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LED burnout Answered

I've have an LED problem that I have done extensive reading and troubleshooting on but I can't seem to solve it. I'm running 4 super bright white LEDs (NTE30071) in parallel from a 24v 600mA DC wall wart. The LEDs are 4V, 25mA. I need that wall wart to also run a solenoid. When a button is pressed a relay powers the lights and the solenoid for a few seconds. The problem is my LEDs keep dying. At first I had a 10K ohms 1/4W resistor on each and I lost 3 LEDs. After more study and calculation just replaced them with 1K ohms 1/2W resistors. After a few more hours I lost another. When I removed the 10K ohms resistors they read 1K ohms. And I just tested the one on the 1K ohms resistor that went out and it reads 200 ohms. I know there is something that I just don't get; something basic that highlights my lack of any electronics experience. I appreciate any help you can give, this is just is just wearing me down.

14 Replies

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robotninjasquid (author)2007-09-04

Two weeks and still burning! Looks like the diode did the trick. Thanks everyone!

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rc jedi (author)2007-08-15

this is the basis of how "hotshot" cattle prods operate. your led's are getting prodded!

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rc jedi (author)2007-08-15

yes the switchin on and off os the relay will make a voltage spike. It is a project from electronics school to put a 70 v neon bulb across the leads of the coil and switch the relay on and off. the 15v supplied cannot light the neon. But the neon flashes briefly as the relay is switched off. a magnetic field that closes the relay collapses and causes a voltage spike in excess of 70 volts. more than enough to kill voltage sensitive semiconductors. like LED's. patric pending is right. you need diodes to shunt the excess voltage spike.

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Patrick Pending (author)2007-08-13

I am fairly sure this has nothing to do with the resistors or the LEDs. This is most likely to be back emf generated as the solenoid is switched off. The back emf generated as an Inductor (e.g., solenoid or relay) is switched off causes a brief reversed voltage spike which can be much higher than the voltage that originally induced it. This might give you say 90 volts reverse bias on a diode that is rated for 4 Vr. The cure is simple (and common practice) you place a reverse biased silicon diode across the terminals of the solenoid. The diode will not be conducting when the solenoid is switched on. When you switch off the solenoid the diode will conduct and prevent a negative going voltage spike from developing. Cheers, Pat. Pending

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user

Makes sense. I knew that about relays, but I didn't give it much though when installing the solenoid. Would a 1N4001 diode be ok?

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user

An IN4001 would be fine, it's not critical any general purpose diode will do. Just remember to reverse bias it so that it is not conducting when the solenoid is powered. Let us know how you get on. Cheers, Pat Pending

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NachoMahma (author)2007-08-09

. The main reason LEDs burn out is too much current/heat. Since the LEDs are "super bright," do they require a heatsink?

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Brennn10 (author)NachoMahma2007-08-14

That is what I was thinking. Are your connections secured properly?

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robotninjasquid (author)2007-08-10

Thanks to everyone for their excellent comments. Actually every LED has its own resistor; similar to option 2 below but 4 resistors. I did notice that I had connected the resistor to the cathode not the anode of the LEDs. I've switched th last one to see how long it will last. Do you think that had an effect on LED life? I'll rework it to match option 1 below. Should I stick with the 1/2W resistor?

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matseng (author)robotninjasquid2007-08-10

Moving the resistor from the cathode or the anode will not affect anything.

Using option 2 with four separate resistors is a perfectly valid option. Use 1.2 K ohm 1/2 watt resistors. You will then have 17.1 mA through the leds. The power dissipation in the resistors will be approx 0.4 watt each so a 1/2 watt resistor is within range.

If you are certain that your power supply is not much above the stated 24 volt you can reduce the resistors to 1 K or even 820 ohm. And still be within the max 25 mA spec of the leds.

Currently the only other thing I can think of that could cause your problem with the leds is that your power supply is not DC but AC. Your leds can, according to the Data Sheet, only withstand 4 vols in the reverse direction. And if you and running them on AC they'll get 24 volts backwards 50 or 60 times each second.

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robotninjasquid (author)matseng2007-08-10

I checked the power supply and it is definately DC. It reads 24.6v when the LEDs and solenoid are on and 30v when the relay is off (which shouldn't be reaching the LEDs). I've still got one burning. I plan on replacing the whole lot with fresh LEDs using these solutions.

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matseng (author)2007-08-10

How have you connected the leds and the resistor? You say that the leds are connected in parallel - are the resistor also parallel (wrong) or is it in series (right) with the leds? In the attached picture there are two correct ways of connecting it. The first is Since you have a wallwart with a very high output voltage I'd recommend the first variant where the leds are dropping more of the voltage. Connect your wallwart to the A and B points. For the first variant using a 24 volt power supply, four leds in series (each having a voltage drop of 3.4 volts) and a target current of 20-25 mA you should use a resistor of 470 ohm. The second variant requires a 1 Kohm resistor, but I really can't recommend using it since the currents through the leds will not be balanced.

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westfw (author)2007-08-10

Are you sure you're reading your resistors right? Your 24 wall wart may be putting out a lot more than that with a light load (just the leds), but the 10k resistors should have been plenty to prevent burnout anyway. Some of the cheap Chinese import white LEDs have pretty awful lifetimes, but I'd think NTE semiconductors would do better...

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lemonie (author)2007-08-09

Well, 10K should limit you to 2.4 mA, and 1 K to 24 Not enought to blow an LED(?) but the rest sounds wierd... L

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