14 Questions with Answer Exam 1 - Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download 14 Questions with Answer Exam 1 - Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Microelectronic Circuits Exam 1 February 7, 2002 Dr. W, Alan Doolittle Print your name clearly and largely: 55 [ uit ; gn S 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. 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 cannot read it, it will be considered 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: 1 DID NOT observe any ethical violations during this exam: l observed an ethical violation during this exam: First 25% Multiple Choice and True/False (Circle the letter of the most correct answer) 1.) (2-points) True of False: in equilibrium, an electric field can NOT exist, otherwise a current would have fa flow: 2.) (2-points) or False: A semiconductor has an energy bandgap greater than a metal and Jess than an insulator. 3.) (2-points) True of ) To obtain a p-type elemental group four semiconductor, you need to dope with an elemigp roup five of the periodic table. 4.) (2-points) True of In equilibrium, if the electron density is increased by doping, the hole density must also increase to maintain charge balance. 5.) (2-points), CTruc pr False: GaP (gallium phosphide with Ga being a group II element and P being from group V) is a binary compound semiconductor. Select the best answer for 6-10: 6.) G-points) The probability that a state is UNOCCUPIED is given by: a.) The Fermi-Dirak integral of order 1/2 b.) The fermi-distribution function c.) The density of states function (4) (1-f(E)) where f(E)=the fermi-distribution function 7.) (Q-points) Ifa given state at energy E=E; has a density of states equal to 2e6 ‘cm? eV the then number of electrons occupying this state (within an energy range dE) is: 1e6 em”? B.} 206 cm? c.) lem? d.) 0.5 cm” ¢.) This question is not fair! 8.) (3-points) The following energy band diagram indicates the material is: a.) p-type e b.) n-type EE ___...------------------------- intrinsic “) Silicon E, 9.) (3-points) For to the following band diagram, what is known from the information given: a.) The device is bent. There is an electric field in this material ¢.)} There is no current flow in this device E d.) There is no diffusion current in this material. Ey 10.) (3-points) In equilibrium: a,) The drift current is equal in magnitude and in the same direction as the diffusion current. (b.) The drift current is equal in magnitude and opposite in direction as the diffusion current. c.) There is always no drift current. d.) There is always no diffusion current. If you read over the previous tests on the web you are very happy right now because this is the same exact question as last semester.  Pulling all the concepts together for a useful purpose: (4 25%) an AN 14.) (25-points) Light is absorbed in a silicon wafer of thickness 500 um. The wafer is p-type and is uniformly doped with 10'* cm® acceptors. The li cee is absorbed throughout the material according to the function Gr=Groe* 3/sec a where the absorption coefficient, «, is 1000 em’. x=0 is the top op surface of t ihe wafer, Qe 7s x=500 um is the back surface of the wafer. Determine the excess electron concentration in the wafer for all positions if the top surface of the wafer is maintained at an excess electron concentration. of 2e14 cm” while the back surface is kept at an excess electron concentration of 0 cm?? Ta Class =O D Gro = = ledlax? Ons LOemag. dAn, An, addhtcona | Taterminsion * 44 a7 Sb=0 Given: 0= Pe General Solution is: An, (x)= de ‘1 4 Be tt, " ax? T, PB d’An, An xf ay Given: 0=D, ae 2 £4G, General Solution is: An, (x)= de 7 + Be Va +G,2, tT, d’An Given: 0=D, z 2 General Solution is: An, (x) =A+ Bx dx @An Given: 0=D, so +G, General Solution is: An, (x)= Ax? + Br tC dx 2 aA Given: 0=D at +Giof(x) General Solution is: An, (x y= Ee £0 [rst] m0 ” dA An - Given: a => £ General Solution is: An, ()= An, ¢ = Oe fi Given: 0=— 1 General Solution is: An, =G,t, Ty, = 1000 cn X90 ot oO Oo - a a bee ** a4 0, vA A ‘ ax Tw Oe On Loe ——_— X2500 are =O,05 em 5 2 bn eH Gre -~ «xX ax> Oy © o Dr Gre ~ aX 7 = Ot e + aX Dy -XX Anlyd = Bet - Gxe e Dn x Extra work can be done here, but clearly indicate which problem you are solving. Anlez0)= Delton? Qeitem ( -— Gee. Poo ~y. lea del c 10 Ltcoo)* C= Belt cn? Anlk=0,08 em) = O Jea} G~ Sloaiv3er ~ tase - 6 (9.05)= 24 ~ pet B=-6 el em? ew n(x) = Belt ~GelFxX - [ele bn A “ es eee vant wenn tninenatennete enced ennai [raneneranonner L Nee’ Due +o Gp? DD y Generat iar profile has liae te etfeet an Andx) pore. for y Pp Zurn Ahis term is negligizle, .