ECE 3040B Microelectronic Circuits Exam 1 - Prof. William Doolittle, Exams of Electrical and Electronics Engineering

Download ECE 3040B Microelectronic Circuits Exam 1 - Prof. William Doolittle and more Exams Electrical and Electronics Engineering in PDF only on Docsity! ECE 3040B Microelectronic Circuits Exam 1 June 7, 2001 Dr. W. Alan Doolittle Print your name clearly and largely: SOLUTIONS 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), your note sheet from the previous exams 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 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! Problem # ==> #1 #2 #3 #4 #5 #6 #7 #8 #9 #10 #11 #12 #13 #14 #15 #16 #17 Bonus Bonus Number of Tests= 45 Point Value of problem= 2 2 2 2 2 3 3 3 3 3 5 5 5 5 5 25 25 10 5 Individual Problem Average= 86.7 100.0 93.3 91.1 75.6 97.8 84.4 95.6 82.2 91.1 92.9 96.9 83.1 88.0 65.8 31.3 41.9 7.6 30.2 Exam Average= 64.5 Exam Standard Deviation= 15.6 Exam Max= 100.0 Exam Min= 34.0 First 25% Multiple Choice and True/False (Select the most correct answer) 1.) (2-points) True or False: Drift current results from movement of electrons and holes from areas of high concentration to areas of low concentration. 2.) (2-points) True or False: Adding acceptors to a semiconductor results in more holes than electrons in the material. 3.) (2-points) True or False: If Carbon (Group 4 element) is used to dope InP (In is group 3, P is group 5), a p-type semiconductor will always result. 4.) (2-points) True or False: The energy band gap in a semiconductor is the energy required to free an electron that normally bonds atoms together, allowing the electron to move through the crystal. 5.) (2-points) True or False: When the fermi-energy is far above an allowed state, the state is probably occupied by an electron. Select the best answer for 6-10: 6.) (3-points) Given Si and Ge are from group 4, In and Ga are from group 3 and P is from group 5, which of the following semiconductors is a binary compound semiconductor? a.) Si b.) Ge c.) In0.47Ga0.53P d.) InP 7.) (3-points) The electron effective mass… a.) …is needed to account for the interaction of the electron with the periodic potentials in the crystal. b.) …is smaller than the mass of an electron in vacuum because the electron is moving at close to the speed of light. c.) …is always larger in compound semiconductor. d.) …is equal to the hole effective mass since they have equal but opposite charge. 8.) (3-points) The following energy band diagram indicates the material is: a.) p-type Ec b.) n-type Ef c.) intrinsic Ei d.) s-type Ev 9.) (3-points) For to the following band diagram, what is known from the information given: a.) The device is leaning on it’s side. b.) There is a non-zero electric field in this material Ec c.) There is no current flow in this device Ev Ei d.) There is no electric field in this material. 10.) (3-points) A plane intersecting the coordinate axes at x=3a, y=3a and z=3a, where a is the lattice constant has which of the following Miller indexes: a.) (100) b.) (111) c.) (666) d.) (333) e.) Forget it, I will just quit school and go sell T-shirts at the beach. Third 25% Problems (3 25%) 16.) (25-points) A semiconductor has the following parameters: Mobility, ,=200 cm?/VSec Substrate relative Dielectric Constant, €; semiconductor Ks=1 1.7 Dielectric Constant of free space, £9 =8.854e-14 F/em Substrate intrinsic concentration, n=1e10 om? Substrate Doping, Na(x)=1e] 5e*68@™/100um) 03 Plot and label (abel the maximum and minimum values) the electric field from x=0 to x=100 um for this material in equilibrium. gs +o solve this, Two way Method AY Almost exactly Ii ke our homework provlen | Ep-E, = -AT L (2) = Ath (MGd)] +47 Ln.) = Arfln(lers) +A (perm rxtored)) «tr, E = p,tAr[ Des) t 46sin(amx/ooun) | ¢4T LCi) €=4 dé =H (2 4.6 cos (ms ) looam AL > 75,86 cos BS) T4986 View Extra work can be done here, but clearly indicate with problem you are solving. Method 2 i Tr equ | ibrium , Jp = Jnr2O, =) Jer q 4p PE ~ 4, De Ve =O Wns 1a” Drift Dit fusion 4 Y vee wT AT. 9 Ejnstein Relationship ~ we é . AT [ elSe 4.6 son (amx/ooum) [cl serena: | y leis @ tein (A mx/ieoum) ty La € = (F) a “(Se Same plots as before. ,,, Also, signiti cand paints given tor a well reasoned qualitative Cro ats) answer arrived at by : Nalx) => Egarlibeiam Band oliag ram = € leesras tat porenvial => E lec#rvc Fre lof Pulling all the concepts together for a useful purpose: (4 25%) 17,) (25-points) Light from two identical light sources is absorbed on BOTH sides of a silicon wafer of thickness 520 um (the wafer is similar to that passed around in class). The wafer is p-type and is uniformly doped with 10’ cm? Steno Sta ‘acceptors. The light has been on since the . dinosaurs walked the earth Light ‘can be approximated as x=-10um > Reston being absorbed uniformly x=0 > within a 10 um region either surface (i.e. no light penetrates past 10 ‘ um into the wafer). The diffusion coefficient for No Light Absorbed i “/Sec.= 0, lates sock inority- Se? cet Gears millisecond (le-3 20, x=510 um > ght Ahsarhed Cletecrons seconds), uw What is the excess 7 electron JO cm hee concentration for all positions in the wafer. Hint: break the problem up into three regions (two of which have identical solutions) and use the x axis pictured above. Your answer should be a numeric expression with x being the only independent variable. (Bonus-10 points) The excess electron concentration changes with position in the wafer. Thus, an electron current density must result. What is the magnitude and direction of this current density at x=125 um from the top surface? (Bonus-5 points) Explain why no net current would flow. d’An, dn - + Given: 0=D, ie Pf General Solution is: An, (x)= Ae th, + Be ty t, @An, An - + Given: 0=D, = 2-—*+G, General Solution is: An, (x)= Ae 4, +Be t +G,T, d’An , Given: 0=D, me ? General Solution is: An, (x)= A+ Bx d’An Given: 0=D, ee P4+G, General Solution is: An, (x)= Ax? +Bx+C dAn An 7) Given: a =-— General Solution is: An, ()=An p(t = Oe 4 f, An Given: 0=-—*+G, General Solution is: An, = G,1, Extra work can be done here, but clearly indicate with problem you are solving. Bonus #2: The only reason why a net Current 15 hot Flowing 15 phat the ligh+ S symetrica ll, applied +o phe water. Thus & current £ lon- $ From the center pewarch rhe. elses and hole current F lous from the edlses powarch the center, prenk More: Unless Mp= Mn, ah Tabet’) A Jn le=e'), (rf Mp > Ah Jaz Tp), Number of Tests Histogram of Test Scores 60 70 Test Score