Here’s a fun experiment – adding 8 very fast output pins via SPI, using just a 74AHC595 shift register:
This is the “spiCycle” test sketch I used to try out the above hardware:
It sends bytes to the shift register, with only bit 0 set, then only bit 1, etc. At the end, a zero byte is sent to clear all bits again, which is why the loop runs to 9 i.s.o. 8. Here’s the output, as seen on the scope via a logic probe:
The yellow trace is SCK. The blue trace is PB0 (Arduino digital 8), which is used as chip select for the shift register. The purple lines are the 8 outputs of the shift register. You can see the “1” bit rippling across all 8 pins.
As you can see, there’s plenty of “ringing” on these signals, which is not surprising, given that it’s a bus running at 8 MHz (and it was all set up on a breadboard using jumper wires). Setting these 8 bits takes just 2 µS!
Note that the number of output bits can easily be extended by “cascading” multiple shift registers, so this is in fact a way to add lots of output pins to an ATmega which can be set very, very quickly.
There’s one oddity in there which I can’t explain – for some reason, the PB0 select pin is being pulled low longer and longer as the loop repeats (see the blue line “low” states).
Tomorrow, I’ll explain the point of this exercise…
Hmmm… what are you up to…?
Are you planning to sequence all your LED strips and host a new year rave?
Heh… that’d be called DMX, I suppose :)
Could the oddity be explained by bit(i) taking longer for larger values of i? More “shift left” instructions to perform, perhaps. I’m too lazy to test it though…
You’re not planning on making a LED cube, aren’t you? ;-)
That’s an idea! (but no, that’s not it)
Try 1<<i instead of bit(i). With the ATmega's barrel shifter that should have a constant runtime.