Quelltext der Seite LED Cube/Column Resistors in Detail

===Resistor===
Now, we are finally able to calculate our resistor value. This is simply done by subtracting all dropout voltages from the supply voltage and then apply U = R * I. So:
 R = (Vcc - Vio - Vta - Vled)/I
   = (5.0 V - 0.5 V - 0.7 V - 1.9 V)/(20 mA)
   = 1.9 V / 20 mA
   = '''95 Ohm'''

Of course, we do not need to use the full 20 mA, which are quite on the limit for the used components. A more reasonable value would be 15 mA:
 R = (Vcc - Vio - Vta - Vled)/I
   = 1.9 V / 15 mA
   = '''127 Ohm'''

or 10 mA:
 R = (Vcc - Vio - Vta - Vled)/I
   = 1.9 V / 10 mA
   = '''190 Ohm'''

For a blue LED (3.0 V drop) at 10 mA we get:
 R = (Vcc - Vio - Vta - Vled)/I
   = (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(10 mA)
   = 0.8 V / 10 mA
   = '''80 Ohm'''

Here is something interesting. We could also calculate with 20 mA:
 R = (Vcc - Vio - Vta - Vled)/I
   = (5.0 V - 0.5 V - 0.7 V - 3.0 V)/(20 mA)
   = 0.8 V / 20 mA
   = '''40 Ohm'''
But now let's assume that the USB voltage increased to 5.5 V for whatever reason, which is well within specs. If we assume that this will result in a voltage increase of 0.3 V at the resistor, our current is now 28 mA. For 9 enabled LEDs and the supply current of 20 mA for the uC, this results in a current through the Vcc pins of 268 mA, which is relatively close to the absolute maximum ratings of 300 mA.