ECE 3040 Microelectronic Circuits

Transcription

1 ECE 3040 Microelectronic Circuits Exam 1 February 17, 2014 Dr. W. Alan Doolittle Print your name clearly and largely: 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 INDICA TED. Write legibly. Ifl 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 True/False (Circle the letter of the most correct answer or answers) 1.) (2-points)~r False: The energy bandgap results from hybridization (intermixing) of sand p orbitals an~es allowed and disallowed energies for electrons in a semiconductor or insulator. 2.) (2-points@ or False: Atoms with weak (small) chemical bond strengths usually result in smaller energy bandgaps and large mobility (due to infrequent collisions resulting from low atomic de~. 3.) (2-points) True r False: The fermi distribution function defines whether a state will be likely to be empty o ed~ 4.) (2-points) True o False lno.3gao.1n 2 is a valid semiconductor formula in standard semicond~ota n. 5.) (2-points True r False: The density of states describes the number of allowed states at a given energy and ar aw~ energy) from the band edges has a decreasing exponential form. 6.) (2-points) True o False To determine th~concentration, one must multiply the density of states by the Fermi 1bution function and integrate that product over all energies. Short Answer 7.) (22 points total) Write in the blanks (on the next page) the answer to the following questions using the information below: Given a semiconductor at room temperature (27 degrees C) has the following parameters and energy band diagram: Hole Diffusion coefficient, Dp=l1.86 cm 2 /Sec Electron Diffusion coefficient, Dn= cm 2 /Sec Substrate intrinsic concentration, ni=lelo cm- 3 Energy-axis a be d e x-axis. f gh I. J

3 Example Answer: '' region c<x<e, region h<x<i and point f' a.) (4-points) Which region(s) contain nonzero electric fields? lo (... X (._ R... ()AJ.. b.) (4-points) Which region(s) are n-type? c.) ( 4-points) Which region(s) are degenerate? d.) (4-points) Which region(s) has zero total (sum) current? e.) (6-points) If the slope ofec for the region h<x<j is 1000 ev/cm what is the electron drift velocity at point i? (Hint: Be careful of your units.) Fenc.,; e"" t~6{ t.ev~ \ ts f (e,. t- K~<.> ""'r".. "t ( I ) ( e I) ) -Itt j ) I -t~t. /ooo,-- (/.b. fo ev.b (o C.. <:.~ (Ooo too o UJ.,.. - f'/" t'\ ~ -= DtJ S ~T('L, -- ( n. b l.:;- '-"'"/r ) ( I ooo '1<,...,) D.D 2)tf V { uol., ::: /. 3 o /o" <-... fs}

4 Short Answer ("Plug and Chug"): For the following problems (8-10) use the following material parameters and assuming total ionization: ni=2e6 cm- 3 N 0 =2e15 cm- 3 donors NA=1e15 cm- 3 acceptors mp*=0.55m 0 mn =0.36mo Ea=l.45 ev Electron mobility, J.ln = 2200 cm 2 Nsec Hole mobility, f..lp = 500 cm 2 N-sec Temperature=27 degrees C 8) (7-points) Where is the Fermi-energy (relative to the valence band which is referenced to zero energy)? II\ -:::: IJ0 ~ Nt<, -r [( Nv:N,.,_)',_ +"'d 'h. ::: ~ tc''i"; ~-~~I) + [ c ID 1 ~ 1 /0 1 <;/' + (L lb' )'-J''' ::: / to"',;: J = "( '- -::: (2. to'- c.~'\) t -:. -l -J P V\ 1 (t> (.{4 Vf,{...(). '' -l 'f ft> C"~-t ;:: -= { "" Q e V p ::- Nv 12. tev-erl/h ~ Ef =. Ev- Jc.T f.) 1...( T~ere. <lre ~ev(/'11 11 ~ )/ a ~ (~(""" ~ hve_ Nv ::.. (Z.)I /D,;. ) (o.~5) 2. = f.jo'',.:-j.-. s 0 ( ""+if) ~) ' EF ::. (0 ev)- (o.o2..s-~ ev\(" ('1~ 1 '"~l' ) ::' f.zq.ev I I o ""' 9) ( 10-points) A 10 urn x 10 urn x 500 urn rectangular semiconductor resistor is made from the semiconductor from problem 8. 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. r ::: T' to c.~ tl -J -1 T.v -== t r-v\ t\ t.,.. (f.~ (o- 11 c,)(nod -~~;)(t-to 1 'c..:: 1 )(/So,f) ~> (;,"?,,";(, ~/,'!} Ttl".f>y.JN:: (/o-\,_,')(b).j{, /t/o.,') lt!: : b). ) c.,... A]

5 Section 3 (more short answer) 10) (I 0-points total) The material in problems 8 and 9 is exposed to a laser light that generates 2e 16-3 em extra mmonty earners. a) (2 points) Is this low level or high level injection? ((, -'l AV\-::. ~p-:::. t.. JD c.m n 0 -= I to 1 r c..""-j ----;> b) (8 points) Draw the 1 dimensional energy band diagram showing the placement of both the quasi-fermi levels (numeric answer) relative to~. Ec, and Ev. 11\::. 11\0 + A"' f + 6f ::.. ) /0 J') ~-J + 2 /01b ("'-~ ::: 2...{ fi:l,, (.f'-\-3 r... <J- ;o (...,4., (0 {.Jio\. -~ lb - s ~ 1... I c/fo c."'.. :3. o.1>~ ev E; ::. o.1 ~ ev &v:; 0 e V -.. _----- Fr::. o.jjl e..v

6 Pulling all the concepts together for a useful purpose: 11.) (39-points) A semi-infinite length of semiconductor in a device flies on a satellite launched in 1966 and has been exposed to constant cosmic radiation and sunlight. The light is absorbed uniformly in the semiconductor. Over time, the radiation damages the front surface. This problem deals with the performance after being exposed to the radiation for a very long time (i.e. not the initial pristine condition). The semiconductor is doped p-type with an acceptor concentration of le15 cm" 3 and has a minority carrier lifetime, of 400 picoseconds (400e-12 seconds). Photons absorbed in the semiconductor generate le19 extra electron-hole pairs per cm 3 per second. It is found that at the surface exposed to the radiation, x=o, the excess electron concentration is smaller than in the rest of the device,!l.n(x=0)=2e9cm 3. If the semiconductor is held at room temperature (27 degrees C), determine the minority carrier diffusion current density at all positions in the semiconductor (x ~ 0 urn). Assume a minority carrier mobility of 100 cm 2 Nsec, 27 degree C operation and the intrinsic concentration is le6cm 3. 1'~ \ ~ (.'tv\ \\t ~ f-~f ~~ rp ~f d 2 ilil Given: 0= Dn 2 P dx ililp Tn General Solution is: d 2!l.n!l.nP Given: 0 = Dn dx2 p --+GL Tn d 2!l.n Given: 0 = Dn dx2 p General Solution is: Given: O=Dn d 2!l.n 2P+GL dx General Solution is: 2 Given: 0 = Dn d dx i'l~p + Gwf(x) General Solution is: MP (x )=[- Gw Jf f(x)dx] + Bx + C DN,t'-1\S v.j)f\1~ d!l.np = _ i'lnp () - t/ Given: General Solution is:!l.np t =!l.np(t = O)e 1 '" dt Tn t~ i r ~!l.~n~ Given: 0 = P + GL General Solution is: f'lji',tj Tn 'f.~.(~e~e.. i~ c... to"\c..t v\'fy"r-h"ta., 0 r~ta;e"'t V\f.Par ft...e >""r.f4c1z t!jf fi,.e- JtviLL, iv\tttttj.. fr"o""- 6. lr(c) 1' b~.-t"'. How.ev.t1, otf J..e..f f\.,s.f"-t".fl.er ~,., ft,e. S"'rfi'-e., f~ll c_,..,._~... fv... h'b... ;)Y"~c~,'eA+ )ec.rt.~t~j A V\ r ( ~) = G(., t.., J'l.A~f{ =0 Jy. x::~

7 Extra work can be done here, but clearly indicate which problem you are solving. r'v\(_\ t~+- \l,_t,+ ~ ~ ~ ;~X (d~pfl, i~<~.f-o ~~vite) 0 [So ).c,l( (.:;>.,.dj fiov\5 '. b ~ f l 0 J :::::- l... ' { D 1 C..M... ~ (,.e~ Q..('"Cf { s & t l.1 +if)~ : o = D ~~~f. -?VI_t_ ~ '=- N )~~ t~ ~ 1::> Vlf l)c.) ~ A, e.... x lt..n -+ g e.jrx.jt.."' 4 l:>l. t.., ~ V\t c 0 ') -::: A -t B -+ tf.. I 0 '~ ':: 2. I D (."" 1 -J A;, + B -= - L '(D' (~ A ( \ 0 ( \..J.. ll I 1 -~ c1 1 ~ 3 uvir' QO J - 0 -t D 1'6) 1 1. o C.!'-) -= 7 fo c.,.., [3;:.0 A = - L [ Gflrl

8 Extra work can be done here, but clearly indicate which problem you are solving. ~ r\1 :: ):1'; t fa~~~. :: (_o. D l.) 1 V) ( l o o c..~ "l.lv s) :: 2!>1 c."" fs Lrv:: JDw't~~~ = J(2.)'1 c.~/s)( ~ {o.. 10 s) =- 3.2'2.. to~r c.tl-, -;: ,..,.."" J rj ;: tprv V"' -=- t 0~ ~~VJ.t (l." fo...,~ (..)[1 51 "~A)(-2. /D1c., -3) -x!o.~',_r-r't -r 3.22 to ("' e