/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 In an n-type silicon, which of t... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 1

In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.

Short Answer

Expert verified
Option (c) is true; holes are minority carriers and pentavalent atoms are the dopants in n-type silicon.

Step by step solution

01

Understanding the Basics of n-type Semiconductors

n-type semiconductors are formed by adding dopants to silicon that have more valence electrons than silicon. This means that the dopant atoms must have five valence electrons.
02

Identifying the Dopant Type

In n-type silicon, pentavalent atoms, such as phosphorus (P), arsenic (As), or antimony (Sb), are used as dopants because they have five valence electrons. This aligns with providing additional electrons.
03

Understanding Majority and Minority Carriers

In n-type materials, the process of doping introduces extra electrons into the conduction band. Thus, electrons are the majority carriers, and holes are the minority carriers.
04

Analyzing Each Option

Let's evaluate each option: (a) This is incorrect because trivalent atoms are used for p-type doping, not n-type. (b) Incorrect because electrons are the majority carriers, not the minority. (c) Correct; holes are minority carriers and pentavalent atoms, which provide extra electrons, are used as dopants. (d) Incorrect because holes cannot be majority carriers in n-type. Thus, the correct statement is option (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Majority Carriers
In n-type semiconductors, electrons serve as the majority carriers. This means that the conduction is primarily carried out by electrons, which are negatively charged particles. When a dopant with five valence electrons (a pentavalent element) is added to silicon, it provides an extra electron that is free to move.
These extra electrons increase the number of free electrons in the material, making them the primary players in conducting electricity.
So, in a nutshell, the electrons outnumber any other type of charge carrier in an n-type semiconductor.
Minority Carriers
Contrary to majority carriers, minority carriers in n-type semiconductors are holes. These are essentially the absence of an electron, creating a positive charge carrier.
Although holes are present, their number is significantly lower than that of electrons.
In n-type materials, the presence of holes comes naturally due to the thermal generation of electron-hole pairs, and they act as the minority carriers.
Pentavalent Dopants
The concept of pentavalent dopants is crucial in forming n-type semiconductors. Elements like phosphorus (P), arsenic (As), and antimony (Sb) are used because they have five valence electrons. Silicon, with four valence electrons, can incorporate these pentavalent dopants by replacing one of its own atoms in the crystal lattice.
This replacement results in one extra electron being added to the silicon's structure because pentavalent elements bring an additional electron.
These extra electrons become available for electrical conduction, effectively changing the electrical properties of the silicon to make it n-type.
Electron Doping
To transform a semiconductor into an n-type, electron doping is performed. This is the process where pentavalent elements are introduced into the crystalline structure of the silicon.
This introduces more electrons that are free to move about, drastically increasing conductivity.
In essence, electron doping infuses the semiconductor material with more free charge carriers, specifically electrons, helping in the majority charge flow.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For transistor action, which of the following statements are correct: (a) Base, emitter and collector regions should have similar size and doping concentrations. (b) The base region must be very thin and lightly doped. (c) The emitter junction is forward biased and collector junction is reverse biased. (d) Both the emitter junction as well as the collector junction are forward biased.

For a transistor amplifier, the voltage gain (a) remains constant for all frequencies. (b) is high at high and low frequencies and constant in the middle frequency range. (c) is low at high and low frequencies and constant at mid frequencies. (d) None of the above.

In an intrinsic semiconductor the energy gap \(E_{g}\) is \(1.2 \mathrm{eV} .\) Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at \(600 \mathrm{~K}\) and that at \(300 \mathrm{~K}\) ? Assume that the temperature dependence of intrinsic carrier concentration \(n_{i}\) is given by \(n_{i}=n_{0} \exp -\frac{E_{g}}{2 k_{B} T}\) where \(n_{0}\) is a constant.

For a CE-transistor amplifier, the audio signal voltage across the collected resistance of \(2 \mathrm{k} \Omega\) is \(2 \mathrm{~V}\). Suppose the current amplification factor of the transistor is 100 , find the input signal voltage and base current, if the base resistance is \(1 \mathrm{k} \Omega\).

A p-n photodiode is fabricated from a semiconductor with band gap of \(2.8 \mathrm{eV}\). Can it detect a wavelength of \(6000 \mathrm{~nm}\) ?
See all solutions

Recommended explanations on Physics Textbooks

Engineering Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Physical Quantities And Units

Read Explanation

Space Physics

Read Explanation

Energy Physics

Read Explanation

Measurements

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.