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Chapter 20: Problem 9

The depletion layer in the \(P-N\) junction region is caused by (A) Drift of holes (B) Diffusion of charge carriers (C) Migration of impurity ions (D) Drift of electrons

Short Answer

Expert verified
(B) Diffusion of charge carriers

Step by step solution

01

Understanding the P-N Junction

A P-N junction is formed when a p-type semiconductor is fused to an n-type semiconductor, creating a junction between the two. The p-type semiconductor has abundant holes, while the n-type has excess electrons.
02

Formation of Depletion Layer

When the P-N junction is formed, because of the concentration gradient, the free electrons from the n-region start diffusing towards the p-region and holes from the p-region towards n-region. This is known as the process of diffusion. As these majority charge carriers cross the junction, they combine with their opposite charge carriers on the other side. As a result of recombination, a layer that's devoid of free charge carriers forms at the junction. This layer is known as the 'depletion layer'.
03

Determining the Cause

Looking at the given options for what causes the depletion layer, based on the explanation above, it is the diffusion of charge carriers that causes the formation of the depletion layer, not drift of holes, migration of impurity ions, or drift of electrons. Hence, the correct answer is (B) Diffusion of charge carriers.

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Most popular questions from this chapter

Two \(P-N\) junctions can be connected in series by three different methods as shown in the Fig. \(20.5\). If the potential difference in the junctions is the same, then the correct connections will be (A) In the circuit (1) and (2) (B) In the circuit (2) and (3) (C) In the circuit (1) and (3) (D) Only in the circuit (1)

In a common emitter transistor, the current gain is 80 . What is the change in collector current, when the change in base current is \(250 \mu \mathrm{A} ?\) (A) \(80 \times 250 \mu \mathrm{A}\) (B) \((250-80) \mu \mathrm{A}\) (C) \((250+80) \mu \mathrm{A}\) (D) \(250 / 80 \mu \mathrm{A}\)

A solid which is transparent to visible light and whose conductivity increases with temperature is formed by (A) Metallic binding (B) Ionic binding (C) Covalent binding (D) Vander Waals binding

The maximum efficiency of full wave rectifier is (A) \(100 \%\) (B) \(25.20 \%\) (C) \(40.2 \%\) (D) \(81.2 \%\)

No bias is applied to a \(P-N\) junction, then the current (A) Is zero because the number of charge carriers flowing on both sides is same (B) Is zero because the charge carriers do not move (C) Is non-zero (D) None of these
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