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Chapter 18: Problem 39

Estimate the electrical conductivity, at \(75^{\circ} \mathrm{C}\) of silicon that has been doped with \(10^{22} \mathrm{m}^{-3}\) of phosphorus atoms

Short Answer

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Question: Estimate the electrical conductivity of silicon doped with \(10^{22}\,\mathrm{m^{-3}}\) of phosphorus atoms at \(75^{\circ} \mathrm{C}\). Answer: The estimated electrical conductivity of silicon doped with \(10^{22}\,\mathrm{m^{-3}}\) of phosphorus atoms at \(75^{\circ} \mathrm{C}\) is approximately \(101.44\, \mathrm{S/m}\).

Step by step solution

01

Understand given information and what we need to find

We are given: 1. Doped silicon with phosphorus atoms 2. Doping concentration: \(N_d = 10^{22} \mathrm{m}^{-3}\) 3. Temperature: \(T = 75^{\circ} \mathrm{C}\) 4. Conductivity: \(\sigma\) (our goal)
02

Write down the formula for electrical conductivity

The formula for electrical conductivity is: \(\sigma = q \cdot N_d \cdot \mu_n\), where: - \(q\) is the elementary charge, \(q = 1.6 \times 10^{-19} \mathrm{C}\) - \(N_d\) is the concentration of phosphorus atoms (donor impurities) which contribute electrons to the conduction band. - \(\mu_n\) is the electron mobility
03

Calculate the mobility of electrons at the given temperature

For silicon, the electron mobility can be estimated using the following equation: \(\mu_n = \frac{\mu_{n0}}{1+ \frac{T}{T_0}}\) Here, we will use the typical values of silicon electron mobility at room temperature (\(\mu_{n0} = 0.13 \, \mathrm{m^2/Vs}\)) and the temperatures \(T = 75^{\circ} \mathrm{C} = 348 \mathrm{K}\), and \(T_0 = 300 \mathrm{K}\). Now, we can calculate the electron mobility: \(\mu_n = \frac{0.13}{1+ \frac{348}{300}} = 6.34 \times 10^{-2} \, \mathrm{m^2/Vs}\)
04

Calculate the electrical conductivity

Now, we can use the formula for electrical conductivity \((\sigma)\) and substitute the values, \(\sigma = q \cdot N_d \cdot \mu_n = (1.6 \times 10^{-19}\,\mathrm{C}) \cdot (10^{22}\,\mathrm{m^{-3}}) \cdot (6.34 \times 10^{-2} \,\mathrm{m^2/Vs}) = 101.44 \,\mathrm{S/m}\) Hence, the estimated electrical conductivity of silicon doped with \(10^{22}\,\mathrm{m^{-3}}\) of phosphorus atoms at \(75^{\circ} \mathrm{C}\) is approximately \(101.44\, \mathrm{S/m}\).

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Most popular questions from this chapter

(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.

(a) In your own words, explain how donor impurities in semiconductors give rise to free electrons in numbers in excess of those generated by valence band-conduction band excitations. (b) Also explain how acceptor impurities give rise to holes in numbers in excess of those generated by valence bandconduction band excitations.

(a) Calculate the number of free electrons per cubic meter for silver, assuming that there are 1.3 free electrons per silver atom. The electrical conductivity and density for Ag are \(6.8 \times 10^{7}(\Omega-\mathrm{m})^{-1}\) and \(10.5 \mathrm{g} / \mathrm{cm}^{3}, \mathrm{re}\) spectively. (b) Now compute the electron mobility for Ag.

Compare the temperature dependence of the conductivity for metals and intrinsic semiconductors. Briefly explain the difference in behavior.

We noted in Section 12.5 (Figure 12.22 ) that in FeO (wüstite), the iron ions can exist in both \(\mathrm{Fe}^{2+}\) and \(\mathrm{Fe}^{3+}\) states. The number of each of these ion types depends on temperature and the ambient oxygen pressure. Furthermore, we also noted that in order to retain electroneutrality, one \(\mathrm{Fe}^{2+}\) vacancy will be created for every two \(\mathrm{Fe}^{3+}\) ions that are formed; consequently, in order to reflect the existence of these vacancies the formula for wüstite is often represented as \(\mathrm{Fe}_{(1-x)} \mathrm{O}\) where \(x\) is some small fraction less than unity. In this nonstoichiometric \(\mathrm{Fe}_{(1-x)} \mathrm{O}\) material, conduction is electronic, and, in fact, it behaves as a \(p\) -type semiconductor. That is, the \(\mathrm{Fe}^{3+}\) ions act as electron acceptors, and it is relatively easy to excite an electron from the valence band into an \(\mathrm{Fe}^{3+}\) acceptor state, with the formation of a hole. Determine the electrical conductivity of a specimen of wüstite that has a hole mobility of \(1.0 \times 10^{-5} \mathrm{m}^{2} / \mathrm{V}\) -s and for which the value of \(x\) is \(0.040 .\) Assume that the acceptor states are saturated (i.e., one hole exists for every \(\left.\mathrm{Fe}^{3+} \text { ion }\right) .\) Wüstite has the sodium chloride crystal structure with a unit cell edge length of \(0.437 \mathrm{nm}\)
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