Yea im buying enough for 2 keyboards do at least 130 do you think I should pick up some extras? Also how do if figure out what resistors I am going to need?
You will definitely need/want spares (incase you blow a few ; p, I've blown like 2-3 in a project using one...5V over a LED is not smart).
As for resistors, it depends on in what fashion you want to hook them up. I think it'd be most efficient in using one/one set of resistors in series with all/lots of LEDs in Parallel. It really depends on how much room you have in your board.
You can always do one resistor per LED (easy to do the maths or if you want to control each LED individually with circuitry).
If you put a ton of LEDs in parallel you may need a very small resistance for your resistors.
Lets just assume you are using you're 2.5V, and you are pulling 30mA.
For 1 resistor in series with 1 LED:
2.5 - 30m*Resistance = 1.85
Reistance ~= 22 ohm
For 1 resistor in series with n LEDs in parallel:
2.5 - 30m*n*Resistance = 1.85
Feel free to substitute 2.5 for 5v if you use power from your USB.
For the 2.5 V you can support like 20ish LEDs off a 1ohm resistor. (Simple resistor circuit).
For 5V you can do like 100 LEDs off a 1ohm resistor.
I'm not sure which one is the best way to do it. There may some pros and cons I'm not aware of when you hook up so many LEDs in parallel. But in the case of using less components you may not need many. If someone else with some good electronics knowledge will step in and comment on this, that'd be great.
Note: you can make resistance of <1ohm by putting resistors in parallel. e.g. placing 2 1ohm resistors in parallel creates a 0.5ohm equivalent resistor.
If you need to test brightness of the LEDs(for all we know it could be way too bright). You might just grab some extra resistors and hopefully can calibrate the brightness. Something cool you could do (but will cost just a bit more, is getting a potentiometer, so you can adjust brightness as you want,without soldering/adding extra resistors).
Background to this stuff (for if you want to learn some electronics):
Voltage = Current * Resistance (Voltage dropped across a component is equivalent to the current going through it multiplied by it's current)
Thats where we get
Vcc - IR = VLed (Kirchoffs voltage law)
We want the Resistor to drop the voltage to what the VLed should operate at optimally (1.85V @ 30mA).
Vcc is our power supply (2.5V/5V)
R is the resistance of the resistor in series.
For n LEDs in parallel:
The current from the resistor will be split/comes from all the LEDs (each 30mA). Therefore I can take I and multiply it by n to get the current through the resistor
Vcc - nIR = VLed.
