Hey friends,
I have a two daisy chained shift registers (74AHC595) which are controlled via an ESP32. I want to set one output to high at a time before switching to the next.
The code seems to work, but the outputs O_9 and O_10 are not staying high (zoom) after setting them, whereas all the other ones are working fine. This is the used code snipped:
pinMode(SHIFT_OUT_DATA, OUTPUT);
pinMode(SHIFT_OUT_CLK, OUTPUT);
pinMode(SHIFT_OUT_N_EN, OUTPUT);
pinMode(SHIFT_OUT_LATCH, OUTPUT);
digitalWrite(SHIFT_OUT_N_EN, LOW);
uint16_t input_bin = 0b1000000000000000;
for(int i=0; i<17; i++){
byte upper_byte = input_bin >> 8;
byte lower_byte = input_bin & 0x00FF;
digitalWrite(SHIFT_OUT_LATCH, LOW);
shiftDataOut(SHIFT_OUT_DATA, SHIFT_OUT_CLK, MSBFIRST, lower_byte);
shiftDataOut(SHIFT_OUT_DATA, SHIFT_OUT_CLK, MSBFIRST, upper_byte);
usleep(10);
digitalWrite(SHIFT_OUT_LATCH, HIGH);
delay(10)
input_bin = input_bin>>1;
}
Is there anything I’m doing wrong, or any idea on where the problem may lie? I’ve already tried looking for shorts and other error sources, but the design was manufactured on a PCB and no assembly issues are noticeable.
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Would you not want to shift out the upper byte first? I could be misinterpreting your setup.
You’re probably right, but that should only change the order of the outputs right?
What does shiftDataOut do? You loop over it but you give the whole byte to it anyway in each loop.
Would it work if you made that delay 1000?
Also try upping the usleep call?
Hmmmm do you want to write to both shift register at the same time? I say this because you’re looping 16 times, but seem to be sending the high and low bytes out 16 times over rather than one bit each time, although you are shifting the input.
Maybe I’m getting ahead of myself, but maybe try using digitalWrite for a single bit instead of shiftDataOut?
Good idea, I’ve tried usleep after all lines, but no change…
The first two lines of the for loop,
byte upper_byte = input_bin >> 8; byte lower_byte = input_bin & 0x00FF;don’t really accomplish anything. The first line is bit shifting to the right 8, and then you just bitwise and it resulting in the same thing. For example, starting with
input_bin:1000 0000 0000 0000 >> 8 0000 0000 1000 0000 & 0xFF 0000 0000 1000 0000So, every time you go through a cycle of the for loop, you’ll just start with the same values in
upper_byte, andlower_byte. To sequentially output each shifted value, you’ll instead want something like:output_value = 0b1 for i = 1 to 16: latch(low) shift_out(output_value) latch(high) output_value = output_value << 1That is, if I interpereted correctly that you want the shift registers to output the following:
output_count, upper_shift_register, lower_shift_register 1, 00000000, 00000001 2, 00000000, 00000010 3, 00000000, 00000100 . . . 16, 10000000, 00000000Note: Lemmy has a bug where it doesn’t format some symbols correctly, so the left angle bracket gets formatted as
. The same issue exists for the right angle bracket, the ampersand, and I would presume others.I think you got and and or switched, first two lines should be fine for shifting the top 8 bits down.
I don’t follow what you mean.
I think what he refers to is that you seem to do a bitwise or for the second line instead of the bitwise and.
Yes that’s what I was thinking
2nd line of what?Oh you are completely right. My bad. Idk why I wrote that. I’ll fix my comment.
You’re 100% right, I’ve lost ‘i’ somewhere in my debugging process
byte upper_byte = input_bin >> (8+i) ; byte lower_byte = (input_bin >> i) & 0x00FF;