Midterm 2_F07_sol

# Midterm 2_F07_sol - MatE 153 Fall 2007 Midterm 2 SOLUTIONS...

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Unformatted text preview: MatE 153 Fall 2007 Midterm 2 SOLUTIONS 65 Points Total 1.) Sketch the band diagrams for a metal and an intrinsic semiconductor. Label on the appropriate band diagram the following things. (10 points) a. EF0 for metal b. EFi for semiconductor c. Point that is defined as 0 energy d. EC e. EV f. EG g. Χ h. Shade where the electrons are at 0K Your diagrams should look like Figure 4.12 a for the metal and Figure 5.1c for the semiconductor with EFi added in the middle of the band gap. 2.) What is the definition (in words, not equations) for n in a metal? And for n in a semiconductor? (5 points) n in a metal is the electron density (concentration) in the valence band. n in a semiconductor is the electron density (concentration) in the conduction band. 3.) Describe sp3 hydridization and how that relates to the bonding of Si atoms. When Si atoms come together in a solid, the 3s and 3p interact to form new, hybrid orbitals. These 4 new orbitals differ in shape from both the traditional s and p orbitals. The new sp3 hybrid orbitals are 4 lobes that are as far spaced out from each other as possible, forming a tetrahedral. A single Si atom has 1 electron in each of these four sp3 orbitals. Si covalently bonds to four other Si atoms. These bonds are made as sp3 bands from neighboring atoms overlaps (each contributing 1 electron to the band). Because the bands are in a tetrathedral shape around the atom, this leads to Si being tetrahedrally bonded to 4 other Si atoms. 4.) Using the attached data, calculate the probability of finding an electron at kT below the Fermi energy at 500K for Au. You can assume that EF is temperature independent. (5 points) From the given table, EF0 = 5.5 eV (though you didn’t even need this as the EF terms subtract from each other in the equation below) eV kT = 8.62 × −5 500K = 0.043eV K E F 0 − kT = 5.457eV f (E = E F0 − kT ) = 1 1 = = 0.73 ⎛ E − EF ⎞ ⎛ E F 0 − kT − E F ⎞ 1 + exp⎜ ⎟ 1 + exp⎜ ⎟ kT ⎝ kT ⎠ ⎝ ⎠ 1 5.) Using the attached data, calculate the effective number of valence electrons per atom for Au. (10 points) 2/3 ⎛ h 2 ⎞⎛ 3n ⎞ 1.6 × 10 −19 J ⎟⎜ ⎟ = 5.5eV E F0 = ⎜ = 8.8 × 10 −19 J ⎜ 8m ⎟ π eV ⎠ e ⎠⎝ ⎝ ⎛ ⎞ 3/ 2 ⎜ −31 −19 ⎟ ⎛ 8m e E F0 ⎞ ⎛ π ⎞ ⎜ 8 ⋅ 9.11 × 10 kg ⋅ 8.8 × 10 J ⎟ n=⎜ ⎟ ⎜ ⎟= 2 2 ⎟ ⎝h ⎠ ⎝ 3 ⎠ ⎜ 6.63 × 10 −34 Js 2 kg ⋅ m ⎜ ⎟ 2 J ⋅s ⎝ ⎠ # valence / atom ⋅ d ⋅ N A n= M at ( n ⋅ M at # valence / atom = = d ⋅ NA 3/ 2 ) ⎛π⎞ 28 electrons ⎜ ⎟ = 5.85 × 10 m3 ⎝3⎠ electrons g ⋅ 196.97 3 mol m = 0.99 3 g ⎛ 100cm ⎞ 23 atoms 19.32 3 ⎜ ⎟ ⋅ 6.02 × 10 mol cm ⎝ m ⎠ 5.85 × 10 28 6.) Ge is doped with 1015 atoms/cm3 of As. Assume all the dopants are ionized. Using the attached data, determine n, p, ni and EF at 500K. (20 points) From the periodic table that was given, you know that As is in Col V (5 valence electrons). This means that when As substitutes for Ge (col IV, 4 valence electrons), there will be an extra electron. As is a donor. ND = 1015 atoms/cm3 (this uses the assumption given that all dopants are ionized) To calculate ni, you need to find NC and NV at 500K In the calculation, you should use me* and mh* b. The subscript b refers to the value to use when calculating density of states (see caption of Figure 5.1). However, I accepted your answer if you used the a values. ⎛ 2πm * kT ⎞ e ⎟ N C = 2⎜ ⎜ h2 ⎟ ⎝ ⎠ 3/ 2 ⎞ ⎛ J ⎜ 2π ⋅ 0.56 ⋅ 9.1 × 10 −31 kg ⋅ 1.38 × 10 − 23 500K ⎟ K ⎟ = 2⎜ 3 ⎟ ⎜ 2 kg ⋅ m 6.626 × 10 −34 Js ⎟ ⎜ 2 Js ⎠ ⎝ ( ) ⎞ ⎛ J 3/ 2 ⎜ 2π ⋅ 0.4 ⋅ 9.1 × 10 −31 kg ⋅ 1.38 × 10 − 23 500K ⎟ * ⎛ 2πm h kT ⎞ K ⎟ ⎟ = 2⎜ N V = 2⎜ 3 ⎜ h2 ⎟ ⎟ ⎜ 2 kg ⋅ m ⎝ ⎠ 6.626 × 10 −34 Js ⎟ ⎜ Js 2 ⎠ ⎝ Note: You needed to use h and k in Joules to get the units to work out. ( ⎛ − EG n i = N C N V exp⎜ ⎝ 2kT 3/ 2 ) ⎞ ⎟ = 2.25 × 10 25 m −3 ⋅ 1.36 × 10 25 m −3 ⎠ = 8.3 × 10 21 m −3 2 = 2.25 × 10 25 m −3 3/ 2 = 1.36 × 10 25 m −3 ⎛ ⎞ ⎜ ⎟ − 0.66eV ⎟ exp⎜ ⎜ ⎟ − 5 eV 500K ⎟ ⎜ 2 ⋅ 8.62 × 10 K ⎝ ⎠ Note that ni is greater than ND. At this temperature, the material is intrinsic. p=n=ni (If ni had been less than ND, than n would equal ND and you would find p from the mass action law. 2 p= ni n EG ≈ E V + 0.33eV 2 ⎛N ⎞ = E Fi + kT ln⎜ D ⎟ ⎜n ⎟ ⎝ i⎠ Because it is intrinsic, E F = E Fi ≈ E V + If ni had been less than ND, than E F n 3 ...
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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.

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