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Exam1Spring2011 - ECE 3040 Microelectronic Circuits Exam I...

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Unformatted text preview: ECE 3040 Microelectronic Circuits Exam I February 21, 20]] Dr. W. Alan Doolittle r- Print your name clearly and largely: 9/} (M ‘l' i a h 5 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: I DID NOT observe an ethical violations durin this exam: 67 E I observed an ethical violation during this exam: First 33% Multiple Choice and TrueiFalse (Circle the letter of the most correct answer or answers) 1.) (3—points) True 0 m e stronger the chemical bond, the smaller the energy bandgap. 2.) (3- points) True 0 W The Pauli- Exclusion principle says that no two electrons, regardless of Spin can have the same energy at the same location in space. 3.) (3~points’or False: If the Fermi-energy were located above the conduction band, there would be mm y occupied states in the conduction band despite the fact the conduction band states are normal] idered to be mostly empty. 4.) (3 —[email protected] False: The intrinsic concentration of electrons/holes results from the extremely un ikely event of multiple phonons (lattice vibrations) adding up to enough energy to break a valence 31v ~ . off the atom. 5.) (3~points) True As doping is added to the semiconductor the average fermi-energy remains constant. 6.) (3~points) True o’l‘he density of states (not the fermi-distribution function) predicts that ' ' e at higher energles j . - lectrons are found. 7.) (3-poiuts) True cufiAuger Recombination occurs mostly at low electron concentrations where electrons have 1‘ e “room” to move around. Select the m answer or answers for 6-10: 8.) (4~points) The minority carrier diffusion equation a. can predict drift currents . é" cannot be used when drift current is present can only determine the majority carrier current. is a simplification of the current continuity equation. e.) is something I really do not understand and thus I might not pass this exam! 9.) (4-points) The probability of occupying a state at energy E=Ef (where E is the fermi-energy) is... a.) ...essentially 1. ...essentially 0. equals 0.5. cl.) not known without knowing the density of states at that energy. 10 )(4- points) The appropriate equation to use for a n—type degenerate semiconductor to determine the electron concentration is: a) . . . b) . . I: 712 Ht? E -i:. ”(1' n=Nce( ’ ‘) nmn,e( ’ ') Second 17% Short Answer (“Plug and Chug”): For the following problems (11-12) use the following material parameters and assuming total ionization: Germanium: ni=lel4 cm'3 (not a mistake) NA=1e15 cm“3 acceptors ml,‘=0.36m0 m,:=0.55m0 EG=0.66 eV Electron mobility, u“ = 1800 cmstee Hole mobility , up = 150 cmzN sec Temperature=27 degrees C 11.)(1 O-points) Where is e fermi energy (relative to the valence band which is referenced to zero new -—~ sé - 55”“ .____--III' _ __. h m 3* ”f ““1. I: p 6“" I; ‘0 “if 0.56 :'§.05fi éel/ Elf 30‘05’3-3- 0,39] l " " '9 ”57,31 R‘CM ”cur—n.1,, um- , z: 3.7 Mgaalthd) / , —-—_ ..._,.___,._« 13.) (20—points total) Defects creating traps: Oxygen makes a donor trap state 0.18 eV below the conduction band, EC, edge in silicon. If the silicon is doped n-tyge and has a _h_ol_e concentration of 2130 cm'3, what is the concentration of occugied oxygen trap states? Assume the concentration of oxygen, [0], is small enough not to affect the overall dOping and is equal to [O]=1el4 cm'3, the energy bandgap is 1.12 eV and the effective density of states for electrons and holes, NC = is 2.83e19 cm'3 and N, is 1.82e19 cm‘3 for silicon, Hint: Since the oxygen defect state energy is referenced from the conduction band, 13.3, it may be helpful to re-write the energy terms in the fermi-distribution fimction referenced from the conduction band energy. l3.)(20-points total) Defects creating traps: Oxygen makes a donor trap state 0.18 eV below the conduction band edge in silicon. If the silicon is doped n-gjpe and has a ELIE concentration of 2130 crn'3, what is the concentration of occupied oxygen trap states? Assume the concentration of oxygen, [0], is small enough not to affect the overall doping and is equal to [O]=1e14 cm'3, the energy bandgap is 1.12 eV and the effective density of states for electrons and holes, NC = is 2.83e19 cm‘3 and NV is 1.82e19 cm‘3 for silicon. -E:/;4r\ — saw"? h;_We 7 )EOT': 5/16 6 1'5 cm" ..,. -4..."qu m_._......-fl--w ”'“‘ l3.)(20-points total) Defects creating traps: Oxygen makes a donor trap state 0.18 eV below the conduction band, EC, edge in silicon. If the silicon is doped n—type and has a m concentration of 2130 cm'3, what is the concentration of occugied oxygen trap states? Assume the concentration of oxygen, [0], is small enough not to affect the overall doping and is equal to [O]=1el4 cm}, the energy bandgap is 1.12 eV and the effective density of states for electrons and holes, NC = is 2.83e19 cm3 and NV is 1.82e19 cm'3 for silicon. Hint: Since the oxygen defect state energy is referenced from the conduction band, EC, it may be helpfiil to re-write the energy terms in the fermi-distribution function referenced from the conduction band energy. 0/ I... a 61/463244 fl WW. /n";’:\ + 5,5 7’ V _. -'5 N0 : 1w 6 r 49/4 fojnemrm QIQ’OflJQV Pulling all the concepts together for a usefiil purpose: 14.) (30-points) In a particular region of a transistor we will study in detail later, called the “base” region, there is a condition established where extra minority carriers are injected (added) at one side while the opposite side has extra minority caniers extracted (removed). If this region is 200 nm in length and the left end (x=0) has a le15 cm"3 more minority carriers than found in equilibrium while the right end ( x=200 nanometers ) has a 1e15 cm' fewer carriers than found in equilibrium, what is the minority carrier diffusion current density in the region. In this “base region”, the minority carrier lifetime is 10 nanoseconds. The device is packaged inside an opaque plastic container so no ago light reaches the device and the device has been operating since Dr. Doolittle was born “'""‘ T. C? (ancient times © ). Assume a minority carrier mobility o 4.0 cm N sec. 3 I +1e15 cm3 An(x=0 nm) Antx=200 nrn)= -1e15 cm3 General Solution is: Anp (x): Ara—y!“ + Be+%” dzAn An _x M Given: 0:1)” 2'” — ‘0 +GIr General Solution is: Anp(x)= A3 A" +Be /"”’ +G1Tn ah: I” ‘ ' dim}, . . Given: 0 = D" 2 General Solution 15: An (x): A + Bx dx P d 2An C II b "a + 5.3 Given: General Solution is: Anp (x) = sz + Bx + C . A" . . G ,1. 14W”) Given: 0=Dn dxzp +Gmf(x) General Solution 13: Anp(x)=[—D—L:Hf(x)dx:|+3x+C E9 dAn An - 1 fl a-tidhfi'ir Given: dt p m — p General Solution is: Anp (t) = Anp (t = 0)e /’ fie I ii? In Anp Given: 0 = — + Gr, T” ‘3 6'“ (5 ~0535 + 1.363 ) z _ 1.5366 Extra work can be done here, but clearly indicate which problem you are solving. , 5 ,x/auae—r :5 X/s,.25~5‘ ,, Ahékawmo'e gmflwo 6 J n 7h ‘3 %[D“] 7% kl \0 5‘04] , 3451919 4/3”“? LINE"; Mme—5 : 0,6e—(fi) Ky, a Bram—Te ‘ wee-56 ___ I; [% €45 #3 (6‘9“; -—- |r “9-196;— 1 X flmm ___~._,,,_fiw____- ,— m __ ' X ‘ .. 1 JA :—1.ll€ Wk? -0516 /352e5 4/013} ...
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