HW2_sol - 1.2 ta) Given: ufl=1fl3mnisnc E = em. = 2Wcrn...

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Unformatted text preview: 1.2 ta) Given: ufl=1fl3mnisnc E = em. = 2Wcrn is? = min? = Ell-ii cmEIV-sec {h} [it Lattice scattering {ii} Instimd impufit} scattering (See the Relatintttthtt'p tn Scattering discussion in Subsection 3.1.3.] in} grinning-1.; is higher than nine“? W Reast‘m: in intrinsic material the scattering is due eitchisiveig.r tn lattice scattering. In heavily doped materials, ionized impurity,r scattering is also important. The fl'ifll'e scattering there is. the latter the mobility. {d}! Given ND: and NA; >:=- in, we knew from Eqs. {3.3} that I l = ...n-type wafer l; p = GMDI €1an In most semicnndnctnts including Gene, III" is greater than lit for it given doping and system temperatute. Since we are given N3] = N32, taking tfle wafer temperatures to be the sense, and with it“ 2» ‘try. we mnclude from the ebnve equatinns that ptwni’er I} :- ptwnrer I]. Here that the cnnclnsinn here is ecmsistent with Fig. 3.8m, teJnN=tinqynn = tnmssinsttn) = ssncmitsee p "gs—type wafer 2 if} tip ere nu, rt Erin ...il' tit-type An «re p", p Epn ...i|' p-tgpe (g) R-G center {11} Innrense. Per Eq. {3.33s}. 11, = liePNT. Since N1 decreased after processing, 1']; increased. M 1 {a} P = ' "Eli-{333} qflnNti = ‘—" = {1.501 ohm-cm {rm pi _ 315 a (1.5 x lfl-Iijtizaismmsj #n g I; j p = [L5 nhm‘cm why inspmtion fmm Fig. 3.3{21} {[3} Since NA : Np. H 2 p = 111' : ifl'flfcmz. Mommies. this Email numhsr nf scattering cantus is N9 + NA = 1 3-: lilmi'cml Thus. from Fig. 3.5(a), ti“ = 1165 CHEN—sec, HP 2 419 cmEN-sec, and P = l = 1 = 3.95 I 105 ohm-cm stun + upjnj {1.6x lfl‘19}{1534}{1fl10} {til-1cm n 2p = rt. 2 lflmicml With NA = I) and ND = i], one has this maximum passihlc. carriar mobilities. From Fig. 3.54;}. pnm= 135$ cmsz—sec and ,ttpmu = 4m cmEN-s=c_ i l = —-—— = = 3.44 x “)5 ultra-cm p stun + mm (1.15:: lfl'wjfilfligmflm} Because of the low ImbilifiE-S in compensated rrttttctisL p{pa.rt h} : pfipait c]. {:1} R = prim ,0 = RAH = ESCIGJIEID'EFU} = Suhm-cm Sinai: this bar is n‘typc, we mnciudt from Fig. 3.31:3} that Np 2 9 x Ifll‘icnfl. it} Far :31 sampic when: NI) 3::- m, p = lfqitflp. Furthermmt, since the Sflrnplt is lightly dopei lattice scattering will dominate anti pin will dim-sass 1.irith increasing T. Fig. 3.7a Wilfil'ms the ptweding ahsmatittn. Thus. with p t: Ifttn, heating up the san‘tplt: causes If": msistivity to increase. 311'. The brief explanation how one arrives at a given answer. an explanation applicable to all the energy hand diagrams-s is given immediater below. Sketches indicating the generoifonn of the expected answers follow the explanatitms. [a] for all cases. The semiconductor is ooncluded to he in equilibrium heeauoe the Penrii level has the same energy value [it is constant} as a function of position. [hi lr' vs. 1' has the same functional form as the "upside down" of E; {or E1 or Ev). The sketches that follow were eonstrueted taking the arbitrary reference voltage to he ltr = I} at x = fl. (ci E vs I is determined by simply.: noting the slope of the energy bands as a function of position. {d} For electrons, PE = Ec—EF and KE 2 E—Eg; for holes PE = Eta—Eu and HE = Eel-4E: {e} The general canierooimnu'arion variation with position can be deduced by noting E's-E; vs. I. Under equilibrium conditions, it = niettpflEF-E mm and t1 = nieap[{Ei~EF}t'fl'] if the semiconductor is nondegcnerate. if] Sine: Ifiidi-ifi = qttfittE , the general variation of Imam with whim can be deduced by conceptually fuming the product or the S vs. xdepcndence sketched in part {c} and the e vs. a dependence sketched in part {c}. Under equilibrium conditions, I“ = Intent: + Juicer = D. This Imam = ‘JNldrift- Diagram {a} Diagram {h} Diagram {I2} V L 1:; L PE BA B. E E} 1-“ L I L .I E3 E L I L J; PE PE J: I I E E I I I {1 {is} ' I)in {:1} Diagram {ell Diagram m 14$ (a) {b} theraliun . "111m is a dcficit 0f can'iars at I = 0* and therefor: the R—G process opal-ales tn eliminate the dtficit by adding {gantrating} flamers. {C} . Ifinlznn= lfldfcmE fi<pfl=Nfi= lfllfificmj an: n—ms—nu Em=ni1flfin= lflmflflm= 1Wkrn3} 2 an (d) ...... .. HEE=DN%fF%£+GL ...minofitycalri¢rdiff.eq. I x n GL = D since light is nut affccting [11: wafer and m: Unfazfinpflxlj term is zcm becaus: there is no cancfintrntifln gradient. Thus dfln Em _|’_ _—_ -_E ' ' ' m In ...31rnphflad nquauon Mptfl} = — m]. "boundary {sandman mpg} = rte—“tn solution — :11] = .4 “applying '31:. 1.1.9 Sinoe Gm is applied for a time I >> tn, Steady state conditions will pm:ij prior to t = fl Analogous to Sample Problem No. l discussed in the text, we can therefore state that oopflli = Gmfit Note that since the light intensity is reduced at t = I]. finfifl} = Mplm. and dilemma; = If}th = it!” x 1545 = 11'.I||3':"i"u:'.ot3 {(WENA = muffin? Low level Mention conditions clearly mail at all times. Also. all other oonditiocne needed for use of the minority eel-tier difiusion countion have been met. Thus. we need to solve 2 SEE =Dflafl" _EE+GL a: 3x1 1'“ which. for the problem at hand, simplifies to flau__éeii Eu: d: — {n+2 ...t>fl This equation is subject to the boundary condition. onptilii = Gmt'n One obtains unit} 2 GL5)“ + ale—m“ ...generel solution and applying the boundary condition. iii-BEEF "Emu = [IDWEfl-W-Wfl = 1.0? 2-: lfl15fcm3 a} "13 = pa = mam—EFF” ={1flm}rfl-3m-U359 = 131 3:10“ch b} n = mew“ 'Efl’” ={1ulmefla3mm59 = 1.15 x 1015:ch p = flig{Ei_FP]fl=T : {mlflmflfimfllfl = x lfllflcmfi A} ND E an s LEE:' 1-: lfl'5fcrn3 NIB—£1 3} Duo: ID illuminafinn, 3,0 n m; and :1 differs significanfly from ma. Fur lflw lewt injemion fine mus: have fip <2: m and n a m} I {Lfix lfl-19)(1345}u.m x 143151 I Mm : QJIuND _ _- _._.———l = _—-—"—'—_I —' pm” qmnn + ppp} {l_fi><lfl'19}[E1345)-E1.15>¢lfl15}+(458}[1fl?x1fl15}1 = 4.34 ohmncm = L35 ohm-cm ii {a} {1!} {cl {:1} {e} [5 m = = masmfflm EMU} = 11.614 1; nl- {1029: = If N “'2 _ . xp [—flq N—D—AWWD} m] _ 3.555xm 5 cm I = Kaila—ML .m- n [ q anflwm. _ 131x1n5cm W = In'l'xpz N {1.6X1fl'19} m1: _5 at} z _. G D In = ___—_[—}Ew _ I. - fl 2 — U-fixm‘mluxmlfi 3.1555x1fl'5 3 mm _ 335m x" _ —'—-—-i-—-i = “as v {21m .8}{E_Ej>{]fl-I4J T‘ L; em =—$=—%%»4%J E #32 JE E 2 5x52 “’11 {b} Poissan's aquarium slams dad: a #fl'san. Thus E «is: “I PU? K353];- {c} Undccr the depiction appmximafim p = gum-NA) “FM? Ex 5 W32 Thus MEI-NA = pm ...—Wr2 s x 5 W2 "flit: ND—NA plot 1 ...
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This note was uploaded on 09/05/2009 for the course ECE 3040 taught by Professor Hamblen during the Spring '07 term at Georgia Institute of Technology.

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HW2_sol - 1.2 ta) Given: ufl=1fl3mnisnc E = em. = 2Wcrn...

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