ECE 3040 Microelectronic Circuits. Exam 1. February 20, 2012

Transcription

1 ECE 3040 Microelectronic Circuits Exam 1 February 20, 2012 Dr. W. Alan Doolittle Print your name clearly and largely: 5tJ(u 'f-:()/0 Instructions: Read all the problems carefully and thoroughly before you begin working. You are allowed to use 1 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. IfI 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. A periodic table is supplied on the last page. 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:

2 First 33% Multiple Choice and True/False (Circle the letter of the most correct answer or answers) 1.) (3-points False: The Pauli Exclusion Principle is the basis of the energy bandgap formatioch energy state in a crysta1 much split into a different energy thus, forming a band of sat call either the valance band or conduction band. 2.) (3-points True r False: Atoms with large chemical bond strengths usually result in large energy ban gaps and low mobility (due to frequent collisions resulting from high atomic density). 3.) (3-points@r False: The fermi distribution function predicts that for a non-degenerate doped semiconductor the states in the conduction band will be mostly empty. 4.) (3-points) True 0@Ino.23Gao.47N0.3 is a valid semiconductor formula in standard semiconductor not 5.) (3-points) True o The law of mass action for a wide bandgap semiconductor with a small intrinsic cation can predict a negative electron concentration. 6.) (3-pointsr False: The density of states far away (in energy) from the band edges has a decreasing exponent" form. 7.) (3-points) True 0 False Impact ionization requires accelerating an electron to higher kinetic energy where it col I es with other electrons forcing a recombination event between three particles (example - 2 electrons and 1 hole). Select the best answer or answer for 8-10: 8.) (4-points) Which of the following are true about the drift velocity it depends on the current flowing in the material... it is proportional to electric field but only at low electric fields c....it plateaus to a constant value at high electric fields... it determines the diffusion current through the use of the Einstein equation c:»... in some semiconductors, it can have a "peak value" at an intermediate electric field. 9.) Moints) Select all of the following that are true. C!0)The material is in equilibrium Ee. The material is in steady state non-equilibrium c. The material has non-uniform doping Ef=FN=Fp The material has uniform doping e.) The material has a non-zero electric field everywhere Ev The material has zero electric field everywhere The material clearly has a net current flow h.) The material clearly has no net current flow Something is wrong since the Fermi-level in above Ec in parts of the material C!f j.) There are too many choices in this problem for only 4 points. 1O.)(4-points) The following energy band diagram indicates the material is: a.) In equilibrium b.) Degenerate and n-type c.) Degenerate and p-type. Non-degenerate n-type e Non-degenerate p-type In low level injection g.) In high level injection E h.) In steady state Et=Fp Ev + 3kT...,...,..., ,

3 t f 12.) Second 17% Short Answer ("Plug and Chug"): For the following problems (11-12) use the following material parameters and assuming total ionization: ;J."e/ nicm-3 No=2e15 cm- 3 donors N A =le15 cm- 3 acceptors m p =0.5511lo m o =0.36m o Ea=0.66 ev Electron mobility, Iln = 2200 cm 2 Nsec Hole mobility, IJp = 500 cm 2 N-sec Temperature=27 degrees C 11.)(7-points) Where is the fermi energy (relative to the valence band which is referenced to zero energy)? -J E./ -: '!&?Y.Clh-S - h? IV" - NIr J[ tv!) - f'lrrj,;)_ I e /'c j-h, ::l E +!;/;,T{/) '). it ith, - CI 'f31ev h ::: le 1'- -::: f? -:;!'I l. ;: ().,If(:' 13), r. Ie 15 :::. 71 f'le II J 1<6 e 13 e (1:'/:'- tj,53f]j.3')/.1.y t.f':: [) (4- '3 e V \ -..-I _ (!J r IfJ ';: 7, e (I -:: J 1'bt2/ E'.. ( Fr)/ '7 II I /:-f'::' 0 /(.7. to V I It./ \..J. 0' r. rv {.!V -F-r} T _ III if-rj/ T k:j0 =5ItrXld/ ce.- p - att )4r (Jr f-=:- I v '., /- t"71 V' - JV. VJ 7..t4e 1 - (j. l't L ) 1..;(9eJ?(7); (10-points) A 6um x 6 urr( rectangular semiconductor resistor is made fr rt9le? v E:f=tP, tfj.'tev ' semiconductor from problem 11. It is biased on two opposing sides (longest dimension) with 9 volts. Determine both the electron and hole current density and currents flowing in the device. If -={(e -+1-=116 ('- 7 c..,"< 6 L -:. O. (JjC fr) 6 f ) n;:! tr;... e -=- f..ian h lq V/b, t7 :y":c "" ) - (i,tje - J1) (';I. ;(170 C '" >-/v-5 )(/<2 I,5D:; 3)( /qo /I-i-.) 1 '-... / /.Ll J 1. "-::-,.1'1 d---;;-> )"'.:::: (f)"7 I? 0 1/IeJ<JI (} r..j..::: Jf1 -=- ;z, lj,pi n '/(t?-11) (i;?uo) 7, tg1+e I r )

4 \0 SectifJ!3 3 (more short answer) 13.)Wpoints total) The material in problems 11 and 12 is exposed to a laser light that generates 2e16 cm- 3 extra minority carriers. a) ('points) Is this low level or high level injection? LX -'7 l' -=2::f >;;> h" ---? f1!? 4, /e vel /11J(;;>[i-/t1 b) ( points) Draw the 1 dimensional energy band diagram showing the placement of both the quasi-fermi levels (numeric answer). J\?: I f IJP +- Je It = ').,i8e13 e (r;, -E/J!a., f. ( fh - f./):: {J,, 7f - I (7,!'te" -I- de 16):::.. ;), $e 1 3 e (E)- FF) =: 0 1 ( 7() ( r:::',...,) b - f - f-p,. I {Jf 2; %fj If',... ''" r:;l /'1\ -O,171e o"o'rev O,( ev ' IF-, /;;;e r 17geV ;'/ 1Y -' t (/f 16 8 foe *,\J.-, ev v \ t/.. i-=., -' - v p

5 ylga,.,."'et,\q/\ 1\0 {l f"dtol' I 0 )[?----J"" o l1 e:: joooo - ift7y''1ir "')' Pulling all the concepts together for 14.) (40-points) t--=0 )('f:;. f5()p 1'*\ 1::. t ()ps ).:::O J( r:y' - up ---- _u J --.-., %'.... I,. _.. t,._ tj.. d 2 I':J.np I':J.np.. 1GIven: 0 == D" General SolutIOn IS: _. p 'H! u_ I C;X Tn - L,,:. O-e. d I':J.n I':J.n. _x/ +;1" J,l Given: 0 = Dn d 2 P ---p +GI. General Solution is: I':J.np(x)= Ae /1-" +Be. I,,, +Gj,T n...: JJ"'(lew)) x r II d 2 I':J.np Given: O=D General Solution is: I':J.np(x)==A+BX dx 2 II d 2 1':J.n Given: O=D", 7 p +G L General Solution is: I':J.n p (x) = AX2 + Bx +C 2 Given: O=D" d Ị':J.n )P+Gwf(x) General Solutionis: I':J.np(x)= [ - GDI.: Hf(x)dx1 +Bx+C di':j.n I':J.n Given: --p == --p General Solution is: I':J.np(t) =I':J.np(t O)e dt r II I':J.n General Solution is: I':J.n =G r p L n JI.6. C. C>n (I<. ::!."}:: Ie '"1 c. 11>\ - ot,et et }tf Igj V\&: 16"41..1(1+\) -::: f e 17 c,",,-'} LA.': SO.A -th.ere /s no jel1. e l"t:if,.y. id l1. x::! 0 >l. -: I b-t?.u tel::. fte-o t- f3 (?11: Ae -,s-o/w.81 t [Je 'll c)/50," '=7 10''1 ('7 I.t' -,.wi6l?,8"f () ;w,,1,c;.<ti /1-:: e-( -8) -: 10 - t.j) e + 'J e /0,7_([2"'/ 7 5"0 :: f3 {,<:(, 0' -tj, ()$' tt J d -=- rcf 6 I; -3 D!1.')(I() Ch-,. A 15" -"!> /) ff= - 5", a63 x 10 C ow\,?) Y\ lx::. r,'3,( (0 I- (e X/S-d." - )(.t1 \ - - e "j c."'" :.

6 Extra work can be done here, but clearly indicate which problem you are solving. - IA 2)" )" t -J..x _"( _) (:J,-rt)!J..: "JGe/f' \ / ;<,/.g, - X tw,sct\ -L' e. lor ' l{ro,g1alo-*)) le - e ) -j _ 0 d (.x/. <6'f.u- -.x/.gq..'(ii, \ A / a,,- q..;l 1} e - e.j rch-,

7 Extra work can be done here, but clearly indicate which problem you are solving. PERIODIC TABLE OF THE ELEMENTS Table of Selected Radioactive Isotopes GROUP fa YIII 2--;: ihe.,rc...ic \ ""1'1"1,1' \ KEY,/ ::: " "",,-.-, "'..."... L"' w'\l30 <,,,.01 _--- *..J"'.....-,,, """"Q::::"z r,/ -"'1);'... / 1."" l)r,mrt... j(f'.il. I (1\ 111'111""'1 J::. --,.. ':'... ""' NQATH UNDFn AV't fxok\f. :lunc)ts ffl)ol! :::;'1!:;':iS= w.-t-';.i