# Burned LED, WTF?



## MikePL (Nov 2, 2007)

Sooo... having a lot of LM317 chips lying around, I've decided to use them for fun. I have found this website that describes the LM317.

http://www.reuk.co.uk/Using-The-LM317T-With-LED-Lighting.htm

Read from the photo of the chip onwards. Anything above it is not important.

From what I've gathered, the LM317 can be used as a constant current regulator. All I have to do is connect the 'out' pin with the 'adj' pin using the proper resistor. Well, I did it per the instructions. I wanted to use a 2ohm resistor to approach 600-700mA, but I couldn't find one so I used a 5ohm resistor. According to the calculation provided on the site, the thing looks like this R=volts/amperes. So R=1,25/0.25 R=5. A 5ohm resistor should give me 250mA of current. Read on...

As far as I understand all the LM317 datasheets, the pinout is the following (assuming the chip lies on it's back, like in the pic):
- left pin is adjustment
- middle pin is out (as well as metal backing)
- right pin is input

So according to the diagram I should short the adj and out pins with my 5ohm resistor and I did it. Then I connected the LED to the LM317 and connected everything to 12V. The LED burned out within 0.1 of a second. Why? It normally operates at 750mA and here I was supposed to have only 250mA. What did I connect wrong?

Afterwards I measured the voltage and the source has 11.9V, while the LM317 output has 11.2V. What's going on?


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## FRANKVZ (Nov 2, 2007)

It regulates current, but not voltage. The way you set it up the led would have seen 250 ma at almost 11 volts! You would need to run several leds in series or a lower voltage power supply.


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## rhuck60 (Nov 2, 2007)

MikePL said:


> Sooo... having a lot of LM317 chips lying around, I've decided to use them for fun. I have found this website that describes the LM317.
> 
> http://www.reuk.co.uk/Using-The-LM317T-With-LED-Lighting.htm
> 
> ...


 
You burned out your LED by applying about 3 times the voltage it was meant to run on.

what you need to do is reconfigure the LM317 as a voltage regulator by using 2 resistors. Or 1 resistor and 1 pot. You need a constant voltage source, not constant current. The current will be determined by the output voltage and the resistance of the LED itself.

If you cant find that circuit I will look it up and post details for you later.

Edit: BTW I looked in the link you provided and you did overlook the fact that the design was to power 3 LED's in series. Check back and you will see it.


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## tonycollinet (Nov 2, 2007)

The info from the previous two responses is not quite correct.

The LM317 WILL regulate current. When it regulates the current to 250mA, the voltage accross the LED will be defined by the LED V/I characteristic, and will be in the region of 3V (assuming you are using somthing like a luxeon, or cree etc).

The rest of the voltage (around 9V) will be dropped accross the regulator - which will now be dissipating about (9*0.25) 2.25W, and WILL need a heatsink.

I suspect what you have done is connect the LED to the output pin of the regulator. However, when you run the regulator in this current source mode, the load (LED) should be connected to the ADJ pin 

(see the diagram about half way down the page - Voltage input, through LM317, through reistor to output (output of the circuit, not of the LM317)

The output of the circuit ("output" of the resistor) is also connected to the ADJ pin.


You can test this, when you have the circuit connected properly, by simply measuring the output current with a meter.

One thing to bear in mind with this type of current limit. It is VERY inefficient. In your case, with 12V input, and around 3V LED, only 1/4 of your input power will be going to the LED. The rest will be dissipated in the regulator. You can improve this by driving more leds in series (so that the regulator needs to drop fewer volts) or by reducing the input voltage so that it is only a couple of volts higher than the LED voltage. (Ideally whatever the dropout voltage of the regulator is, above the LED voltage - and if possible use a low dropout voltage regulator)

Most of the current regulators sold for use with high power leds are of the "switching" variety (switching buck, or switching boost regulators). This means they have a very high efficiency (of the order of 90%) regardless of the input or load voltage.


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## evan9162 (Nov 2, 2007)

rhuck60 and frankvz are wrong. 

tonycollinet is almost certianly correct. You connected the LED to the output pin and not to the other end of the resistor (which happens to be connected to the ADJ pin). If this is the case, you just didn't follow the circuit diagram on that page.


your configuration should have looked something like this:







The LED is not connected directly to the output pin, it's connected to the other side of the resistor.

BTW - you need to make sure your 5 ohm resistor is capable of dissipating the appropriate power. In this setup, it will be dissipating 0.3W. You need at least a 1/2W resistor. If you use a normal 1/4W resistor, it will likely overheat and burn up.

When messing with circuits you are unfamiliar with, or diving into electronics, it's best to use something besides an expensive LED as a test load when first powering up. You can simulate an LED with 4 1N4001 diodes in series, which cost like $0.10 each. Use those as a test load, and verify all the parameters (current and voltage, and expected voltage drops across appropriate components) before connecting an LED to it. That should save some heartache and $.


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## tonycollinet (Nov 2, 2007)

evan9162 said:


> You can simulate an LED with 4 1N4001 diodes in series, which cost like $0.10 each. Use those as a test load, and verify all the parameters (current and voltage, and expected voltage drops across appropriate components) before connecting an LED to it. That should save some heartache and $.


 
Good advice for anyone building LED lighting - even if you buy a regulator off the shelf.


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## MikePL (Nov 2, 2007)

Yup guys, that's right. I was unnecessarily messing around with it in the middle of the night. Probably was too sleepy to see the diagram correctly. Now I see everything clearly. I've connected things the right way and it works.

I measure 11.3V on the adj pin, which proves that LEDs are definitely curent driven, no matter the voltage.

The resistor and the LED are cold but the chip does heat up. I know that it's inefficient but I need a simple setup that will give constant current as I need it to test various LEDs at various currents. I have 20 pieces of the LM317 so I have lots of options.

I am simply getting a little bit bored while waiting for 15 days for my order from Kaidomain. I didn't get a confirmation e-mail, I didn't get a tracking number. I only got a message from PayPal that my payment has been processed. There was also a tracking number there, but it doesn' work when typed into Kaidomain's tracking number field. 
Is this normal for them? This lack of any confirmation...

I wrote to them a week ago and also two days ago but neither time got a reply.
Was I hosed?


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## tonycollinet (Nov 2, 2007)

How are you measuring the 11.3V? (where are you connecting the meter?) You shouldn't have 11.3V accross the LED.


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## Handlobraesing (Nov 2, 2007)

If you're doing 0.7A and you're running one Luxeon from 12v, you need to consider thermal dissipation of the regulator.

~4V for Luxeon, 8v dropped. 
Dissipation = 5.6W in regulator + series resistor.
Regulator efficiency = 33%

You're better off to do 12v input with two LEDs in series and 350mA current to get similar total lumen output from efficiency point of view. The LEDs are more efficient at 350mA and you also raise the regulator efficiency to 66%.


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## laserblue (Nov 27, 2007)

sometimes these LM317 are pain in the neck.


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