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Chapter 5: Problem 5

A silicon sample is \(2.5 \mathrm{~cm}\) long and has a cross-sectional area of \(0.1 \mathrm{~cm}^{2}\). The silicon is n type with a donor impurity concentration of \(N_{d}=2 \times 10^{15} \mathrm{~cm}^{-3} .\) The resistance of the sample is measured and found to be \(70 \Omega\). What is the electron mobility?

Short Answer

Expert verified
The electron mobility is approximately 1116 cm虏/V路s.

Step by step solution

01

Calculate the Volume of the Silicon Sample

The volume of the silicon sample can be calculated using the formula for volume: \( V = A \times L \), where \( A \) is the cross-sectional area and \( L \) is the length. Here, \( A = 0.1 \text{ cm}^2 \) and \( L = 2.5 \text{ cm} \). So, \( V = 0.1 \times 2.5 = 0.25 \text{ cm}^3 \).
02

Calculate the Conductivity

The conductivity \( \sigma \) is related to the resistance \( R \) by the equation \( \sigma = \frac{1}{R} \times \frac{L}{A} \). Plugging in the given values: \( R = 70 \Omega \), \( L = 2.5 \text{ cm} \), and \( A = 0.1 \text{ cm}^2 \), we find \( \sigma = \frac{1}{70} \times \frac{2.5}{0.1} = \frac{2.5}{7} \text{ S/cm} \).
03

Use Conductivity to Find Electron Mobility

Conductivity \( \sigma \) is also related to electron mobility \( \mu_n \) by the formula \( \sigma = q \times N_d \times \mu_n \), where \( q \) is the charge of an electron \( (1.6 \times 10^{-19} \text{ C}) \) and \( N_d = 2 \times 10^{15} \text{ cm}^{-3} \) is the donor impurity concentration. Solve for \( \mu_n \): \( \mu_n = \frac{\sigma}{q \times N_d} = \frac{\frac{2.5}{7}}{1.6 \times 10^{-19} \times 2 \times 10^{15}} \).
04

Solve for Electron Mobility

Substitute the values into the formula: \( \mu_n = \frac{2.5/7}{1.6 \times 10^{-19} \times 2 \times 10^{15}} = \frac{0.3571}{3.2 \times 10^{-4}} \). Calculating this gives \( \mu_n \approx 1116 \text{ cm}^2/\text{V}\cdot \text{s} \).

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Key Concepts

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

Understanding Electron Mobility
Electron mobility is a key parameter in semiconductor physics. It reflects how quickly electrons can move through a material when an electric field is applied. In simple terms, it's like understanding the speed limit for electrons. Here's how it works:
  • Higher mobility means electrons can travel faster through the material, enhancing the semiconductor's performance.
  • It's influenced by factors such as temperature and material purity.
A semiconductor's electron mobility directly impacts its conductivity. If electrons can move freely, the material conducts electricity better. This is why understanding electron mobility is crucial for designing efficient electronic components.
In the original exercise, knowing the impurity concentration and calculating the mobility helps determine the silicon's effectiveness for specific applications.
Conductivity Calculation Explained
Conductivity, expressed as \(\sigma\), is a measure of a material's ability to conduct electric current. In essence, it shows how easily electrons can move through a semiconductor. The higher the conductivity, the more conductive the material.
  • The formula for conductivity is \(\sigma = \frac{1}{R} \times \frac{L}{A}\), where \(R\) is resistance, \(L\) is the length, and \(A\) is the cross-sectional area of the sample.
  • This formula originates from the equation \(\sigma = q \times N_d \times \mu_n\), linking conductivity to both electron mobility and donor concentration.
In the given exercise, after finding the volume of the sample, we calculate the conductivity using the measured resistance. This step is essential in determining electron mobility, showing the interconnected nature of these concepts in semiconductor physics. Understanding conductivity calculations can help predict how materials will behave in real-world electronics.
Exploring N-Type Silicon
N-type silicon is a type of semiconductor where extra electrons are introduced into the silicon crystal, which increases its conductivity. This is done by doping the silicon with donor impurities, typically elements like phosphorus or arsenic.
  • N-type silicon has more electrons available for conduction, making it a better conductor compared to its pure form.
  • The extra electrons come from the donor atoms that fit into the silicon lattice, giving it its n-type characteristics.
In the context of the exercise, the donor concentration \(N_d\) is specified as \(2 \times 10^{15} \text{ cm}^{-3}\). This value is important for calculating the electron mobility because it tells us how many carriers (electrons) are available to move through the material. Understanding n-type silicon is fundamental in electronics as it forms the backbone of many semiconductor devices.

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Most popular questions from this chapter

(a) A GaAs semiconductor resistor is doped with donor impurities at a concentration of \(N_{d}=2 \times 10^{15} \mathrm{~cm}^{-3}\) and has a cross- sectional area of \(5 \times 10^{-5} \mathrm{~cm}^{2}\). A current of \(I=25 \mathrm{~mA}\) is induced in the resistor with an applied bias of \(5 \mathrm{~V}\). Determine the length of the resistor. \((b)\) Using the results of part \((a)\), calculate the drift velocity of the electrons. \((c)\) If the bias applied to the resistor in part \((a)\) increases to \(20 \mathrm{~V}\), determine the resulting current if the electrons are traveling at their saturation velocity of \(5 \times 10^{6} \mathrm{~cm} / \mathrm{s}\)

A silicon crystal having a cross-sectional area of \(0.001 \mathrm{~cm}^{2}\) and a length of \(10^{-3} \mathrm{~cm}\) is connected at its ends to a \(10-\mathrm{V}\) battery. At \(T=300 \mathrm{~K}\), we want a current of \(100 \mathrm{~mA}\) in the silicon. Calculate ( \(a\) ) the required resistance \(R,(b)\) the required conductivity, (c) the density of donor atoms to be added to achieve this conductivity, and \((d)\) the concentration of acceptor atoms to be added to form a compensated p-type material with the conductivity given from part \((b)\) if the initial concentration of donor atoms is \(N_{d}=10^{15} \mathrm{~cm}^{-3}\)

The electron concentration in silicon at \(T=300 \mathrm{~K}\) is given by $$ n(x)=10^{16} \exp \left(\frac{-x}{18}\right) \mathrm{cm}^{-3} $$ where \(x\) is measured in \(\mu \mathrm{m}\) and is limited to \(0 \leq x \leq 25 \mu \mathrm{m}\). The electron diffusion coefficient is \(D_{n}=25 \mathrm{~cm}^{2} / \mathrm{s}\) and the electron mobility is \(\mu_{n}=960 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\). The total electron current density through the semiconductor is constant and equal to \(J_{n}=-40 \mathrm{~A} / \mathrm{cm}^{2} .\) The electron current has both diffusion and drift current components. Determine the electric field as a function of \(x\) which must exist in the semiconductor.

Germanium is doped with \(5 \times 10^{15}\) donor atoms per \(\mathrm{cm}^{3}\) at \(T=300 \mathrm{~K}\). The dimensions of the Hall device are \(d=5 \times 10^{-3} \mathrm{~cm}, W=2 \times 10^{-2} \mathrm{~cm}\), and \(L=10^{-1} \mathrm{~cm} .\) The current is \(I_{x}=250 \mu A\), the applied voltage is \(V_{x}=100 \mathrm{mV}\), and the magnetic flux density is \(B_{z}=500\) gauss \(=5 \times 10^{-2}\) tesla. Calculate: ( \(a\) ) the Hall voltage, \((b)\) the Hall field, and ( \(c\) ) the carrier mobility.

The hole concentration in p-type GaAs is given by \(p(x)=10^{16}(1+x / L)^{2} \mathrm{~cm}^{-3}\) for \(-L \leq x \leq 0\) where \(L=12 \mu \mathrm{m}\). The hole diffusion coefficient is \(D_{p}=10 \mathrm{~cm}^{2} / \mathrm{s}\). Calculate the hole diffusion current density at (a) \(x=0,(\) b) \(x=-6 \mu \mathrm{m}\), and \((\) c) \(x=-12 \mu \mathrm{m}\).
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