/*! 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 43 In GaAs, the donor impurity conc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In GaAs, the donor impurity concentration varies as \(N_{d 0} \exp (-x / L)\) for \(0 \leq x \leq L\), where \(L=0.1 \mu \mathrm{m}\) and \(N_{d 0}=5 \times 10^{16} \mathrm{~cm}^{-3}\). Assume \(\mu_{n}=6000 \mathrm{~cm}^{2} / \mathrm{V}\) -s and \(T=300 \mathrm{~K}\). (a) Derive the expression for the electron diffusion current density versus distance over the given range of \(x\). \((b)\) Determine the induced electric field that generates a drift current density that compensates the diffusion current density

Short Answer

Expert verified
The electron diffusion current density is \( J_{diff} = q \cdot \frac{N_{d 0}}{L} \cdot D_n \cdot \exp(-x/L) \). The electric field that compensates this is \( E = \frac{1}{\mu_n} \times \frac{D_n}{L} \).

Step by step solution

01

Understand Diffusion Current Density

The electron diffusion current density, \( J_{diff} \), is given by \( J_{diff} = -q D_n \frac{d n}{d x} \), where \( q \) is the charge of an electron, \( D_n \) is the electron diffusion coefficient, and \( \frac{d n}{d x} \) is the gradient of electron concentration. To find \( \frac{d n}{d x} \), differentiate \( n(x) = N_{d 0} \exp(-x/L) \).
02

Calculate the Derivative

Differentiate \( n(x) = N_{d 0} \exp(-x/L) \) with respect to \( x \). We have \( \frac{d n}{d x} = -\frac{N_{d 0}}{L} \exp(-x/L) \).
03

Find Electron Diffusion Coefficient

Use the Einstein relation \( D_n = \mu_n \frac{kT}{q} \). Given \( \mu_n = 6000 \, \text{cm}^2/\text{V} \cdot \text{s} \), \( k = 1.38 \times 10^{-23} \, \text{J/K} \), and \( T = 300 \, \text{K} \), calculate: \[ D_n = 6000 \times \frac{1.38 \times 10^{-23} \times 300}{1.6 \times 10^{-19}} \].
04

Expression for Electron Diffusion Current Density

Substitute \( D_n \) and \( \frac{d n}{d x} \) into \( J_{diff} = -q D_n \frac{d n}{d x} \). The expression becomes: \( J_{diff} = q \cdot \frac{N_{d 0}}{L} \cdot D_n \cdot \exp(-x/L) \).
05

Determine Induced Electric Field

The drift current density \( J_{drift} \) is given by \( J_{drift} = q n(x) \mu_n E \), where \( E \) is the electric field. For steady-state, \( J_{drift} = J_{diff} \). Solve for \( E \): \( E = \frac{J_{diff}}{q n(x) \mu_n} \). Substitute \( J_{diff} \) and \( n(x) \) into the formula.

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 91Ó°ÊÓ!

Key Concepts

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

Electron Diffusion
Electron diffusion in semiconductor physics is a fundamental process where electrons move from high concentration regions to low concentration regions. This movement is driven by the natural tendency to achieve uniform electron distribution. In the context of GaAs semiconductors, understanding this phenomenon helps examine how charge carriers behave within the material.

To quantify electron diffusion, we employ the diffusion current density. The formula for electron diffusion current density, denoted as \( J_{diff} \), is:
  • \( J_{diff} = -q D_n \frac{d n}{d x} \)
Here, \( q \) is the electron charge, \( D_n \) is the electron diffusion coefficient, and \( \frac{d n}{d x} \) represents the gradient of electron concentration. A negative sign indicates the flow from high to low concentration.

In GaAs, where donor impurities follow a specific distribution \( n(x) = N_{d 0} \exp(-x/L) \), it becomes crucial to differentiate this function to find \( \frac{d n}{d x} \). This derivative is \( -\frac{N_{d 0}}{L} \exp(-x/L) \), representing how the electron concentration changes with position. Understanding this change allows us to predict how electrons will behave in any localized region.
Electric Field
In semiconductor physics, the electric field plays a crucial role in balancing diffusion to maintain equilibrium. The electric field is the driving force that causes charge carriers to move in a specific direction, resulting in what is known as drift current.

The relationship between the drift current density \( J_{drift} \) and the electric field \( E \) is given by:
  • \( J_{drift} = q n(x) \mu_n E \)
Here, \( q \) is the electron charge, \( n(x) \) is the electron concentration, \( \mu_n \) is the electron mobility, and \( E \) is the electric field.

To achieve balance, the drift current must match the diffusion current, meaning \( J_{drift} = J_{diff} \). By setting these equal, we solve for the electric field \( E \):
  • \( E = \frac{J_{diff}}{q n(x) \mu_n} \)
This formula highlights how the electric field adjusts to counteract the diffusion process, keeping the semiconductor in a steady state by ensuring no net motion of charge carriers.
GaAs Semiconductor
Gallium Arsenide (GaAs) is a compound semiconductor with unique properties that make it highly valuable in electronics. GaAs is particularly noted for its high electron mobility, meaning electrons can travel through it with relative ease, making it advantageous for high-speed applications.

In doping processes, impurities like donor atoms are added to control the electrical properties. In the exercise, the donor impurity concentration in GaAs varies with depth as described by \( N_{d 0} \exp(-x/L) \). This indicates a non-uniform distribution where concentration decreases with distance.

This material is highly prized in fields such as:
  • High-frequency and microwave electronics
  • Optical communication systems
  • Solar cells
Because of its ability to efficiently manage electron flow, GaAs is often used in applications where performance and speed are critical factors. Understanding how electrons diffuse within GaAs and how electric fields are managed allows engineers to design better and more efficient electronic devices.

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

The steady-state electron distribution in silicon can be approximated by a linear function of \(x\). The maximum electron concentration occurs at \(x=0\) and is \(n(0)=2 \times\) \(10^{16} \mathrm{~cm}^{-3}\). At \(x=0.012 \mathrm{~cm}\), the electron concentration is \(5 \times 10^{15} \mathrm{~cm}^{-3} .\) If the electron diffusion coefficient is \(D_{n}=27 \mathrm{~cm}^{2} / \mathrm{s}\), determine the electron diffusion current density.

Consider silicon at \(T=300 \mathrm{~K}\). A Hall effect device is fabricated with the following geometry: \(d=5 \times 10^{-3} \mathrm{~cm}, W=5 \times 10^{-2} \mathrm{~cm}\), and \(L=0.50 \mathrm{~cm} .\) The electri- cal parameters measured are: \(I_{x}=0.50 \mathrm{~mA}, V_{x}=1.25 \mathrm{~V}\), and \(B_{z}=650\) gauss \(=\) \(6.5 \times 10^{-2}\) tesla. The Hall field is \(E_{H}=-16.5 \mathrm{mV} / \mathrm{cm} .\) Determine \((a)\) the Hall voltage, \((b)\) the conductivity type, \((c)\) the majority carrier concentration, and \((d)\) the majority carrier mobility.

The conductivity of a semiconductor layer varies with depth as \(\sigma(x)=\sigma_{o} \exp (-x / d)\) where \(\sigma_{o}=20(\Omega-\mathrm{cm})^{-1}\) and \(d=0.3 \mu \mathrm{m}\). If the thickness of the semiconductor layer is \(t=1.5 \mu \mathrm{m}\), determine the average conductivity of this layer.

(a) Assume that the electron mobility in an n-type semiconductor is given by $$ \mu_{n}=\frac{1350}{\left(1+\frac{N_{d}}{5 \times 10^{16}}\right)^{1 / 2}} \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} $$ where \(N_{d}\) is the donor concentration in \(\mathrm{cm}^{-3}\). Assuming complete ionization, plot the conductivity as a function of \(N_{d}\) over the range \(10^{15} \leq N_{d} \leq 10^{18} \mathrm{~cm}^{-3} .\) (b) Compare the results of part ( \(a\) ) to that if the mobility were assumed to be a constant equal to \(1350 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .(c)\) If an electric field of \(E=10 \mathrm{~V} / \mathrm{cm}\) is applied to the semiconductor, plot the electron drift current density of parts \((a)\) and \((b)\).

Consider a gallium arsenide sample at \(T=300 \mathrm{~K}\). A Hall effect device has been fabricated with the following geometry: \(d=0.01 \mathrm{~cm}, W=0.05 \mathrm{~cm}\), and \(L=0.5 \mathrm{~cm}\). The electrical parameters are: \(I_{x}=2.5 \mathrm{~mA}, V_{x}=2.2 \mathrm{~V}\), and \(B_{z}=2.5 \times 10^{-2}\) tesla. The Hall voltage is \(V_{H}=-4.5 \mathrm{mV}\). Find: \((a)\) the conductivity type, \((b)\) the majority carrier concentration, \((c)\) the mobility, and \((d)\) the resistivity.

See all solutions

Recommended explanations on Physics Textbooks

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.