I built an LED cube (8x8x8) with some cheap LEDs. I am going to control it with 8 74HC595 on the columns. Can I skip resistors and power the shift registers with the LEDs voltage spec (2v)? I pulled some LM317 off of an LED strips so I can do that. Because it would make thinks alot easier.
Thanks for any help,
ematson5897
sure, it will eventually kill your led's and your shift registers, but you can
I havent given any thought to it, but it might be better to set the 317 to act as a constant current device
Why, no current limits?
yep
Yea. For cheap leds, they have survived a lot though. I have had them on a 1/8 duty cycle on 5v through a shift register.
This is why you always need something to limit the current with an LED.
http://www.thebox.myzen.co.uk/Tutorial/LEDs.html
Yes I have seen your website. For some reason I was under the impression that if an LED is being fed its fwd voltage, it will only take the current it needs
Yes that is a common misconception that is why I made that page. 
Ok we'll thanks for the help. So I guess I'll just go buy 64 resistors
No for an 8 by 8 matrix you only need 8 resistors.
its a cube though
It depends then on how many LEDs are going to be on at any one time and how you wire them.
An Led requires a CONSTANT CURRENT for proper operation, regardless of Duty Cycle the current must be limited. Above the turn-on point (voltage ~2V for red and ~3V for green/blue) the led is a short circuit and current MUST be limited. Short Duty cycles apparently work... They DON"T, instead you wind up spiking the 5V source every time you enable a non current limited LED. If you limit the current you don't NEED to shorten the Duty Cycle. Don't forget to BY-PASS, BY-PASS, BY-PASS. Cheap insurance and often times the answer to what looks and acts like a bad sketch. Nice thing about a Bread board is that you can just PLUG them in... IMO
Doc