20085ee2_1_hw5sol

20085ee2_1_hw5sol - Problem set 5 solution Q1 Na= 1015...

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Unformatted text preview: Problem set 5 solution Q1 Na= 1015 cm‐3, Nd=4×1015cm‐3. therefore n type with overall Nd=3×1015 cm‐3. from figure 18 µ=1300. then 1 1.6 Ω Q2 From figure 19, n0_27=1015 cm‐3 at 27 degrees centigrade. Also Therefore normalizing to 17 degrees, we get 10 / / 1.02 10 cm‐3 Q3 1 1 1 1 500 1 900 3.11 10 Therefore µ=321.4 cm2V‐1S‐1. Q4 equating this to zero to get total current of zero we get where E is the built in electric field under thermal equilibrium. Q5 a) Location of EF at T=300k. Two key equations: / Taking ni=1010cm‐3 at 300k, we get EF=0.345eV + Ei. b)Location of EF at T=200k. _ 10 200 300 Therefore ni_200=3.15×109. Therefore EF=0.249eV + Ei. In the above notice that in the equation for EF in terms of Ei both ni and T of kT depend on temperature. Q6 From problem 2, n0=1015 cm‐3 therefore µn=1350. 3 2 . 225.1 / In the above notice that mn=0.26m where m is mass of free electron 9.11×10‐31 kg. 450 Q8 FCC has 4 atoms in one unit cell therfore n=N/V=4/a3=4/(3.61×10‐8)3=85×1021 cm‐3. 43.25 cm2V‐1s‐1. Therefore 64 10 where mn=0.26m where m is the mass of free electron. ...
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