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I’m aware that LEDs are doped with impurities on each p-type and n-type material to increase the charge carriers of each semiconductor.

My curiosity now is, how does each side continue to have extra charge carriers after a certain amount of time?

In other words, I imagine each side wants to be balanced and neutral and when power is applied they recombine at the depletion zone where photons are released. So, why does it keep happening and not ever stop? Is it the power source that keeps it going or am I missing something?

My confusion could be stemming from how a battery works and once depleted that’s it.

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    \$\begingroup\$ A battery doesn't run out of electrons, it runs out of energy. An LED keeps getting energy put into it which it converts to light. \$\endgroup\$ – Finbarr Apr 3 '18 at 15:47
  • \$\begingroup\$ The wires that connect to the LED are chock full of electrons. \$\endgroup\$ – The Photon Apr 3 '18 at 15:49
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    \$\begingroup\$ N- and P-type semiconductor is still valently neutral at rest (assuming no external force); there are no "extra" electrons. \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 3 '18 at 16:12
  • \$\begingroup\$ @IgnacioVazquez-Abrams Even though each is doped to have more charge carriers? \$\endgroup\$ – ohmmy Apr 3 '18 at 16:20
  • \$\begingroup\$ The charge carriers still belong to atoms. \$\endgroup\$ – Ignacio Vazquez-Abrams Apr 3 '18 at 16:21
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I find the whole concept of extra electrons and holes is sort of a poor description of doped semiconductors and leads to questions like yours.

In reality there are no extra or missing electrons (Holes).

What actually happens is the doping process modifies the semiconductor material, which by itself is an insulator, such that is becomes conductive either in the conduction band with free electrons, or in the valance band.. with what we term holes.

Conduction band and valance band are in fact two different types of current carrying methods.

An N-type semiconductor is conductive because the doping material will easily release electrons. A P-Type is conductive because the doping material will easily catch electrons.

NOTE: There are no "extras", just a tendency to expel or attract electrons.

You can think of a PN junction as a team of base-ball pitchers on one side and a team of catchers on the other. One side wants to throw, the other side wants to catch.

Of course once a pitcher has thrown his ball, he needs a replacement and will quickly catch another from the air. The catcher on the other hand wants to pass on that extra ball.

Why don't the pitchers run out of balls when biased?

As you surmised, because the applied voltage is continually supplying fresh electrons (balls) at the cathode, the process is never ending.

Batteries are a different beast entirely and the charges are produced by half-reactions at each terminal which provide the voltage.

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    \$\begingroup\$ Thank you! The doping explanation helped a lot as every source I’ve read does say it adds/takes away electrons. Is there a reason no one explains it the way your answer does? Would have eliminated my confusion \$\endgroup\$ – ohmmy Apr 3 '18 at 18:13
  • \$\begingroup\$ @ohmmy the extras explanation it's so entrenched into the classic street description it's hard to get away from. \$\endgroup\$ – Trevor_G Apr 3 '18 at 18:18
  • \$\begingroup\$ pitcher with fresh electrons. nice analogy. The girls would like it. \$\endgroup\$ – Sunnyskyguy EE75 Apr 3 '18 at 19:50
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A battery is more like a charged capacitor with a high Er dielectric with a chemical cell base potential. An LED is a PN semiconductor that also has dielectric properties but it is the recombination effect of electrons and holes that makes electroluminescence.

Once the applied voltage or forward current is removed, the recombination dies out but in theory there may be still a charge potential with a very small xx pF capacitance (depends on power rating /size) , but in practice leakage resistance decays this voltage to 0.

  • The charge carriers recombine in a forward-biased P-N junction as the electrons cross from the N-region and recombine with the holes existing in the P-region.

  • Free electrons are in the conduction band of energy levels, while holes are in the valence energy band.

  • Thus the energy level of the holes is less than the energy levels of the electrons. Some portion of the energy must be dissipated to recombine the electrons and the holes.

  • This energy is emitted in the form of heat and light.
    • Any supplier who says LED's don't emit Infrared or heat, don't know the Black Body Physics of conducted heat.

This is a continuous flow of recombination and current that controls the emission of light to maintain the E field in the gap and recombination of charges.

enter image description here Ref

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  • \$\begingroup\$ So the energy from the power source keeps the electrons jumping from the valence band to the conduction band in a continuous cycle? Once power is removed each semiconductor remains “unbalanced”? \$\endgroup\$ – ohmmy Apr 3 '18 at 16:06
  • \$\begingroup\$ static charge bleeds out thru leakage resistance (1uA aprox in small junctions) \$\endgroup\$ – Sunnyskyguy EE75 Apr 3 '18 at 16:09

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