Calulating BJT base resistance for PIC led switching?

[xnihilo]

Hi,

Hoping this is not too off topic.
I need my PIC to switch an ultrabright led.

Using a BC182L general purpose NPN transistor from futurlec
(http://www.futurlec.com/Transistors/BC182L.shtml)
and an ultrabright led from futurlec
(http://www.futurlec.com/LED/LED5YULB.shtml)

I need to use a BASE resistor for my BJT.

I'm using the ultrabright led at 2.4V 20mA with a 9V power supply, so I will be using a 330 ohms limiter resistor (9V-2.4V/0.02A)
Then I've checked the web to find clues to calculate RB for the BJT but there are parameters I don't understand.

according to
http://www.kpsec.freeuk.com/trancirc.htm
I should calculate the RB value with a pretty simple formula but when I have to find the Ic for the load, I'm stuck. How can I determine the right load resistance? Does it mean the total resistance of the current limiter for the led AND the resistance of the led? If so how can I calculate the resistance?
I did not find the value of the led resistance in the specs of the led.
Maybe using:
U=R*I -> R = 2.4V/0.02A = 120 ohms?
Then I should add the 330 ohms of the led current limiting resistor to have the Ic value using: 9V/.450Kohms = 20 mA?
If so, then the RB should be (if HFE min is 100 and PIC supply is 5V):
(5V*100)/(5*0.02A) = 5KOhms

Am I right?
Thanks for any hints...


PS: I should add some voltage drop value somewhere for the transistor?


In the same idea, I also have to switch an LM555 times but I don't know what value I should consider as the LM555 (the 'load') current and resistance...