electronics

Although the serial-in, parallel-out shift register, aka. SN74AHC595, SN74AHC594 or just “595″ / “594″, is an easy component to hook up and use, it might not be immediately obvious what’s going on inside the chip. At least it was not to me. Therefore I found Mike Szczys’s demonstration very enlightening. He shows how to drive the shift register manually, all without any micro-controller. The functional pins of the register is hooked up to buttons, so he can slowly go through the sequence of setting the data bit, clocking or shifting that onto the register, and then closing the latch to show the result on a line of LEDs.

Combined with the data sheet for the 595, it becomes more clear how the chip works: There are two separate registers on the chip; the visible storage register, and the internal shift register. When the latch (aka. “RCLK”, or “ST_CP”) goes high, the bits from the shift register is copied over to the storage register, and becomes visible on the external pins.

So that makes for 16 potentially different bits of information. However, as the title suggest, there’s one more available: It’s the output pin, typically used to daisy chain multiple shift register chips together. Even though it is part of the internal shift register, which thus consists of 9 bits, it is externally visible, and retains its state after the latch is closed. The drawing below shows this, with nine yellow external dots indicating visible bits, and eight blue dots indicating internal bits, for a total of 17 bits.

In the figure I’ve marked the QH’ as externally visible, even though it is part of the internal shift register. There probably aren’t many applications in which you’d use that as an active output, but one that comes to mind is a 3×3x3 LED cube. In a typical setup, you’d have three layers of 9 LEDs, for 12 wires altogether. Instead of controlling each LED in a layer individually, you could use a shift register, however you’d be one pin short (or 7 pins wasted if using two chips). So why not use the output pin for the 9th LED?

The programming would work as normally when addressing a shift register, but instead of shifting out 8 bits, you’d shift 9. The only problem would be that the 9th LED connected to the output pin would see the previous bits flying past as new bits are shifted onto the register. This might not be a big problem, as cycling the layers will probably be too fast to notice. If it is an issue, one could just disable all layers while bits are shifted out, and turn on only the relevant layer when done. I think this is a small experiment I’ll have to make, so stay tuned.

On the heals of the success with the Arduino kit last week, I went ahead and bought some extra parts: a 74HC595, 8 bits Shift Register, some red/green/orange LEDS (TLUV5300), and a bigger bread board. I also got some resistors, however, the proved to be wrong. The guy in the shop suggested 15 Ohm for 5 V, however, all the LED resistor calculators I looked at suggested 100-120 Ohm, I I’ll get some new ones tomorrow.

So for the application. Not very fancy so far, but great fun to see the register chip working so easily. Based on this code example by Carlyn Maw and Tom Igoe, and the tutorial at SparkFun (for their 74HC595 offering), I hacked away and came up with a "binary counter" of sorts. Here’s some pictures, and the code that went with it. I modified the example to use the shiftOut Arduino library function.

On the breadboard, yellow is +5V, and brown in ground. The extra breakaway at row 25 does nothing but support the 4 wire sound cable I used hook up to the Arduino. From left to right, Arduino to 74HC595: 12 -> SH_CP (clock), 11 -> ST_CP (latch), 9 -> DS (data). Finally, don’t be confused by the series of resistors to the right. I wanted to see the difference between 15 and 90 Ohm on one of the LEDS; it was almost not noticeable.

// Based on:
// http://arduino.cc/en/Tutorial/ShftOut13

//Pin connected to ST_CP of 74HC595
int latchPin = 11;
//Pin connected to SH_CP of 74HC595
int clockPin = 12;
////Pin connected to DS of 74HC595
int dataPin = 9;

//holders for infromation you're going to pass to shifting function
//byte data;

void setup() {
//set pins to output because they are addressed in the main loop
pinMode(latchPin, OUTPUT);
pinMode(clockPin, OUTPUT);
pinMode(dataPin, OUTPUT);

//function that blinks all the LEDs
//gets passed the number of blinks and the pause time
blinkAll(2,500);
}

void loop() {
for (int j = 0; j < 255; j++) {
write(j);
}

}

void write(byte data) {
digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST, data);
digitalWrite(latchPin, HIGH);
delay(1000);
}

//blinks the whole register based on the number of times you want to
//blink "n" and the pause between them "d"
//starts with a moment of darkness to make sure the first blink
//has its full visual effect.
void blinkAll(int n, int d) {
write(0);
for (int x = 0; x < n; x++) {
write(255);
write(0);
}
}