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In an LED the electric circuit causes the electrons and holes in the P-N Junction to neutralize each other to produce light energy. Whereas in a rectifier diode, if I am not wrong, the electric circuit does not cause the electrons and holes to neutralize each other. Why is this so? Also, do the electrons in a circuit flow through the diode? If they do, why dont the electrons neutralize the hole?

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up vote 14 down vote accepted

In a rectifier diode, there will still be some e-h pairs that recombine. However there are reasons why they don't produce light

Firstly, Silicon is an indirect band-gap semiconductor. What this means is that the energy levels in the band-gap don't line up nicely making it very inefficient at emitting photons. Generally recombination will take place in a way which results in either vibration/heating of the silicon lattice, or other means (e.g. transferring the energy to other carriers). It is possible for silicon to emit light, however it is very inefficient.

Secondly, the band-gaps in rectifier diodes will not have the right energy level for photon production. Remember that the frequency of a photon is proportional to its energy (\$E_{ph}=h\nu\$). To produce a photon of a given colour, the semiconductor band-gap energy must be approximately equal to this energy (i.e. \$E_g = E_{ph}\$).

Thirdly, the doping levels of rectifier diodes will be much to low. Lower doping levels means fewer carriers which in turn means there will be less recombination of e-h pairs than, say, a heavily doped LED p-n junction. This is evident from the fact that the forward voltage of the rectifier diodes is much less than an LED (doping levels determine the built in potential).

Finally, LEDs are fabricated in a way that there is an emission surface from which photons can be emitted (a so-called "heterostructure"). Without this, the emitted photons would just be re-absorbed. This surface is then encapsulated in a transparent lens to get the light out efficiently - and to minimise the number of photons which reflect off the semiconductor/air interface back into the device. Rectifier diodes do not have any of this - they are sealed in opaque epoxy packages.

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Things get more complicated and blurry when you go into the labs. Schottky barrier LEDs exist there, and electroluminescence effects have been observed in schottkys too. I think the most important point is that LEDs are specifcially manufactured to produce as many photons as possible in a wavelength band as narrow as possible, wherein other types of diodes nobody cares what happens to the energy, and the few photons that are produced in wavelengths all over the place get absorbed anyways. – PlasmaHH Feb 12 at 11:24
Does the re-absorption of the photons cause the efficiency of the diode to decrease? "nobody cares what happens to the energy" doesn't seem right in these days of very low powered devices - but if the energy gets re-absorbed with a net of not really any change, then I guess it doesn't matter... – Steve Feb 12 at 13:18
@Steve Reabsorption will basically just produce new e-h pairs. – Tom Carpenter Feb 12 at 13:36
All current results in e-h recombination in the junction. And, I will point out that III-V transistors light up quite nicely in operation, you just don't normally notice because they are usually in a package without a window. – Jon Custer Feb 12 at 15:28
Thanks for the help. However I would also like to ask: Do electrons in a circuit flow through a rectifier diode? If so why do these electrons not recombine with the holes in the P-N junction? – Raphael Low Feb 14 at 12:06

All diodes essentially work the same, but different kinds of diodes have different forward voltages. Diodes with higher forward voltage generate electromagnetic radiation with a smaller wavelength.

LEDS generate visible light (and are constructed so that the light can escape). 'Normal' diodes generate infrared radiation, so they just heat up.

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