Use interrupts with analog pins in arduino or another possible solution?

Question

I have a very simple circuit, but my program is some what complicated.

Here is the circuit diagram:

Here is the code:

/*
  01 - GND -------- GND
  02 - +5V -------- 5V
  03 - V0  -------- Potentiometer Middle
  04 - Reset ------ 9
  05 - Read/Write - GND
  06 - Enable ----- 8
  07 - Data0 ------ NC
  08 - Data1 ------ NC
  09 - Data2 ------ NC
  10 - Data3 ------ NC
  11 - Data4 ------ 10
  12 - Data5 ------ 11
  13 - Data6 ------ 12
  14 - Data7 ------ 13
  15 - +LCD ------- 5V
  16 - -LCD ------- GND
*/

//include LCD library
#include <LiquidCrystal.h>

//Initialize an LCD object
/*Pins should be mentioned in this order:
  Reset
  Enable
  Data4
  Data5
  Data6
  Data7
*/
LiquidCrystal lcd(2, 3, 4, 5, 6, 7);
int onTimePin = A0;
int onTime = 0;
int offTimePin = A1;
int offTime = 0;
int ledPin = 13;

void setup()
{

  pinMode(onTimePin, INPUT);
  pinMode(offTimePin, INPUT);

  pinMode(ledPin, OUTPUT);

  //Begin the LCD interface
  lcd.begin(16, 2);

  lcd.print("ON  TIME : ");
  lcd.setCursor(0, 2);
  lcd.print("OFF TIME : ");

}

void loop()
{
  onTime = map(analogRead(onTimePin), 0, 1023, 1, 3);
  lcd.setCursor(11, 0);
  lcd.print(onTime);

  offTime = map(analogRead(offTimePin), 0, 1023, 1, 9);
  lcd.setCursor(11, 1);
  lcd.print(offTime);

  digitalWrite(ledPin, HIGH);
  delay(onTime * 1000);

  digitalWrite(ledPin, LOW);
  delay(offTime * 1000);
}

The above code works. But there is a small problem:

When I turn the potentiometer wiper, there is a delay.

It's obvious because when the loop completes, the changed value of potentiometer is displayed. So, I decided to do the same thing using interrupts. But only digital pins have interrupts.

I would like to know the solution for using interrupts on analog input.

JimmyB · Accepted Answer · 2018-12-06T13:12:47.973

Avoid delay.

Very simplistic alternative:

unsigned long lastLedToggleTime;
unsigned long currentLedToggleInterval;

void loop() {
  ...

  unsigned long now = millis();

  if ( (now - lastLedToggleTime) > currentLedToggleInterval) {
    lastLedToggleTime = now;
    // We need to toggle the LED now.
    if ( digitalRead( ledPin ) == LOW ) {
      // LED was LOW, so we set it HIGH and setup next toggling after onTime.
      digitalWrite(ledPin, HIGH);
      currentLedToggleInterval = (onTime * 1000);
    } else {
      // LED was HIGH, so we set it LOW and setup next toggling after offTime.
      digitalWrite(ledPin, LOW);
      currentLedToggleInterval = (offTime * 1000);
    }
  } 
  ...
}

Notice that we're dealing with 32 bit unsigned integer values here, which is why this code will work even if/when millis() rolls over to 0. Specifically, not only will now - lastLedToggleTime always return a positive value (because it is cast into the unsigned data type), it will also return the correct value due to two's complement representation and modulo arithmetic base 2^32.

Example: Let's just look at an 8 bit value for simplicity:

Say, lastLedToggleTime was at the last tick of the 8 bit timer value, i.e. lastLedToggleTime == 255. In the next timer tick, now will overflow to 0, so next we have now - lastLedToggleTime = 0 - 255 = -255. In two's complement, this value would need (at least) 9 bits to represent (binary: 1 0000 0001), but because we only have 8 bits of storage, the sign bit (MSB) is just discarded resulting in 0000 0001 binary = 1. We can see that 1 is the correct value for the number of ticks to get from 255 to 0; it's exactly one tick.

score 3 · Answer 2 · answered Oct 21 '16 at 14:08

3

You are looking for the bad solution for a simple problem. As Ignacio Vazquez-Abrams puts it, the problem is you using delay(). The simplest solution is to manage the timings with millis(). See the Blink Without Delay Arduino tutorial.

answered Oct 21 '16 at 14:08

Edgar Bonet

45,094
4
42
81

Use interrupts with analog pins in arduino or another possible solution?

2 Answers2