• ### Code

Max-Felix Müller05/10/2019 at 22:33 0 comments

Here is the completed working code for a random pattern. Nothing special, not efficient or fast at all. But it needed to be done quick since it's a gift for my fathers birthday which is now...

```#define MOS1 8
#define MOS2 9
#define MOS3 10
#define MOS4 11

#define DAT1 0
#define DAT2 1
#define DAT3 2
#define DAT4 3
#define DAT5 4
#define DAT6 5
#define DAT7 6
#define DAT8 7

#define OE1 A0
#define OE2 A1
#define LE1 A2
#define LE2 A3

void setup() {
// mosfets
pinMode(MOS1, OUTPUT);
pinMode(MOS2, OUTPUT);
pinMode(MOS3, OUTPUT);
pinMode(MOS4, OUTPUT);

// data
pinMode(DAT1, OUTPUT);
pinMode(DAT2, OUTPUT);
pinMode(DAT3, OUTPUT);
pinMode(DAT4, OUTPUT);
pinMode(DAT5, OUTPUT);
pinMode(DAT6, OUTPUT);
pinMode(DAT7, OUTPUT);
pinMode(DAT8, OUTPUT);

// latch
pinMode(LE1, OUTPUT);
pinMode(LE2, OUTPUT);

// disable outpputs
pinMode(OE1, OUTPUT);
pinMode(OE2, OUTPUT);
digitalWrite(OE1, HIGH);
digitalWrite(OE2, HIGH);
}

int data1[] = {0, 0, 0, 0};
int data2[] = {0, 0, 0, 0};

void loop() {
// make new random data
for (int i = 0; i < 4; i++) {
data1[i] = int(random(256));
data2[i] = int(random(256));
}

for (int x = 0; x < 10; x++) {
for (int i = 0; i < 4; i++) {
// enable outpputs
digitalWrite(OE1, LOW);
digitalWrite(OE2, LOW);

if (data1[i] % 2 == 0) {
digitalWrite(DAT1, HIGH);
}
if (data1[i] % 3 == 0) {
digitalWrite(DAT2, HIGH);
}
if (data1[i] % 4 == 0) {
digitalWrite(DAT3, HIGH);
}
if (data1[i] % 5 == 0) {
digitalWrite(DAT4, HIGH);
}
if (data1[i] % 6 == 0) {
digitalWrite(DAT5, HIGH);
}
if (data1[i] % 7 == 0) {
digitalWrite(DAT6, HIGH);
}
if (data1[i] % 8 == 0) {
digitalWrite(DAT7, HIGH);
}
if (data1[i] % 9 == 0) {
digitalWrite(DAT8, HIGH);
}

// latch
digitalWrite(LE1, HIGH);
delay(1);
digitalWrite(LE1, LOW);

// reset
digitalWrite(DAT1, LOW);
digitalWrite(DAT2, LOW);
digitalWrite(DAT3, LOW);
digitalWrite(DAT4, LOW);
digitalWrite(DAT5, LOW);
digitalWrite(DAT6, LOW);
digitalWrite(DAT7, LOW);
digitalWrite(DAT8, LOW);

if (data2[i] % 2 == 0) {
digitalWrite(DAT1, HIGH);
}
if (data2[i] % 3 == 0) {
digitalWrite(DAT2, HIGH);
}
if (data2[i] % 4 == 0) {
digitalWrite(DAT3, HIGH);
}
if (data2[i] % 5 == 0) {
digitalWrite(DAT4, HIGH);
}
if (data2[i] % 6 == 0) {
digitalWrite(DAT5, HIGH);
}
if (data2[i] % 7 == 0) {
digitalWrite(DAT6, HIGH);
}
if (data2[i] % 8 == 0) {
digitalWrite(DAT7, HIGH);
}
if (data2[i] % 9 == 0) {
digitalWrite(DAT8, HIGH);
}

// latch
digitalWrite(LE2, HIGH);
delay(1);
digitalWrite(LE2, LOW);

// reset
digitalWrite(DAT1, LOW);
digitalWrite(DAT2, LOW);
digitalWrite(DAT3, LOW);
digitalWrite(DAT4, LOW);
digitalWrite(DAT5, LOW);
digitalWrite(DAT6, LOW);
digitalWrite(DAT7, LOW);
digitalWrite(DAT8, LOW);

// enable mosfet
if (i == 0) {
digitalWrite(MOS1, HIGH);
delay(1);
digitalWrite(MOS1, LOW);
} else if (i == 1) {
digitalWrite(MOS2, HIGH);
delay(1);
digitalWrite(MOS2, LOW);
} else if (i == 2) {
digitalWrite(MOS3, HIGH);
delay(1);
digitalWrite(MOS3, LOW);
} else if (i == 3) {
digitalWrite(MOS4, HIGH);
delay(1);
digitalWrite(MOS4, LOW);
}

// disable outpputs
digitalWrite(OE1, HIGH);
digitalWrite(OE2, HIGH);
}
delay(1);
}
}```

If you want only one led to light up, you have to use modulo with prime numbers. Like this, when modulo 4 is zero, modulo 2 will always be zero too.