Question
Explain the working of a LED.
✓
Answer
Working :
An LED is forward-biased with about $1.2\ V$ to $3.6\ V$ at $12\ mA$ to $20\ mA$ . Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p -layer, some of these excess minority carriers electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of diferent energies.
In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy hv.
[Note : The photons originate primarily in the p-side of the junction which has a bandgap $E _{ Gp }$ narrower than that of the $n$-side, $E _{ Gn }$. Thus, with hv < $E _{ Gn }$, the photons are emitted through the wide-bandgap n-region with essentially no absorption.]
An LED is forward-biased with about $1.2\ V$ to $3.6\ V$ at $12\ mA$ to $20\ mA$ . Majority carriers electrons from n-type layer and holes from p-type layer are injected into the active layer. Electrons cross the junction into the p-layer. In the active p -layer, some of these excess minority carriers electrons, recombine radiatively with majority carriers, holes, thereby emitting photons. The resulting photon has an energy approximately equal to the bandgap of the active layer material. Modifying the bandgap of the active layer creates photons of diferent energies.
In the energy band diagram this recombination is equivalent to a transition of the electron from a higher energy state in the conduction band to a lower energy state in the valence band. The energy difference is emitted as a photon of energy hv.
[Note : The photons originate primarily in the p-side of the junction which has a bandgap $E _{ Gp }$ narrower than that of the $n$-side, $E _{ Gn }$. Thus, with hv < $E _{ Gn }$, the photons are emitted through the wide-bandgap n-region with essentially no absorption.]
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