Simple Arduino - 74hc595 shift register construction [tmli5]

Question

I'm quite new to Arduino world and I'm trying to understand how shift registers work. I built this construction :

In fact, it is an Arduino Duemilanove with AtMega on it, but afaik it doesn't matter.

The tutorial I try to follow use SPI library so I try to do it too. I kind of understood that data input, latch and clock had to be pins 7, 4 and 11.

Basically, all I try to do is sending a byte to light the LEDs. As for me now, the sequence should light all LED, then no one, then one after another. I know there is one more input ont the right side of the SR but I didn't plan this model so I ran out of LEDs...

This is the code I use :

#include <SPI.h>
#include <StandardCplusplus.h>
#include <vector>
#include <iterator>
#include <size_type>

using namespace std;


//pin du shift register
#define DATA_PIN 13
#define LATCH_PIN 4
#define CLOCK_PIN 11
//à utiliser plus tard
#define INPUT_PIN 2


void setup() {  
  pinMode(DATA_PIN, OUTPUT);
  pinMode(LATCH_PIN, OUTPUT);
  pinMode(CLOCK_PIN, OUTPUT);
  //pinMode(INPUT_PIN, INPUT);

  SPI.setBitOrder(MSBFIRST);
  SPI.setDataMode(SPI_MODE0);
  SPI.setClockDivider(SPI_CLOCK_DIV2);
  SPI.begin();

  //SPI.transfer(getBit(3));
  //digitalWrite(LATCH_PIN, HIGH);
  //digitalWrite(LATCH_PIN, LOW);

  Serial.begin(9600);
  Serial.println("start !");
}


void loop() {
  shift(B11111111);
  shift(B00000000);
  shift(B10000000);
  shift(B01000000);
  shift(B00100000);
  shift(B00010000);
  shift(B00001000);
  shift(B00000100);
  shift(B00000010);
  shift(B00000001);
}


void shift(int n){
  SPI.transfer(n);
  digitalWrite(LATCH_PIN, HIGH);
  digitalWrite(LATCH_PIN, LOW);
  Serial.print("shift : ");
  Serial.println(n);
  delay(1000);
}

When I upload this code, LEDs sequence has no sense, but serial output was totally OK.

What I see (without the output 0 of course) is : 1111111, 0111111, 1011111, 1110111, 1111101 and then 1111111 without further changes, whereas serial output is 255, 0, 128, 64, 32, 16, 8, 4, 2, 1, 255, 0, 128 and so on.

I assume I didn't understand something important, and it prevent me from finding answers in google.

Thanks for helping, it's really disturbing for me.

EDITS :

changed the title, it is a 74HC595.

Strangest behavior too (not gonna create another topic for that now), I dont have to connect both the gnd and VCC from my arduino for this to light LEDs. It changes a little the brighness of LEDs, but surprisingly VCC< GND< both

This is the tutorial I use to understand shift registers : https://youtu.be/6fVbJbNPrEU?t=195, and the one that uses SPI is nXl4fb_LbcI

I am now aware that I should get one resistor for each LED but I don't have that much resistors and as far as I can see this don't harm my LEDs nor underpower them.

Answer 1

You have a number of problems here. The first is the wiring.

My wiring diagram above is for the Uno / Duemilanove. Since that is what you have, you should follow that.

The pins are fixed in hardware, except for the slave select pin. Thus D13 on the Duemilanove should go to pin 11 on the 595. D11 on the Duemilanove should go to pin 14 on the 595. And your slave select (which may as well be pin 10 on the Duemilanove) should go to pin 12 on the 595.

---

Your code looks more complex than it needs to be. This code will exercise the 8 LEDs:

#include <SPI.h>

const byte LATCH = 10;

void setup ()
{
  SPI.begin ();
}  // end of setup

byte c;
void loop ()
{
  c++;
  digitalWrite (LATCH, LOW);
  SPI.transfer (c);
  digitalWrite (LATCH, HIGH);
  delay (20);
}  // end of loop

---

As other responders have said, you should have a current-limiting resistor for each LED. You can't share it like that.

Also the 595 ground (pin 8) should be wired to the Duemilanove ground, not via a resistor.

---

Reference

Using a 74HC595 output shift register as a port-expander