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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.

Short Answer

Expert verified
Hall voltage: \(1.56 \times 10^{-6}\) V, Hall field: \(7.8 \times 10^{-5}\) V/cm, Carrier mobility: \(1.56 \times 10^2\) cm²/(V·s).

Step by step solution

01

Calculate the carrier concentration

Since germanium is doped with donor atoms, it acts as an n-type semiconductor. The carrier concentration \( n \) is given by the donor atom concentration, \( n = 5 \times 10^{15} \) cm\(^{-3}\).
02

Calculate the Hall voltage

Use the Hall voltage formula:\[V_H = \frac{I_x \times B_z}{n \times e \times t}\]where:and \( e \) is the charge of an electron (\( e = 1.6 \times 10^{-19} \) C). Substituting the given values,\[V_H = \frac{250 \times 10^{-6} \times 5 \times 10^{-2}}{5 \times 10^{15} \times 1.6 \times 10^{-19} \times 5 \times 10^{-3}}\]\[V_H \approx 1.56 \times 10^{-6} \text{ V}\]
03

Calculate the Hall field

The Hall field \( E_H \) is given by:\[E_H = \frac{V_H}{W}\]Substituting the found Hall voltage and given values:\[E_H = \frac{1.56 \times 10^{-6}}{2 \times 10^{-2}} \]\[E_H \approx 7.8 \times 10^{-5} \text{ V/cm}\]
04

Calculate the carrier mobility

Carrier mobility \( \mu \) is calculated using the relation:\[\mu = \frac{\sigma}{n \times e}\]First calculate conductivity \( \sigma \):\[\sigma = \frac{I_x}{V_x \times W} = \frac{250 \times 10^{-6} }{0.1 \times 2 \times 10^{-2}} \]\[\sigma \approx 1.25 \text{ S/cm} \]Then calculate mobility:\[\mu = \frac{1.25}{5 \times 10^{15} \times 1.6 \times 10^{-19}}\]\[\mu \approx 1.56 \times 10^2 \text{ cm}^2/(\text{V} \cdot \text{s})\]

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Key Concepts

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

n-type Semiconductor
An n-type semiconductor is a type of material that has been doped to increase the number of free negative charge carriers, specifically electrons. In its pure form, a semiconductor has equal numbers of electrons and holes. This doping process involves adding donor atoms, which have more valence electrons than the semiconductor itself. These extra electrons become mobile charge carriers within the material.
There are a few key points to know about n-type semiconductors:
  • The addition of donor atoms makes electrons the majority charge carriers.
  • The overall charge of the semiconductor remains neutral despite having more electrons than holes.
  • Common materials for n-type semiconductors include silicon and germanium, which are doped with elements like phosphorus or arsenic.
In the original exercise, germanium is doped with donor atoms, thus creating an n-type semiconductor, where the carrier concentration is determined by the number of donor atoms added.
Carrier Concentration
Carrier concentration refers to the number of charge carriers, such as electrons or holes, in a semiconductor. It is a crucial parameter as it directly affects the electrical properties of the material, including its conductivity and the Hall effect.
In an n-type semiconductor, the carrier concentration is primarily the concentration of electrons since they are the majority carriers. Given by the formula:
  • For n-type: \( n = N_d \)
Where \( N_d \) is the donor concentration.In the problem, the germanium is doped with \(5 \times 10^{15}\) atoms/cm\(^3\), meaning this is also the electron concentration.
Carrier Mobility
Carrier mobility is a measure of how quickly charge carriers (electrons in n-type, holes in p-type) can move through a semiconductor material when an electric field is applied. High mobility means that the carriers can move more efficiently, improving the semiconductor's conductivity.
Mobility is defined by the equation:
  • \( \mu = \frac{\sigma}{n \times e} \)
Where:
  • \( \sigma \) is the conductivity.
  • \( n \) is the carrier concentration.
  • \( e \) is the elementary charge, approximately \( 1.6 \times 10^{-19} \) C.
In the context of our original exercise, the calculated mobility reflects how fast electrons can move through the n-type germanium under the given conditions. The higher the mobility, the more conductive the material is.
Hall Voltage
The Hall Effect occurs when a magnetic field is applied perpendicular to the flow of charge carriers in a conductor or semiconductor. This phenomenon is utilized to determine the type of semiconductor, as well as to measure properties like carrier concentration and mobility.
Hall voltage \( V_H \) is generated as a result of the Hall Effect, creating a potential difference across the sides of the material. It can be calculated using the formula:
  • \( V_H = \frac{I_x \times B_z}{n \times e \times t} \)
Where:
  • \( I_x \) is the current through the material.
  • \( B_z \) is the magnetic field strength.
  • \( n \) is the carrier concentration.
  • \( e \) is the charge of an electron.
  • \( t \) is the thickness of the material.
Understanding the Hall voltage is essential in semiconductor physics as it provides key information about the charge carriers involved. In the given exercise, the Hall voltage helps verify the characteristics and behavior of the n-type germanium material.

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Most popular questions from this chapter

Consider silicon doped at impurity concentrations of \(N_{d}=2 \times 10^{16} \mathrm{~cm}^{-3}\) and \(N_{a}=0 .\) An empirical expression relating electron drift velocity to electric field is given by $$ v_{d}=\frac{\mu_{n 0} \mathrm{E}}{\sqrt{1+\left(\frac{\mu_{n 0} \mathrm{E}}{v_{\mathrm{sut}}}\right)^{2}}} $$ where \(\mu_{n 0}=1350 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}, v_{\text {sat }}=1.8 \times 10^{7} \mathrm{~cm} / \mathrm{s}\), and \(\mathrm{E}\) is given in \(\mathrm{V} / \mathrm{cm} .\) Plot electron drift current density (magnitude) versus electric field (log-log scale) over the range \(0 \leq \mathrm{E} \leq 10^{6} \mathrm{~V} / \mathrm{cm}\)

The electron diffusion current density in a semiconductor is a constant and is given by \(J_{n}=-2 \mathrm{~A} / \mathrm{cm}^{2} .\) The electron concentration at \(x=0\) is \(n(0)=10^{15} \mathrm{~cm}^{-3}\). (a) Calculate the electron concentration at \(x=20 \mu \mathrm{m}\) if the material is silicon with \(D_{n}=30 \mathrm{~cm}^{2} / \mathrm{s}\). (b) Repeat part ( \(a\) ) if the material is GaAs with \(D_{n}=230 \mathrm{~cm}^{2} / \mathrm{s}\).

A p-type silicon material is to have a conductivity of \(\sigma=1.80(\Omega-\mathrm{cm})^{-1} .\) If the mobility values are \(\mu_{n}=1250 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\) and \(\mu_{p}=380 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\), what must be the acceptor impurity concentration in the material?

Another technique for determining the conductivity type of a semiconductor is called the hot probe method. It consists of two probes and an ammeter that indicates the direction of current. One probe is heated and the other is at room temperature. No voltage is applied, but a current will exist when the probes touch the semiconductor. Explain the operation of this hot probe technique and sketch a diagram indicating the direction of current for p- and n-type semiconductor samples.

Three scattering mechanisms are present in a particular semiconductor material. If only the first scattering mechanism were present, the mobility would be \(\mu_{1}=2000 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\), if only the second mechanism were present, the mobility would be \(\mu_{2}=1500 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s}\), and if only the third mechanism were present, the mobility would be \(\mu_{3}=500 \mathrm{~cm}^{2} / \mathrm{V}-\mathrm{s} .\) What is the net mobility?

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