I don't really see any serious flaw with your test code.
So how have you connected the '595s together? You chained them serially, right? The serial out of the first shiftregister, to serial in on the next, etc. But having common clock and latch.
Maybe you have tristated or reset two of them?
The 595 is pretty quick, you can just latch it high then low immediately after, especially using digitalWrite().
digitalWrite(LATCH_PIN, HIGH);
digitalWrite(LATCH_PIN, LOW);
here is my quick and dirty fritzing mockup (left out power supply and some arduino stuff, just an illustration)
i have left out the outputs and this one only has 2 595s
imagine this being blue and green
would having the MR connected on all three create an issue?
Diagram seems fine for the 74HC595s. I would suggest having a small capacitor across the power supply near them though, like 100nF or so.
What do you mean by "MR"?
left out power supply and some arduino stuff, just an illustration
It would be better to see what you actually did rather than what you think you did.
I used to think fritzing was a good idea, that Illustration changed my mind, I cant hardly tell what's going on
It would be better to see what you actually did rather than what you think you did.
this IS what i did, don't have a picture but was able to recreate an image from fritzing
I used to think fritzing was a good idea, that Illustration changed my mind, I cant hardly tell what's going on
It's only as good as the designer, especially in the breadboard mode. Just like a real breadboard, I'm afraid...
this IS what i did,
No it's not you left off lots of stuff where you might have gone wrong like the power supplies.
If the last shift register is working and they are actually chained then data is flowing through the other two but not appearing on the output. This could be that output enable is not wired up right or the shift / load is not wired right.
So the red is the first byte you sent out? That ends up in the last shift register.