Individual pin control on shift register (74HC595)

Using Arduino Programming Questions
1

I am not familiar with using shift registers. I am using this code to control 8 leds.

int tDelay = 500;
int latchPin = 11;      // (11) ST_CP [RCK] on 74HC595
int clockPin = 9;      // (9) SH_CP [SCK] on 74HC595
int dataPin = 12;     // (12) DS [S1] on 74HC595

byte leds = 0;
// this first function to turn them off, from right to left (least significant)
void updateShiftRegister()
{
   digitalWrite(latchPin, LOW);
   shiftOut(dataPin, clockPin, LSBFIRST, leds); //LSBFIRST starts from the least significant Byte, that corresponds to 8th pinout 
   digitalWrite(latchPin, HIGH);
}
// this second function is to turn them off
void updateShiftRegister2() 
{ 
   digitalWrite(latchPin, HIGH);
   shiftOut(dataPin, clockPin, LSBFIRST, leds); //if we start with MSBFIRST in this function, then it would start from the most significant, that is the 1st pinout.
   digitalWrite(latchPin, LOW);
}

void setup() 
{
  pinMode(latchPin, OUTPUT);
  pinMode(dataPin, OUTPUT);  
  pinMode(clockPin, OUTPUT);
   Serial.begin(9600);
}


void loop() 
{
   leds = 0;
  updateShiftRegister();
  for (int i = 0; i < 8; i++)
  {
    delay(tDelay);
    bitSet(leds, i); //bitset sets the byte of each led to on
    updateShiftRegister();   
   }   
}

1757×634 130 KB

It turns on the led's from right to left. The code is working well. But I want to individually control an led, to control led 1 or 2 or any led of this project, how can I do so ! I couldn't figure it out.

Please help me...
Thanks in advance :slight_smile:

2

Hello,

Use bitSet and bitClear, or use a library ( http://bildr.org/2011/08/74hc595-breakout-arduino/ )

3

I want to individually control an led, to control led 1 or 2 or any led of this project, how can I do so ! I couldn't figure it out.

  for (int i = 0; i < 8; i++)
  {
    delay(tDelay);
    bitSet(leds, i); //bitset sets the byte of each led to on
    updateShiftRegister();   
   }

This for loop sets each bit of the variable i to HIGH in turn and sends the byte to the SR each time one is set

EDIT : corrected the sentence above
This for loop sets each bit of the variable leds to HIGH in turn and sends the byte to the SR each time one is set

As an experiment set i to say 0b10101010 and call the updateShiftRegister() function.. What do you see ?

You can use bitSet() to turn any bit in a byte HIGH and you can use bitClear() to set any bit LOW

4

UKHeliBob re read your post and fix it :slight_smile:

5

guix:
UKHeliBob re read your post and fix it :slight_smile:

Yes Sir !
Done :slight_smile:

6

My intention was, to control any specific output of shift registers. As there, 8 led's are connected, How I could control any specific led, if I want to.
And I found the simplest way to do so:

int tDelay = 500;
int latchPin = 11;      // (11) ST_CP [RCK] on 74HC595
int clockPin = 9;      // (9) SH_CP [SCK] on 74HC595
int dataPin = 12;     // (12) DS [S1] on 74HC595

byte leds = 0;
void updateShiftRegister()
{
   digitalWrite(latchPin, LOW);
   shiftOut(dataPin, clockPin, LSBFIRST, leds); 
   digitalWrite(latchPin, HIGH);
}

void setup() 
{
  pinMode(latchPin, OUTPUT);
  pinMode(dataPin, OUTPUT);  
  pinMode(clockPin, OUTPUT);
   Serial.begin(9600);
}


void loop() 
{
  updateShiftRegister();
  bitSet(leds, 0);
  updateShiftRegister();
  
  delay(2500);
  
  updateShiftRegister();
  bitClear(leds, 0);
  updateShiftRegister();
  
  updateShiftRegister();
  bitSet(leds, 2);
  updateShiftRegister();
  
  delay(1000);
  
  updateShiftRegister();
  bitSet(leds, 4);
  updateShiftRegister();
  
  delay(200);
  
  updateShiftRegister();
  bitClear(leds, 2);
  updateShiftRegister();
  
  updateShiftRegister();
  bitSet(leds, 1);
  updateShiftRegister();
  
  delay(500);
  
  updateShiftRegister();
  bitClear(leds,1);
  updateShiftRegister();
  
  delay(4000);
  
  updateShiftRegister();
  bitClear(leds, 4);
  updateShiftRegister();
  
  
  updateShiftRegister();
  bitSet(leds, 3);
  updateShiftRegister();
  
  delay(1000);
  
  updateShiftRegister();
  bitClear(leds, 3);
  updateShiftRegister();

  delay(5000);
   
}

Thank's for everyone's support :slight_smile: I managed to complete whatever I was wishing for :slight_smile:

1 Like
7

You don't need to call updateShiftRegister before modifying 'leds'. And, you could do it like this :

void updateShiftRegister( const uint8_t value )
{
    digitalWrite(latchPin, LOW);
    shiftOut(dataPin, clockPin, LSBFIRST, value ); 
    digitalWrite(latchPin, HIGH);
}

void loop() 
{
    // this is equivalent to your code
    updateShiftRegister( 0b00000001 ); // led 0 on, other leds off
    delay(2500);
    updateShiftRegister( 0b00000100 ); // led 2 on, other leds off
    delay(1000);
    updateShiftRegister( 0b00010100 ); // leds 2 and 4 on, other leds off
    delay(200);
    updateShiftRegister( 0b00010010 ); // leds 1 and 4 on, other leds off
    delay(500);
    updateShiftRegister( 0b00010000 ); // led 4 on, other leds off
    delay(4000);
    updateShiftRegister( 0b00001000 ); // led 3 on, other leds off
    delay(1000);
    updateShiftRegister( 0 ); // all leds off
    delay(5000);
}
1 Like