Solved Questions for Exam Two - Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download Solved Questions for Exam Two - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B Microelectronic Circuits Exam 1 September 20, 2001 Dr. W. Alan Doolittle Print your name clearly and largely: Sala Hong ne Instructions: Read all the problems carefully and thoroughly before you begin working. You are allowed to use 1 new sheet of notes (1 page front and back) as well as a calculator. There are 100 total points in this exam plus 5 point bonus. 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. If1 cannot read it, it will be considered 4 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: LDID NOT observe any ethical violations during this exam: T observed an ethical violation during this exam:  0% SUA rex oor sxe wexg 68t SUOpEAIC prEpuNs WEXy 9"8s =oSusoay weXy 0S o's. |zbh Joos [rox |9°s9 |9'28 |r'08 |9's6 098 Jo'r9 |0'96 |o'89 Jo'z9 Jo’86 "06 Jove |o'96 ors =eSesoay wa]qod TEMPLAIPUT Is s ee ISS AS SS IS Oe ee ee eee ee le ie fe ee fe =mraiqoad Jo onpe, yuOd os 8989, Jo Joquinyy snuog |snuog | L1# | 914 | st# | pr# | ere | zi# | 11H | ore | 6 | gH se _| ve | ce | cw | te <= #woiqo S110, 88419} 3109 }S01 S$9J09S JO JOqUINN 14.)(5-points) If 10,966Volts is placed across the resistor in part 13, what are the electron current density, hole current density and the electron drift velocity? (If you found no answer in part 13, then use a resistor length of 1 cm in your calculation). Myre: No Abasion | Je + ) Jez 4 eE)n n E)n NANT” Er cto ov 0, Bem J.- (\.6e-14) (300 (75.75 7)(6.46e15) Tn= 62,6» Shen? Tp = (1be-!4) (900) C78.787) (1. Hele) Je = 37.3 Alen? 15.}(5-points) A semiconductor is doped with 5¢19 cm” very deep acceptors (large binding energy) which are only partially ionized at room temperature. If the fermi energy is 0.leV above the valence band, and the acceptor energy (Ea) is 0.16 eV above the valence band, what is the number of ionized acceptors in this material? You can assume the degeneracy factor, g,=4. Hint: do not make this problem harder than it is. - Na Na = TV hone tone MAT ~ sel __ — Tha gos 0.) /0.0a54 Ne = |. eg cm? Third 25% Problems (3" 25%) 16.) (25-points total) A semiconductor has the following parameters: _ Hole Diffusion coefficient, D,={cm’/Sec b f Op Hole Mobility, 1,=200 cm?/VSee “+ §.1% ——- so Substrate relative Dielectric Constant, &;.semiconduoto=Kg=11.7 y 4p Dielectric Constant of free space, &o =8.854e-14 F/em Substrate intrinsic concentration, n=1e10 cm? The hole concentration in NON-EQUILIBRIUM in the material is maintained at p(x}FlelSe@!™™ om? «for X20 +O Xz Odum a.) (20 points) Plot and label (label the maximum and minimum values) the hole current density if an electric field of 108% V/cm is applied across the material. b.) (5 points) Explain why we cannot determine the electron concentration as a function of position. a) p> 4 #0 ~ 4 Dp Wp = (.6e-19) x) (\0 d[ ese (.6e19) (58) | 2 | = [ora ~ 0.083% | ¢ X/a.o\em = O.2.37e x /o.0\ tm Alen* x/0.0) “] _ x20 Kz 0,0fFem b) Ln non~ equilibrium, hp J nt Pulling all the concepts together for a useful purpose: (4 25%) 17,) (25-points) Light is absorbed in a silicon wafer of thickness 500 um (the wafer is similar tg that passed around in class). The wafer is p-type and is uniformly doped with 10 cm? nccepiors. The Hight The light light has been on for a very lon; time and can be approximated as being “ASTER Ue TG T the cess HOLE generation rate is 10 cm/sec and the minority jer lifetime is 1 milliseconds (1e-3 seconds); mM, = 1000 om Meee NMS uniform a.) (4 points) Should the absorption coefficient be large or small for such a condition to @ Ssorption occur? . . wee Ww un form b.) (5 points) What is the excess electron concentration for all positions in the wafer. ©.) (16 points) What would be the electron concentration as a function of time for all An times after the light is turned off? Bonus: (5 points-no partial credit on this bonus} Determine what electron current density flows in the material. Support your answer thoroughly for credit. @An, An, - + Given: 0= D,—>* a OF General Solution is: An,(x)= de 7 + Be’? dAn, dn - Given: O= D, me —4+6,° General Solution is: An,(x)= de” + Be’ +.G,1, d’An Given: 0=D,— 5" General Solution is: An, (x)= A+ Bx . d’An, ., 2 Given: 0= D, 5" +G, General Solution is: An, (x)= Ax? + Bx+C dAn An - Given: a = 7 General Solution is: An, (t)= An, (= O)e 4 An, Given: 0=-—*+G, General Solution is: An, =G,1, Almost Toleniseee | te the problen, done: in class} | = FR 2° a) T= Toe Vis x ert fe inal for, wea k absorption ae 3 COA DE,, fav by 3 ou Oh 1 - Se +6. Arp= & B, dng = 10 "len en? )(le- 3 sec) G) See next (age