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Chapter 18: Problem 25

An \(n\)-type semiconductor is known to have an electron concentration of \(3 \times 10^{18} \mathrm{~m}^{-3}\). If the electron drift velocity is \(100 \mathrm{~m} / \mathrm{s}\) in an electric field of \(500 \mathrm{~V} / \mathrm{m}\), calculate the conductivity of this material.

Short Answer

Expert verified
Question: Determine the conductivity of an n-type semiconductor with an electron concentration of \(3 \times 10^{18} \mathrm{~m}^{-3}\), an electron drift velocity of 100 m/s, and an electric field of \(500 \mathrm{~V/m}\). Answer: The conductivity of the n-type semiconductor is \(9.6 \times 10^{-3} \mathrm{~S/m}\).

Step by step solution

01

Write the expression for current density

The current density (J) can be expressed as the product of the charge of the electron (e), electron concentration (n), and electron drift velocity (vd), as shown below: $$ J = n e v_d $$
02

Write the expression for the electric field

Using Ohm's Law, we relate the voltage across the material (V) to the current density (J) and the conductivity (σ), as shown below: $$ J = \sigma E $$ where E is the electric field.
03

Write the expression for the conductivity

From Steps 1 and 2, we can now express the conductivity (σ) in terms of the electron concentration (n), electron drift velocity (vd), and electric field (E): $$ \sigma = \frac{J}{E} = \frac{n e v_d}{E} $$
04

Substitute the given values

Now, we can substitute the given values for electron concentration (\(3 \times 10^{18} \mathrm{~m}^{-3}\)), electron drift velocity (100 m/s), and electric field (\(500 \mathrm{~V} / \mathrm{m}\)) into the expression for the conductivity (σ). $$ \sigma = \frac{(3 \times 10^{18} \mathrm{~m}^{-3})(1.6 \times 10^{-19} \mathrm{~C})(100 \mathrm{~m/s})}{500 \mathrm{~V/m}} $$
05

Calculate the conductivity

Now that all the values are plugged in, we can solve for the conductivity (σ): $$ \sigma = 9.6 \times 10^{-3} \mathrm{~S/m} $$ The conductivity of this n-type semiconductor is \(9.6 \times 10^{-3} \mathrm{~S/m}\).

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Most popular questions from this chapter

Consider a parallel-plate capacitor having an area of \(2500 \mathrm{~mm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\), and with a material of dielectric constant \(4.0\) positioned between the plates. (a) What is the capacitance of this capacitor? (b) Compute the electric field that must be applied for \(8.0 \times 10^{-9} \mathrm{C}\) to be stored on each plate.

(a) For each of the three types of polarization, briefly describe the mechanism by which dipoles are induced and/or oriented by the action of an applied electric field. (b) For solid lead titanate \(\left(\mathrm{PbT} \mathrm{O}_{3}\right)\), gaseous neon, diamond, solid \(\mathrm{KCl}\), and liquid \(\mathrm{NH}_{3}\), what kind(s) of polarization is (are) possible? Why?

A charge of \(3.5 \times 10^{-11} \mathrm{C}\) is to be stored on each plate of a parallel-plate capacitor having an area of \(160 \mathrm{~mm}^{2}\left(0.25 \mathrm{in} .^{2}\right)\) and a plate separation of \(3.5 \mathrm{~mm}(0.14 \mathrm{in}\).). (a) What voltage is required if a material having a dielectric constant of \(5.0\) is positioned within the plates? (b) What voltage would be required if a vacuum were used? (c) What are the capacitances for parts (a) and (b)? (d) Compute the dielectric displacement for part (a). (e) Compute the polarization for part (a).

Is it possible for compound semiconductors to exhibit intrinsic behavior? Explain your answer.

Using Equation \(18.36\) and the results of Problem \(18.33\), determine the temperature at which the electrical conductivity of intrinsic germanium is \(22.8(\Omega \cdot \mathrm{m})^{-1}\)
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