According to page-1268-69 of Halliday, Walker & Resnick's Fundamentals of Physics (10th edition),

To emit enough light to be useful as an LED, the material must have a suitably large number of electron-hole transitions....What we need is a semiconductor material with a very large number of electrons in the conduction band and a correspondingly large number of holes in the valence band. A device with this property can be fabricated by placing a strong forward bias on a heavily doped p-n junction, as in Fig. 41-16. In such an arrangement the current I through the device serves to inject electrons into the n-type material and to inject holes into the p-type material. If the doping is heavy enough and the current is great enough, the depletion zone can become very narrow, perhaps only a few micrometers wide. The result is a great number density of electrons in the n-type material facing a correspondingly great number density of holes in the p-type material, across the narrow depletion zone. With such great number of densities so near each other, many electron-hole combinations occur, causing light to be emitted from that zone.

Now, if electron-hole pairs are ceasing to exist due to recombination and resulting in a greater number of "gridlocked" electrons and light, who will continue to conduct electricity? Won't current flow stop after a while due to the absence of electron-hole pairs?

  • 2
    $\begingroup$ would not the current replace the electrons and holes due to the lattice structure?T the current must be coming from outside the LED lattice, a battery or power supply. $\endgroup$
    – anna v
    Jun 2, 2021 at 10:32

2 Answers 2


You won't run out of electrons and holes. There are two main processes through which you will get more.

Thermal energy will naturally generate electron/hole pairs in your semiconductor. That's where they come from in a semiconductor at thermal equilibrium. However, this process is slower than we would like for an LED, or any diode. Fortunately, we can help it along by injecting carriers from outside the p-n junction. We can imagine a simple p-n junction with metal contacts set back a bit from the p-n junction. You can directly inject electrons into the n side, replenishing the lost electrons. At the p side you will have a greatly increased electron/hole generation rate allowing you to effectively inject holes into the p-side (by generating an electron/hole pair and removing the electron into the metal).

  • $\begingroup$ Is my answer (below this comment) correct? $\endgroup$ Jun 3, 2021 at 4:10
  • $\begingroup$ Are electrons/holes being injected with the help of the battery? $\endgroup$ Jun 3, 2021 at 4:15
  • $\begingroup$ @user545735 Yes, the battery or whatever else is sourcing the current. Your answer uses nonstandard terminology so I cant really tell what you meant and therefore cant determine if it is correct or not. $\endgroup$
    – Matt
    Jun 3, 2021 at 10:37

No, the current won't stop flowing as the battery maintains constant electromotive force. Some free electrons will lose their kinetic energy and become gridlocked. You can view this phenomenon, that is, free electrons losing their kinetic energy, as free electrons losing kinetic energy to a resistance, say a light bulb.


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