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Exam1_sol - ECE 3040 Microelectronic Circuits Exam 1...

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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam 1 February 5, 2010 5 o ”(fish/L Print your name clearly and largely: Instructions: 1. Closed book, closed notes. You are allowed to use 1 sheet of notes (single side, letter size) as well as a calculator. 2. Show all work in order to receive full credit for your answers. Incorrect answers may be awarded partial credit when appropriate. 3. CIRCLE YOUR FINAL ANSWER WITH THE PROPER UNITS INDICATED. Write legibly. If I cannot read it, it will be considered a wrong answer. 4. Do all work on the paper provided. Clearly indicate which problem you are solving on each page. Turn in all scratch paper, even if it did not lead to an answer. 5. There are 10 pages to this exam, including the cover sheet and 3 blank pages. There are 100 total points in this exam. Observe the point value of each problem and allocate your time accordingly. Good luck! Part l:( 30 points) Multiple Choice and TrueIFalse Circle the lctte ' the most correct answer: 1.) (2-points) ru r False: Ionized donors are positively charged. 2.) (2-points) True or-Diffusion current results from movement of electrons and holes in response to an applie Electric field. 3.) (2—points) True or False: A zincblende crystal structure has more atoms in the unit cell than the diamond crystal structure does. 4.) (2-points) .r False: If Magnesium (Mg is group 2 element) is used to dope GaN (Ga is group 3, N is group 5 a p-type semiconductor will result if the Mg replaces a Ga atom. 5.) (2-points) True [email protected] A p—type Silicon semiconductor has more electrons than holes AND has more hole than an intrinsic silicon semiconductor. 6.) (2—points) True ortFalse) The probability of occupying a state located at the fermi-energy is always 1. Select the best answer: 7.) (3-points) A heavily doped p-type semiconductor block is connected to a battery. Which of the following is true: a.) There is no net current in the block. b. The block has a flat Fermi energy 0.) Majority carriers carry most of the current . Electrons carry most of the current 8.) (3-points) Effective mass a. ...is a fundamental constant that does not change from material to material. @. . .takes into account many difference forces acting on the carrier that would otherwise not be present in free space. c.) ...is always less than the mass of an electron in vacuum. d.) .. .of an electron is always smaller than for a hole. 9.) (3- oints) The following energy band diagram indicates the material is: ' . p-type Ec b.) n-type c.) intrinsic El- """""""""""""""""" d.) Silicon Br 13.. 10.) (ii—points) In steady state: a.) The number of electrons equals the number of holes. b. No net current can flow. c. The time rate of change of the system is zero. d.) The minority carrier concentration can not change from position to position. 11.) (3-points) For the electron and hole shown in the following band diagram circle all that are left true. The electron will move to the right b.) The hole will move to the right c. The electron will move to the left ‘The hole will move to the left c.) There is net current flow in this device f.) The device is not in equilibrium rit _.gi+ 12.)(3-points) The conduction band... at) ...is mostly filled with electrons. ®...is mostly empty. c.) ...is lower in energy than the valence band. d.) ...is mostly filled with valence electrons. Part 2: (25 points) Short Answer (“Plug and Chug") For the following problems (13-15) use the following material parameters: ni=1e14 cm'3 ND=8e17 em'3 donors NA:6e12 crn'3 acceptors. 1nn*={).21no Inpi=12mo Effective density of states: Nc=3.75324 cm‘3 Nv=5.51625 cm'3 Electron mobility, in. = 900 cmstec Hole mobility, [JP = 10 cmZNsec Temgemturer- 150 degrees C 13.)(lO-points) Assuming total ionization, what is the electron and hole concentrations and is the material p or n-type? Np 77 NA ND 77m]; I h 2; ND : 8917 CM”3 3 f 1 -— H” WW) P L ~C -—/2~ . - h 8617 _ Sew cm The We'lLE/Vtck/t {,3 thjfe; 14.)(I 0-points) What is the bandgap of the semiconductor material? 14.: '3 INC NV 83 bé/ZKT V11: EC, 1: “ZKT [KW v _ ., m IE/V ‘* Z XQ‘GHX/U $XU§0+27 ii‘ 3) L : L87 6V 15.) (S-points) What is the conductivity of the semiconductor? <7 ‘3 ‘1 (/LLHM‘L MW) ‘: i6 XIDFIWOIW X 99}? “f )0 x;,25‘€/0) : is mfixczooxieii : 219.2 1“ (JAM-CM Part 3: (20 points) 16.) (20-points total) A semiconductor has the following parameters: Hole Mobility, W500 cm2/VSec Substrate relative Dielectric Constant, e,_semimdum=Ks=l1.8 Dielectric Constant of free space, so =8.854e-l4 F/cm Substrate intrinsic concentration, ni=1e10 cm'3 The hole concentration in NON-EQUILIBRIUM in the material is maintained at p(x)=2616e(xno0 ““0 sin-3 a.) (15 points) Calculate the hole current density if an electric field of 5 V/cm is applied across the material. b.) (5 points) Can we determine the electron concentration as a function of position? Why? 0i.) Jr: MFFEUEZPF VF :Siws ~t531-Mp-VF v A? 6 X0) .i Mm xswomxro' EM, x5 " L6 Madly 0‘025‘7 X900 XZWDMQ (2%”) ...L 3(0.0/ :( gmww) QWW Part 4: (25 points) Pulling all the concepts together for a useful purpose 17.) (ZS—points) A 500 pm thick InP semiconductor was put under a light at room temperature (27 degree C) a long time ago. The light uniformly generates 1020 additional holes per cm3 per second. At atime, we call FD, a student trips on the power cord and the light is shut off. Determine the excess electron concentration in the InP for all positions at both KB and t>0. Assume a minori . carrier diffiision length of 0.1 pm and minority carrier diffusion coefficient at room temperature of 3.2 cm lse'c. (my: An -x- x Given: 0: Dn dxz p — F General Solution is: Amp (x): Ag Ln +Be+/L" T" dzAn An -x or Given: 0 = 1),, dxz ” — p + I General Solution iszAnp (x): A8 A +Be A" +61. rn . T" 2 G. . 0 _ D d An!’ - ' . _ wen. — General Solution 15. An (20- A + Bx n dxz p d 2An Given: 0 = D p + G General Solution is: An (x) = sz + Bx+ C n dxz L ' p (1A7! An _ I Given: P = -— P General Solution is: An (t): An (I = 0).? /r' d! r” p F An Given: 0 = -— p + GL General Solution is: Anp = G L 2'” r at I 3123* 65*”) 56C. Extra work can be done here, but clearly indicate with problem you are solving. ._ 20 A“? ‘— 30 x3423 gem) :3’13‘6? CMUB [:37 Jc >0 ...
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