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Shift register missing bits (discuss.tchncs.de)

submitted 1 year ago* (last edited 1 year ago) by [email protected] to c/[email protected]

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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&lt;17; i++){

    byte upper_byte = input_bin >> 8;
    byte lower_byte = input_bin &amp; 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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[–] [email protected] 2 points 1 year ago* (last edited 1 year ago) (1 children)

Would you not want to shift out the upper byte first? I could be misinterpreting your setup.

[–] [email protected] 1 points 1 year ago

You're probably right, but that should only change the order of the outputs right?

[–] [email protected] 1 points 1 year ago (1 children)

Would it work if you made that delay 1000?

[–] [email protected] 1 points 1 year ago (1 children)

Also try upping the usleep call?

[–] [email protected] 1 points 1 year ago* (last edited 1 year ago) (1 children)

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.

[–] [email protected] 1 points 1 year ago (1 children)

Maybe I'm getting ahead of myself, but maybe try using digitalWrite for a single bit instead of shiftDataOut?

[–] [email protected] 2 points 1 year ago

Good idea, I've tried usleep after all lines, but no change..

[–] [email protected] 1 points 1 year ago* (last edited 1 year ago) (2 children)

The first two lines of the for loop,

byte upper_byte = input_bin >> 8;
byte lower_byte = input_bin &amp; 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
&amp; 0xFF
0000 0000 1000 0000

So, every time you go through a cycle of the for loop, you'll just start with the same values in upper_byte, and lower_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 &lt;&lt; 1

That 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, 00000000

Note: 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.

[–] [email protected] 1 points 1 year ago (1 children)

I think you got and and or switched, first two lines should be fine for shifting the top 8 bits down.

[–] [email protected] 2 points 1 year ago (1 children)

I don't follow what you mean.

[–] [email protected] 2 points 1 year ago* (last edited 1 year ago) (2 children)

I think what he refers to is that you seem to do a bitwise or for the second line instead of the bitwise and.

[–] [email protected] 1 points 1 year ago* (last edited 1 year ago)

~~2nd line of what?~~ Oh you are completely right. My bad. Idk why I wrote that. I'll fix my comment.

[–] [email protected] 1 points 1 year ago

Yes that's what I was thinking

[–] [email protected] 1 points 1 year ago

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;

[–] [email protected] 1 points 1 year ago

What does shiftDataOut do? You loop over it but you give the whole byte to it anyway in each loop.