Name the p-n junction diode which emit spontaneous radiation when forward biased. How do we choose the semiconductor, to be used in those diodes, if the emitted radiations, is to be in the visible region?
Answers
Answer:
The p-n junction diode, which emits spontaneous radiation when forward biased, is known as the light emitting diode or LED. The visible light is from 0.45 µm to 0.7 µm and corresponding energy is between 2.8eV to 1.8 eV .therefore, the least band gap of the semiconductor to be used in LED, in order to have the emitted radiation to be in the visible region, should be 1.8 eV. Phosphorus doped gallium arsenide and gallium phosphide are two such semiconductors.
Explanation:
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Answer:
Explanation:
(Light emitting diode (LED) gives the spontaneous radiation.) At the junction boundary, the concentration of minority carriers increases (due to forward biasing where electrons are sent from n→p and holes are sent from p→n). There excess minority carriers recombines with majority charge carriers near the junction and energy is released in the form of photons. Visible LED's must at least have a band gap of 1.8 eV.Read more on Sarthaks.com - https://www.sarthaks.com/968905/explain-with-help-of-circuit-diagram-the-action-of-a-forward-biased-p-n-junction-diode