Under these conditions, RL must drop 3V across it at 20mA, since the voltage must be the supply (5V) minus the voltage drop across the LED. The value of RL can then be easily obtained from Ohm's Law (R = V/I):
Hence: R = 3/0.02 = 150R Current gain, , is Ic/Ib so:
Ib = Ic/gain = 0.02/100 = 0.2mA
Assuming 0.7V across the b / e junction, Rl must drop (5 – 0.7)V = 4.3V. Given that we have already calculated base current as 0.2mA, the value of Rl is:
Rl = V/Ib = 4.3/0.2 X 10-3
= 21,500 = 22K (preferred value)
In practice, the value is by no means critical, and in this simple circuit any value between 1K0 and 33K would probably work all right. It is common to see values of either 10K or 15K used for the purpose.