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Semiconductor Electronics Materials Devices And Simple Circuits

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semiconductor-electronics--materials-devices-and-simple-circuits Question 36

Question: Q. 15. Give reasons for the following :

(i) A photodiode, when used as a detector of optical signals is operated under reverse bias.

(ii) The band gap of the semiconductor used for fabrication of visible LED’s must at least be $1.8 \mathrm{eV}$.

A&E [CBSE SQP 2014]

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Solution:

Ans. (i) When operated under reverse bias, the photodiode can detect the changes in current with the changes in light intensity more easily. $\mathbf{1} \frac{1}{2}$

(ii) The photon energy, of visible light photons varies from about $1.8 \mathrm{eV}$ to $3 \mathrm{eV}$. Hence for visible LED’s, the semiconductor must have a band gap of $1.8 \mathrm{eV}$.

[CBSE Marking Scheme 2014] 1½

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Semiconductor Electronics Materials Devices And Simple Circuits

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Semiconductor Electronics Materials Devices And Simple Circuits

semiconductor-electronics--materials-devices-and-simple-circuits Question 36

Long Answer Type Questions

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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