We just calculated that a 68 Ω resistor will give us the maximum forward current from a 3.3 V supply, but when we hooked up the LED to the 3.3 V GPIO pin earlier, we used a 330 Ω resistor. This is because the GPIO pins of BeagleBone's processor are only rated to source a maximum current of 4-6 mA. Using a 330 Ω resistor gives 1.3 V / 330 Ω = 3.9 mA.
There are a few ways we can source more current than the 4-6 mA maximum of the GPIO pins; one simple way is to use an NPN Bipolar Junction Transistor (BJT).
For this circuit, you will need:
Breadboard
1x 5 mm LED
1x 4.7 kΩ resistor
1x 68 Ω resistor
1x 2N3904 NPN transistor
Jumper wires
Wire it up on your breadboard as shown here:
Note
If you have a resistor kit like the one from SparkFun (https://www.sparkfun.com/products/10969) you might not have a 68 Ω resistor handy. If this is the case, you can use one 220 Ω resistor in parallel with one 100 Ω resistor for a combined resistance of 68.75 Ω.
Let's take a quick look...