Calulating BJT base resistance for PIC led switching?

[mister_e]

let's see..

CASE 1:
normal red led driven by pic
I need 2V, 15mA, driven by 5V from PIC
I will need 200 Ohms current limiter (5-2/0.015) between led and PIC.

OK

CASE 2:
ultrabright led supplied by 9V battery, controled by a BJT (BC182L, HFE(min) = 80) driven by PIC (5V)
Current limiter to get 2.1V 20mA with 9V => (9-2.1/0.02) = 345 Ohms
Then for the Base current limiter I will have: 80/5 = 16 => 20/16, 5-0.7/1.25 = 3.44 KOhms resistor or less

OK!

CASE 3:
infrared led supplied by 9V battery, controled by a MOSFET (IRLD110) driven by pic (5V)
Current limiter to get 1.4V 250mA: 9-1.4/0.25 = 30.4 Ohms

Right too... BUT
P=I*I*R = 250mA*250mA * 30.4 = ~2Watt
sometime it's nice to place some led in serie to drop the current a little and avoid to brown your PCB... nice to heat an enclosure though

CASE 4:
One Pic pin controls a BJT (BC182L) to pull to GND a pin of another PIC.
The pins of the two pic are connected via a 1K resistor, so current between pics will be 5/1000=5mA
5/16 = 0.3125, 5-0.7/0.3125 = 13.76K resistor or less to be used as base current limiter

OK

CASE 5:
LCD backlight controled by pic (5V) via a BJT (BC182L). Backlight need 5V 70mA.
70/16 = 4.375, 5-0.7/4.375 = 0.98K or less to be used as base current limiter.

OK

CASE 6:
A 5V, 30mA relay controled by a pic (5V) via a BJT (BC182L).
30/16 = 1.875, 5-0.7/1.875=2.29K or less to be used as a base current limiter

OK

CASE 7:
A PIC (5V) is used to pull to GND the TRIGGER pin of a LM555 via a 1K current limiter (so we will have 5V, 5mA) and a BJT (BC182L).
5/16 = 0.3125, 5-0.7/0.3125 = 13.76K resistor or less to be used as base current limiter

OK

Sounds like you understood correctly what i said... easy huh?