L.E.D's and the photoelctric effect

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SUMMARY

The discussion centers on the relationship between light from filament lamps and voltage production in Light Emitting Diodes (L.E.D.s) through the photoelectric effect. Participants explain that when light strikes the semiconductor material of an L.E.D., it provides sufficient energy for valence electrons in the n-type semiconductor to transition to the conduction band, creating a flow of electrons toward the p-type semiconductor. This movement generates a voltage difference, enabling the L.E.D. to emit light. The low band gap energy of the semiconductor allows this process to occur even with lower energy white light.

PREREQUISITES
  • Understanding of semiconductor physics, specifically n-type and p-type materials
  • Knowledge of the photoelectric effect and its principles
  • Familiarity with band gap energy concepts in semiconductors
  • Basic principles of how L.E.D.s operate
NEXT STEPS
  • Research the band gap energy of various semiconductor materials used in L.E.D.s
  • Explore the principles of the photoelectric effect in greater detail
  • Learn about the differences between n-type and p-type semiconductors
  • Investigate the impact of photon energy on electron excitation in semiconductors
USEFUL FOR

Students and professionals in physics, electrical engineering, and anyone interested in the principles of semiconductor technology and L.E.D. functionality.

ryan750
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hello - i feel confident that i understand the principles of how L.E.D's work. I can grasp the concept of the depletion zone and how this leads to conduction in one direction.

But how does light from an ordinary filament lamp cause a voltage in an L.E.D to be produced when the L.E.D has no obvious power supply?

I know of the principles of the photoelectric effect and realize it has something to do with this but need a concise explanation of how the voltage is produced.

i have been discussing this with sum1 else and have come up with this explanation:

light from the lamp when incident on the semiconductor in the L.E.D causes valence electrons in the n type semiconductor to be released into the conduction/ depletion layer. these are attracted to the p type semiconductor and they move into the 'holes'. this causes a drop in energy level for the electron which is called relaxation and causes light to be emitted.
this process means that a current is flowing from the n to the p type semiconductor which means that a voltage is present in this mini - circuit.

now the fact that white light has a low photon energy means that if it were the photoelectric efect that causes this process - the material in the L.E.D would need to have a low work function or sum other proerty that allows this to happen.

anyone who can correct any details - add more information - or completely change the explanation to the question - please feel free

thanks
 
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for the explanation! You are correct in your understanding of the photoelectric effect and how it relates to the production of voltage in an L.E.D. The key factor here is the band gap energy of the semiconductor material used in the L.E.D. This band gap energy is the minimum amount of energy required for an electron to move from the valence band to the conduction band.

When light from the filament lamp hits the semiconductor material, it provides enough energy for some of the electrons in the valence band to jump to the conduction band. This creates a flow of electrons from the n-type to the p-type layer, resulting in a voltage difference. This voltage difference is what powers the L.E.D.

The reason why white light, which has a lower photon energy, can still produce this effect is because the semiconductor material used in L.E.D.s is specifically chosen to have a low band gap energy. This allows the electrons to easily jump to the conduction band with the energy provided by the light.

I hope this helps clarify the process of how L.E.D.s use the photoelectric effect to produce light and power. Let me know if you have any further questions or need more clarification.
 

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