How to read and use the voltage of 5V from the sensor as a signal and extend that signal and output it to some pin

Using Arduino Sensors
1

Hello everyone,
I am beginner to using the Arduino, I use Arduino Uno.
We have one project in which I need to read a voltage of 5V from a sensor that is short about 3ms. When the sensor does not read anything I have 0V and when it reads I have 5V voltage. I need to read that voltage on some pin as a signal and to output 5V as a signal on the other pin and extend it to 50ms.
I need the output signal to be 5V when the sensor gives 5V, and to be 0 when the sensor reads nothing.

Can someone help me?

Thanks in advance.

2

3ms is short for a human, but very slow for an Arduino.
Just digitalRead the input pin, and write a HIGH to an output pin that you set LOW again after a 50ms delay.

Let's see some code.
Leo..

3

Hi Leo,

Thank you for your response.

This is code I made, but it doesn't work.

int sensor = 13;
int PHD = 7;

void setup() {
pinMode(PHD, OUTPUT);
pinMode(sensor, INPUT);
}

void loop() {
digitalRead(sensor);
if (sensor=HIGH);
{
digitalWrite(7, HIGH);
delay(50);
digitalWrite(7, LOW);
}
}

4 
if (sensor=HIGH);

"=" is the assignment operator. "==" is used for comparisons.,

5

You're not storing the outcome of your digitalRead anywhere. Apart from the problem pointed out by @jremington, you seem to compare the pin value (int sensor = 13) with 'HIGH', which doesn't make sense.
Try something like

if (digitalRead(sensor))
{
  //Rest of code here
}

Another issue is the use of delay() to keep your output high for 50ms. During the delay period, you will miss any 3 ms signals that you might have wanted to capture. Try something creative with millis() (many tutorials on this I understand, but it's not rocket science anyway) and/or Google on 'non-blocking code Arduino' etc.

6

Hello guys,

Thank you for your reply.

In the meantime, I changed the code.

int sensorState = 0;

void setup()
{
pinMode(7, INPUT);
pinMode(13, OUTPUT);
}

void loop()
{
sensorState = digitalRead(7);
if (sensorState == HIGH) {
digitalWrite(13, HIGH);
} else {
digitalWrite(13, LOW);
}
}

This should work.

For the test I used 5V from the Arduino, when I bring 5V to pin 7 the LED "L" on the Arduino starts to light up. When I remove the voltage of 5V LED "L" stops lighting. This confirms that the Arduino is reading the signal. But even if I touch PIN 7 very quickly, the LED "L" remains lit. It lights up for about seven seconds and then goes out. I also read a long signal on the oscilloscope. It seems to me that the Arduino is reacting slowly to change.

Do any of you perhaps know why the Arduino behaves like this?
I tried something with another Arduono Uno device and it behaves the same.

Thank you in advance.

7

Try to avoid using pin13 as input or you will have issues with small boards like a Nano.
Basic sketch attached. If you want it to do more than that, then the delay() should be replaced with millis() timing.
Leo..

const byte sensorPin = 2;
const byte outPin = 13; // also flashes the built-in LED

void setup() {
  pinMode(13, OUTPUT);
}

void loop() {
  if (digitalRead(sensorPin)) {
    digitalWrite(outPin, HIGH);
    delay(50);
    digitalWrite(outPin, LOW);
  }
}
8

Hard to say without a schematic. In principle the response of the arduino should be a few microseconds with the code you posted. A possible cause is that you have some amount of capacitance at pin 7 that charges as soon as you touch it with 5V and that discharges slowly through the leakage current of pin 7.

9

An input pin is 'floating' when nothing is connected to it, and can take any digital value in this state.
You must have a 'bleed' resistor from input pin to ground. Any high value would do (100k-1Meg).

Where does that '5volt' come from.
It's ok if it comes from the Arduino itself, but it could be wise to connect it via a (10k) protection resistor if it comes from a device not powered from the Arduino.
Leo..

10

Hi guys,

I did it.

This code works:

int sensorState = 0;

void setup()
{
pinMode(7, INPUT);
pinMode(13, OUTPUT);
}

void loop()
{
sensorState = digitalRead(7);
if (sensorState == HIGH) {
digitalWrite(13, HIGH);
delay(50);
} else {
digitalWrite(13, LOW);
}
}

The problem arose when I used 5V directly from the Arduino itself to pin 7 for the test. For some reason, it did not want to break the voltage quickly, even if I just touched pin 7.
When I connected the Arduino to the circuit I made to reduce the 12V from the photo cell to 5V to use as an input signal. Then Arduino started working properly.

Thank you so much for help!

11

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