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Chapter 18: Problem 30

Germanium to which \(10^{24} \mathrm{m}^{-3}\) As atoms have been added is an extrinsic semiconductor at room temperature, and virtually all the As atoms may be thought of as being ionized (i.e., one charge carrier exists for each As atom). (a) Is this material \(n\) -type or \(p\) -type? (b) Calculate the electrical conductivity of this material, assuming electron and hole mobilities of 0.1 and \(0.05 \mathrm{m}^{2} / \mathrm{V}\) -s, respectively

Short Answer

Expert verified
Answer: Doping germanium with As atoms results in an n-type semiconductor. The electrical conductivity of this semiconductor is approximately \(1.6 \cdot 10^5 \mathrm{S/m}\).

Step by step solution

01

Determine if the material is n-type or p-type

As atoms form Group V elements in the periodic table. When As is added to a germanium semiconductor, it will donate electrons to the semiconductor's conduction band, making it an n-type semiconductor.
02

Write the formula for electrical conductivity

The electrical conductivity (σ) of a semiconductor material can be calculated using the formula: $$\sigma = q \cdot (n \cdot \mu_n + p \cdot \mu_p)$$ where: - q is the elementary charge (\(1.6 \cdot 10^{-19} \mathrm{C}\)) - n and p are the concentrations of electrons and holes, respectively - \(\mu_n\) and \(\mu_p\) are the mobilities of electrons and holes, respectively
03

Calculate the concentrations of electrons and holes

Since the semiconductor is n-type, and virtually all As atoms are ionized, the concentration of electrons (n) will be approximately equal to the concentration of As atoms, which is \(10^{24} \mathrm{m}^{-3}\). Since the material is extrinsic and n-type (due to As doping), the hole concentration (p) can be calculated using the mass action law: $$np = n_i^2$$ Where \(n_i\) is intrinsic concentration of carriers. In extrinsic semiconductors, the relation \(n \gg n_i\) and \(p \approx \frac{n_i^2}{n}\), which means the concentration of holes will be negligible compared to the concentration of electrons in this case.
04

Calculate the electrical conductivity

Now we can plug in the values into the electrical conductivity formula: $$\sigma = (1.6 \cdot 10^{-19} \mathrm{C}) \cdot (10^{24} \mathrm{m}^{-3} \cdot 0.1 \mathrm{m}^{2} /\mathrm{V}\mathrm{s})$$ $$\sigma \approx (1.6 \cdot 10^{5} \mathrm{S/m})$$ The electrical conductivity of this germanium semiconductor doped with As atoms is approximately \(1.6 \cdot 10^5 \mathrm{S/m}\).

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Most popular questions from this chapter

The polarization \(P\) of a dielectric material positioned within a parallel- plate capacitor is to be \(4.0 \times 10^{-6} \mathrm{C} / \mathrm{m}^{2}\) (a) What must be the dielectric constant if an electric field of \(10^{5} \mathrm{V} / \mathrm{m}\) is applied? (b) What will be the dielectric displacement \(D ?\)

For each of the following pairs of semiconductors, decide which will have the smaller band gap energy, \(E_{g},\) and then cite the reason for your choice. (a) \(C\) (diamond) and Ge, (b) AlP and InSb, (c) GaAs and ZnSe, (d) ZnSe and CdTe, and (e) \(\mathrm{CdS}\) and \(\mathrm{NaCl}\)

A parallel-plate capacitor using a dielectric material having an \(\epsilon_{r}\) of 2.2 has a plate spacing of \(2 \mathrm{mm}(0.08\) in.). If another material having a dielectric constant of 3.7 is used and the capacitance is to be unchanged, what must be the new spacing between the plates?

(a) Compute the electrical conductivity of a \(7.0-\mathrm{mm}(0.28 \text { -in. })\) diameter cylindrical sil icon specimen \(57 \mathrm{mm}(2.25\) in.) long in which a current of 0.25 A passes in an axial direction. A voltage of \(24 \mathrm{V}\) is measured across two probes that are separated by \(45 \mathrm{mm}(1.75 \text { in. })\) (b) Compute the resistance over the entire \(57 \mathrm{mm}(2.25 \text { in. })\) of the specimen.

At temperatures near room temperature, the temperature dependence of the conductivity for intrinsic germanium is found to cqual $$\sigma=C T^{-3 / 2} \exp \left(-\frac{E_{g}}{2 k T}\right)$$ where \(C\) is a temperature-independent constant and \(T\) is in Kelvins. Using Equation \(18.36,\) calculate the intrinsic electrical conductivity of germanium at \(175^{\circ} \mathrm{C}\)
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