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Question

Physics Question on Semiconductor electronics: materials, devices and simple circuits

In a reverse biased diode when the applied voltage changes by 1V1\, V, the current is found to change by 0.5μA0.5\, \mu\,A . The reverse bias resistance of the diode is

A

2Ω2 \Omega

B

200Ω200 \Omega

C

2×106Ω2 \times10^6 \Omega

D

2×105Ω2\times10^{5} \Omega

Answer

2×106Ω2 \times10^6 \Omega

Explanation

Solution

Reverse resistance =ΔVΔI= \frac{\Delta V}{\Delta I} =10.5×106= \frac{ 1}{0.5 \times10^{-6}} =2×106Ω = 2 \times10^{6} \Omega