Super simple high power LED driver

Step 2: How it works

The LM317 regulator gives out a constant voltage of 1,25 volts between ADJ and Vout, so by adding a resistor between these two outputs, you'll get a constant current.

Ohm's law says that U/I=R, which means that Voltage divided by Ampere makes resistance.

so if you want to connect one or more luxeon 1W LEDs, which has a power consumption of 350mA, the calculation should look like this: 1,25 (the constant reference voltage of the LM317) divided by 0,350 (the LEDs power consumption) makes 3,57. So if the resistor is 3,57, constant current will be 350mA. The closest E12 value is 3,9 ohms, it will give you a constant current of 321mA. However you can't see any difference in the light output.

If you use 3W LEDs, which has a current consumption of 700mA, the calculation should be: 1,25 divided by 0,7 makes 1,78. The closest E12 value is 1,8 ohms, the output will be 694mA

the resistor must be at least 1W in both calculations.

Although the LM317 is rated for 1,5 Ampere, I wouldn't recommend it for applications that need more than 1 Amperes, because it gets very, VERY hot. the LM350 is equal to the LM317, but it's rated for 3 Amps
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hamtons says: 1 year ago
Hi, I was trying to design a driver with LM1084IT-ADJ.

I noted that the reference voltage is 1.5V, I need 3 Amps so 1.5/3 = 0.5 ohms for the resistor value right?

Then another thing is how do I know what watts the resistor should be?

I'm running 3 LEDs 3.2V each at 3Amp in series, for a total of 9.6V.

Is it 1.5V*3amps = 4.5W or is it 9.6V*3amps for nearly 30W?

Thanks (:
PedroDaGr8 in reply to hamtons1 year ago
the first one. You are measuring the voltage drop across the resistor.
hamtons in reply to PedroDaGr81 year ago
4.5W it is. ok thanks (:
PedroDaGr8 in reply to hamtons1 year ago
Don't forget to give yourself some headroom on the W. I wouldn't use a 5W resistor. Its going to get quite hot and there just isn't much headroom should things go haywire. 7.5W or 10W resistor would be better.
TheGreatS says: 1 year ago
Ohm never forgot his dying uncles advice.

"Remember: with great power comes great current squared times resistance"

Excellent instructable by the way, bravo.
chse720 says: 2 years ago
i am wanting to power an LED that requires 16.2V and between 1.5-2.5A. I would like to use a current of 2A for this project, how would i go about doing this and power source would you suggest? i would like to use a dewalt rechargeable 18V battery
eugenehaller says: 3 years ago

I am trying to build this, but I am using two 10w LEDs with a Vf of 3.6 and a current consumption of 2800 mA. For the life of me I can't figure out what regulator to use. Any ideas?
yugang in reply to eugenehaller2 years ago
Hi Eugenehaller,

could you help me in designing a circuit for lighting 10 w power LED or 1*10 watt leds..

Thanks,

Yugang
valveman in reply to eugenehaller3 years ago
Use an LM338 it is rated to 5A.
seethasub in reply to eugenehaller3 years ago
you have to add a PNP power transistor with IC to incease its current rating.
(Transistor 2N6111 On Heat Sink with LM317 )

diy_bloke in reply to seethasub1 year ago
you are correct in adding a transistor, but by then ofcourse it is not a ' simple' regulator anymore and it would be better to build a different circuit then, not based on a LM317 because that has a big voltage drop. Use a FET or power transistor with 2 diodes or a simple signal transistor
Artificial Intelligence (author) in reply to eugenehaller3 years ago
Hi, sorry for not replying. I must have missed your comment. I have not tried using 10W LEDs with the LM350, but I don't think it's a very good idea, because when I use the LM317 (1500mA max) to drive a LED that just uses 700mA, it gets very hot and needs a pretty big heatsink. This happens because both the LM350 and LM317 are pretty inefficient, so they "burn off" the power as heat. For very high power LEDs, you might be better off with a more efficient regulator. Try the LM1084IT-ADJ. It costs more than the LM350, but it's more efficient and can handle up to 5 Amps. It has the same reference voltage and pinout as the LM350/317, so you can use my instructable with it.
doransignal in reply to eugenehaller3 years ago
try using a LM350 it is rated at 3 amps or 3000 ma

Lee
broxlin says: 2 years ago
Hi!
I have 50 blue and 50 white leds, here are the spec:
" http://i1104.photobucket.com/albums/h329/broxlin/Proiect%20Diamant/Blue.jpg " and " http://i1104.photobucket.com/albums/h329/broxlin/Proiect%20Diamant/White.jpg ".
This will be the light for my reef tank.
I want to make 2 or max 4 lamps, if it's possible, so one circuit for 25 / 50 leds.
If I put less leds, the number of sources will be higher and the cost the same.
How can i make it better?
Thanks!
Jkirk3279 says: 2 years ago
Hi.

I've been studying this subject.

This regulator seems just what I'm looking for. Not tuned in yet on the "constant current" concept though.

I want to drive twenty 1 watt LED's in series. Each uses 2.79 volts and 350 mA.

So if I drove them in parallel I'd need 2.79v but a staggering 7 amps !

I asked an electronics teacher and he said "run them in series".

So then you need 28 volts but only 350 mA for 20 watts, plus a balancing resistor.

Very little power wasted.

As I understand it each LED sees 350mA, but the voltage pressure drops progressively until finally it hits the resistor.
mjhilger in reply to Jkirk32792 years ago
The voltage drop across any LED is dependent on the junction temprature. The temp will rise as they are driven in higher power. And, the voltage change between dim and fully bright is small (compared to the drive voltage). However the current to light output is relatively consistent. One should drive LED's via current to protect longevity and operation. Because each LED of the same part number, from the same manufacturer batch will vary its exact voltage per current, you cannot run LED's in parallel; there is no way to regulate the current to each in this manner. So you must drive them in series to maintain control to drive near peak output.
The constant current concept is a consistent way to safely drive power through the LED's one or many. You are correct the voltage goes up on your supply demands, but the constant current (or current regulator) provides the proper power control.
Hope that helps to explain constant current concept
bbas8 says: 3 years ago
Just would like to say thankyou for all your time and  replies to everyone.  I am learning a lot by how well you explain the different set-ups required by reading thriugh all the comments and links.  Hope to build my own soon when i understand enough..:-)

Thanks again.
matthewkhoury75 says: 3 years ago
Hi again, newtonn2 told me to use this regulator for everything http://cgi.ebay.com/5-x-LM350-LM350T-Adjustable-Voltage-Regulator-3A_W0QQitemZ250529410749QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item3a54b76ebd#ht_1542wt_939.  I just wanted to make sure it is going to work for my LED, more information is in the comments if you have questions
Artificial Intelligence (author) in reply to matthewkhoury753 years ago
You could use that regulator, but for your LED, I won't recommend it it's similar to the LM317. The only difference is that the LM350 is capable of delivering 3 Amps instead of 1.5, but the LM350 requires the input voltage to be at least 3 Volts higher than the output voltage, so it's pretty inefficient. I recommend the LM1084IT-ADJ. It works the same way, but doesn't heat up that much.
Is there any cheaper regulator?? the LM1084IT-ADJ is really expensive.
Artificial Intelligence (author) in reply to matthewkhoury753 years ago
The LM350 has a dropout voltage of 3 Volts. Since you're powering you LED with 15V (1 volt more than the LED voltage drop), a 3 volt dropout will result in too low output voltage. The LM1084IT-ADJ has a dropout of 1 volt, which makes it perfect for your application.
i dont know if this matters, but I am also powering other things with the same power supply.  I have a 12v lcd a 12v fan and a 2w speaker, so would it matter if the LED did drop 3v even though it is not going to get the full 15v??
matthewkhoury75 says: 3 years ago
Hi, I need some help. I dont know much about electronics and am making an LED projector.  I have a 14v 1.5a LED and need a driver for it.  I also need to know the output so I know how much the power supply needs to be.  my question is What is the difference between a regulator and a resistor. Also because my LED is different than yours can I still use the LM317 Regulator?? I did the math for the resistor and it says I need a resistor that is 9.333333... Is this correct??

Artificial Intelligence (author) in reply to matthewkhoury753 years ago
The LM317 is rated for a maximum of 1.5 amps, so I wouldn't recommend using that! You could go for the 3 Amp version of the LM317 (The LM350), but choosing the right regulator also depends on what voltage you're using to power your circuit. There are more efficient regulators out there. More efficiency means lower voltage drop. And lower voltage drop means less heat. The higher the resistance, the lower the current, so you must have gotten something wrong. The resistor should have a resistance of 0.83 ohms. A resistor with that value is a little hard to obtain, so you could try with a 1 ohm. This will run the LED cooler, but I don't think the brightness would be much lower.

What voltage do you use to power the circuit?
I think I am going to use 15v to power the whole circuit but other things are included.  Here is a schematic that Newtonn2 made me. Maybe this will help.
matthewkhoury75 in reply to matthewkhoury753 years ago
I just found a 2 small heatsinks.  Will these work for the regulators?? I am using two of them and I really don't know how big the heatsinks need to be.  Here are some pictures in comparison to a quarter which is 1" in diameter.
matthewkhoury75 in reply to matthewkhoury753 years ago
I just found some resistors on ebay. I just wanted to make sure these are the right ones so here is a link http://cgi.ebay.com/6-Resistors-1-4-Watt-1-ohm_W0QQitemZ170357518698QQcmdZViewItemQQptZLH_DefaultDomain_0?hash=item27aa19716a
Artificial Intelligence (author) in reply to matthewkhoury753 years ago
If you're planning to run it at 15 Volts, you'll definately need a more efficient regulator. The LM1084IT-ADJ would be a good choice in your case. You can get them from eBay. The resistors won't work. They have too low wattage, so they will probably smoke. also to get closer to the right value, you can connect two 1.8 ohm resistors in parallel. You can connect two of these resistors in parallel for the most accurate results. The heatsink looks ok.
by parallell do you mean one on the adj and one on the out or do you mean two on the out?? Also can I use the LM1084IT-ADJ for my 12v regulator also??
Artificial Intelligence (author) in reply to matthewkhoury753 years ago
I mean two 1.8 ohm resistors connected in parallel, where you would usually use one. Since the LM1084IT-ADJ is adjustable, you can adjust to give out 12 Volts.
This may be a stupid question, but how do you adjust the regulator?? Also, what should the regulator be adjusted to for the LED??
Artificial Intelligence (author) in reply to matthewkhoury753 years ago
How you adjust the regulator depends on whether you want a constant current (like we want with the LED) or a regulated voltage. For an explanation of the different hookup methods, please take a look at the datasheet page 7 to 11. It contains schematics and explanations. For the LED, we will like to have a constant current across it, so the current should be set to 1500mA (preferably a little less, to keep it from overheating. Therefore a resistor of 0.9 to 1.2 ohms would be appropriate.
Thanks so much for the help so far.  Sorry, but the datasheet is foreign to me.  I need a regulator to regulate the voltage to 12v and I need another regulator to regulate the voltage to 7.5v, Do I need to buy anything for this to happen, or do I just have to wire it up a certain way??
lightime says: 4 years ago
If I have three led's that I'd like to run at 700mA each will this work with Lm317? Or would the 700mA be divided into each of the three for 233mA each?
Artificial Intelligence (author) in reply to lightime4 years ago
If you are connecting the LEDs in series, the resistor should be 1.8 ohm to give a constant current of 700mA.
lightime in reply to Artificial Intelligence4 years ago
Thanks very much for the reply. So each of the three LED's would get 700mA a piece for a total of 2100mA? or would each get 233mA for a total of 700mA?
Artificial Intelligence (author) in reply to lightime4 years ago
Are you connecting them in serial or parallel?
lightime in reply to Artificial Intelligence4 years ago
I would be connecting them in series. Thanks
Artificial Intelligence (author) in reply to lightime4 years ago
Then you should use a 1.8 ohm resistor the current through each LED will be 700mA, and the current for the whole circuit would be 700mA, since you are connecting them in serial.
lightime in reply to Artificial Intelligence4 years ago
Thanks for clearing that up. I appreciate it very much!!
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