Exam1SolutionsSpring2001

Exam1SolutionsSpring2001 - ECE 3040 Dr. Alan Doolittle...

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Unformatted text preview: ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 ECE 3040B Microelectronic Circuits amp/in: Cm—ern‘a Exam] t . [27% errar' Eu!- [fewer it“; Februaryg’2001 T,‘Me_ RQQMMMé}! lofaéiflr. me-wmzu {err-(.97: C 3 m ad m.nw+ej [Va Clue. = 61002,) Dr. W.Alan Doolittle Print your name clearly and largely: k6 ‘1 Instructions: a Read all the problems carefully and thoroughly before you begin working. You are allowed to use 2 sheets of notes (1 page front and back) as well as a calculator. There are 100 total points in this exam. Observe the point value of each problem and allocate your time accordingly. SHOW ALL WORK AND CIRCLE YOUR FINAL ANSWER WITH THE PROPER UNITS INDICATED. Write legibly. If I can not read it, it will be considered to be a wrong answer. Do all work on the paper provided. Turn in all scratch paper, even if it did not lead to an answer. Report any and all ethics violations to the instructor. Good luck! Sign your name on ONE of the two following cases: I did not observe any ethical violations during this exam: —————————————__—__—__—__—_ I observed an ethical violation during this exam: N6 2:20 59505... E mem mowommom moumufim Em ESQ SGJOQS 1391 JO JaqwnN mm @038 8 $530.8 5% 5% s8 Sm Q8 8 E08 BF 8* 9.: “m2 3 2 NF : S N w mm {mm m3 m2 v.8 93 Q8 3m 58 3m 08 0%“ 5v . 3mm Em $3 EN gmw E awaimmso 639:3? mfoBmEo... v6 mama—.33.". Nd ; i_ f n .358. t: 3 ed #6 _ ,ly wd COBmGDO \_®Q 0.50m fix”. “ or 222m 50> 6395:: mfoimEo; =m @500 E 8 on 3. cm om on ow 2.5. + AquEwum {952:3 *o wmwacwohwavmmmsr n Boom “mm... _ cm 02‘ . ,icgofioflwicn ac Baum Haggai E .53. 91003 1391 ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 First 25%, T rue/F alse and Multiple choice: Circle the correct answer (only one) 1. (2 points) True cw An insulator has a smaller energy bandgap than does a semiconductor. 2. (2 points) True 0 A p-type semiconductor has more electrons than holes. 3. (2 points) True 0 An intrinsic semiconductor has a higher number of donors than acceptors. 4. (2 points) @or False: Antimony (symbol Sb from group 5 of the periodic table) would act as a donor in 51 Icon (symbol Si from group 4 of the periodic table). 5. (2 points) True 0@ Silicon Carbide, SiC, is an example of an elemental semiconductor. (Both Si and C are from group 4 of the periodic table). Select the best answer for 6-10. 6. (3 points) For a plane intersecting the coordinate axes at x=a, y=2a and z=3a, where a is the lattice constant, the Miller indexes are: a.) (123) b. 321 t. enough information to answer this problem. 7. (3 points) When a formerly intrinsic material is converted to an extrinsic material by doping with acceptors, a.) There are more electrons than holes b. The fermi level is above the intrinsic energy @There are fewer electrons than when the material was intrinsic. . None of the above 8. (3 points) The electrons in an intrinsic material come from: a.) Donors b.) Acceptors c Photogenerated by breaking the bonds that hold the atoms in the crystal together. Thermally generated by breaking the bonds that hold the atoms in the crystal together. e.) The electron factory. f.) None of the above 9. (3 points) The effective mass... a.) of an electron is always smaller than the mass of an electron in a vacuum. b.) of an electron is always larger than the mass of an electron in a vacuum. is different from the mass of an electron in a vacuum because it is physically smaller. .. is different from the mass of an electron in a vacuum because it sees different electrostatic potentials. f.) is the same for electrons as it is for holes. ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 10. (3 points) A semiconductor doped with donors has it’s fermi energy, E, above it’s conduction band energy, E. This semiconductor a.) is non-degenerate. b. is a compound semiconductor. $5.. is a degenerate semiconductor. d.) obeys the minority carrier diffusion equations. e.) is really messed up! Second 25%, Short answer 11. (5 points) A semiconductor has 12.5% of the atomic concentration Indium (In) with the rest of the material made up of Aluminum (Al) and Arsenic (As). What is the properly reduced semiconductor formula for this material? (By reduced, I mean put the formula in the format discussed in class, not the standard chemical equation/i) I"n.5"/a Alum. 55W. \fl, ’Calucea/ I»; ’4 Ions A 5 12. (10 points) The figure below and to the right is the zincblende unit cell of the semiconductor in question (1 1). If the smaller atoms are arsenic, label the larger atoms numbered 1, 2, 3, and 4 as Al or In such that the structure is consistent with your answer in question 11. Note, your answer may or may not be unique and atoms 1, 2, 3, and 4 are all completely contained inside the unit cell. Atoml: I,‘ Atomz; or any , 8 a+om5/una'* ‘9'” Cambihafitfi’“ #416045 Muff Atom 3: ,‘kvalvt'nj ZINCBLENDE ya a Txese Atom4' i In + .5 are a” on cuée 3' Li Larje I'h‘l‘em‘df a +0 13. (5 points) In equilibrium, what restriction is placed on the product of the electron,” m (w + be 3 “M, 3 concentration, 11, and hole concentration, p regardless of doping concentrations? “47: "fr; I ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 14. (5 points) If a material is doped with donors, draw the energy band diagram, labeling the conduction band energy E, the valence band energy E, the intrinsic energy E, the fermi energy Ef and the donor energy Ed at (a)... room temperature assuming total ionization (b) ...0 degrees Kelvin . c_,a_____ ,~___.,__~Ea mg, war: 0 E- _ p” _’_ __ _. _ 4‘ h “*9 A~~ E4, Third 25%, Problem Solutions 15. You are given the following information about a semiconductor sample at 27 degrees Celsius: N3 = 1.6615 cm‘3 Nd: 2615 cm‘3 m*n=0.01m0 m*p=5mo Eg=0.5 eV Nc=2.51x1019 (m"‘n / m0)3/2 Nv=2.51x1019 (m"‘p / m0)3/2 un=2000 cmz/V second pp=500 cmz/V second Cross sectional area 0.01 cm2 by 1 cm long a.(8 points) Find the intrinsic, electron and hole concentration (assume total ionization) — _. 0540.03513 3) 14:: New e E’AAL; n; : Erma”(a.m)”°][2.s'lxw'YrWfl e m.“ :- 2.6 5613 6 a 5—)17L7ew):L : 96114 + @el‘t)a #— 4.704): ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 b.(5 points) Find the intrinsic energy. m f ,1. Ex. 52 + gaérw a) e.(4 points) Find the fermi energy position relative to the intrinsic energy. (5,2- 15;) =4TA(%~) %éwfl[% : 0,029? A “'5 3’“ 566444.53 Np "’ "a' d.(4 points) The resistance of a piece of material. .3 I I , F ?(9~6’00 (4.681%) 3‘- 505(613813D féchh + ,affl) __ 6L. 6,570) ‘ R” A 2 m9: e.(4 points) The hole drift velocity of the sample when 1000 volts are applied along the sample’s length. _3 E: VA“ : twav lam A /V}['—' Mao E -= 5790 “709 A111: 5'xl05 Lin/{co ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 Fourth 25%, Problem Solutions 16. (25 points) A single “pixel” from semiconductor camera (a CCD camera) has a 10 pm long region of silicon that is at 27 degrees C, is in non-equilibrium and is uniformly doped. The material is p—type, is doped with 1016 cm'3 acceptors, and has a minority carrier lifetime that is 4 long enoungh to ne lect recombination when com ared with the deminsions of the material. ;’> A C 0 The electron mobility is 500 cmz/V Second and the intrinsic concentration is 1610 cm'3. The material is continu Minority carrier extraction Minority carrier extraction 0 illuminated with light AA for a very long time. The light is uniformly I“ absorbed throughout the volume of the semiconductor. The light generates 1017 cm‘3 per second extra holes. Minority carriers are extracted from the sample at x=0 and x=10 pm maintaining the minority carrier concentration at a constant value equal to the equilibrium minority carrier concentration at both edges of the material. Write an numerical expression (no unknowns) for the minority carrier concentration as a fimction of position. ( The distance variable, “x” should be the only alphabetic variable in your answer. I.E. no unsolved constants. ) AA 0 )lAhp _ A” a 22;: Dn ’Sx—a +61. X=0 X=10um 4A ._ (lei-7 IV?! —7,7;e '9 ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 Bonus: (5 points) In problem 16, could you solve the problem if the equilibrium carrier concentration was given as, p (X): 1 018 e(-x/(2 um)) cm-3 Support your answer. ...
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Exam1SolutionsSpring2001 - ECE 3040 Dr. Alan Doolittle...

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