Shift register question.

Can the 74HC595 source and sink current on its output pins? I have 7 Bi-colored LEDs,the 2 lead type, and I connect all 14 leads up to 2 74HC595 shift registers (with appropriate resisters). I only want/need 1 LED on at a time, so set one pin high will light that LED, correct? It's other pin, connected to a 74HC595 pin set to LOW, would be the same as connected to ground, correct?

I believe that is correct, yes. As always be sure there is a current limiting resistor between the LED and one of those pins on the shift register (doesn't matter which)

Also for another take on coolness, since LEDs are after all diodes (only conduct current in one direction usually), you could stack 2 LEDs on the same set of pins, just orient them backwards. One current-limiting resistor should be all you need. So set the pins one way, one LED lights, set them the other way and it's the other LED that lights.

Set both pins 1 or 0 and none of them light.

Thanks. That is how they will be as they are bi colored…

as already noted yes a 595 will source or sink, its just not a bunch of current, check your data sheet, and if you need some cheat space you can stack one on top of another basically putting 2 in parallel to help

that being said I have yet to really kill one, though thats random luck over short periods of time

While individual outputs can sink or source up to 35mA, the total current for the chip is usually around 70mA.

They are generally pretty robust, but keep those limits in mind.

yea thats why I said check the data sheet, some say 20ma per pin and 70 total, some say 35 depending on who's making it