2){
   time2 = m;

   if (value2 == LOW)
     value2 = HIGH;
   else
     value2 = LOW;

   digitalWrite(led2, value2);
 }

   Serial.println(interval2);
}

Answer 1

尝试这样的事情：

void setup()
{
 Serial.begin(9600);
 randomSeed(analogRead(0));
 pinMode(led1, OUTPUT);      // sets the digital pin as output
 pinMode(led2, OUTPUT);
}

long interval2 = random(100,1500);

void loop()
{
 unsigned long m = millis();

 if (m - time1 > interval1){
   time1 = m;

   if (value1 == LOW)
     value1 = HIGH;
   else
     value1 = LOW;

   digitalWrite(led1, value1);
 }

 if (m - time2 > interval2){
   time2 = m;

   if (value2 == LOW) {
     value2 = HIGH;
     interval2 = random(100, 1500);
   } else {
     value2 = LOW;
     interval2 = random(100, 200);
   }

   digitalWrite(led2, value2);
 }

Answer 2

这总是需要不到一秒的时间才能运行，这样您就可以回到主循环并确保不会错过另一个LED的1秒开/关：

void flicker(){
    boolean state = false;
    int r = random(20, 175);
    for(int i = 0; i < 5; i++){
        digitalWrite(led2, state);
        state = !state;
        delay(r);
        r = random(20, 175);
    }
    digitalWrite(led2, HIGH);
}

顺便说一下。我正在替换这个切换代码：

if (value2 == LOW)
  value2 = HIGH;
else
  value2 = LOW;

digitalWrite(led2, value2);

用这个：

state = !state;
digitalWrite(led2, state);

现在，随机调用flicker();也许每15-45秒或任何你觉得合适/现实的东西。

LED同时闪烁，图案分离且不一致

2 个答案: