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Unformatted text preview: 1.2
ta) Given: uﬂ=1ﬂ3mnisnc
E = em. = 2Wcrn
is? = min? = Ellii cmEIVsec {h} [it Lattice scattering {ii} Instimd impuﬁt} scattering (See the Relatintttthtt'p tn Scattering discussion in Subsection 3.1.3.]
in} grinning1.; 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
= ...ntype 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' tittype
An «re p", p Epn ...i' ptgpe (g) RG 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 ohmcm {rm pi _ 315 a
(1.5 x lﬂIijtizaismmsj #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 cmENsec, and P = l = 1 = 3.95 I 105 ohmcm
stun + upjnj {1.6x lﬂ‘19}{1534}{1ﬂ10} {til1cm 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 cmENs=c_ i l = ——— = = 3.44 x “)5 ultracm
p stun + mm (1.15:: lﬂ'wjﬁlﬂigmﬂm} Because of the low ImbiliﬁES in compensated rrttttctisL p{pa.rt h} : pﬁpait c]. {:1} R = prim
,0 = RAH = ESCIGJIEID'EFU} = Suhmcm 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 dimsass 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 diagramss 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 = Eel4E:
{e} The general canierooimnu'arion variation with position can be deduced by noting
E'sE; vs. I. Under equilibrium conditions, it = niettpﬂEFE mm and t1 = nieap[{Ei~EF}t'ﬂ'] if the semiconductor is nondegcnerate. if] Sine: Iﬁidiiﬁ = 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
opalales 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'.I3':"i"u:'.ot3 {(WENA = mufﬁn? Low level Mention conditions clearly mail at all times. Also. all other oonditiocne needed for use
of the minority eeltier 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. iiiBEEF "Emu = [IDWEﬂWWﬂ = 1.0? 2: lﬂ15fcm3 a} "13 =
pa = mam—EFF” ={1ﬂm}rﬂ3mU359 = 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 lﬂ19)(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 ohmcm 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 — Uﬁxm‘mluxmlﬁ 3.1555x1ﬂ'5 3
mm _ 335m x" _ —'——i—i = “as v {21m .8}{E_Ej>{]ﬂI4J 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 = gumNA) “FM? Ex 5 W32
Thus MEINA = 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.
 Spring '07
 HAMBLEN

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