hey guys, im wondering if the super bright LED can be use a an head light for cars ? maybe 10 or 20 LED's?

if anyone knows how, im kinda interested coz ordinary halogen (100 - 150w)wont last long and it melts the plastic connector

 

If you can keep the junctions from overheating, it is probably possible. There are state and federal regulations aout headlamps that may prevent you from actually using such illumination. I would want to be awfully sure about the equivalence of illumination and reliability before actually using LED headlamps.

 

hello sir, i the only possible way i think for it not to overheat is changing the bulb to a lower wattage but it has a poor light.

im not sure about regulations in headlamps but ive seen a few headlights already that uses LED, maybe 2 or 3..

and maybe if there is a law about it, they wont allow to use those type of headlights (i forgot the name) the one that uses ballast and electronic type, and it goes around 6000 watts. thats a cool one but very expensive for me

 

It will not give nearly enough light yet. A 5W LED can give about 65 lumens.
Contrast this with 1550 lumens ±15% for a H1 Halogen and you'd need
23 of them X 2 so about 46-50 of those painfully expensive chips plus the heat sinks, a pretty massive current regulator and something to focus the light where it needs to go.
Look at HID lighting. It' s far more efficient than filament and approaches florescent
lamps with a stunning capacity for high intensity light.
This has some links to discharge lamps, there is another link to automobile HID lights
http://members.misty.com/don/contxe.html

 

ok sir, thx for the informations. well thats sad, anyways can u guys teach me how to wire a super bright LED.

i need to wire 4 pcs of them as a light for my PSU. (i can get a source of 12V, i think thats 10A from that same PSU or 5V but i think its 30A)

thanks again

 

The other thing about super bright LED's are that they are nearly impossible to look into. I would hate to meet an oncoming car that might be using them.... kinda like looking directly into a laser....

 

ok thank u guys about the replies.

i hope someone can teach me how to wire a 4 pcs led direct to a PSU 12v (or same as where the PC fan molex connector is connected)

 

...i hope someone can teach me how to wire a 4 pcs led direct to a PSU 12v (or same as where the PC fan molex connector is connected)

I can, if you wish.

 

cume pls :>

heres what i remember but i dunno if its gonna work

series connect D1, D2, D3, D4 and a 270k 1/4 watt resistor. is it correct sir?

how can i calculate the resistor i need (for future LED projects ^^)

 

hmm, no one wants to share?

 

Add together all the voltage drops of the series LEDs. Subtract that total from the supply voltage to find drop across the resistor. Apply both Ohms Law and algebra: R=E/I

Or, visit: http://ourworld.compuserve.com/homepages/Bill_Bowden/led.htm

 

ty sir. nice site

 

cume pls :>

heres what i remember but i dunno if its gonna work

series connect D1, D2, D3, D4 and a 270k 1/4 watt resistor. is it correct sir?

how can i calculate the resistor i need (for future LED projects ^^)

Sorry, did not saw that. But is is simple. You will need to know that is the voltage drop for an individual LED for a given current. It is better to have a voltage/current curve to get this information. Then you use the formula:
Rs = (Vsupply - 4 x Vled) / Iled

Rs is the series resistor resistance, Vsupply is the supply voltage, Vled is the voltage drop for each led and Iled is the current across the LED, which is the same across the series. If you are using white LEDs for an automotive application, I would expect a voltage drop from 3V to 4V, so I yould advice you to use series of two LEDs (since voltage will vary greatly, and can achieve 18V). You can use the two series of two LEDs in paralel. You can even use a single resistor for both series, although some others might think that this is prejudicial to the LEDs (I think it might, but in the remote case of one series of LEDs being interrupted). If you use the same resistor consider the currnt needed for the association equal to the double of the current across each LED, or:
Rs = (Vsupply - 2 x Vled) / (2 x Iled)