Shift Register 74HC595 issue

Other Hardware LEDs and Multiplexing
1

Hi , I am trying to use two Shift Registers as described here :

lets say i wanna shift out this binary " 0000010000000001" which is 1025 Decimal.
when i use this code :

digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST,1025);
digitalWrite(latchPin, HIGH);

the output is "0000000000000001" or one in decimal and when i change the code to:

digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST,1025);
shiftOut(dataPin, clockPin, MSBFIRST,1025);
digitalWrite(latchPin, HIGH);

I could get the desired output. However, this wont work for different output. for example if i wanna have 1024 decimal or "0000010000000000" i couldn't get the output even by one time "shiftOut(dataPin, clockPin, MSBFIRST,1024);" or repeating it twice .
I would be really appreciate it if someone could help me.
Thanks in advance.

2

Shiftout only sends 1bytes worth of data to the IC.

8bits worth, the next time you use shiftout, 8 more bits get shifted into the first register any bits previously store get shifted out to the 2nd register.

So the trick is to get the high/low values and send it in 2 lots (16bits)

3

Thanks a lot .. That works .. didn't know that .
i wanna have : 1000000000111111 or 32831 in the output and i used this code which works great:

digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST,128);
shiftOut(dataPin, clockPin, MSBFIRST,63);
digitalWrite(latchPin, HIGH);

Thanks for you help.

4

could also have:

int dataOut; // declare in pre-setup part of sketch

// then update in loop as needed:
dataOut = 0b1000000000111111;
// or 
dataOut = 32831;

// then send it out:
digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST, highByte (dataOut);
shiftOut(dataPin, clockPin, MSBFIRST, lowByte (dataOut);
digitalWrite(latchPin, HIGH);
5

Thanks a lot for your response CrossRoads.
I have a question .. if i use more than two 595, how should i change your code? for example for three 595 should it be like this?:

digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST, highByte (dataOut);
shiftOut(dataPin, clockPin, MSBFIRST, highByte (dataOut);
shiftOut(dataPin, clockPin, MSBFIRST, lowByte (dataOut);

digitalWrite(latchPin, HIGH);
6

That will send the same byte twice, is that what you want?
If the three bytes are to be unique, you could put the data into a 32-bit unsigned long, and break out the 8-bit chunks from there. Or just use three unique bytes:

digitalWrite(latchPin, LOW);
shiftOut(dataPin, clockPin, MSBFIRST, dataOut3;
shiftOut(dataPin, clockPin, MSBFIRST, dataOut2;
shiftOut(dataPin, clockPin, MSBFIRST, dataOut1;

digitalWrite(latchPin, HIGH);

7

well i meant 32bit output generally(or even more). this solution is the one which cjdelphi also suggested. However, what interested me about your code which makes me to ask a question was this part of it :

highByte (dataOut);
lowByte (dataOut);

that how this form could be used for more than 16 bit. so in this case for any length of bit i don't need to separate them in individual 8bit .

8

highByte and lowByte are just functions that someone wrote
http://arduino.cc/en/Reference/HighByte

There may be something that splits up unsigned long too.
Can also do it yourself, maybe something like this:
unsigned long 32bitWord;

byte bits7to0 = 0x000000FF & 32bitWord; // mask off 8 bits
32bitWord = 32bitWord >>8; // shift away 8 bits, upper 24 left
byte bits15to8 = 0x0000FF & 32bitWord; // mask off 8 bits
32bitWord = 32bitWord >>8; // shift away 8 bits, upper 16 left
byte bits23to9 = 0x00FF & 32bitWord; // mask off 8 bits
32bitWord = 32bitWord >>8; // shift away 8 bits, upper 8 left
byte bits31to24 = 0xFF & 32bitWord; //

Can mix that in with the SPI.transfers