HW4_Fa11_solutions - MatE 153: HW 4: Solutions Reading...

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Unformatted text preview: MatE 153: HW 4: Solutions Reading Kasap: 4.1-4.8, 5.1 and 5.2 Be sure to look over the reading review notes on D2L. This will help you prepare for the final. Total Points: 80 1. Why is the bandgap of GaAs larger than that of Si? (5 points) GaAs is predominantly ionically bonded (where Ga gains an electron from As). This ionic bonding is stronger than the covalent bonding in Si so it takes more energy to move an electron away from this bonding (out of the valence band to the conduction band). 2. What assumptions are different between Boltzmann and Fermi-Dirac probability functions? (5 points) The key difference in assumptions is that Boltzamm statistics assumes that the site the electron is moving into is empty. Fermi-Dirac has more complicated math because it checks to make sure the site is empty (by using the probability of it being empty). 3. Using Excel, generate a plot of the density of states, g(E), in the valence band for copper from the bottom to the half way point in band. Assume the Fermi energy is the half way point in the band. (5 points) This would be a plot of g(E) from E=0 eV to E= 7eV. You need to have your units in Joules in order for everything to cancel out. ⎛m ⎞ g(E) = 8π 2 ⎜ 2e ⎟ ⎝h ⎠ 3/ 2 ⎛ 9.1 × 10 −31 kg E(J) = 8π 2 ⎜ ⎜ 6.626 × 10 −34 Js ⎝ ⎛ kg ⎞ Here' s a check of the units : ⎜ 2 2 ⎟ ⎝J s ⎠ = ( 3/ 2 J 1/ 2 kg 3 = 33 Js kg kg m kg 2 m 1J 2 s 4 1 kg = 3 4 = 33 2 4 s Js J s 1kg m J ⋅ m3 h 6.63E34 m 9.10E31 ⎞ ⎟ 2⎟ ⎠ ) 3/ 2 E(J) kg 3 (Nm ) = 3 3 Js ⎛ kgm 2 ⎜2 ⎜s ⎝ ⎞ ⎟ ⎟ ⎠ E(ev) E(J) g(E) 0 0 0.00E+00 0.5 8E-20 3.00E+46 1 1.6E-19 4.24E+46 1.5 2.4E-19 5.20E+46 2 3.2E-19 6.00E+46 2.5 4E-19 6.71E+46 3 4.8E-19 7.35E+46 3.5 5.6E-19 7.94E+46 4 6.4E-19 8.48E+46 4.5 7.2E-19 9.00E+46 5 8E-19 9.49E+46 5.5 8.8E-19 9.95E+46 6 9.6E-19 1.04E+47 6.5 1.04E18 1.08E+47 7 1.12E18 1.12E+47 8 7 E(eV) 6 5 4 3 2 1 0 0.E+00 2.E+46 4.E+46 6.E+46 8.E+46 1.E+47 1.E+47 g(m^-3*J^-1) 4. Calculate the average effective speed of the conduction electrons in Au at 0K and 300K. Since Au is valency I metal, its electron concentrations, n are then the atomic concentrations multiplied by the group number, or: Nd 6.022 ×1023 mol−1 19300 kgm −3 nAu = (Valency) A Au = (1) = 5.9 ×1028 m −3 −3 M Au 197 ×10 kg/mol At 0 K, ( 2 ( )( ) h 2 ⎛ 3n ⎞ 3 6.626 × 10 −34 Js ⎛ 3 × 5.9 × 10 28 m -3 ⎞ ⎜ ⎟ E F 0 (Au ) = ⎜ ⎟= ⎟ π 8me ⎝ π ⎠ 8 × 9.1× 10 −31 Kg ⎜ ⎝ ⎠ -19 = 8.863×10 J or 5.54 eV 2 ) 2/3 At 300 K, ⎡ π 2 ⎛ kT ⎞ 2 ⎤ ⎡ π 2 ⎛ 0.02585eV ⎞ 2 ⎤ ⎟ ⎥ = 5.54eV ⎢1 − EF (Au) = EF 0 (Au) ⎢1 − ⎜ ⎜ ⎟⎥ ⎟ ⎜ ⎢ 12 ⎝ EFO (Au) ⎠ ⎥ ⎢ 12 ⎝ 5.54eV ⎠ ⎥ ⎣ ⎦ ⎣ ⎦ EF(Au) = 5.54 eV (essentially unchanged due to the very small temperature ∴ dependence) Therefore, the average speed of the electrons would be the same at both temperatures. You should have used the mean speed of conduction electrons: 1 3 me ve2 = Eav = E 0 2 5 1/ 2 ∴ v e = (6 E F 0 / 5me ) and ve(Au) = (6EF0(Au)/5me)1/2 = ((6×5.54×1.6×10-19 J)/(5×9.1×10-31kg))1/2 ∴ ve(Au) = 1.08×106 ms-1 Or if you solved it using speed at EF: 1 me ve2 = EF0 2 1/ 2 ∴ v e = (2 E F0 / me ) and ve(Au) = (2EF0(Au)/me)1/2 = ((2×5.54×1.6×10-19 J)/(9.1×10-31kg))1/2 ∴ ve(Au) = 1.4×106 ms-1 5. Calculate the density of electrons in the conduction band (n) in the intrinsic semiconductor Ge at 500K. In an intrinsic material, n=p=ni ⎛ − EG ⎞ n i = N C N V exp⎜ ⎟ ⎝ 2kT ⎠ ⎛ 2πm * kT ⎞ e ⎟ N C = 2⎜ ⎜ h2 ⎟ ⎠ ⎝ 3/ 2 ⎛ 2πm * kT ⎞ h N V = 2⎜ ⎜ h2 ⎟ ⎟ ⎝ ⎠ 3/ 2 The values for me* and mh* can be found in Table 5.1 (relative to me). I used the b values because we are calculating a density. To make my calculations easier, I used a spreadsheet (to repeat the calculation quickly at each temperature). For Ge T in K 500 NC in m^-3 2.254E+25 me*/me 0.56 NV in m^-3 1.360E+25 mh*/me 0.4 ni in m^3 8.280E+21 ni in cm^-3 8.280E+15 me in kg 9.10E31 k in J/K 1.38E23 h in Js 6.63E34 Eg iv eV 0.66 6. GaAs is a compound semiconductor used for mobile communications applications. Calculate the difference in the intrinsic carrier concentration (ni) of GaAs at room temperature and operating temperature (around 100ºC). (10 points). I solved this equation in Excel as well (making it very easy to adjust for temperature and answer part b. With mh*/me=0.4: For GaAs T in K 300 me in kg 9.10E-31 NC in m^-3 4.334E+23 me*/me 0.067 NV in m^-3 6.323E+24 mh*/me 0.4 ni in m^-3 1.973E+12 k in J/K 1.38E-23 ni in cm^-3 1.973E+06 h in Js 6.63E-34 Eg iv eV 1.42 With mh*/me=0.5: For GaAs T in K 300 me in kg 9.10E-31 NC in m^-3 4.334E+23 me*/me 0.067 NV in m^-3 8.836E+24 mh*/me 0.5 ni in m^-3 2.332E+12 k in J/K 1.38E-23 ni in cm^-3 2.332E+06 h in Js 6.63E-34 Eg iv eV 1.42 b.) Repeat the calculation at 100°C (a possible operating temperature of your computer chip accounting for outside heat and internal Joule heating of the chip.) (3 points) With mh*/me=0.4: For GaAs T in K 373 me in kg 9.10E31 NC in m^3 6.009E+23 me*/me 0.067 NV in m^-3 8.765E+24 mh*/me 0.4 ni in m^-3 5.897E+14 k in J/K 1.38E23 ni in cm^-3 5.897E+08 h in Js 6.63E34 Eg iv eV 1.42 With mh*/me=0.5: For GaAs me in kg T in K 373 9.10E-31 NC in m^-3 6.009E+23 me*/me 0.067 NV in m^-3 1.225E+25 mh*/me 0.5 ni in m^-3 6.971E+14 k in J/K 1.38E-23 ni in cm^-3 6.971E+08 h in Js 6.63E-34 Eg iv eV 7. 1.42 Using the values of the density of states effective masses me* and mh* in Table 5.1, find the position of the Fermi energy in intrinsic Si and GaAs with respect to the middle of the bandgap (Eg/2). (10 points) ∗ ⎛ me ⎞ 1 3 ⎜ ∗⎟ EFi = Ev + E g − kT ln⎜ ⎟ 2 4 ⎝ mh ⎠ ∗ ∗ For Si, me = 1.08 me and m h = 0.6 me ∴ EFi = Ev + ⎛ 1.08 me ⎞ 1 3 E g − (8.62 × 10 −5 eVK −1 )(300 K ) ln⎜ ⎟ ⎜ 0 .6 m ⎟ 2 4 e⎠ ⎝ ∴ E Fi = Ev + 1 E g − 0.011 eV 2 So intrinsic Fermi level of Si is 0.011 eV below the middle of the bandgap (Eg/2). For Ge, ∗ me = 0.56 me and ∗ mh = 0.4 me ∴ E Fi = Ev + ⎛ .56 me ⎞ 1 3 E g − (8.62 × 10 −5 eVK −1 )(300 K ) ln⎜ ⎜ 0 .4 m ⎟ ⎟ 2 4 e⎠ ⎝ ∴ E Fi = Ev + 1 E g − 0.0065 eV 2 So intrinsic Fermi level of Ge is 0.0065 eV below the middle of the bandgap (Eg/2). For GaAs, ∗ ∗ me = 0.067 me and mh = 0.5 me ∴ E Fi = Ev + ⎛ 0.067 me ⎞ 1 3 E g − (8.62 ×10 −5 eV K −1 )(300 K ) ln⎜ ⎜ 0 .5 m ⎟ ⎟ 2 4 e⎠ ⎝ ∴ E Fi = Ev + 1 E g + 0.039 eV 2 So intrinsic Fermi level of GaAs is 0.039 eV above the middle of the bandgap (Eg/2). 8. Define n-type and p-type dopants. List an element that is a possible n-type dopant in GaAs and an element that is a possible p-type dopant in GaAs. (5 points) A n-type dopant is an impurity that is added to a semiconductor that has one (or more) extra valence electrons than the semiconductor atom they substitute for. The n-type dopants can be ionized to have this extra electron excited to the conduction band. A p-type dopant is an impurity that is added to a semiconductor that has one (or more) less valence electrons than the semiconductor atom they substitute for. The p-type dopants can be ionized to have this hole filled, creating a hole in the valence band. Si (or anything from Col 4) susbtituting for Ga is a possible n-type dopant in GaAs and Zn (or anything in Col 2) substituting for Ga is a possible p-type dopant. ...
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This note was uploaded on 03/13/2012 for the course MATE 153 taught by Professor Staff during the Fall '08 term at San Jose State University .

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