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Chapter 38: Problem 2914

The electrical resistance of depletion layer is large because (A) it contains electrons as charge carriers (B) it has holes as charge carriers. (C) it has no charge carriers. (D) It has large number of charge carriers

Short Answer

Expert verified
The correct reason for the large electrical resistance in the depletion layer is option (C) because it has no charge carriers.

Step by step solution

01

Analyze the first option

Option (A) states that the electrical resistance of the depletion layer is large because it contains electrons as charge carriers. This is not the correct reason because the presence of electrons as charge carriers would actually lead to electrical conduction, not an increase in resistance.
02

Analyze the second option

Option (B) states that the electrical resistance of the depletion layer is large because it has holes as charge carriers. This is also not the correct reason because, like electrons, the presence of holes as charge carriers would also lead to electrical conduction, not an increase in resistance.
03

Analyze the third option

Option (C) states that the electrical resistance of the depletion layer is large because it has no charge carriers. This is the correct reason because, in the depletion layer, the recombination of electrons and holes leaves behind a region with no charge carriers, thus resulting in a large electrical resistance.
04

Analyze the fourth option

Option (D) states that the electrical resistance of the depletion layer is large because it has a large number of charge carriers. This is not the correct reason because having a large number of charge carriers would lead to an increase in electrical conduction, not an increase in resistance.
05

Conclusion

Based on the analysis of each option, the correct reason for the large electrical resistance in the depletion layer is option (C) because it has no charge carriers.

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Most popular questions from this chapter

A P-N junction is said to be forward based when (A) a magnetic field is applied in the region of junction. (B) a potential difference is applied across \(\mathrm{P}\) and \(\mathrm{N}\) regions making \(\mathrm{P}\) region negative and \(\mathrm{N}\) region positive. (C) not potential difference is applied across \(\mathrm{P}\) and \(\mathrm{N}\) regions. (D) a potential difference is applied across \(\mathrm{P}\) and \(\mathrm{N}\) regions making P region positive and N region negative.

The reverse biasing in junction diode (A) increase the potential barrier (B) increases the number of minority change carriers (C) increases the number of majority change carriers (D) decreases the potential diode

The number of minority carriers crossing the junction of diode depends primarily on the (A) magnitude of potential barrier (B) magnitude of the forward bias barrier. (C) rate of thermal generation of electron hole pair. (D) concentration of doping impurities.

What is the resistance of \(\mathrm{P}-\mathrm{N}\) junction diode in forward biasing ? (A) zero (B) high (C) infinity (D) a few ohms

In a semiconductor diode, barrier potential offers opposition to only (A) free electrons in \(\mathrm{N}\) region (B) holes in the P region (C) minority carriers in both regions (D) majority carriers in both regions.
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