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I read somewhere that LEDs could self limit current up to a certain point and had a few lying around in my box so I decided to test this statement. Its true for roughly the voltage drop across the LED. But the question comes is what happened when I applied 20V to the LED? It made a loud popping sound. What happened to the internals of the LED?

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    \$\begingroup\$ Um - They popped - blown, gone, killed.... \$\endgroup\$ – Linker3000 May 24 '11 at 21:50
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    \$\begingroup\$ I also noticed that my white LEDs went from white to blue and then to purple before they popped. @Linker3000 Define "popped", "blown", "gone", "killed". They are all different words with different meanings and each of the words has large number of meanings which are not related to process that occurred here. To me it looks like the question is about the physical process of LED destruction, which is very important to understand so that its early stages could be detected in a circuit that appears to be running fine. \$\endgroup\$ – AndrejaKo May 24 '11 at 21:55
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    \$\begingroup\$ You turned it into a DED, a Dark Emitting Diode. Although you might have gotten lucky and turned it into a SED, or Smoke Emitting Diode. \$\endgroup\$ – user3624 May 25 '11 at 2:49
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    \$\begingroup\$ You released the Magic Smoke! \$\endgroup\$ – Connor Wolf May 26 '11 at 23:32
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    \$\begingroup\$ What happened was approximately the same thing that happened to a neon bulb I put accross 230 Vac as a kid at age 8, without the > 100k series resistor. It made a bang. Also, your parents might have been running towards your work bench, checking if you were still o.k. \$\endgroup\$ – zebonaut Jun 3 '11 at 22:40
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Increasing heat from power dissipation causes a failure of the LED die.

The change in colour, e.g. red and green LEDs going yellow at high currents, is probably because the die is actually glowing hot, i.e. near failure. Note that the red LED has a fall in wavelength, but the green one has an increase in wavelength.

White LEDs going blue could be explained by the yellow-emitting phosphor in the LED being less effective at high currents. White LEDs are often constructed from a blue LED coated with a special phosphor which emits yellow light when blue light hits it creating a fairly even white light. So perhaps you are seeing more blue than the yellow phosphor can convert.

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    \$\begingroup\$ Maybe the phosphor already died at that point. The orange/red glow is probably from heat alone. Once upon a time I had an EPROM glow red when I accidentally reversed the polarity of the power supply. \$\endgroup\$ – starblue May 25 '11 at 19:28
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An LED is a Light Emitting Diode. The key part of that name is "Diode." An LED is a diode.

Diodes do not limit forward current very well. The extremely steep current/voltage curve (an exponential curve) is probably their second most important functional characteristic which results in sales of diodes.

When you put a forward voltage on a diode (an LED, BE junction of a BJT, or whatever), it's current DOUBLES with each incremental 26mV of voltage. So if you applied 0.7V (700mV) and got 50mA, then you should get about 100mA at 726mV, 200mA at 752mV, 400mA at 778mV, and so on. So what would you expect at 20000mV? The theoretical answer is about 7 times 10 to the 222nd power amps. That is a '7' with about 222 zeros after it. But your home's circuit breaker (thankfully) limits current draw to something less than Quadra-Bazillions of times the total of all power plants on planet Earth. Your LED draws maybe 20 amps for a few microseconds, turning a very small volume inside it about as hot as the Sun, and then it is all over.

If you do this with larger parts, the shrapnel can kill you. An electrian, at a factory I worked at, had the misfortune of crossing two phases of industrial strength AC with his screwdriver. As far as could be figured out, he was not electricuted: the plastic screwdriver handle protected him from that. However, the short instantly vaporized the metal in the screwdriver. The resulting explosion killed him.

So provide external current limiting in line with your LED, or... wear safety goggles.

EDIT: I got carried away with the point I was trying to make and erroneously said current doubles with each 25 mV across the diode junction. The actual factor is 'e', where 'e' = the base of the natural logarithm = about 2.7. So the current would increase by a factor of 2.7, not 2, for each 25mV. The damage to the device looks pretty much the same for large voltages, though....

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  • \$\begingroup\$ A friend of a friend was working in a busbar cabinet, bolting down something with a spanner when he was startled by a spider - so he threw the spanner at it! As you said, the tool vaporized, but the resulting induction field twisted half the box off the wall. The engineer survived. No word on the spider. \$\endgroup\$ – Linker3000 May 27 '11 at 13:48
  • \$\begingroup\$ You are describing the Wikipedia topic Shockley diode equation. \$\endgroup\$ – Richard Chambers Jan 7 '18 at 3:42
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The weakest electrical portion of the LED fused. That part would vary based on LED construction. I've had this happen and blow off the tip of a 3mm LED with surprising energy, when I was young and messing with such questions. It hit the ceiling pretty hard.

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