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hw sol ch12

# hw sol ch12 - 12.1 This problem calls for us to compute the...

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Unformatted text preview: 12.1 This problem calls for us to compute the electrical conductivity and resistance of a silicon specimen. {a} We use Equations (19.3) and (19.4} for the conductivity, as _l_L_ I_' G— ‘ ‘ 2 p A d VTE(§) . —3 _ 0.1a asxlc m :.I4_g{ﬂ_mj-l _ -3 {125 ijf {b} The resistance, R, may be computed using Equations (19.2) and (19.4), as 50.3 x10'3 m 3 =166.9 5.1! 5.1 x10‘ m 2 — [14srrz-mj'1](n) 12.4 Let us demonstrate that, by appropriate substitution and algebraic manipulation, Equation {12.5} may be made to take the form of Equation (12.1). Now, Equation {12.5} is just J = (E But, by definition, J is just the current density, the current per unit cross—sectional area, or J = l.-'A. Also; the electric field is defined by E : WI. Andi substituting these expressions into Equation (12.5) leads to _ v I_ A But, from Equations {12.2} and (12.4} and Solving for V from this expression gives ‘v‘ 2 IR, which is just Equation (12.1). 12.6 When a current arises from a flow of electrons, the conduction is termed electronic; for ionic conduction, the current results from the net motion of charged ions. 12.10 The drift velocity of a free electron is the average electron velocity in the direction of the force imposed by an electric ﬁeld. The mobility is the proportionality constant between the drift velocity and the electric field. It is also a measure of the frequency of scattering events (and is inversely proportional to the frequency of scattering}. 12.14 {a} This portion of the problem asks that we calculate, for gold, the number of free electrons per cubic meter {n} given that there are 1.5 free electrons per gold atom, that the electrical conductivity is 4.3 x 107 {o—mﬂ and that the density (p Au} is 19.32 g.-"cm3. (Note: in this discussion, the density of gold is represented by pAu in order to avoid confusion with resistivity which is designated by p.) Since n : 1.5N, and N is defined in Equation (4.2}, then p' N h = 1.5N 21.5[%] Au _ 1 5 (19.32 g.-’cm3}(8.023 x 1023 atoms-moi} ‘ - 196.9? gi‘mol = 8.86 x1022 cm'3 2 sea x10;28 m'3 {b} Now we are asked to compute the electron mobility, tie. Using Equation {12.8} _ U LLEt—nle _ 4.3 x 107 (ﬂ—mf' _ {8.86 x 1028 m'3}{1.602 3410— .19 = 3.03 x 10'3 mZN—s C} 12.17 We are asked to determine the electrical conductivity of a Cu—Ni alloy that has a yield st ength of 125 MPa (18,000 psi}. From Figure '8.16(b}, the composition of an alloy having this yield 8 strength is about 20 wt% Ni. For this composition; the resistivity is about 2? x 10— Q—m Figure 19.9}. And since the conductivity is the reciprocal of the resistivity; Equation (12.4), we have ‘I ‘I 6 o=—=—_8:3.T0x10 1’ 27x10 Q—m {L'J-m}_1 1223 These semiconductor terms are defined in the Glossary. Examples are as follows: intrinsic—— high purity (undoped) Sl, GaAs, CdS, etc; extrinsic--P-doped Ge, El-doped Sl, S-doped GaP, etc.; compound-GaAs, lnP, CdS, etc; elemental--Ge and Si. 12.28 Nitrogen will act as a donor in Si. Since it (N) is from group VA of the periodic table (Figure 2.6), an N atom has one more valence electron than a Si atom. Boron will act as an acceptor in Ge. Since it (B) is from group IIIA of the periodic table, a B atom has one less valence electron than a Ge atom. Zinc will act as an acceptor in GaAs. Since Zn is from group IIB of the periodic table, it will substitute for Ga; furthermore, a Zn atom has one less valence electron than a Ga atom. Sulfur will act as a donor in lnSb. Since S is from group VIA of the periodic table, it will substitute for Sb; also, an 8 atom has one more valence electron than an Sb atom. Indium will act as a donor in CdS. Since In is from group IIIA of the periodic table, it will substitute for Cd; and, an In atom has one more valence electron than a Cd atom. Arsenic will act as an acceptor in ZnTe. Since As is from group VA of the periodic table, it will substitute for Te; furthermore, an As atom has one less valence electron than a Te atom. 12.31 Using the data in Table 12.2 we are asked to compute the electron and hole concentrations in intrinsic lnSb at room temperature. Since the conductivity and both electron and hole mobilities are provided in the table, all we need do is solve for n and p using Equation (1215). Thus, 0 n: : p leltuewhl : 2 x104 {rt—mg"l {1.602 x 10'19 out? + 0.0?) m2N-s 21.61:»:1O;22 m"3 12.34 This question asks that we compare and then explain the difference in the temperature dependence of the electrical conductivity for metals and intrinsic semiconductors. For a pure metal, this temperature dependence is just [This expression comes from Equations [12.4) and (1210).] That is, the electrical conductivity decreases with increasing temperature. By way of contrast, for an intrinsic semiconductor [Equation (12.18)] E moEc-—1 Or, with rising temperature, the conductivity increases. The temperature behavior for metals is best explained by consulting Equation (12.8] o = n| el Me As the temperature rises, n will remain virtually constant, whereas the mobility (Lie) will decrease, because the thermal scattering of free electrons will become more efficient. Since |e| is independent of temperature, the net result will be diminishment in the magnitude of o. For an intrinsic semiconductor, Equation (19.15) describes the conductivity; i.e., U=dd0g+um=ddu%+um Both n and p will increase with rising temperature. rather dramatically, since more thermal energy becomes available for valence band-conduction band electron excitations. The magnitudes of tie and uh will diminish somewhat with increasing temperature (because of the thermal scattering of electrons and holes), which effect will be overwhelmed by the increase in n and p. The net result is that 0 increases with temperature. “12.41 This problem asks that we estimate the temperature at which GaAs has an electrical conductivity of 3.? x 10—3 (Q—m)‘1 assuming that the conductivity has a temperature dependence as shown in Equation (12.3%). From the room temperature [298 K) conductivity [‘ID43 (Q—mﬂ] and band gap energy (1.42 eV) of Table 12.2 we determine the value of C" [Equation (12.3%)] as E .._ 2 g InC —lno+2|nT+2kT : In [10‘6 {Q—mﬂ] + g In [298 K) + + (2)(8.62 x 10 eV.tK](298 K) = 22.3? Now we substitute this value into Equation (12.3913) in order to determine the value of T for which 6 = 3.? x10'3(Q—m)'1: E _ .. 2 a Ino—lnC —2lnT—2kT In [3] x 10'3 (ta—myl] = 22.37 _ 3 In T _ 1.42 {2)(8162 x 10 eWK)T This equation may be solved for T using the E-Z Solve equation solver. entered into the workspace The following text is |n[3.7‘10"‘—3) = 22.37 - 1.5*|n(T) — ‘I.42f(2*8.ﬁ2*‘|D“—5*T) And when the "Solve new run" button in the toolbar is clicked, the value of T = 43? appears in the data grid. This value is the temperature in K which corresponds to 154°C. ...
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hw sol ch12 - 12.1 This problem calls for us to compute the...

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