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Let us suppose we have an LED with the allowed voltage range of 1.5V to 4.5V and we gave it somewhat 12V. Now we know that the LED will instantly stop forever. But I want to know what will happen to the material (the semiconductor, etc) when this happens.

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what will happen to the material?

The LED may experience an incident of Electrical Overstress (EOS). This depends upon if the supply can source more current than the LED's maximum allowed current. And if the LED can tolerate the supply's maximum current whether the supply's voltage output will reduced itself to the LED's forward voltage. Over current and over voltage (i.e. over driven) will result in an EOS event.

EOS damage in LEDs can be catastrophic in that the LED is permanently non-functional immediately after the EOS event; alternatively there may be just partial damage whereby a significant degradation in performance or the complete failure of the LED only occurs later on. Partial EOS damage in LEDs might manifest itself as reduced light output, poor thermal performance, and/or a shorter service life.

When an LED is overdriven the failures here include the thermal overstressing of the semiconductor or a fused wire bond. Notice in the photo below the stress is greatest where the bond wire is attached to the semiconductor material.

Example of a fused wire bond
In this experiment, an LED with a 250 mA max was driven with 1000 mA, four times the data sheet maximum current. Failure occurred in about 10–20 seconds.
enter image description here


X-Ray of the above LED failure mode

enter image description here

In the image below the same LEDs were driven with a single pulse of 3000 mA for 300 ms. Due to the higher power load, the bond wire melted.

enter image description here

Source: The Basic Principles of Electrical Overstress (EOS) App Note

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An LED can be modeled with a very complex circuit, but for the purposes of answering this question, it can be modeled as a voltage drop with a series resistor in parallel. The allowed voltage range is determined by keeping this series resistor in mind. More specifically, the power dissipated across this resistor should be less than what the physical package of the LED can dissipate into the environment. When the supply voltage is more than the rated value, the resistor generates more heat than what the package can dissipate into the environment thus heating the LED gradually. Once a particular temperature is crossed, the junction can burn down in several different ways depending on the construction of the junction.

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  • \$\begingroup\$ TL;DR: It melts. \$\endgroup\$ – Dampmaskin Nov 21 '18 at 13:00
  • \$\begingroup\$ ^that^ or the bondwire melts like in a fuse, or the semiconductor material evaporates, or a combination of all that is mentioned here. \$\endgroup\$ – Bimpelrekkie Nov 21 '18 at 13:42
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    \$\begingroup\$ The magic smoke (that makes it work) leaks out. \$\endgroup\$ – Edgar Brown Nov 21 '18 at 18:00
  • \$\begingroup\$ Ohh. Well, that clears my doubt. Thanks everyone! 😁 \$\endgroup\$ – Ishaan Pathak Nov 22 '18 at 13:05
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    \$\begingroup\$ An LED cannot be modeled as a resistor. An LED can be damaged from over voltage and a resistor cannot. You do not model a voltage drop, its the result of the modeling of a resistance. An LED could be modeled as a variable resistor where the resistance is a function of the current flowing through it. \$\endgroup\$ – Misunderstood Nov 22 '18 at 19:00

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