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प्रश्न
With the help of a neat circuit diagram, explain the working of a photodiode.
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उत्तर
a. A photodiode is a special purpose P-N junction diode fabricated with a transparent window to allow light to fall on the diode.
b. When the photodiode is illuminated with light (photons) with energy h greater than the energy gap Eg of the semiconductor, then electron-hole pairs are generated due to the absorption of photons.
c. The diode is fabricated such that the generation of electron-hole pairs takes place in or near the depletion region of the diode.
d. Due to electric field of the junction, electrons and holes are separated before they recombine.
e. The direction of the electric field is such that electrons reach N-side and holes reach Pside. Electrons are collected on N-side and holes are collected on P-side giving rise to an e.m.f.
f. When an external load is connected, current flows. The magnitude of the photocurrent depends on the intensity of incident light.
g. It is easier to observe the change in the current with change in the light intensity, if a reverse bias is applied. Thus, photodiode can be used as a photodetector to detect optical signals.
संबंधित प्रश्न
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|
LED is a heavily doped P-N junction which under forward bias emits spontaneous radiation. When it is forward-biased, due to recombination of holes and electrons at the junction, energy is released in the form of photons. In the case of Si and Ge diode, the energy released in recombination lies in the infrared region. LEDs that can emit red, yellow, orange, green and blue light are commercially available. The semiconductor used for fabrication of visible LEDs must at least have a band gap of 1.8 eV. The compound semiconductor Gallium Arsenide – Phosphide is used for making LEDs of different colours.
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OR
iii. Draw V-I characteristic of a p-n junction diode in
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