1
const int Trigger = 6;
const int Echo = 7;
const int LEDpin1 = 4;
const int LEDpin2 = 5;
const int Parameter = 400;
float Time;

void setup()
{
Serial.begin(9600);
pinMode(Trigger, OUTPUT);
pinMode(Echo, INPUT);
pinMode(LEDpin1,OUTPUT);
pinMode(LEDpin2,OUTPUT);
}
void loop()
{

digitalWrite(Trigger, HIGH);
delayMicroseconds(10);
digitalWrite(Trigger, LOW);
Time = pulseIn(Echo, HIGH);
Serial.print("Distance: ");
Serial.println(Time);
delay(10);
if (Time < Parameter)
digitalWrite(LEDpin1, HIGH);
else
digitalWrite(LEDpin1, LOW);

if (Time < Parameter)
digitalWrite(LEDpin2, HIGH);
else
digitalWrite(LEDpin2, LOW);

//**When I include -delay(5000)- right here, it also delays the Ultrasonic sensor to only detect every 5 seconds**
}
Coder9390
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Leonardo Morales
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    The answer is using non-blocking code involving the `millis()` function in a pattern like the `BlinkWithoutDelay` example shows. There are many tutorials about that on the web. When you learn that coding principle you will be able to solve your problem – chrisl Nov 20 '21 at 18:56
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    Please format code as code (you can edit your post) so it's easier to read. – Dave Newton Nov 20 '21 at 20:10

0 Answers0