1

Yesterday I asked a question about de-bouncing this rotary encoder and Jsotola suggested that I should use a method like This to detect the previous state and then do the counting. using that method each turn is counting twice in one turn so I came up with my version of de-bouncing the rotary encoder:

int s1, s2;
int count = 0;
boolean right = 0;
boolean left = 0;
int rotary = 0;
unsigned long timer = 0;
unsigned int bounce = 50;

void setup() {
  Serial.begin(9600);
  pinMode(6, INPUT);
  pinMode(7, INPUT);

}

void loop() {

  s1 = digitalRead(6);
  s2 = digitalRead(7);
  if (s1 == 1 && s2 == 1) {
    right = 0;
    left = 0;
    rotary = 0;
  }
  if (s1 == 0 && s2 == 1) {
    left = 1;
  }
  if (left == 1 && (s1 == 0 && s2 == 0)) {
    rotary = 2;
  }
  if (s1 == 1 && s2 == 0) {
    right = 1;
  }
  if (right == 1 && (s1 == 0 && s2 == 0)) {
    rotary = 1;
  }

  if (rotary == 1) {
    if ((millis() - timer) > bounce) {
      count++;
    }
    right = 0;
    left = 0;
    timer = millis();
  }
  if (rotary == 2) {
    if ((millis() - timer) > bounce) {
      count--;
    }
    right = 0;
    left = 0;
    timer = millis();
  }
  Serial.print(s1);
  Serial.print(s2);
  Serial.print(", ");
  Serial.print(count);
  Serial.println(" ");
  delay(15);
}

The code is working just fine. When turning the rotary this are the steps that I get:

10-00-01 < (11) > 10-00-01

problem is if I turn the knob quickly rotary encoder will skip one or two steps like: 01-01-01 < (11) or 10-00-00 < (11).

This problem exist in all methods/libraries (I tested them all), is there anyway to fix this with software/hardware or this rotary encoder is just junk and I should get a better one?


Another method of de-bouncing as Jsotola requested:

 int pinA = 3;  // Connected to CLK on KY-040
 int pinB = 4;  // Connected to DT on KY-040
 int encoderPosCount = 0; 
 int pinALast;  
 int aVal;
 boolean bCW;

 void setup() { 
   pinMode (pinA,INPUT);
   pinMode (pinB,INPUT);
   /* Read Pin A
   Whatever state it's in will reflect the last position   
   */
   pinALast = digitalRead(pinA);   
   Serial.begin (9600);
 } 

 void loop() { 
   aVal = digitalRead(pinA);
   if (aVal != pinALast){ // Means the knob is rotating
     // if the knob is rotating, we need to determine direction
     // We do that by reading pin B.
     if (digitalRead(pinB) != aVal) {  // Means pin A Changed first - We're Rotating Clockwise
       encoderPosCount ++;
       bCW = true;
     } else {// Otherwise B changed first and we're moving CCW
       bCW = false;
       encoderPosCount--;
     }
     Serial.print ("Rotated: ");
     if (bCW){
       Serial.println ("clockwise");
     }else{
       Serial.println("counterclockwise");
     }
     Serial.print("Encoder Position: ");
     Serial.println(encoderPosCount);

   } 
   pinALast = aVal;
 } 
  • 1
    Did you delete your 1st question? I can not find it. 1) It would be best if you kept a 4 state machine such that you could tell left from right based on the current code and the previous quad-code. I don't think I see that in your code above. 2) When dealing with firmware sampling one can assume some samples will be missed. So the software needs to compensate or you need better hardware. Usually cheap hardware and good software wins out because of cost. So, I would have the software make up left & right base on history for illegal code sequences. – st2000 Oct 14 at 13:23
  • @st2000 deleted first question because i couldn't explain what the problem actually was and the question was a mess. | How can i detect one step from another when like the example in the question it just skipping all 3 or 2 steps? i need at least the first step... – Hamed Oct 14 at 13:34
  • My guess is the that all those serial.prints are really slowing down your code, making it too slow to detect fast rotary movements. You could for example only print the count, then rotate right fast, and then rotate back slowly, till it reaches the starting position. Then check if count is back to 0. – Gerben Oct 14 at 15:21
  • @Gerben removed delay() and serial prints too, no difference what so ever... – Hamed Oct 14 at 17:31
  • you almost have it ... the program detects the transition, but you are jumping to trying to determine the direction of rotation .... forget about the direction and focus on consistently detecting either the rising or the falling edge of the signal ... when you see a transition, wait 50 ms and check again ... if a transition is confirmed, then check the signal level .... if it is high, then a rising edge occured ... if it is low, then a falling edge occured .... at this point, do not worry about the direction ... simply increment the counter only on rising edge ... observe the results – jsotola Oct 14 at 19:44
1

I applaud you for your tenacity with the "debouncing" idea, but there is an excellent library for this rotary encoder, and the author has also written an article about it: Ben Buxton - rotary encoders done properly. Here is a link to his library: GitHub - buxtronix.

I’ve used an older version of the library before with interrupts, and it worked perfectly for my application. I see the author has updated the library and has 2 example sketches; one uses interrupts, and the other uses polling. I tried both of them today on an Uno and I was very impressed with the polling sketch’s performance. It seemed to equal that of the interrupt method.

If you take the important bits out of the library files, then add them to a sketch, you can see how little code it takes for this "method" of processing the encoder’s output. Here is a test sketch with the library bits in it.

#define R_START 0x0
#define R_CW_FINAL 0x1
#define R_CW_BEGIN 0x2
#define R_CW_NEXT 0x3
#define R_CCW_BEGIN 0x4
#define R_CCW_FINAL 0x5
#define R_CCW_NEXT 0x6
#define DIR_NONE 0x0
#define DIR_CW 0x10
#define DIR_CCW 0x20

const byte pin1 = 5;  // DT
const byte pin2 = 6;  // CLK
int state = R_START;
int counter = 0;

const unsigned char ttable[7][4] = {
  // R_START
  {R_START,    R_CW_BEGIN,  R_CCW_BEGIN, R_START},
  // R_CW_FINAL
  {R_CW_NEXT,  R_START,     R_CW_FINAL,  R_START | DIR_CW},
  // R_CW_BEGIN
  {R_CW_NEXT,  R_CW_BEGIN,  R_START,     R_START},
  // R_CW_NEXT
  {R_CW_NEXT,  R_CW_BEGIN,  R_CW_FINAL,  R_START},
  // R_CCW_BEGIN
  {R_CCW_NEXT, R_START,     R_CCW_BEGIN, R_START},
  // R_CCW_FINAL
  {R_CCW_NEXT, R_CCW_FINAL, R_START,     R_START | DIR_CCW},
  // R_CCW_NEXT
  {R_CCW_NEXT, R_CCW_FINAL, R_CCW_BEGIN, R_START},
};

unsigned char process(){

  // Grab state of input pins.
  unsigned char pinstate = (digitalRead(pin2) << 1) | digitalRead(pin1);

  // Determine new state from the pins and state table.
  state = ttable[state & 0xf][pinstate];

  // Return emit bits, ie the generated event.
  return state & 0x30;
}

void setup(){
  pinMode(pin1, INPUT_PULLUP);
  pinMode(pin2, INPUT_PULLUP);
  Serial.begin(9600);
}

void loop(){
  unsigned char result = process();
  if(result == DIR_CW){
    counter++;
    Serial.println(counter);
  }
  else if(result == DIR_CCW){
    counter--;
    Serial.println(counter);
  }
}

Heres a simplified sketch (It works the same but I didn't like the hex numbers and I don't use define):

const byte pin1 = 5;  // Connected to DT on KY-040
const byte pin2 = 6;  // Connected to CLK on KY-040
int state = 0;
int counter = 0;

const unsigned char ttable[7][4] = {
  {0, 2, 4, 0},
  {3, 0, 1, 0 | 16},
  {3, 2, 0, 0},
  {3, 2, 1, 0},
  {6, 0, 4, 0},
  {6, 5, 0, 0 | 32},
  {6, 5, 4, 0},
};

void setup(){
  pinMode(pin1, INPUT_PULLUP);
  pinMode(pin2, INPUT_PULLUP);
  Serial.begin(9600);
}

void loop(){

  unsigned char result = process();

  // Clockwise rotation.
  if(result == 16){
    counter++;
    Serial.println(counter);
  }

  // Counter clockwise rotation.
  else if(result == 32){
    counter--;
    Serial.println(counter);
  }

}

unsigned char process(){

  // Grab state of input pins.
  unsigned char pinstate = (digitalRead(pin2) << 1) | digitalRead(pin1);

  // Determine new state from the pins and state table.
  state = ttable[state & 0xf][pinstate];

  // Return emit bits, ie the generated event.
  return state & 48;
}
  • DUDE! it's magic! why i couldn't find this library! Thank you. – Hamed Oct 14 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.