HW6_Fa11_solutions - MatE 153: HW 6 Solutions Reading...

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Unformatted text preview: MatE 153: HW 6 Solutions Reading Kasap: 5.5-5.6, 8.1-8.2 Be sure to look over the reading review notes on D2L. This will help you prepare for the final. Total Points: 1. Owing to a concentration gradient of donor impurities in a Ge sample there is a built in voltage at room temperature between the ends of the Ge sample of 0.125V. The local resistivity at the high impurity end is 10Ωm. Find the local resistivity at the other end. (10 points) To solve this problem, you need to recognize that the voltage comes from a dn dV dn = −enμ e + eD e =0 dx dx dx kTμ e dn dV e = enμ e e dx dx V2 kT n 2 dn ∫V1 dV = e ∫n1 n kT ⎛ n 2 ⎞ V2 − V1 = ln⎜ ⎟ e ⎜ n1 ⎟ ⎝⎠ J e = enμ e Ε x + eD e ⎛ 1.6 × 10 −19 C(0.125V ) n2 1eV 1J ⎞ ⎛ e(V2 − V1 ) ⎞ ⎟ = 125.9 = exp⎜ ⎟ = exp⎜ −19 ⎜ 0.02585eV n1 kT 1.6 × 10 J C ⋅ V ⎟ ⎝ ⎠ ⎝ ⎠ Let’s call the known (high doped, low res.) end (#1) and the unknown, (low doped, high res) end (#2). Recognizing that there is a relationship between doping concentration and resistivity we have: ρ eμ n2 = 1 e1 n1 ρ 2 eμ e 2 Now if we assume that the mobility is the same at each end (may or may not be a good assumption), then: V2 − V1 = kT ⎛ ρ1 ln⎜ e ⎜ ρ2 ⎝ ⎞ ⎟ = −0.125V ⎟ ⎠ ρ1 ⎡ eV ⎤ = exp ⎢ 2−1 ⎥ ρ2 ⎣ kT ⎦ ⎡ 0.125eV ⎤ = 125.9 ⋅ 10Ωm = 1.26kΩm ⎣ 0.02585eV ⎥ ⎦ ρ 2 = ρ1 exp ⎢ Our assumption that the mobility was unchanged probably wasn’t very good! With two orders of magnitude change in doping, there probably would be a change in mobility as seen in Figure 5.19; there could be as much as an order of magnitude change in the mobility in fact.; but it depends on what the actual doping was so to get a more accurate answer you would need to solve this iteratively. 2. You have two samples: Sample A is Si at 300K and Sample B is GaAs at 300K. Assume equal minority carrier lifetimes for electrons and holes in Si and GaAs (τnSi=τpSi=τnGaAs=τpGaAs=10-7 s). Note: each one of your answers below should have two parts: one for Si and one for GaAs a. If both Sample A and Sample B are p-type, what are the minority carrier diffusion lengths in Si and GaAs? (5 points) Assume a doping concentration if 1015 atoms/cm3. (Stated in part c) Assume GaAs acts like Si with 1015 dopants , that is it gives a μ close to intrinsic value. In p-type samples, the minority carriers are electrons. D e (Si ) = kTμ e = e 1.38 × 10 − 23 cm 2 J CV ⋅ 300K ⋅ 1350 2 Vs = 34.9 cm KJ s 1.6 × 10 −19 C cm 2 ⋅ 10 −7 s = 1.87 × 10 −3 cm L Si = D e τ e = 34.9 s D e (GaAs) = kTμ e = e 1.38 × 10 − 23 L GaAs = D e τ e = 34.9 J CV cm 2 ⋅ 300K ⋅ 8500 2 KJ Vs = 219.93 cm s 1.6 × 10 −19 C cm 2 ⋅ 10 −7 s = 4.69 × 10 −3 cm s The mobility was approximated from Figure 5.19. approximately the intrinsic value. At such low doping, the value is b. If 1014 electrons/cm3 are injected at x=0, what are the diffusion current densities in Si and GaAs at x=0? Hint: You need to differentiate Δn(x) to get an expression for dn/dx. Solve this expression at x=0 to get J. (10 points) dΔn p J e,diffusion = eD e dx ⎛−x⎞ Δn p = Δn n (0) exp⎜ ⎜L ⎟ ⎟ ⎝ n⎠ dΔn p dx =− ⎛−x⎞ Δn n (0) exp⎜ ⎜L ⎟ ⎟ Ln ⎝ n⎠ dΔn p ( x = 0) −0 1014 electrons / cm 3 ⎛ ⎞ 16 4 (Si ) = − exp⎜ ⎟ = −5.35 × 10 electrons / cm −3 −3 dx 1.87 × 10 cm ⎝ 1.87 × 10 cm ⎠ dΔn p cm 2 J e,diffusion (Si) = eD e = 1.6 × 10 −19 C ⋅ 34.9 ⋅ (− 5.35 × 1016 electrons / cm 4 ) s dx A C = −0.299 = −0.299 2 2 cm s ⋅ cm 14 3 dΔn p ( x = 0) −0 ⎞ 16 4 (GaAs) = − 10 electrons / cm exp⎛ ⎜ ⎟ = −2.13 × 10 electrons / cm −3 −3 dx 4.69 × 10 cm ⎝ 1.87 × 10 cm ⎠ dΔn p cm 2 J e ,diffusion (GaAs) = eD e = 1.6 × 10 −19 C ⋅ 219.93 ⋅ (− 2.13 × 1016 electrons / cm 4 ) dx s A C = −0.75 = −0.75 2 cm s ⋅ cm 2 Due to the increased mobility, the diffusion current is much higher in GaAs. c. If there are 1015 holes/cm3, what electric field would be required to give the same current density in Si and GaAs as in part b? (5 points) J h ,drift (Si) = epμ h E = 1.6 × 10 E = 4.15 −19 ( ) A cm 2 E = −0.299 C ⋅ 10 holes / cm 450 Vs cm 2 15 3 V cm J h ,drift (GaAs) = epμ h E = 1.6 × 10 −19 C ⋅ (10 )400 cm Vs 2 15 holes / cm 3 E = −0.75 A cm 2 V cm 3. InGaN is famous as a blue LED E = 11.7 a.) What is it’s bandgap range? (5points) I calculated this from the range of blue light. ΔE = hc λ = 6.6 ×10 −34 Js ⋅ 3 ×108 m / s 6.6 ×10−34 Js ⋅ 3 ×108 m / s to 492 ×10 −9 m 455 ×10 −9 m = 4.05 × 10 −19 J to 4.37 × 10 −19 = 2.53eV to 2.74eV b.) What are some possible n-type and p-type dopants in this material? (5 points) There are many answer from this based on the periodic table. In and Ga are Col III (3 valence electrons). Substitutional dopants for In/As from Col II (such as Hg, Cd) would be p-type dopants. Substitutional dopants for In/As from Col IV (such as Si) would be n-type dopants. N is Col V (5 valence electrons). Substitutional dopants for N from Col IV (such as Si) would be p-type dopants. Substitutional dopants for N from Col VI (such as Te) would be n-type dopants. c.) In your own words, describe how the built-in voltage gets set up. How do you think you could reduce the space charge area? (5 points) Majority carriers in the n-type (nn) diffuse down a concentration gradient into the p-type region. Holes from the p-type (pp) go in the opposite direction. When the carriers leave the dopant atoms, they leave behind an ionized dopant. This electric field causes a drift of carriers back in the opposite direction. Diffusion continues until the drift current is equal an opposite to it. At this point, no more carriers diffuse away so the voltage (from the ionized dopants) remains constant (the built in voltage). The area in the junction where the dopants are ionized (depletd of carriers) is called the space charge region or depletion region. This area can be reduced if the dopant density is higher (so it would take less volume to ionize the same number of dopants and create the same voltage). d.) Read about InGaN LEDs and summarize, in your own words, one interesting fact about how they are fabricated. Reference your source ( 5points) Your answers of course will vary here. There are many, many papers out there about fabricating InGaN LEDs. Here is a paper by Shuji Nakamura, inventor of the blue LED: http://140.120.11.121/~lcli/file/periodical/GaN/Bychikhin/ref2NakamuraJJAP1995.pdf 4. What class of magnet are refrigerator magnets. Briefly explain why you chose that class. (5 points) Refrigerator magnets have a permanent magnet (which is why the stick to your fridge). The classes of permanent magnets are ferromagnets or ferrimagnets. Refrigerator magnets are most likely made of ferromagnets because they are more common (cheaper). ...
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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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