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This Instructable will show you how to built a Constant Current for high power LEDs, using only two components.

High power LEDs are getting cheaper and cheaper, however the constant current drivers, to drive them are pretty expensive.

Here, I'll show you how to built a simple and cheap, yet very effective constant current source.

The image shows the constant current driver hooked up to a 1W white Luxeon LED.

EDIT: This LED driver supports PWM, which means that you can control the brightness of the LED(s). Those fancy and expensive drivers doesn't support that. I'll post some schematics and applications as soon as i have time.
 
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Step 1: Get the Parts.

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Here is a list of the the things you'll need.

a LM317 Regulator.
a Resistor (see next step).
a Heatsink for the LM317 (you don't need one as big as mine, I just took one i had laying around).
some Luxeon, or other brands of high power LEDs (see next step too).
some Wire to hook it up.
it will be a good idea to use a heatsink for the LED as well.
source LED says: May 12, 2013. 7:40 PM
It is a great article! We are a US based company in China, Jiangsu Allray inc. I am specialized in 1W high power LED. Any questions on 1W LED, please ask me, I will reply you all the time. :)
Chips of LED are from bridgelux and epistar. Luminous is 130lm/w(avg.), forward current 350mA, voltage 3.0-3.3V, the price with Star MCPCB is $0.4/pcs.
My email: hup@allrayinc.com, call me Richard.
gizmo13 says: Apr 26, 2013. 9:16 AM
I tried to use this for a 3w led and the only thing that happened was the regulator
got hot.
I first tried to hook it up to 2 3w led's in series then I removed one of them.
Neither time did any of them light up.
I think I need a different regulator. Which one should I use?????
Great aritcle in any case!!!!!!!!!!
gizmo13 in reply to gizmo13Apr 26, 2013. 9:18 AM
Forgot to mention the specs of the led's.
forward voltage: 2.9v-3.3v
forward current: 750 mA
hope that helps.
petebarchetta says: Apr 5, 2013. 10:30 AM
I'm building a bike light version of this running 3 Cree MC-E LEDs, I'm curious if this circuit would run them, also would it interface with a pwm 555 chip if so what would the correct component values be?
santmich says: Feb 21, 2013. 9:38 AM
another question excuse me for using that resistor 1W LED
santmich says: Feb 21, 2013. 9:29 AM
hello I congratulate you for what you share only one question in your experience that other component or components you can include it stays cool for both thanks
arashagha says: Nov 29, 2012. 9:40 PM
hi, if i wanna use 3, 1W leds in series and waNNA connect them to 12-14 volts in my car, what would be the correct R with what Watt??

thx for article
tdool says: Oct 21, 2012. 1:57 PM
Have a few questions on setting this up for my reff tank and possibly simple Dimming ability

OK I want to do a string of 9x 3w LEDs with fv of 3.6v and run them at 700ma.

So if I am understanding I would run them in series add a Power Supply that delivers 35.5volts { (9x3.6)+3 for LM317 }

For R1 I would use a 1.8ohm resister that can take 1amp. is that correct or do I need 5-10amp?

Now if I want to dim it without any fancy PWM circuit can I just add a Pot after R1 to increase resistance which will lower the current in effect dimming the LED's. will this work?? if so what size Pot will I need?

Thanks
tdool in reply to tdoolOct 21, 2012. 3:04 PM
I meant 1w or 5-10w resistor not amp sorry
svb93 says: Sep 25, 2012. 9:49 AM
hi
i have mobile battery (3.7 volt , 900ma)
Can I plug my battery directly to the 1 watt or 3 watt power led ?
please help me
thanks
spanda5 says: Sep 19, 2012. 11:43 PM
http://sevenpanda.com/100w-warm-white-high-power-led-light-lamp-100-watt-p-1415.html#.UFq6bFFfb9Y
spanda5 says: Sep 19, 2012. 11:41 PM
I don't know much about of electronics, But I want to recommend you guys some low price LEDs.
http://sevenpanda.com/100w-warm-white-high-power-led-light-lamp-100-watt-p-1415.html#.UFq6bFFfb9Y

http://toyswill.com
vallka says: Aug 31, 2012. 12:31 AM
Hi all, Just leaning about this stuff and want to try to buld up a nice headlamp or two.
Will this driver work for 3 of this Cree XLamp XR-E Q5 White 3W LED Light Emitter, or should I build one for each like?
http://www.ebay.ca/itm/5PCS-Cree-XLamp-XR-E-Q5-White-3W-LED-Light-Emitter-mounted-20mm-Star-PCB-/320941810262?pt=LH_DefaultDomain_0&hash=item4ab99f6256

I would like to power the 3 with a Ryobi 18V drill battery, I also want to build a couple and power them with 3.7V li-ion (I work with computers and have a pile of 18650 batteries from laptops) so i can put together a few battery sizes.
This is the light design.
http://www.instructables.com/id/Improved-high-power-LED-bike-head-light-with-integ/
PedroDaGr8 says: Apr 3, 2012. 10:22 AM
Instead of the LM317, I like to use the Sharp PQ12RD21. It is a four pin voltage regulator, similar in nature to this one but the fourth pin is an on/off pin which can be connected to a PWM for easy dimming. Additionally, because there is a separate pin for PWM dimming, the PWM doesn't need to handle large amounts of current. A simple 555 PWM circuit can do it.
Maragnus in reply to PedroDaGr8Jun 18, 2012. 8:11 AM
It looks like the PQ12RD21 doesn't have an ADJ pin, and is available as separate models of four voltages. Is that still usable in this application?
jaki001002 in reply to MaragnusJul 15, 2012. 5:04 PM
I think thats On/Off control terminal , pin 4. When you put a PWM signal in that pin , the output signal (V) response will be almost the same as input , about frequency . So , you'll get PWM signal with high current for dimming LED brightness . If you want to gain more brighness with the same power , this is good way to do.
1234.jpg
placidus_69 says: Jul 15, 2012. 3:11 AM
HI Team,
Can i use a 12V DC running 3x1w led ?
panchavanparivendhan says: May 19, 2012. 1:49 AM
but.......u still did'nt say the value of resistance needed for this project.
ptacnik says: May 11, 2012. 1:22 PM
I Have a 90W LED that works at 21V and consumes 700mA, if i have lots of 3.7 Li-Po 5000 ( 1 Cell ) how many would i need to power this baby, i know i can use a Voltage doubler with capacitors and diodes so Voltage is not my main concern. How can I know the amperage on my custom battery pack ? just by measuring the Resistance and multiplying by voltage ? if this is too high how can i regulate it :(
I'll install a micro piezo blower to cool it down ;-)
bjcryss says: Apr 11, 2012. 7:15 AM
Can i put the LED with the LM317 on the same heatsink? Won't it shortcut or something ?
PedroDaGr8 in reply to bjcryssApr 23, 2012. 3:30 PM
Both should be electrically isolated from the heatsink (the 317 by a silpad or similar and the LED by the star). That being said, putting both on one heatsink is asking for trouble.
aleixoar says: Apr 3, 2012. 11:45 AM
Excellent but if power source has highest voltage you can't regulate with a solid resistence, i suggest a LA7812 it's cheaper and will work in a range between 13 and 16v with a good heat sink.
PedroDaGr8 in reply to aleixoarApr 3, 2012. 2:38 PM
I assume you mean LM7812. It isn't much different than the LM317. LM317 has a dropout voltage of around 2.5V and the LM7812 has a dropout voltage of around 2V. That being said, its 6 of one half dozen of the other. Both will dissipate the same amount as heat because dropout voltage only matters when Vout is getting close to Vin-Vdo.
hamtons says: Mar 26, 2012. 6:32 AM
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 hamtonsApr 3, 2012. 10:10 AM
the first one. You are measuring the voltage drop across the resistor.
hamtons in reply to PedroDaGr8Apr 3, 2012. 11:38 AM
4.5W it is. ok thanks (:
PedroDaGr8 in reply to hamtonsApr 3, 2012. 2:23 PM
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: Apr 3, 2012. 11:56 AM
Ohm never forgot his dying uncles advice.

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

Excellent instructable by the way, bravo.
zami says: Nov 5, 2011. 12:23 AM
Thnx for the gr8 article.It really helps.I have a question.
I m making a 90W Led panel,Using 1W led(rating 350mA,3.2V)
In which design shud i make it??

And for Led lights panel,Constant current source is better or Constant Voltage source??????
n do the series or parallel combinations of LEDs affect the efficency or temperature of the panel....


Please Email me the design at
good_boy_9211@yahoo.com
PedroDaGr8 in reply to zamiApr 3, 2012. 10:16 AM
For high wattage LEDs, ONLY use Constant current. Additionally, series is better if you have the voltage overhead. You will need a constant current driver for every string of LEDs you make. so for example
          |--CCDriver--LED-LED-LED-|
          |--CCDriver--LED-LED-LED-|
12V--|--CCDriver--LED-LED-LED-|--Ground
          |--CCDriver--LED-LED-LED-|
          |--CCDriver--LED-LED-LED-|

enLighten says: Mar 7, 2012. 4:53 PM
I am looking for a low cost method to adjust the intensity of a single 3W Cree LED using a 12 VDC or 24 VDC input power. It appears that the LM317 design would require an expensive 2 Ohm Rheostat to make the adjustment. I need the light from the LED to be fairly stable (i.e. not fluctuating). Any idea's that can be tested?
PedroDaGr8 in reply to enLightenApr 3, 2012. 10:09 AM
Use a PWM for dimming. Voltage dimming is spotty at best on LEDs; additionally it tends to be non-linear and causes color shift. Personally, instead of the LM317 I would use the Sharp PQ12RD21. It's a 2A 12V reg like the 317 but it has a fourth pin dedicated to PWM. So you can use a simple 555 based PWM circuit.
nigel cox says: Dec 19, 2011. 12:50 PM
What input voltage are you using in this circuit please, thank you Doc Cox
chse720 says: Dec 6, 2011. 10:32 AM
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
ironsmiter says: Oct 24, 2011. 10:40 AM
So, when it comes to circuit design, I can be a bit addlebrained.

What's the advantage(besides power savings) of the modded PWM version ov, say, using a variable resistor in THIS circuit.

To control brightness.

Or am i confusing voltage and amperage yet again?
(sorry, bast my normal bedtime)
markee2 says: Sep 10, 2011. 6:41 AM
Nice instructables. No doubt this is one of the simplest LED driver. Its only limitation is current is limited only up to 1.5A
I found a  very cheap LED driver that claims can drive up to 10A current
I dont know know and never tried building it building it, here is the circuit
copied from a site
http://www.simple-electronics.com/2011/09/cheap-power-led-driver-circuit.html

here is the sites circuit




cheap led driver circuit.jpg
PabloMDP says: Jul 8, 2011. 11:48 AM
Hi, i have 4 leds 10w (9 - 11v 1amp) and i have a 12v 10amp power supply. Can i make a driver only with one LM1084IT-ADJ? I want to dimmer the leds with one control. Thanks!!!
Avaruushirvi in reply to PabloMDPSep 4, 2011. 9:56 AM
From datasheet ==> "The LM1084 is a series of low dropout voltage positive regulators with a maximum dropout of 1.5V at 5A of load current. It has the same pin-out as National Semiconductor's industry standard LM317."

So you cannot use full bright!
PabloMDP says: Jul 8, 2011. 10:23 PM
Sorry, i have another question. What happend if i put 4 leds in parallel, 1 amp each led, all with only 1 regulator LM1084IT-ADJ, (0.3 ohm resistor) and 1 led is damaged. In this case, the other leds work at 1.33 amps? Thanks, and sorry for my english.
Avaruushirvi in reply to PabloMDPSep 4, 2011. 9:54 AM
IF that one led shorts, then 4 apms go through it... And If it open then it's go just as you suspected!
burton2663 says: Jul 18, 2011. 3:23 AM
im looking for a 1 watt luxeon led driver to maximize brightness when connected to the outputs of the vu meter lm3915, an actual schem would be nice tnx
eugenehaller says: Feb 3, 2010. 12:22 PM
 AI, I am amazed that you are still diligently answering comments two years after you you made your instructable.

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 eugenehallerJun 6, 2011. 3:14 AM
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 eugenehallerSep 9, 2010. 1:55 PM
Use an LM338 it is rated to 5A.
seethasub in reply to eugenehallerMar 11, 2010. 1:40 AM
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 seethasubNov 19, 2012. 12:30 PM
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
led-driver5.jpg
Artificial Intelligence (author) in reply to eugenehallerMar 2, 2010. 7:20 AM
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 eugenehallerMar 1, 2010. 3:01 PM
try using a LM350 it is rated at 3 amps or 3000 ma

Lee
aekara says: Jan 27, 2011. 11:33 PM
yugang in reply to aekaraJun 6, 2011. 3:13 AM
Hi,

could you please help me in designing to light 10w power led or 1w * 10 led?

I appreciate your help!

Yugang
macnomad84 in reply to aekaraFeb 9, 2011. 10:31 AM
DATA SHEET
http://www.national.com/ds/LM/LM117.pdf

"The LM117 series of adjustable 3-terminal positive voltage
regulators is capable of supplying in excess of 1.5A over a
1.2V to 37V output range."

P = V * I = 16 Volts * 1.5 Amp = 40 W

I think you can. Go for it. Someone correct me if I'm wrong. Gonna keep reading this DS...
ngwanhoe says: May 18, 2011. 9:30 AM
I anyone could show me a simple circuit for :-

Input voltage : DC 4.8V - 5.6V
Led : 3 X 3w Leds, working voltage 4v - 5v each led, so had to wire in parellel.
Prefered current : 0.8A - 0.9A

Now I am just using resistor to do the job.

Please email to ngwanhoe@yahoo.com

Thanks a million.
Jayvis Vineet Gonsalves in reply to ngwanhoeMay 20, 2011. 10:44 AM
You can wire the LEDs in Series and for the resistor, you can use a 1.56 ohm resistor if you desire 0.8A of current. Use the schematic given in the instructable.
ngwanhoe in reply to Jayvis Vineet GonsalvesMay 22, 2011. 5:52 PM
Thanks Vinet,

If wire the Leds in series then I would have to put in 8v - 15v power supply. Don't have space for so many batteries.
pandyaketan says: May 9, 2011. 1:40 AM
My projects might provide simple answers to your complex issues...

reg
ketan
--------------------------------------------------------
"May the good belong to all the people in the world.
May the rulers go by the path of justice.
May the best of men and their source always prove to be a blessing.
May all the world rejoice in happiness.
May rain come in time and plentifulness be on Earth.
May this world be free from suffering and the noble ones be free from fears"
---- Vedic blessing
broxlin says: May 1, 2011. 3:44 PM
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: Feb 5, 2011. 10:38 AM
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 Jkirk3279Feb 22, 2011. 3:56 PM
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
TheNuclearWatermelon says: Feb 13, 2011. 8:22 AM
Ugh! The LM317 is the most common current regulator circuit ever! I hate it's 3v dropout! Why can't anyone show us a simple boost circuit like the LM3410?
ovven says: Jan 7, 2011. 9:48 AM
Great guide, really useful. I am trying to rig up a system of 11 high power LEDs, with a forward current of 350 mA and a forward Voltage averaging 3.5 V.
would it be best to run these all with different drivers or all in series or a mixture and also what would be the best regulator for the system. I'm nearly at the point of comprehending all of this but thats not the same as understandingit to a competant enough level to make it, just yet.

Cheers
morphious69 says: Nov 29, 2010. 10:41 PM
hi. I am wondering if you could use a couple of these for higher power like what is consumed by the 10 watt or higher LEDs? can they be adjusted to put out the higher voltages needed like 12 or 24 volts and just be expected to supply the amount of amperage they can handle but multiply this by having several of them in parallel so as to add more amps?
Electroinnovation says: Nov 12, 2010. 5:09 PM
Could this handle 2 1W white leds? and if so would i put them in a parallel or a series circuit?
LEDismylife says: Oct 2, 2010. 3:21 AM
To anyone who knows, how can I extend my LED output from 700mA to 1000mA? should i change just the regulator?
oli1122 says: Sep 9, 2010. 8:46 AM
Could i do this for a 10W Led of 1000ma???
SRWitt says: Sep 7, 2010. 8:12 PM
Lets see if you're still answering questions. So I'm building an array of 3w Leds in red, yellow, and blue.

I'm planning on using the LM1084it-adj vs the LM317.

So far I've figured based on datasheets @ 700ma, I should have about 2v total drop with the 1084 (1.25adj+ ~.8@700mA). Is this correct?

My source is going to be a computer PSU converted to a benchtop PSU, using the 12V out. Based on my LED's datasheets, I should be at about 2.2V per Red or Yellow @700ma, based on this, the 1084 will be dissipating 1.2V@700mA with 4 red or yellow LEDs wired in series (total of .84W). For the blue I've figured ~3.6V for 700mA, with 2 in series the 1084 would have to drop ~2.8V@700ma (1.96W).

Forgive me, I am a relative newb to electronics, does my math seem correct? Am I misinterpreting the datasheets when it comes to how much voltage the 1084 drops under varying loads?
peterbeihoffer says: Aug 22, 2010. 1:45 AM
Hi I need some help. I am trying to design a high powered LED light. I want to put together 10X 1W high power LEDs. I was thinking of using 10 topledlight # BY1W-WH they have a DC Forward Voltage (VF): 3.0V ~ 3.8V, and a DC Forward Current (IF): 350mA ~ 400mA. I was wondering what the simplest and easiest way to wire this. Could I just use 1 10w driver? any help or tips would be much appreciated. Thanks
WingmanSVT says: Aug 8, 2010. 5:44 PM
Im a bit confused about using this circuit with LEDs wired in series. I have 2, 3 watt LEDs wired in series, each has a forward voltage of 3.8 max and 700 ma. What resistor would i need to have between pin 2 and 3 on the LM317?
jon1888 says: Jul 17, 2010. 4:02 PM
does anyone know what it would take to power a 50w 24v-26v
TheWhatnought says: Jun 30, 2010. 1:12 AM
Hey, is this usable for some of the really high current emitters? I'm trying to find a cheap, only on/off CC driver for a 3.86V 2800mA LED. I'm also trying to drive it off of a 12VDC supply. Any suggestions?
pedroalaray says: May 22, 2010. 10:00 AM
Hi could you take a look at this circuit I've planned out? I think it would work but not sure if I should order the bits thank you Peter

http://farm4.static.flickr.com/3339/4629373728_186cbc5e3f_o.jpg
Joshus says: Jun 11, 2009. 10:32 PM
Hi, I have been messing with DC power supplies for some time. Can I just use a bridge rectifier with my 110V ac at home followed by your design? How could I acomplish this? Thank you so much for all of the information! Reading your instructable and answers to questions has helped me ALOT!
Rusdy in reply to JoshusJan 3, 2010. 9:47 PM
 After many googling and forums, I've found this inexpensive current source, with wide AC inputs (110-240VAC). Link is here

(If the link doesn't work, just type "current source" on the linked website, www.dealextreme.com, and browse for the correct input voltage)

I just received it today, so haven't managed to tried it myself, but googling the driver's datasheet (AP3706), definitely a true current source! The product is rated to 350mA, but if you're good with electronics, you can change the Rsense (which I only can guess at the moment which one on board) to suit
lightartist1 in reply to RusdyApr 30, 2010. 6:03 PM
 hey rusdy , i came across your comment while researching powering a led 'star'. i have a couple of questions: how are the drivers from deal extreme working out for you?  also, what type of  power rated wall wart would i need to power just one star safely?  there are good deals on surplus sites for wall warts, i just want the type that would complement the 350 mA star as cheaply as possible.  thanks!
Rusdy in reply to lightartist1May 1, 2010. 6:03 AM
I believe this one (I linked above from DealExtreme) will be the best option for powering a single white LED. Simple, cheap, and really small. Just note that there are 2 types of driver offered: 350mA and 700mA option, so they need to be matched with your LED (or else the driver can fry your LED!)

Also note that this driver only have max output voltage of around 4V, so only good for powering a single white LED. Obviously, you have to be comfortable with the hand skill for installing such a small driver (with live voltage!). Hope that helps...
lightartist1 in reply to RusdyMay 1, 2010. 7:34 PM
 hey rusdy. thanks for your promt reply. i appreciate the info very much.
Artificial Intelligence (author) in reply to RusdyJan 4, 2010. 5:53 AM
I have that exact same driver, and it works quite well, and its very compact. Doesn't appear to support PWM from an external source though.
Artificial Intelligence (author) in reply to JoshusJun 12, 2009. 11:44 AM
Unfortunately, the maximal input voltage of the LM317 is 37V, so hooking it up to mains voltage wouldn't be a good idea. There are two things, you can do. 1. connect it to a wall wart adapter or another DC supply, that puts out less than 37 Volts. 2. Buy a preassembled LED Constant current Power supply that runs directly on mains power, like the ones they sell at dealextreme. Their website is currently down, so I can't post a link.
tkjtkj in reply to Artificial IntelligenceAug 21, 2010. 6:41 AM
my intended voltage source varies from 32v to 41.3 .. i wonder if that high limit would destroy the regulator .. or is the referenced 3amp regulator perhaps more tolerant of applied voltages a lil above its specified max ?? tkjtkj@gmail.com
Joshus in reply to Artificial IntelligenceSep 21, 2009. 12:17 AM
What if I took a bridge rectifier from 110VAC followed by several LM317 regulators in parellel. Each LM317 would "feel" a reduced voltage right?
Artificial Intelligence (author) in reply to JoshusSep 23, 2009. 1:40 PM
Well. If they are connected in parallel, they will all take the 110V and you will be letting out some of that magical blue smoke. They will most likely explode.
cdousley in reply to Artificial IntelligenceJul 28, 2010. 10:43 PM
blue?
Joshus in reply to Artificial IntelligenceJun 13, 2009. 2:00 AM
Could I just us a transformer to reduce the ac voltage to a smaller value, then us a bridge rectifier follower by the voltage ragulator and resistor?
Artificial Intelligence (author) in reply to JoshusJun 13, 2009. 2:35 AM
Yes you can.
aleysth says: Apr 27, 2010. 11:39 PM
hi, can this circuit use to power a Bridgelux LED ARRAY, BXRA-C0400-00000? voltage = 10.6, constant current 600mA, Power = 6.4w.
jfenwick says: Apr 13, 2010. 8:37 PM
I'm confused about how the voltage regulator behaves with only one resistor.
The voltage of the LM317 can usually be modified by changing the ratio of the two resistors going into ADJ and Vout.
This is usually based on the equation Vout = 1.25(1 + R2/R1).
If R1 = R2, then doesn't the equation give you 1.25(1 + 1) = 1.25 * 2 = 2.50V?
If that's the case, I would think your computation for how to determine what resistor you need for limiting current would come out differently.
For instance, when you have a 1W LED, which requires 350 mA, you use Ohm's Law to come up with 1.25/0.350 = 3.57.
Wouldn't you instead end up with 2.50/0.350 = 7.14?
jfenwick in reply to jfenwickApr 13, 2010. 8:39 PM
The last two lines were a mistake.
I'm having a hard time understanding how you can even use Ohm's Law here since any time you change your resistor, you're going to change how much voltage you're getting because you'll be changing the voltage going to the ADJ pin.
cdousley says: Apr 8, 2010. 3:36 PM
would this work for a laser ? if it does what would i have to do to get about 4v
and 30 millamps
Jayvis Vineet Gonsalves says: Mar 19, 2010. 1:04 AM
Thanks AI I got the LEDs up and running with 12v power supply. Thank you once again for all your replies. Soon I will post an instructable about the headlights.
Jayvis Vineet Gonsalves says: Mar 14, 2010. 11:52 PM

I am about to use this circuit and 3 LEDs in a motorcycle headlight. So the input voltage could be at around 12v to 14v. Is so much voltage sufficient to drive the LEDs. By the way I have not tested the circuit with the LEDs. I have only tested the circuit using a multimeter and with an input voltage of 9v.

Jayvis Vineet Gonsalves says: Mar 14, 2010. 10:47 AM
Thanks AI your circuit works. I measured it using a multimeter. It read 310mah. I used it with a 9v battery. The voltage was 8.38v. Is this much voltage and current ok for using 3pcs of 1watt LED?
Artificial Intelligence (author) in reply to Jayvis Vineet GonsalvesMar 14, 2010. 2:14 PM
 Did you test the circuit with 3 LEDs at 9 Volts??? If you use white LEDs, the LED voltage is likely around 3.5V, so 3.5*3=10.5 + 3 Volts dropout from the regulator is 13.5, so you'll want a voltage of at least 13.5Volts, however you can usually go a little less than one volt lower than that, so you might get it to work successfully down to around 12.5 volts.
Jayvis Vineet Gonsalves says: Mar 14, 2010. 5:40 AM
Also AI for a 1watt LED is 3.9ohms resistor OK.
Jayvis Vineet Gonsalves says: Mar 14, 2010. 3:34 AM
The sign of a diode(LED) consists of a triangle and a line. So is the line the negative terminal of the LED and the base of the triangle its positive terminal? By the bar I meant the line.
Jayvis Vineet Gonsalves says: Mar 14, 2010. 1:04 AM
Hey AI is the bar like thing on the diode sign(LED) signify the negative terminal of the LED?
Artificial Intelligence (author) in reply to Jayvis Vineet GonsalvesMar 14, 2010. 1:51 AM
 Bar like thing?
Jayvis Vineet Gonsalves says: Mar 12, 2010. 9:00 AM
Hi AI don't you require any heat sink for the LED? 
Artificial Intelligence (author) in reply to Jayvis Vineet GonsalvesMar 12, 2010. 9:22 AM
 Hi,
Yeah, You should always use a heat sink for both your LED and regulator. And yes, it will work with  varying voltage as long as the voltage doesn't exceed the maximum input voltage of the regulator (35V for the LM317).
Jayvis Vineet Gonsalves says: Mar 12, 2010. 9:03 AM
Also can I use this circuit if I have an input voltage that keeps on changing between 12v to 14v. I will be using a 1watt LED.
bbas8 says: Feb 18, 2010. 4:55 AM
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.
pravardhan says: Feb 9, 2010. 6:48 AM
Hi AI, I tried your circuit and its very simple.

But, The LM317 and the LED's Star Board is getting very hot for just 3 to 5 Minutes!?! The LED is 1W from a local company (may be Chinese too). I am from Bangalore, Karnataka, India.

I used a 12V, 500mA DC Wall Adapter. The resistor value is 4.7Ohm of 0.5W. I wanted less than 350mA, as i don't have any datasheet for this LED.

So, why does both the LED and IC gets Very Hot soon?
cahillicus says: Jan 29, 2010. 6:32 AM
A I, you seem very helpful.  I hoping you would help me also.  I would like to build dual motorcycle headlights, each with 6 CREE XR-E P4 emitters.  I would like to use your driver but am bad with the calculations.  
Artificial Intelligence (author) in reply to cahillicusFeb 2, 2010. 1:52 PM
Your motorcycle battery is on 12 Volts, right?
cahillicus in reply to Artificial IntelligenceFeb 7, 2010. 9:12 PM
Yes, but voltage increases to near 14v at driving RPM's.  I finally ended up order Cree Q5's and am considering running them somewhere between 700 and 1000 ma.  Any ideas how to run all this?
cahillicus in reply to cahillicusFeb 7, 2010. 9:13 PM
The are 3.7vf ps.
matthewkhoury75 says: Jan 29, 2010. 6:04 PM
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 matthewkhoury75Jan 30, 2010. 6:10 AM
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.
matthewkhoury75 in reply to Artificial IntelligenceJan 31, 2010. 8:00 AM
Is there any cheaper regulator?? the LM1084IT-ADJ is really expensive.
Artificial Intelligence (author) in reply to matthewkhoury75Feb 2, 2010. 1:49 PM
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.
matthewkhoury75 in reply to Artificial IntelligenceFeb 2, 2010. 2:36 PM
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: Jan 21, 2010. 3:26 PM
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??

Thanks in advance
Artificial Intelligence (author) in reply to matthewkhoury75Jan 22, 2010. 10:00 AM
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?
matthewkhoury75 in reply to Artificial IntelligenceJan 22, 2010. 1:56 PM
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.
wiring_schematic_LED.jpg
matthewkhoury75 in reply to matthewkhoury75Jan 22, 2010. 7:30 PM
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.
DSCN1626.JPG
matthewkhoury75 in reply to matthewkhoury75Jan 22, 2010. 2:16 PM
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 matthewkhoury75Jan 23, 2010. 3:29 AM
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.
matthewkhoury75 in reply to Artificial IntelligenceJan 23, 2010. 1:56 PM
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 matthewkhoury75Jan 23, 2010. 3:15 PM
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.
matthewkhoury75 in reply to Artificial IntelligenceJan 23, 2010. 6:28 PM
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 matthewkhoury75Jan 25, 2010. 1:24 PM
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.
matthewkhoury75 in reply to Artificial IntelligenceJan 26, 2010. 6:41 PM
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??
oneupload says: Dec 8, 2009. 5:47 PM
hi I have a white 20 watt
Intensity Typ. : 1100Lm
Viewing Angle : 160°
Forward Voltage : 13.5V-15V
Forward Current : 1500mA
LED could I use this drivr whit it? I have tryed it on 12v power supply rated @ 12v dc 4.16A and that almost cooked it! I would like to run it in a car 12-14V and also use it with my powr supply at home what resistor will I need and were can I get the compnents at a good price in the uk thnks
Artificial Intelligence (author) in reply to oneuploadJan 4, 2010. 6:04 AM
Hi,
Sorry for not replying. The problem with this driver is that it isn't that efficient. It has a voltage drop of 3 Volts, meaning that the input voltage needs to be 3 Vollts higher than the LED forward voltage.
spartakus says: Oct 3, 2009. 1:00 PM
hello AI. i was wondering if you can help me..i am trying to build your driver for a cree xre Q5 LED. it is powered by 7.4 volt lion battery 1500Mah. i used a 1.2k ohm 1watt resistor to get a constant current of 1000mah.. Led lights up but nowhere near the brightness of where it should be....can u tell me what i am doing wrong...
Peregrine7 says: Jul 12, 2009. 9:33 PM
First of all, thank you very much for this instructable. This has helped me a lot in understanding needs for my project. However, I still have few questions. Let's start with explaining exactly what I want to do.

I'm going to convert 4 of my taillights in my car so that each of the 4 lights would use 8 LEDs (I'm constricted to that number due to physical dimensions). Since all 4 lights will be identical, let's use the following as an example of 1 light only.

LED specs:
Red 1 Watt
Forward Voltage : 2.2v-2.4v
Forward Current : 350mA
example: ebay item # 310154192168

Input voltage == regular automotive 12V batt. which I think would vary the voltage from about 10v to 14v.

So if I use 1 driver per light ( 8 LEDs) I could connect 2 series (4 LEDs in series) in parallel (2 strings)? The way I'm looking at it according to your examples would be using LM317 for 2.3+2.3+2.3+2.3+3 = 12.2 at 700mA, so I'd need 1.8 ohm resistor for that? But what exactly will happen in the input voltage from battery drops below 12.2V ? Will the LEDs just become a little dimmer, or would they not light up at all ?

2nd question: These lights are supposed to be dual stage (park and stop light). So for stop light, I'd use the full intensity, but for park light, I'd want to limit the intensity to probably about 70% maybe (?). So can I just, for parklight input, add a resistor to the Vin of the LM317?

3rd question: I realize I need a heatsink for the LM317, although since I'll be using all of the input voltage, there probably won't be much heat generated to dissipate anyway. But I don't get the LED heatsink comments. I can't just mount these LEDs on a regular PCB?? What kind of heatsinks should I use for them and how do I attach them?

Thank you very much in advance for helping!
Peter
Artificial Intelligence (author) in reply to Peregrine7Jul 13, 2009. 4:53 PM
You're right. You can use a 1.8 ohm resistor when running the two strings in parallel. As the input voltage drops, so does the brightness. It really doesn't drop noticeably until the voltage drops to less than 2 volts above the LED voltage. It it gets dimmer as the voltage drop, but it doesn't suddenly turn off. There are different ways of turning down the brightness. You could perhaps use a transistor or relay to switch from the 1.8 ohm resistor (for full brightness) to a higher value resistor (for lower brightness). You could also use a microcontroller to dim the LEDs using PWM. As you mentioned, you probably don't need a very big heatsink for the LM317, however, I think it would be a good idea to use one. Just in case. Are the LEDs, you use mounted on a star?
Peregrine7 in reply to Artificial IntelligenceJul 13, 2009. 7:27 PM
Thank you :)
So following my example above, the brightness of the LEDs would not be noticeably worse unless the input voltage dropped below 11.2V, right? That's perfectly within my expectations :)

As far as turning down the brightness, let me get this straight. The way I thought I could do it was to leave the way you had the 1.8ohm resistor wired between the Vout and ADJ and add another higher resistor at Vin. If Vin was getting 12-14v, the LEDs would be at full brightness, but if the voltage was routed through the higher rated resistor to Vin instead, the LEDs would be dimmer. And I was gonna separate the 2 inputs to Vin using 1amp directional diodes. So you're saying I can't limit the voltage that way to lower the brightness? Any more detailed clues on how to accomplish this goal the easiest and cheapest way? I don't have to change the brightness levels once I find the right brightness drop.

As far as heatsinks, I think attaching the regulators directly to car's chassis should serve as sufficient heat dissipation, right? Although it will be in trunk area, and in summer the metal gets really hot... but so would pretty much anything in the trunk. As for the LEDs, the ones I was looking for are being sold without the stars. Are stars the heatsinks? Unfortunately, all those with the stars that are being sold have different parameters (less lumens, higher forward voltage). So I'd really like to use those here: http://cgi.ebay.com/ws/eBayISAPI.dll?ViewItem&item=310154192168&ru=http://shop.ebay.com:80/%3F_from%3DR40%26_trksid%3Dp3907.m38.l1313%26_nkw%3D310154192168%26_sacat%3DSee-All-Categories%26_fvi%3D1&_rdc=1
What would you suggest about the LEDs then?

Artificial Intelligence (author) in reply to Peregrine7Jul 14, 2009. 4:44 PM
The brightness drop is based on my experience with white LEDs. Perhaps it might be more visible with other colors.

As far as dimming the LEDs, you should be able to connect a resistor on the input of the LM317. I don't know if you'll get problems with the LM317 trying to maintain a constant current, and can't because of the resistor though, but I don't think so.

I don't think mounting the regulator directly on the chassis is a very good idea, because the Vout pin on the regulator also is connected to the tab of the regulator, so if you do, you'll need to insulate it. I suggest mounting the LEDs on an aluminium plate with thermal glue.

Hope that helps :)
Peregrine7 in reply to Artificial IntelligenceAug 22, 2009. 10:12 AM
I'm still having a problem figuring out how to properly do the dual brightness on this. I was wondering if you could perhaps help me a bit more. I found this circuit made by someone else http://www.instructables.com/id/EUSOVGN9AYEWOF32SN/
and I posted the question there, but the author doesn't seem to respond anymore. Maybe you can help me figure that out?
Artificial Intelligence (author) in reply to Peregrine7Aug 26, 2009. 1:22 PM
Hi, Sorry for not replying. I don't have any experience with Dan's circuit, But I have successfully tried dimming LEDs using this driver with PWM (Pulse Width Modulation) from a microcontroller. I think, I'll do some experiments with dimming high power LEDs tomorrow.
Peregrine7 in reply to Artificial IntelligenceAug 27, 2009. 12:10 PM
Here's another suggestion I got from someone from another forum: Problem is, I've been told that the extreme ranges of automotive battery voltage can damage this circuit and that it's safer to do PWM, and I'm not sure how to modify it to use PWM with it. He has also used only one string per regulator because he said that if there are multiple series in parallel, the regulator may be dividing the amperage unevenly and it may burn out some LEDs. With PWM regulation, I guess this would not be a problem anymore, so we could then use 2 strings in parallel. Please let me know what you think and if you can maybe change this to include PWM there.
LM337 Brakelight.PNG
Peregrine7 in reply to Artificial IntelligenceAug 26, 2009. 3:09 PM
Thank you very much. Please let me know how it goes. I don't have any experience with PWM. Someone said I can use IC555, but I'm not sure how to connect it to my circuit. If you could possibly be able to do some helpful diagram and some remaining parts specs, that'd be really greatly appreciated. Also, since my power will be switched by a positive wire, someone suggester I should use LM337T instead of the LM317. Does that make sense to you?
Peregrine7 in reply to Artificial IntelligenceJul 14, 2009. 5:36 PM
Thank you very much!! Yes, that helps a ton :)
hlem says: Jun 21, 2009. 12:31 PM
Hello AI, Can you give me input on this design here? is it correct? thanks Ha
hlem in reply to hlemJun 21, 2009. 12:50 PM
Artificial Intelligence (author) in reply to hlemJun 21, 2009. 1:09 PM
Sorry, I can't sse the image. You have to be a registered member to view the image.
hlem in reply to Artificial IntelligenceJun 21, 2009. 1:57 PM
Artificial Intelligence (author) in reply to hlemJun 22, 2009. 5:38 AM
That picture just shows a fish and says "Reef Central".
hlem in reply to Artificial IntelligenceJun 22, 2009. 7:51 AM
Sorry, i dont know why linking wont work, but i just found the option here to upload image directly. Also, to add to my question, how efficient is your design?
218872LED.jpg
Artificial Intelligence (author) in reply to hlemJun 22, 2009. 9:40 AM
Your setup looks fine to me. As far as efficiency goes, I don't know. Remember using heatsinks for your regulators and of course LEDs.
ac-dc in reply to Artificial IntelligenceAug 26, 2009. 11:04 PM
It is extremely inefficient, among the least efficient ways possible to drive an LED.

There is no point to doing it this way. Simply using a series resistor will do just as well, just as efficiently. The only thing nice about this circuit is an easy way to calculate the current regardless of (an unknown mystery LED) what the forward voltage of the LED is.

This doesn't inherently "support PWM" either.

Above all, LM317 is designed to be a voltage regulator. The best way to drive an LED, whether it be by PWM or not, is by more directly regulating the current but ideally, not using a linear voltage drop stage which is what the LM317 is.

It is a good teaching circuit to learn about LM317, but for powering an LED, not so much.

Check out http://www.dealextreme.com as they have a selection of dirt cheap LED drivers for most 1W to 5W applications. And good prices on the most efficient 3W Cree LEDs too.

To put it another way, usign a $4 driver and $6 LED, you would have over triple the light output using less than 1/3rd the power of running this circuit with one LED off more than roughly (too lazy to do the math) 12V, and eliminate the need for a massive heatsink.

LED drivers are purpose designed to do the job best. DIY projects are great but also appreciate when professional EEs pick the right parts, when a factory can get it all made, and delivered to you at a good price ready to do exactly what you need.

Modern tech combined with the internet for finding and ordering products is very very great, even if you want a DIY project there isn't a need to do everything from scratch anymore and with LEDs, to regain the efficiency is very desirable for not only efficiency but to not have to deal with all that heat in a project casing.
hlem in reply to Artificial IntelligenceJun 22, 2009. 11:39 AM
ok thanks, wonder why people would spend $20 or so for a constant current Buck...when they can easily make this for a few $.
Also, since i have 6 in serial, the total voltage drop is 6x3.2(LED)+3(LM) = 22.2V, but my source is 24V, what do i do with the extra 1.8V? same goes for the 6.5A output from the source, my parallel config is only 4.2A (6x700mA), what happen to the other 2.3A?
thanks
Artificial Intelligence (author) in reply to hlemJun 23, 2009. 6:22 AM
I think, people buy the expensive constant current LED drivers, because they think, it will somehow increase the lifetime of the LED, even though that is not the case. The LM317 will dissipate the 1.8V by turning it into heat. you can use the extra 2.3 Amps to drive more LEDs or other electronics, if you want to.
hlem in reply to Artificial IntelligenceJun 23, 2009. 8:37 AM
thanks AI. I was told that the other constant current drivers is more efficient because they waste very little power, using the LM317 in my config will waste over 1W of power. And since there is only 1.8V left, I cant really run more LEDs,
diy_bloke in reply to hlemNov 1, 2012. 4:37 AM
thge effficiency is easy to calculate and thus to compare (though in practice there may be slight variations): if you are using an LED that is say 3 Watt at a forward voltage of 3 Volts and 1 amp current and you are putting in X Volt, well you do the math.
But right off one can already say that when using one 3 Volts LED the efficiency will never be more than 50% (because of the 3 Volt Voltage drop), whereas other circuits claim to have a 70-90% efficiency. For a 3 Volt LED, you would need at least 6Volt input, whereas other circuits will let you use 5 or even 4 Volts input. Ofcourse the efficiency of the LM317 circuit will get better when using more LED's in series it will always at least lose that 3 Volt x LEDcurrent
Artificial Intelligence (author) in reply to hlemJun 23, 2009. 9:11 AM
Yes you can, because your power supply is capable of delivering 2.3 Amps more than you need and so you can use those excess Amps to power more electronics if you want.
StupidPig says: Jul 28, 2009. 5:15 PM
Will this work for 4 x 3W (2 white in 3.5v and 2 amber in 2.2v, all 700mA) in series with 12v supply in car? I know the total Vf probably too high (11.4v), but will it still work if I'm OK to not running the LED in full 700mA? If not, should I run two sets of white+amber+LM317, or one LM317 with two parallel string of white+amber? Finally, if I run the dual LM317 or single LM317 with parallel string, should I use a separate voltage regulator to drop the supply to 9V before go into the LM317, to minimize the heat? Thanks.
StupidPig in reply to StupidPigJul 29, 2009. 1:11 PM
Sorry, I mean to say LM350 for the 4x scenario. BTW, will it be even better to do a step-up to 14.4v supply and drive the LED in the 4x scenario?
Artificial Intelligence (author) in reply to StupidPigAug 9, 2009. 8:39 AM
I'm sorry for not replying, I've been on vacation for a couple of weeks. I would definately run a circuit with two LM317s. One for the amber ones and one for the white. Use two heatsinks, One for each regulator, because one of the pins are connected to the tab.
StupidPig in reply to Artificial IntelligenceAug 10, 2009. 11:17 AM
No probelm, thanks for the help. One mounting question for the LM317. Do you have any suggestion on how the heatsink should be mounted? I think the tab for LM317 is linked to Vcc, right? So if I have the heatsink connected dirrectly to it, will the heatsink also carry the voltage? I may wrap the whole think in heatsink tube to insulate it, but I guess that defeat the purpose for the heatsink, right?
Artificial Intelligence (author) in reply to StupidPigAug 10, 2009. 12:12 PM
I would suggest mounting the LM317 directly on the heatsink, and then make sure the heatsink doesn't touch any metal, that conducts electricity. The Vout (middle pin of the LM317) is connected to the tab.
ashleyps says: Jul 20, 2009. 3:02 PM
Hi, Thank you so much for this instructable, I'm thinking about to use this circuit, I'm going to use the LM350 instead of LM317 because I have to connect 36 Led, and I decide to split in two 18 Led modules. One module with 18 white led at 950mA each one (1.3 Resistor) and another with 18 blue led at 700mA each one (1.8 Resitor). I want to use the PWM version of you circuit, but I don't know what Q1 use because the BC337 is only 800mA and I'm going to need 2.1A for one module and 2.85A for the another module. Thanks in advanced and sorry for my basic english
circuit.jpg
yeagerxp in reply to ashleypsAug 5, 2009. 3:49 AM
Don't apologize for your english. It is better than most
Artificial Intelligence (author) in reply to ashleypsJul 20, 2009. 3:36 PM
In your schematic, the LEDs will only get around 233mA each, Because the LM350 will give out a constant current of 700mA with that 1.8ohm resistor. The current will be divided by 3, because you have 3 strings in parallel, so actually if you want to achieve full brightness, you'll have to use a 0.59 ohm resistor to get around 700mA for each LED.
ashleyps in reply to Artificial IntelligenceJul 20, 2009. 4:29 PM
You are right, I'm going to use .59 ohm resistor instead of 1.8 ohm, now I'm thinking about to use a TIP31 or BC681 for Q1, what do you think?
Artificial Intelligence (author) in reply to ashleypsJul 21, 2009. 2:12 AM
I can't find any information about a transistor called BC681, but the TIP31 should work.
ashleyps in reply to Artificial IntelligenceJul 21, 2009. 5:26 AM
Sorry is my mistake, is BD681
Artificial Intelligence (author) in reply to ashleypsJul 21, 2009. 6:36 AM
Should work too.
ashleyps in reply to Artificial IntelligenceJul 21, 2009. 5:06 PM
Finally I had to work with TIP31C because I didn't find the BD681, I'm test the circuit with the tip31, 1.2ohm resistor because I only connect 2 strip of led in parallel and everything work fine, the PWM from a picaxe 08M dim the led from 0 to 100%, thanks you so much for you help, you are the best.
Artificial Intelligence (author) in reply to ashleypsJul 21, 2009. 10:28 PM
I'm glad I could help :-)
lightime says: Jun 30, 2009. 6:11 PM
I am trying to buy an LM317 but there appear to be many different versions or brands...which is the correct one or can any LM317 be used??? Thanks VERY much!!!
Artificial Intelligence (author) in reply to lightimeJul 5, 2009. 1:47 PM
There are many different manufacturers of the LM317, but they are all pretty much the same, except for the maximum current, they can handle. Some are 100mA, Some are 500mA, but the 1.5 Amp version is the most common.
baja.pescador in reply to Artificial IntelligenceJul 8, 2009. 10:43 AM
With a 3v drop from the LM317 or 350 and a 2.2 Vf red LED, is there any chance that a 5v computer power supply would light up a parallel array? I'm planning to use the 12v supply but I hate to waste the 20 amps. Thanks for the great, simple project.
Artificial Intelligence (author) in reply to baja.pescadorJul 9, 2009. 2:36 PM
I think it will work with a red LED on 5 volts. It seems to me that the voltage drop is more like 2.5V than 3V. If you're connecting them to a 12V power supply, you can connect some in series. That way you won't waste all the Amps.
lightime says: Jul 1, 2009. 3:18 AM
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 lightimeJul 5, 2009. 1:43 PM
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 IntelligenceJul 7, 2009. 1:03 PM
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 lightimeJul 8, 2009. 2:26 AM
Are you connecting them in serial or parallel?
lightime in reply to Artificial IntelligenceJul 8, 2009. 6:25 AM
I would be connecting them in series. Thanks
Artificial Intelligence (author) in reply to lightimeJul 8, 2009. 8:21 AM
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 IntelligenceJul 8, 2009. 4:11 PM
Thanks for clearing that up. I appreciate it very much!!
avillegas says: Jul 6, 2009. 4:27 PM
Hi everybody, I would like use this led driver on a SSC P7 led, that runs with 3.6vf, and 2800mA with 10W of power, I guess that I need the LP350, but can somebody help me to calculate the resistor value? ( ohms and power)... Thank You. Regards
Artificial Intelligence (author) in reply to avillegasJul 7, 2009. 4:39 AM
NOTE: I haven't tested this driver with LEDs that are more than 3 Watts, so I don't know how hot the regulator will get. If you do it, then please let me know how well it works.
Artificial Intelligence (author) in reply to avillegasJul 7, 2009. 4:34 AM
Yes. You will need the LM350. The calculation is simple: 1.25V divided by 2.8 Amps = 0,44 ohm resistor. You'll need a power resistor. It should be at least 3.5 watts. Remember a heatsink for both the regulator and the LED.
avillegas in reply to Artificial IntelligenceJul 7, 2009. 7:23 AM
Dear A I: Thanks. It's done, I did use the P7 led + LM350t + R47@2W (didn't find .44Ω) and works well, all parts with heatsink. My test was made in my 12V car battery and heatsink turns hot fast. Today I'll test with two and three P7 leds for check how much heat is generated... Best Regards AV
Artificial Intelligence (author) in reply to avillegasJul 8, 2009. 2:40 AM
Thanks for letting me know, that it works. Actually when you connect 3 LEDs, the regulator shouldn't get as hot as when you use just one LED, because when you use one LED, the regulator has to drop more voltage, and that heats it up. I'm not sure if you will get the full brightness, if you connect 3 LEDs though, because the LM317 and LM350 has a voltage drop of 3 Volts. 3.6+3.6+3.6+3=13.8. It might work, but I'm not sure. Please let me know.
jimbowen says: May 31, 2009. 6:09 AM
Hi Artificial Intelligence I have 4 x 1.8a 20w leds do I use the same calculation to get resistor needed for LM350 ? Cheers Jimbo
Artificial Intelligence (author) in reply to jimbowenJun 1, 2009. 1:32 PM
Yes the LM350 works the exact same way as the LM317, so the calculation should be the same. Remember a heat sink for your LM350. Good luck with your project.
alah in reply to Artificial IntelligenceJul 3, 2009. 7:31 AM
Hi, everybody. If I use 12v to power the LM317 and use it as a current regulator, I should have 9 to 10v volts at the output due to the input-output voltage difference. Shouldn´t this destroy the LED?
Artificial Intelligence (author) in reply to alahJul 5, 2009. 1:41 PM
The LM317 maintains a constant current over the LED(s) and adjusts the voltage to a suitable level.
jimbowen in reply to Artificial IntelligenceJun 2, 2009. 12:53 AM
Thanks Artificial Intelligence Do You know if there is a way to get the led to work AC & DC without polarising ,.preferably a very small circuit ? Thanks again A.I Jim
Artificial Intelligence (author) in reply to jimbowenJun 2, 2009. 8:43 AM
You can use the classic bridge rectifier. No matter whether you feed it with AC or DC, it will always put out DC.
jimbowen in reply to Artificial IntelligenceJun 3, 2009. 10:24 AM
Sorry Me again! I have a 350w 13A Switched-mode power supply Do I stil need the LM350's ?? Cheers Jim
Artificial Intelligence (author) in reply to jimbowenJun 3, 2009. 12:53 PM
The amperage of your power supply has nothing to do with the current limitation by the LM350, so yes, you'll still need it. I hope this information is useful. Good luck with your project.
jimbowen in reply to Artificial IntelligenceJun 3, 2009. 5:15 AM
Thanks again A.I Very Helpfull ,.. muchas apritiated !!! : ) Jim
bribby.bribbs says: Jun 22, 2009. 1:52 PM
Thank you for the instructable. I've a question--is there a larger regulator anywhere that will be able to change the voltage used for an arc lamp (27 V, 7 A) for a 50 W LED?
Artificial Intelligence (author) in reply to bribby.bribbsJun 23, 2009. 6:29 AM
There probably is, but I don't know any.
alpesh.masarani says: Jun 21, 2009. 4:02 AM
how to connect in single 1w high power led in 12v power supply.
Artificial Intelligence (author) in reply to alpesh.masaraniJun 21, 2009. 10:01 AM
You should use a 3,9 ohm resistor.
moonlit24x7 says: Jun 18, 2009. 1:43 AM
Hello Aritifical Intelligence, I am currently using the ciruit described by you & it works perfectly fine, however I would like to know can I control the brightness of the LED using a microcontroller that has 3.3V at its output pins using the same circuit but with the addition of the microcontroller? I would really appreciate if you could provide me with some insights on how to achieve this. Regards, moonlit
Artificial Intelligence (author) in reply to moonlit24x7Jun 18, 2009. 11:32 AM
Take a look at this comment.
lond in reply to moonlit24x7Jun 18, 2009. 10:15 AM
You can use the info from my Instructable: IKEA Jonisk LED

// lond
kim gregersen says: May 13, 2009. 2:33 AM
Hello! First off all, great circuit, thanks for showing. I have a question. I want to build a string of 5 3w led. Could i use a sistem like yours? If sow, what power suply would i need?? and what components should i buy? The leds run at 700ma and have a forward Voltage of 3.5V~4.0V Sorry about all the questions but i'm not all that good in electronics! Thanks!!!
Artificial Intelligence (author) in reply to kim gregersenMay 13, 2009. 7:02 AM
If you've read step 2, you would have seen which resistor was the right one for 700mA LEDs, but anyways it's 1.8 ohms. You can calculate the required voltage of the circuit by adding the LEDs forward voltage together (in your case 3.6+3.6+3.6+3.6+3.6) and then add the voltage drop of the LM317 (always 3 Volts). That makes 21 volts. So you'll need to power the circuit with 21 or more volts. If that's too much for your application, you can connect some of the LEDs in parallel, but then you'll need another resistor. Good luck.
kim gregersen in reply to Artificial IntelligenceMay 13, 2009. 12:09 PM
thansk for your reply! I read step 2, but as i said before, i don't understand much about eletronics and i didn't know if i had to change resistor when i had more leds to it. (i realy, realy don't know much about electronics, i just like to trie new things). Once again, thanks for your help and for the great instructable.
Artificial Intelligence (author) in reply to kim gregersenMay 13, 2009. 2:11 PM
you don't have to change the resistor if the LEDs are connected in serial. Only if they're connected in parallel.
kim gregersen in reply to Artificial IntelligenceMay 14, 2009. 6:00 AM
ok, i see! So let's say that i have a 24 v 2 amp power supply, and i want to attach 2 strings of 5 led's each. I would have to change the resistor once i have 2 strings in parallel, right? am i seeing things right? Thanks for your time
Artificial Intelligence (author) in reply to kim gregersenMay 16, 2009. 2:01 PM
Yes. The resistor has to be 0,9 ohms in this case, because the circuit will draw 1400mA Are you from Denmark, because Kim Gregersen sounds like a danish name?
kim gregersen in reply to Artificial IntelligenceMay 17, 2009. 2:11 PM
I again. More or less! My father is Danish, he lives in Skive. I'm Portuguese, and i live in Portugal. Thats why the name lookes familiar to you. I have just one more question! Should i use one driver to the two strings or should i use one driver to each string? Thanks for your help.
Artificial Intelligence (author) in reply to kim gregersenMay 20, 2009. 1:49 PM
I'm sorry for being so slow to answer. You can do both, but if you're using a LM317 on two strings at once, you'll be pushing it's limits, so I suggest using one driver for each string. If you're using one driver for each string, the resistor should be 1.8 ohms.
lavert31 says: May 11, 2009. 7:17 AM
Hey, Say, if i want to connect this star to a simple 9V battery. do i need the regulator? can i just connect a 30 ohm resistor in order to maintain 350mA const? Etay
Artificial Intelligence (author) in reply to lavert31May 13, 2009. 7:03 AM
When using a 9 volt battery, I think a single resistor would be enough.
fallenspirit123 says: Oct 20, 2008. 7:43 PM
Great instructable but I am utterly confused what resistor should I use for this led? or similar?
Artificial Intelligence (author) in reply to fallenspirit123Oct 21, 2008. 5:28 AM
For a 1W LED like this, you'll need a 3.9 ohm resistor capable of at least 0.4W. I'll recommend a 1W though, cause it gets hot.
lux_manish in reply to Artificial IntelligenceMar 20, 2009. 12:22 PM
Can you please suggest any way out for reverse polarity protection,without wasting power?
lux_manish in reply to Artificial IntelligenceNov 12, 2008. 10:00 AM
I have come to know that drop out voltage for LM317 is 3volt.Kindly suggest me how can I use this ckt with a 6volt lead acid battery which typically gives around 5.6volt after a bit discharge.
Artificial Intelligence (author) in reply to lux_manishNov 12, 2008. 11:52 AM
Even though the manufacturers of the LM317 says, that it has drop out voltage of 3V, I've tested this circuit with a white 1W luxeon led, and there was almost difference in the light output compared to a 12V power supply. it might not be that efficient though. I hope you find this information useful.
lux_manish in reply to Artificial IntelligenceNov 13, 2008. 10:04 AM
True,Thanks ,but can suggest me some way out?
fallenspirit123 in reply to Artificial IntelligenceOct 22, 2008. 1:54 PM
ok thanks
angyalati says: Mar 16, 2009. 9:25 AM
I have a constant 12V/10A power supply, and I'd use 3 red 3w led. It have 2,5V forward voltage.
So, 3x2,5V(leds)+3v(lm317)=10,5V
What I have to do with the rest 1,5V voltage?
Forward resistor before the leds by the ohm-law?
Apologize me. My english is so weak.
lusorg says: Jan 26, 2009. 12:21 PM
Hi again, What type/model of transitor do you use ? I have try with 2n2222 and a need a 50 ohms resistor to pass full current, the strange thing is the max current is increase and not because the Ib current (very low). Thanks.
Artificial Intelligence (author) in reply to lusorgJan 27, 2009. 6:13 AM
I used a BC337 NPN transistor. What do you mean by "need a 50 ohms resistor to pass full current"?
lusorg says: Jan 20, 2009. 1:19 AM
Hi, where do you have connected the npn transistor to have pwm control ? Thanks a lot !!!
Artificial Intelligence (author) in reply to lusorgJan 20, 2009. 6:51 AM
Here is a little schematic that shows you how to do it. I'll post some more notes and a video soon (hopefully)
led.jpg
lusorg in reply to Artificial IntelligenceJan 20, 2009. 7:09 AM
Thanks :)
newtonn2 says: Jan 3, 2009. 7:09 AM
WOW this is really simple, do you think i could use this sistem to power a high power led of 20w, 13.5-15.0 Vcd, 1400mA-1800ma? like this one LED 20W
I made a LED projector, but i just bought the 20w Led.

Thank you for your intructable and your help.
Artificial Intelligence (author) in reply to newtonn2Jan 3, 2009. 8:46 AM
Yes. You should be able to do that, but then you should use the LM350, which is a 3 Amp version of the LM317. You'll also need a 2,25W resistor. Keep in mind that the input voltage should be at least 3V higher than the LED voltage. Good luck with your projector.
newtonn2 in reply to Artificial IntelligenceJan 3, 2009. 9:53 AM
Thanks a lot!
short_stack says: Dec 9, 2008. 12:18 AM
so what regulator and resistor would i need to connect a 3W led (white) connect to 12v dc and 700MA? i would appreciate it thanks
Artificial Intelligence (author) in reply to short_stackDec 10, 2008. 10:23 AM
You'll need a LM317 and a 1.8 ohm 1W resistor.
short_stack in reply to Artificial IntelligenceDec 20, 2008. 10:21 PM
so if i wanted to connect a few more more would i need a different resistor? and how many could i actually connect without it being to much. thanks in advanced
THURGOBOB says: Dec 10, 2008. 9:32 AM
Sorry if this has been covered. I want to power 3 x Cree 3 watt led's in series from say a 12 cell nimh which gives 14.4v (I can change this) Can you simplify the calculation to give the ma required, I think I need 700ma.
Artificial Intelligence (author) in reply to THURGOBOBDec 10, 2008. 10:19 AM
3W LEDs usually needs 700mA and when you connect them in series, they'll still only require 700mA, so you're right.
rappa says: Dec 7, 2008. 11:36 AM
(removed by author or community request)
rappa says: Nov 25, 2008. 1:33 PM
I bought 3 LM317T http://www.st.com/stonline/products/literature/ds/2154/lm217.pdf to drive a RGB LED with a Vf of 3.4V (green/blue) and about 2.8 (red). Now I'm wondering if my 5V power supply has enough voltage to drive it. I couldn't find the low dropout voltage on the datasheet.
Artificial Intelligence (author) in reply to rappaNov 26, 2008. 8:12 AM
It might work, but it will not be very efficient and bright. Don't you have a power supply with a higher output voltage?
rappa in reply to Artificial IntelligenceDec 1, 2008. 6:05 AM
I have a 9V power supply but it's only 650mA. The parts are coming today so I will at least try this circuit with the 5V supply. Now I just need to remove some excess solder as the blue and green contacts are bridged.
Artificial Intelligence (author) in reply to rappaDec 1, 2008. 12:24 PM
What is the wattage of the LED(s) you're using
rappa in reply to Artificial IntelligenceDec 1, 2008. 1:16 PM
It's a 3W RGB, with 1W/350mA per color.
codesuidae says: Mar 20, 2008. 9:26 PM
For more current you can add some beefy pass transistors and take the load off of the regulator. See http://www.rason.org/Projects/vreg/vreg.htm for details.
GordieGii in reply to codesuidaeNov 29, 2008. 8:48 PM
I agree with AI, if you want to add a few components I saw a circuit diagram for a switch-mode current regulator for driving LEDs with only half a dozen components. Much more efficient and cool (temperature, not social value). But _two_ components, that's cool (social value).
Artificial Intelligence (author) in reply to codesuidaeMar 21, 2008. 1:48 AM
Thanks for that, but if I've added these extra components, the circuit wouldn't have been the simplest LED driver.
rappa says: Nov 22, 2008. 4:00 PM
Will this circuit work with PWM, using a MOSFET and microcontroller?
Artificial Intelligence (author) in reply to rappaNov 25, 2008. 12:02 PM
Yes, I've just tried it and it works great. I just used a NPN transistor.
rappa in reply to Artificial IntelligenceNov 25, 2008. 1:27 PM
Thanks
lux_manish says: Nov 18, 2008. 7:04 AM
I want to use LM 1117 ,as my input voltage is 5volt and as drop out voltage of LM317 is 3volt approx.I want to have 3.6-3.8 volt for 1watt LED.Kindly guide me which resistances should be used.
silverback013 says: Nov 13, 2008. 7:57 AM
hi Ai, i'd like to run 15 luxeonrebel 2.5w(3.14fv) leds and 1 ledengin 10w(16.64fv) led on the same series circuit,....can you help re:power supply??drivers?resistors etc.appreciate your help
poppymann says: Oct 15, 2008. 10:05 AM
I have a stupid question: If I'm hooking up one Cree XR-E and want to use the LM 317 and a resistor, which LM317 do I use? There about a billion.
Artificial Intelligence (author) in reply to poppymannOct 20, 2008. 12:11 PM
As long as they say LM317, they're all the same.
totally_screwed in reply to Artificial IntelligenceNov 3, 2008. 10:12 PM
NOT all LM317 regulators are the same! LM317L can supply up to 100 mA LM317M can supply up to 500 mA LM317T can supply up to 1500 mA So for a 350 mA LED you need a regulator capable (and appropriately heatsinked) for at least 350 mA. The LM317L can't do that on its own!
Artificial Intelligence (author) in reply to totally_screwedNov 5, 2008. 9:30 AM
I know but I mean the different brands has different names for it. Some manufacturers call it LM317, some KA317, L317 and so on, but they're all the same.
totally_screwed says: Nov 3, 2008. 10:15 PM
LM317 / LM338 current calculator
Might be useful.
http://www.reuk.co.uk/LM317-Current-Calculator.htm
totally_screwed says: Nov 3, 2008. 10:00 PM
Any suggestions for suitable Low Drop Out versions of the LM317 regulator? In constant current mode the LM317 wastes a few volts which can be crucial in battery applications, where a larger battery has weight & cost implications. Thanks.
Jamez20 says: Aug 27, 2008. 6:50 PM
Yo dude! i tried your circuit using a 1 Watt Luxeon Led but when i checked the current flow to the LED the reading from my ammmeter was 40mA what happened to my circuit? i was hoping for a 350mA..when i simulated it in my program your circuit was correct...but the actual reading was different...please help...
totally_screwed in reply to Jamez20Nov 3, 2008. 9:56 PM
Jamez20, Assuming that you haven't made a mistake (reversed the LED, or miswired the regulator, used the wrong resistor, dry solder joint) and that all the parts are working properly (not destroyed the LED (by wiring it backwards or cooked it by running it without a heatsink) or the LM317), then one possibility is that the supply voltage isn't high enough. As I understand it you need to allow for the forward voltage plus the dropout of the regulator. Another possibility Your power supply needs to be capable of supplying sufficient current to drive the LED plus a bit, (Old dying battery perhaps, could supply volts but no current). Hope this helps.
Artificial Intelligence (author) in reply to Jamez20Aug 28, 2008. 7:01 AM
That sounds really weird. Are you sure you used a 3.9 ohm resistor and not a 39 ohm?
ounvme says: Oct 16, 2008. 7:20 PM
Could this be used and set to run at 2200ma and 3.7v? Iam not understanding how the voltage is regulated with constant current....
bravebaker says: Oct 3, 2008. 1:45 PM
This circuit was described a few years ago over at linear's led blog. Are you, by any chance, Rob?

Funny story, my local surplus store just printed out linear's instructions, including his diagram, and replaced "Luxeon" with their own name for them. They even put together a little kit with his instructions packed with the parts.
Artificial Intelligence (author) in reply to bravebakerOct 3, 2008. 2:10 PM
Nope, I'm not Rob, my name is Christian. And you're right, this circuit is very common, because it's so cheap and you don't need any special parts to make it.
gpf67 says: Sep 25, 2008. 7:12 AM
At the risk of sounding stupid(because I actually am), where do I buy resistors of this value. I am very new to this. thanks Ahead of time
Artificial Intelligence (author) in reply to gpf67Sep 25, 2008. 8:11 AM
You should be able to get them in every store that sells electronic parts. Or online. Radioshack and digikey. These resistors are very common.
moris_zen says: Apr 28, 2008. 2:19 PM
Nice first instructable. waiting for more...
Artificial Intelligence (author) in reply to moris_zenApr 29, 2008. 6:23 AM
There is more.
nerobro says: Apr 23, 2008. 2:59 PM
We should say something about why the LM317 gets hot when your voltage gets to high. Because this is constant current, it's output voltage will rise untill the the load passes the proper current. (this is to our advantage...) So we set the current to 300ma. Power disapation for a device is defined by it's voltage drop, times the current. Say our LED's end up dropping 4.2v. The current sense resistor will be dropping 1.25v. So total voltage drop for the load is 5.75v. IF we had a 6 volt input, we'd be looking at a .5v drop across the lm317, at 300ma. Or roughly 0.150w. The LM317 would hardly even get detectably warm. However with a 20 volt input, it's a different story entirely. The voltage drop across the lm317 would be 14.25v at 300ma. Or 4.25 watts. In other words you'd be looking at letting the magic blue smoke out. This is a lesson I learned while working with some 7805's and trying to power LED's. .... Here's hoping my explanation is correct. :-)
Artificial Intelligence (author) in reply to nerobroApr 24, 2008. 2:15 AM
OK, thank you very much, that was helpful.
computerwiz_222 says: Mar 19, 2008. 7:44 AM
I have never had the chance to experiment with a luxeon LED. Just how bright are they? Do they have the power to light up an entire small sized room? Are they a "spot" light or do they diffuse to the entire room? Does the LED get very hot?
callmeshane in reply to computerwiz_222Mar 21, 2008. 5:55 PM
I had a 5W Luxeon LED in a torch, on my bike helmet - and it was SO intense, that the light looks like a solid rod of white going into the night.... Much like a light saber...... A few years back... but VERY VERY bright and strong. I am led to believe that the more moderna ones are much brightera again.
milo0is0hot0 in reply to callmeshaneApr 3, 2008. 7:19 AM
where did you get a 5w luxeon from? i had difficulty sourcing a 3w luxeon.
Artificial Intelligence (author) in reply to milo0is0hot0Apr 3, 2008. 7:50 AM
The reason why you can't find 5W Luxeons anymore, is that they're out. Use Luxeon K2s instead. The K2 are also more effective. The white K2 has a light output of 120 lumens, and a power consumption of 3,42W, while the white 5W Luxeon only has a light output of 105 Lumens and has a power consumption of 4,79W.
Scott_Tx in reply to computerwiz_222Mar 20, 2008. 1:47 PM
the old luxeon stars are out of date, now we have luxeon rebels/ k2s and Cree Q5/R2s that use 3 watts and put out 100+ lumens. Very bright LEDs.
Artificial Intelligence (author) in reply to Scott_TxMar 21, 2008. 1:44 AM
you're right, but this circuit can also drive Rebels and K2s.
Artificial Intelligence (author) in reply to Artificial IntelligenceMar 21, 2008. 1:51 AM
Plus Cree and Seoul P4s
Artificial Intelligence (author) in reply to computerwiz_222Mar 19, 2008. 11:42 AM
here is a picture of the led, in my room. it's hard to see, but it actually lights up the wall
IMGP5882.JPG
Artificial Intelligence (author) in reply to computerwiz_222Mar 19, 2008. 11:31 AM
The 1W LED, that I used is on 45 lumens, and 250000mcd, but the 3W types is on 65 lumens. The one, that I used could weakly light up my room. The LEDs diffuse to the intire room, but you can buy a lense for it, so that it becomes a bright spot. the LEDs gets hot, but you'll only need a small heatsink. If you'll want to light up your room I'll recommend the Luxeon K2. Why didnt you had the chance to experiment with a luxeon LED.
tfj in reply to Artificial IntelligenceSep 14, 2010. 4:45 AM
I need help.
I want to make a diving lamp. I have a battery pack 14.4 v and 9amp.
I would like to use as many sst-50 high power LEDs as possible for better light.
Can I use the simple driver with LM1084IT-ADJ since each sst-50 uses 5 amp.
Driver must be connected in series or parallel?
computerwiz_222 in reply to Artificial IntelligenceMar 19, 2008. 12:31 PM
i work in an electronics shop and none of their suppliers carry them yet
Artificial Intelligence (author) in reply to computerwiz_222Mar 19, 2008. 12:37 PM
Can't you just buy them from online shop, like Led supply
computerwiz_222 in reply to Artificial IntelligenceMar 19, 2008. 1:09 PM
Oh ya, but I don't actually need them for a specific project, to make an order just to monkey around... I just don't need them that bad. Thanks for the site. This store is pretty good though, they have nearly the entire NTE semiconductors, capacitors, resistors, leds, connectors and everything!
jimmixan says: Mar 22, 2008. 12:04 PM
This is a great low cost circuit. It is a very simple to use and make into a constant current source. I sell the LM317 and the 3.9 Ohm Resistor to make a slightly under 350 milliamp output. Perfect for driving 1 watt Luxeons or other Leds that needs this current level. Input voltage can be from inexpensive 9 volt or 12 volt wall wart transformers. With a 12 volt input three Luxeons can be driven in series with no problem. Contact me at Solarlightandpower.com check out the new Luxeon circuit boards that have 3 Luxeon Rebels premounted for your use. Insanely bright!!!
BlindTreeFrog says: Mar 19, 2008. 6:03 PM
I think your math in step 2 is wrong, but it's been a while.

Current draw of the LED == 350mA
Power consumption would be in watts

Current isn't going to drop across the LED, so no matter how many you have in series that resitor won't change. The voltage requirements will, but the resistor won't.

Now, on the chance that I'm wrong and it really is power not current, the equation isn't V=IR like you are using. The equation is P=IIR (that's current squared). and P=VVR (voltage squared)
Artificial Intelligence (author) in reply to BlindTreeFrogMar 20, 2008. 12:31 AM
You got to read it again, because I didn't wrote that that the resistor should change, when you drive more LEDs
BlindTreeFrog in reply to Artificial IntelligenceMar 20, 2008. 11:00 AM
You are correct, you didn't. I missed that you switched to a 3W LED rather then multiple LED's. That said, I needed to read it 6 times to figure out what you were talking about.
Artificial Intelligence (author) in reply to BlindTreeFrogMar 20, 2008. 11:16 AM
6 times? It can't be so hard to understand.
BlindTreeFrog in reply to Artificial IntelligenceMar 20, 2008. 12:44 PM
Well you keep using power consumption when you mean current. You randomly state that 1W resistors are needed without ever actually showing how to calculate what the power dissapation across the resistors actually is (.4 and .8 Watts respectively). Mostly though, your wording is throwing me.
GorillazMiko says: Mar 19, 2008. 4:57 PM
Wow, great job! Done very nicely, and everything, I like your image on your profile, but enough of that, this Instructable is SO cool! Nice job! +1 rating. (added to favorites)
gmjhowe says: Mar 19, 2008. 12:20 PM
Nice! i think the massive heatsink adds to the coolness
backcountry says: Mar 19, 2008. 10:14 AM
Nice trick, great instructable. Assuming you're powering this by battery, how does it fail when the batteries run out and the voltage drops? Does the LEDjust turn off, or does it fade slowly?
Artificial Intelligence (author) in reply to backcountryMar 19, 2008. 11:19 AM
The LED fades down, and when the voltage is really low (3V), the LED begins to flash very fast, I don't know why
Artificial Intelligence (author) in reply to backcountryMar 19, 2008. 11:07 AM
I havent tried it yet, but I'll find out. I used a 12V power supply
GitarGr8 says: Mar 19, 2008. 9:11 AM
This is really cool! Great first instructable. The only thing I think missing was what you used to power the current regulator. I would imagine that you would hook it up to a voltage regulator, but it would be nice to show.
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