Why are 'modern' high-efficiency LEDs easier to damage than

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Why are 'modern' high-efficiency LEDs easier to damage than

Postby mkioiuq » Dec 14, 2016 4:03 am

I have a few 'old time' LEDs from the late 80s and early 90s. The red and green 5mm LEDs (amber was a rarity, blue was 'impossible' back then). Not being very smart, I used to test them with a 9V battery without any resistor and, surprisingly enough, they always outlived the experience.

Fast forward to the third millennium: I bought several dozens transparent 'high-efficiency' (or should I call them 'high brightness', hard to tell without a datasheet) LEDs on the Web. They are superbright, but the one time I tried to "Oh, here's a 9V battery: let's see what color this is" one of those, they almost instantly died after a faint flash that told me "I was blue, you #@@#!"

(Like this: https://www.youtube.com/watch?v=7IoyYj6BJlc )

Now, it is clear to me that a resistor is required to limit the current, but my question is about what exactly kills the LED, or put in another way: why do 'old technology' LEDs survive?

Is it related to the fact that 'old' LEDs exploited recombination between conduction and valence band of a sturdy PN junction, while 'new' LEDs are based on more exotic etherostructures that create quantum wells? Or is it because the 'old' manufacturing process used bigger dies, or thicker bonding wires, or materials that were so lossy that they provided enough series resistance by themselves?

I think I owe an answer to my two dead blue LEDs. :banana_yay:

EDIT: just re-did the experience with an 'old' red LED: I can leave it on for seconds without a problem using the same battery that zapped the 'new' LED.

New EDIT : While I can let the LEDs light up for a few seconds, I managed to blow one up when trying to measure the current. So, they are harder to damage, but not immortal after all. Tried three - four more old LEDs and I can confirm that for at least one second they survive (appearently) unscathed. New LEDs die almost instantly. I will try later to measure the current in a more controlled setup, possibly with short pulses.

I love the smell of burning GaAs in the morning.
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Re: Why are 'modern' high-efficiency LEDs easier to damage t

Postby NZcaver » Dec 16, 2016 1:21 pm

mkioiuq wrote:...Or is it because the 'old' manufacturing process used bigger dies, or thicker bonding wires, or materials that were so lossy that they provided enough series resistance by themselves?

I think this. But it's just a wild guess.

Interesting question/observation!
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Re: Why are 'modern' high-efficiency LEDs easier to damage t

Postby batrotter » Dec 16, 2016 10:49 pm

"Is it related to the fact that 'old' LEDs exploited recombination between conduction and valence band of a sturdy PN junction, while 'new' LEDs are based on more exotic etherostructures that create quantum wells? Or is it because the 'old' manufacturing process used bigger dies, or thicker bonding wires, or materials that were so lossy that they provided enough series resistance by themselves?"

Dude, I'm not really sure what you said there? This is a caving forum. Maybe some other electronic geeks will chime in. Or is this just a bit of sarcasm?
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Re: Why are 'modern' high-efficiency LEDs easier to damage t

Postby caver.adam » Jan 4, 2017 3:10 pm

I suspect it is a matter of material and heat. More efficient LEDs need less material to get the same (or higher) light output. So when you put too much current through them (by over-volting them without a series resistor) they heat up to burnout levels faster. In addition to dissipating less heat internally due to less metal being inside the LED, the PN junctions are likely smaller and more susceptible to overheating. My gut reaction guess anyway.
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