Solved Exam 1 for Microelectronic Circuits | ECE 3040, Exams of Electrical and Electronics Engineering

Download Solved Exam 1 for Microelectronic Circuits | ECE 3040 and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 ECE 3040B Microelectronic Circuits Grading Criveriqa Exam I ~bY: ‘ . Mark errar but proper we C1) an 9, 2001 Time Regus reek by fotege Almast correct seryas “€ 5) Ae 30 minutes Me Clee = &é 100% ) Dr. W. Alan Doolittle Print your name clearly and largely: ke e V f Instructions: 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: pel 00} 06 08 oz 09 0s or o£ 0z oF S9109§ }Sa] Jo JOqUINN weJ6ojsiH }# Wexy goroeao] sonsners [4 Wexy o o 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. .. is a degenerate semiconductor. d.) ... obeys the minority carrier diffusion equations. e.) ... is really messed up! Second 25%, Short answer 11. 6 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, | mean put the formula in the format discussed in class, not the standard chemical e ation) Triasm Tl yry,"6 50%, Ub reduced Ty, A lars A $ 12. (10 points) The figure below and to the right is the zincblende unit cell of the semiconductor in question (11). 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. Atom 1: Ty : roan Atom 2 Al 0 vi 3 atoms Junie cell Al Com hart a Yn ef avams Muse Atom 3: pavely hg ZINCBLENDE These be GOP ” Al 4 Dn + % Al are all on cube Atom 4: fages and are As. OU Large ‘ntertor ate: 13. (S points) In equilibrium, what restriction is placed on the product of the electron” 5 “must be 3 ranf? 3 concentration, n, and hole concentration, p regardless of doping concentrations? np= ny 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 E, and the donor energy E, at ... (a)... room temperature assuming total ionization _(b) ...0 degrees Kelvin . ——— cl DL OT é SS ee OOO Br o Es - -- 2. - ee 7 ert er mr te E v Third 25%, Problem Solutions 15. You are given the following information about a semiconductor sample at 27 degrees Celsius: N, = 1.615 cm? Ne 2e15 cm? m*,=0.01m, * 5m, E,=0.5 eV N=2.51x10" (m*, / m,)*” N. .=2.51x10" (m*, / m, .2 1,=2000 cm?/Vsecond 1,=500 cm’/Vsecond Cross sectional area 0.01 cm’ by 1 cm long a.(8 points) Find the intrinsic, electron and hole concentration (assume total ionization) ~ 0.5/f0.09592) Nizf NM e Pe, nj = [psi (a0) JL asixo'(s)) e qcsege 5 Gs ae o)+ d. Jeu) - pele t @el)* + (i. Ze 14)* 7 ECE 3040 Dr. Alan Doolittle Microelectronic Circuits Exam #1 b.(5 points) Find the intrinsic energy. m x af e- & ,2g4rh(% as 10.05 (0.0254) Bn (4) EE 037 ev | Ff Es A, ¢.(4 points) Find the fermi energy position relative to the intrinsic energy. Ee - Es) = AT Ly (ss) ) # bt be ( = 0,0259 Ly (eben Fr-Ex = 0.0258 eV d.(4 points) The resistance of a piece of material. = | - | ? 4 (2000 (4.620) + 500 (6,3613)) gan + 4pf) = 657 Jb- cm _ pb 667t) 4 R A 0.01 e.(4 points) The hole drift velocity of the sample when 1000 volts are applied along the because Nyx any sample’s length. E> Vi, 2 1000V | crn V7 2 Ap E = 500 (1900) VI > 5 x10” om [see