Originally Posted by nharkey85
I know LEDs usually need resistors on them, or at least in the circuit to prevent too high of voltage. would it be safe to assume that if on a curcuit board, incandescent bulbs are replaced by LEDs (component not PLCC or such), that resistors would be needed. this is NOT on a tacoma let alone a toyota. only info i could find for this vehicle was a completely custom setup instead of using factory locations for bulbs. essentially this vehicle uses what looks like the mini twist sockets in our fog light switch and A/C panel (2005+), but they are soldered to the surface of the PCB where the lead of the bulb wraps around the socket. using or not using the socket looks like would make no difference
Im not asking you guys how to do what i want to do. i already know how to do it. just would like to know bout the resistors
oh and no i do not have multimeter to test. found my old one stuffed away with batteries all corroded.
yes you will need resistors. Heres how to figure it out. First in the specs of the LED you will see Vf (forward voltage) and IF (forward current).
Example PLCC2 White LED. Vf = 3.2V IF = 20mA
First you need to know how much voltage to drop across the resistor.
V = Voltage at the bulb 12-14V for a car.
V - Vf = drop across resistor.
14V - 3.2V = 10.8V (the voltage needed to drop across the resistor)
Now you need the resistance of the resistor needed to drop that voltage at 20mA (the rating of the LED)
R = E (V) / I (If of the led)
R = 10.8V / 0.020A (mA converted to A) = 540 Ohms
You will need a 540Ohm resistor to get 20mA on the LED at max voltage of 14V.
Now you need to know what power rating resistor you need.
P = E (voltage across resistor) x I (Current through resistor)
P = 10.8V x 0.020A = .216W or 216mW you will need a 1/4 (.250)W resistor.
Now If you want to put 2 LEDs in series. just add two Vfs together and use that value as Vf in the first equation.
3.2V + 3.2V = 6.4V Now use 6.4V as the Vf in the first equation.
Note: always use the high end of your voltage range for this calculation. If you use the low end then at higher voltage you will overcurrent the LED.