Cree 10w

[vimal]

Hello everybody. I wanted to make a Led headlight for my car using 4X cree 10w. each led operates at 3A and 3.3-3.5 V max. i used 4 in parallel connected to my car headlight socket  which is 14.5-14.6v.

but unfortunately everytime, 1 led bulb get burnt.

I would like to get some help in looking for the resistor i need to use for each led and capacitor as well.

any help or advice is welcomed.

thanks

[kam]

Hello vimal.

You want to say that you connected 4 in series right? The positive of the battery goes to the anode of the first led, then the cathode of this led to the anode of the other led and so on, and then the cathode of the 4th led back to the negative of the battery. is that correct? We avoid connecting leds in parallel. It is a bad practice

So i trust that you connect them in series. worst case scenario: Battery voltage 14.6V. Which means that each led gets about 3.65V which is already too much! One of the LEDs will always be weaker than the others and it will always be the first to die.

So, i give you 2 simple solutions. First is the (crappy) resistor (as you mentioned). Play it safe, under-power the LEDs, after all 40W are already enough! Say that you want to provide not more than 3.2V for each LED, a total of 12.8V in series. You connect the resistor to "dump" the extra 14.6-12.8=1.8V... Here is the calculation:

R=V/I = 1.8/3 = 0.6 Ohms

3 amperes will go through this resistor, so it has to be able to dissipate:
P=I² x R = 3² x 0.6 = 5.4W

So you want a resistor around 0.6 ohms (i would go for a little bit higher though, say 0.8 or 1 Ohm) rated 10 Watts. Remember that dissipating 5.4W is enough to give you a burn if you touch this resistor. Keep it away from plastics, heat-sink it if necessary.

---------------

Solution #2 - a transistor linear driver. You can use such a driver to "simulate" a resistor. The benefit is that with the transistor you automatically regulate its resistance so that the current through the LEDs is kept constant. The previous solution with the resistor has the disadvantage that any fluctuation on the voltage will appear as change in brightness on the LEDs. A constant current always keeps the current through the LEDs constant (hence the name) regardless of the voltage.

Here is a good constant driver for your case:
Transistor - MOSFET Constant Current Driver

For mosfet you can use the IRF520 or 540. Any NPN transistor will do. RG can be around 470 ohms. RS will control the current. In your case I'd keep it as low as 2.6 amperes (keep leds underpowered):

RS = 0.6 / 2.6 = 0.23 ohms

The 0.23 ohms resistor must be able to dissipate that much power:
P = 3² x 0.23 = 2.07 watts

So, choose a resistor around 0.2 ohms (or higher) at 5 watts

Again, the mosfet HAS TO BE HEAT-SUNK because it will be called to dissipate all the power required to keep the LEDs properly biased, that will be more than 5W 9as we calculated before). The tab will BURN!!!

good luck

[vimal]

hello. thanks for the reply. it really helped.

is the mosfet for only 1 led? or can i use 4 in parallel or in series?

[kam]

4 in series, never in parallel with one driver (even if this driver is a resistor)

[cheerio]

i have a "small" project coming where i have to drive 2x 50W RGB LEDs. i will hijack this thread when it happens ^^

[vimal]

Hello, what is the reference of the transistor used?

thanks

[kam]

you can use any NPN transistor

[vimal]

Hello .I want to know what are the modifications that should be done to the  circuit if I want to add 2 strings of led or a total of 6amps at 12.8v .the input voltage is 14.5v.

I have a mosfet irf840 which is rated 8amps.is it ok if I use it?

thanks

[kam]

That's a huge wattage. better make 2 similar circuits

[vimal]

Thanks a lot for your help.

[vimal]

Hello Kam,

I would like to add a capacitor to the circuit to stabilise the voltage for the 4 led's in series so that when i brake, the light does not dim.

how do i calculate the value of the capacitor?

the output voltage after the mosfet would be 12.8v at 2.6A

thanks

[cheerio]

i think it was:
uF = (mA * mS)/V
so:
uF = (2600 * mS)/12.8

so if you want to be prepared for a 3Second brake you should go:
uF = (2600 * 3000)/12.8 = 609375uF = 0.61Farad

Please confirm Kam, i am very rusty

edit:
after thinking a bit more about it.... why do the leds dim when you brake??

[vimal]

Hello,thanks for your reply. The headlight might dim because when I brake.the stop light will light up. It will only dim a bit.not much.

[cheerio]

But the Capacitor will supply the energy to the whole lighting system i guess. Not just your headlights. You somehow need to stop the energy drain from the other lights. Maybe add some big diodes? That way you could make sure the Caps are only drained by the LEDs

[kam]

With 4 less in series this is a problem that you will have to live with. You already push the supply voltage to the limit so you do not have extra voltage to cope with the drop on the wires due to higher current. As cheerio calculated, you will need a huge capacitor which is practically not possible.