So am I right to say: Total wattage including everything would be…

2.59watts + 0.98watts + (4* 3W) = 15.57W?

Where the (4* 3W) is the wattage of the four LEDs…

Have I done some stupid calculations like before again?

I am getting the cheaper ProLight Luxeons, so driving at 700mA is what they recommend…. 1400mA would probably burn those LEDs badly! :P

And I will certainly go for the LM1085 option. But is the adjusting resistor’s value calculated the same way with it? And is a huge heatsink required on the LM1085 itself?

And what happens if the voltage of the car falls to about 10V (which could happen sometimes in my old car, especially during start up), will the LEDs fail to light at all?

I will mount these 4 LEDs on an old Pentium II processor forced air heatsink, which measures 5cm x 11cm x1.5cm, do you think it will be enough to keep the LEDs cool?

Actually, how hot does these LEDs actually gets when running? I have never bought them before, so I am just putting a bet on making something that would work….

Thanks so much for helping!

You have the right idea – wire as many luxeon in series as you have voltage available (remember the lm317 costs you around 3v in overhead). A separate regulator for each LED = lots more waste. Since the LM317 has a dropout voltage of approximately 3v under load and heated up, you can use this factor along with the input and output voltages to calculate the power dissipation in the regulator:

Iout(Vin – Vout) + Iout * Vdo = 0.7 * (14 – (2.95 * 3)) + 0.7 * 3 = 3.6 + 2.1 = 5.7 watts.

Next we do the resistor:

I^2*R = 0.7 * 0.7 * 2 = 0.98 watts

This means you have to dissipate six watts on the regulator and one watt on the resistor. Seven watts is the reason I hate linear regulators … the numbers just get much worse if you try to drive your luxeons at their full rated power … 700mA will be under-powering your luxeon to about 50% of their 1400mA rating. You can help reduce waste in the regulator by matching the output voltage closely to the input voltage; in that case, a low dropout regulator helps a great deal. The LM1085 is pin compatible with the LM317 but handles up to 3a and offers a 1.5v dropout voltage, half of the LM317. Let’s try this:

0.7a * (14v – (2.95v * 4)) + 0.7a * 1.5v = 1.54 + 1.05 = 2.59 watts

So now you only have 2.6 watts on the regulator … less than half the waste of the above example, just by adding one more LED and using a more modern regulator!

Another thing to remember, the LED (like all light bulbs) is a heater than happens to give off some light, make sure you keep those bad boys cool!