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Unformatted text preview: CHAPTER 1 S E I (a) In theory the two quantities are numerically idenn‘cal. (b) The MUS—C or MUSFET under test is heated to an elevated temperature and a bias is
applied to the gate of the device. Typical conditions for a bias—temperature stress to deteca sodtutn ion contatrtination would be T = lSﬂ'C, VG such that 59; <2 106 Vicar. and t = 5
minutes. to) The ﬁxed oxide charge is thought to be due to excess ionic silicon that has broken away
from the silicon proper and is waiting to react in the vicinity of the SiﬁSiﬂz interface when
the oxidation process is abruptly terminated. (d) BIT is greatest on {111] Si surfaces. smallest on {IUD} Si surfaces, and the ratio of
midgap states on the two surfaces is approximately 3: l. {e} MOS device structures exhibit both an increase in the apparent ﬁxed charge within the
oxide and an increase in the interfacial trap concentration. {f} In reaponse to—BT stressing, the negative—bias instability causes a shifting of the (3—1”
curve toward negative biases. Alkali ion contamination leads to a (34" curve voltage
nanslation in the direction opposite to the applied bias. {g} The VT = VT' + Vpg relationship was derived assuming er changes little over the range
of surface potentials between dis = ﬂ and d5 = 2955:. This becomes a poor assumption if the
device contains a large density of interfacia! traps — if the device is unanoealed. for
example. (h) A depletion—mode n'ansistor is a MDSFET that is "on“ or conducting when VG = 0. (i) The ﬁeld—oxide lies outside the active region in MUS devices and integrated circuits; the
gateoxide lies directly beneath the MOS gates. The ﬁeldoxide is typically much thicker
than the gate—oxide. (j) Simply stated, the "body effect" refers to the deep depletion condition that is created
beneath the gate when the back or body of a MDSFETis reverse biased relative to the
source. The body effect is utilized to adjust the threshold voltage. 18—] i (a) {b} ¢Ms = {ILWHS} = [x'—(EF—Ec}poly5i]“[I'HEcEFkamsmlirasi] = #[tEc  EFJpolySi — (Ec — EF)FB. mystalljm—Si] = ﬂ.4 V
(Now that the computational equation developed horn: is the same: as E]. 13.24.} (c) biased. When VG = a the: polysilioon side of tilt: part {a} diagram is
Imamd, yielding 1.8.3
In general
this = glen — r' — (EC—Enm]
where
‘ noses! ...nt (See Fig. 13.3 caption.)
(ﬁn' — r} = 41.13%? ...n+ poly (See Fig. 13.3 caption.)
l (Ea—Esiw pals  x'= (x'+Eo}I' = 56 = 111” P+ P01?
Also
(EC—EF)FB = {EC—El.) + (El—EFlpg E £6” + (Ei—EF}FE
or .
Eﬁﬂ — {kaqjlﬂNDIni} ...ntype crystalline Si
{ECEFJFB = r
sen + (nonhuman ...prypc orpstallme Si The results of the pd polycrystallinegate computation based on the above relationships are presented in the following plot. The MATLAB program script used to generate the plot is
also listed on the next page. Although it leads to onlyr a minor difference, it should be
mentioned that, instead of employing E:— E1 5 E532, the more accttrate value of Ec—Ei =
ﬂ.57e"t’ was used in constructing Fig. 13.3. 1.2 HMS Imus}
I:
I'll 5::
1h “'2 .'.. .E” ......E....'_..1..I:._§ L :1 ....:.......:.:. to 1a" to" 1o” 10" Mot ND [tan3} 183 LJAJIJu3pnuyenlscﬁpL"
$HEta1Semicenductor ﬁbrkfunetion Difference $Initializatien
clear: close ECUnstElﬂtS and Parmneter 5
ni=1.ﬂe1ﬂ:
EG=1.12;
kT%0.ﬂ255:
semenut‘ﬁpecify the gate material','hl','n+ poly','p+ poly‘l:
if 51, A;ﬂ.03: elseif s==2, A=ﬂ.1B; else A=EG:
end ECaleulete MS Workfunction Difference
%EeEF=(Ee—EF}FB NE=1ogspaee114,13};
BEEanEGIEkT.*leg{HB.!ni]:
EcEFp=EGI2+kT.*1og(NB.!ni]:
ﬁM5n=AECEFn: ﬁHSp;AFECEFpI %Pletting result
semilegxlNB,HHSn,HB,auﬁpl; grid
xlahelt'NA or ND [emr31‘1: ylabe1{'eﬂs [veltsJ'J 18—4 i .
5
I
! 18.3
(a) Given mm, = m = constant. 10 2
ﬁvﬁ(ﬂﬂbﬂ¢] = ___'_'1_I 3:pr = _ MD 10115 [1.6X1ﬂ'193{1ﬂ13}(l[r5)2
(2}(3.9}(3.35 x 1014} =  212‘? (b) Here
pm“ = QMEUG}
where In
QM =1 Piun{x}dx = We
0 Substituting Pier: = QMEUG} intoEq.{lS.13} gives 2
nubile _ _. D
:2.le l: ’— ‘ ‘mﬁ‘WD ' ‘m Clearly the MFG here' 13 twiee that' 111 part (a). AVG = —4ﬁ.‘IV 1.3.41
(a) AVGIIﬁxed charge) = — prﬂu ...Eq (18.15)
(b) We: are. givcn
or niaﬂimnaﬁcally
'0 ...ﬂ' 5 I S Icy—ﬁx
PM i E r ...u s r 5 ix, when: I": xxuﬁu
m2 Substituting pox into Eq.[lB.l I) gives
ﬁx .t‘ﬁx’+xo—M)dx' (a) Mayan m = 1A;
ﬁVGILpazt a) 3x0 If Ax = IU'TGII‘J and 10 = 195cm —> AVGWNﬁVmH] = 11.99? If six = 107cm and x0 = lﬂ'ﬁcm —1r dVGﬂmﬁVG {a} = “.96"! Prmriddd x0 3:: dx, it is assaulially impossible to distinguish bemwn charge disuibuted a
short distance. into die. oxide. and chargc right at the interim For very thin oxides the
difference becomes dctgctablc, but not all that signiﬁcant. even when In is only lﬂdx. 13—6 lﬂé
(allfﬁtr: MOSEisidﬁaicmeptfarmsatﬂandﬂpiﬁﬁhcn
vFB = ans—Q = writ—Qt:
Ca K05!) A plot of Vii5 vmus 10 data should be a straight lint: with an cxnapoiated VFEaxia
internept equal to ms and a. slope of—QFfKoag. (h) The given VF]; versus 19 data is plotted below. A 1:35! squares fit ﬂtmugh the data
jfitlds Thus VF]; = —ﬂ+596—{3.ﬂ2x1041:0 "in in cm
{01.15 = 41.596 V (3.9)(335xm14)(3.t}2xm4)
1.6x1tr19 1:25": = $080810qu = = 6.52x1umtcm1 Lr. I} [L] 0'3 0'3 0.4 I15 13—}; Lil (a)  Ifthcrc is no charge in the oxide, if pm = c. then £0, = constant and the
oxide energy bands are a linear function of position. However, if pm at 13, EM becomes a function of position and the oxide energy bands in turn exhibit curvature. A concave
curvature as pictured in Fig. P133? is indicative of a signiﬁcant positive charge, alkali ions.
in the oxide. ' (In) . The normal component of the D—ﬁeld, where D = £8 , must becontinuous
if there is no plane of charge at an interface between two dissimilar materials {see Sub
section 16.3.2). When a plane of charge does exist, there is a discontinuity in the Dﬁeld
equal to the charger‘cm2 along the interface. Note from Fig. P133? that the slope of the
bands is zero and therefore 8 =(1fq){dchde = {i on the oxide side of the interface. 0n
the semiconductor side of the interface 8 is decidedly nonzero and positive. Thus, there
must be a plane of charge at or near the interface. For the picnirecl situation we in fact
require (gm =ngn£5 and the interface charge must be positive. The interfacial charge
could arise from alkali ions, iterfacial traps, or the ﬁxed charge. In real devices, alkali ions
typically give rise to a spreadout volume charge, making alkali ions an unlikely source of
ﬂingrm. Moreover, the interfacial trap charge is assumed to be negligible in the statement
of the problem. That leaves the ﬁxed charge which closely approximates a plane of
positive charge at the Si—SiO; interface. We conclude {25 at: t}. Although a conclusion has been reached, we need to address an apparent inconsistency.
In this problem and in Exercise 13.3, we have indicated that the ﬁxed charge will cause a
discontinuity in the interfacial D—ﬁeld at the 3i3i02 interface. However, in deriving
Eq.(i 3.1 l), the fit—ﬁeld was explicitly assumed to be continuous across the Si—SiOg
interface. Eq. (13.11) in turn was used to establish the dVGIIﬁxed charge] expression.
This apparent inconsistency is resolved if the mathematical development is examined
carefully. To be precise, by including (2;: in pm in the Eq. {13.1 1) derivation, we actually
took the ﬁxed charge to be slightly inside the oxide. The LIIﬁeld discontinuity then occurs
at x = is; instead of exactly at: =xg. Whether the discontinuity occurs exactly at the
interface or an imperceptible distance into the oxide cannot be detected physically, and
clearly does not affect the mathematical results. 13—3 1.3...3 (a) In an itieal version of an
MUSC. flat hand always ooeurs
at vi] = D, with the ideal tleviee
exhibiting the same value of C at
ﬁat band as the non—ideal device
Because Q” w l], the ideal C—V
curve is obtained by simply translating the given C—V curve
along the voltage axis until the flat band point is at VG = i}. (h) Given Cm = C0 =m In
weeonelude
143
:rc. = M: (3 we. Bjxiﬂ )(2 9x1ﬂ' } : smxtné cm
Co zooms12
Also
CHIN = J)—
l + KMW
K510
making 11 snsxmﬁm ,
=KE1CC_ 1;): i = . 105 =o.3
WT K0 —1 (——3 9) ——%ﬂ( —1) SW em Ittrn Referring to Fig. 16.9, the plot of WT versus Hg or ND, vve oonelude :1 WT = [1.3mm results when
N [1 E lﬂl‘ﬁicma' (e) Sinee QM =.0 and Qu = ﬂ,
QF
ﬂvdfhtbnnd= VFB: “15—76:;
FB = —0.71 in the statement of the problem. Also. for an N3 = lﬂlﬁfemg' ﬁlm—Si} device,
we conclude from Fig. 133 that thus = —ﬂ.24‘v'. Thus zooms'2 ——{—o.24+o.v1}
sexier3 as = cutouts— stJ= %(ms—VE)= = 3.14X1ﬂ'3 eouli‘em2 1.83 We infer from the C4! characteristics that the MOS—C is a phulk device. Alan. W kmw
that acceptor—like traps are negatittelpr charged when ﬁlled with an electron and neutral when
empty. For apthulk MUS—C the effect of biasing on the occupation and charge state of the
acceptor—like traps is summarized in the following figure. Ee
t'}
{J
Es
_ Ev
Ace Dept — II'W {strong} We also note EVE = —% Thus, relative to the "aﬂef‘ or negligible QIT situation, the "before" characteristics will be
shifted pttsitiuelyr 11er is negative) and the displacement will s3;staetnah'call3Ir increase as one
progresses front accumulation. through depletion. to inversion. The deduced "before"
characteristics are pictured below. C Before
After From the answerto Ptobletn 1.5 {see Solutions Manual pages 1—2 and 1‘3), we know that
there are 6.18 X It]14 atotttst'cn'tla and 9.59 X It)” atotnstr‘ctn2 on the (100) and {110) surface planes. respectively. If one assumes the number f  t :_ sorting bonds" is ptomrtional to the number of Si surface atoms, then the (1 to} surface should exhibit
the higher density of residual "dangling bonds" or interf: = traps. xperintents conﬁrm
the above conclusion.) 13—13 r‘+—————
I 1.3.1.]. {at} We note that the interfacial maps will be neutral when the MOS—C is aeeumulatinn nr
lightlyr depletion biased, but bemne positively charged when the deﬁne is ltngI : lﬁﬂ
depletion biased er inversion biased. Filled
Emu—like Clearly, there is no shift in the C—V curve when dtedeviee is aecuntuladnn and ¢3l s: hm
depletion biased. Hawdvcr. when t¢5l : lﬁd depletinn biased at inversinn biased, the
chnmterisﬁcs are traitslaled $11: = WC“ = constant negative value along the mltage
EKIE. C From the preceding one concludes, {e} If diestates are very close to En they retain the same charge om the nonconstant
capaeitanoe portion of the C—V characteistie. Sinoe the states are donorulike and always
empty for all depletion biasing, one expects a positive {211‘ and a negative shifting for the
entire depletion part of the C—Vehataeteristie. \Qﬁcﬂﬂl’lﬂlﬁeﬂm
WWW ﬂ“
becomea neutral—
it is ion in the level . aeoumlaﬁon pcu'tion M eitlteeurve Vo (cl) A donorlike level VET}? close to E, is always ﬁlled and neutral for the nonconstant capac
itance pmtion of the C—Veurve. There will be no observable C—V shift due to such states.
C One cannot see {3n
become positive — it is
lost in the ﬂat inversion
portion of the curve. 3. Sameasidw Note: This problem points out the tilil‘l'leultyr of detecting Err states that are veryr close to
the band edges. 13—12 I 3, 12
The two halves of the MUS—C may be viewed as separate capacitors. irradiate {r MUSE _Elj— % MOS—C Before irradiation, each half will eonnihute precisel onehalfot‘ the observed capacitance,
each yielding a C—Vcharacteristic like that labeled ". before" in the ﬁgure below. After
irradiation, the C—V characteristic of the affected half {labeled "t after“ in the ﬁgure below)
will be shifted toward negative voltages due to the apparent Qp. Graphically combining the
"% before” and "i after" curves yields the total expected "after" curve. (a) The shift in the gd— VG characteristic after +BT stressing is symptomatic of while ions
in the oxide. (1}) Gsneeptually extrapolating the gd — VG curves into the V5 axis, we conclude that the
turn—on voltage has shifted negatively ~ 2V after +BT stressing. The device is now obviously "off" when VG =  2V and V5 = — 3?. Moreover, the Pa = 4V state after
stressing is equivalent to the V5 = — 2"." state before stressing. Thus, IfD
VG = 4v Z w; =—2v, —3v 13*13 Jim
(a) VT will shift in the —V5 direction. Since Qp is positive, an apparent Qp causes a
negative shift in the threshold voltage. ——'—_p +VG (apparent Qp is D) VT 4 1IH‘rc (b) The gate material affects ths. 1i'lIFitl‘t (IIML I: = —G.{}3 eV for Al (see the Fig. 13.3
caption) and th'm II = (163 eV for Cu (from Table 13.1), ms and hence VT will increase
by (1.65? in going from an Al to a Cu gate. ———> we {Al ~aCu gate) V'n} t VT
{c} The substrate doping affects both (bus arid VT'. As given by Eq.(l$.22), VT = 2¢F+£ixo We at: where m: = Ella Ea)
Kg K5130 it '11
“5° (En—Em *5 Eon — (Hi—EFJFB = sac + wt.
this = (Helmet!  I'  (EC—EF}FE] = (IIqJIth'—x'—Egt2] — 19F
Since {pp increases with doping, ¢MS decreases and VT' inereases with increasing NA. However, the increase in VT" is greater than the decrease in MS: and VT = VT' + this
increases with an increase in substrate doping. —__> +VG fmcteaaed Na) V'In —} VT and {d} In generai,xu enters into the determination of both Vf and Vpg. However, because the
MUSFET is speciﬁed to be ideal except for MS at t}. In in this problem affects onlyr VT'.
Inspecting the VT' expression quoted in part (b), one rapidly,r concludes Vf. and therefore
VT. decrease with decreasing In. ———> we (decreased x0} Fri—“Io {e} To ﬁrst order, the implantation of Boron into the near surface region of the Si is
equivalent to adding a negative fixed charge to the system. — The threshold voltage shifts
in the +VG direction. I __p. +543 (Boron implantation} Vm—t VT 13.13 (a) Adding the voltage shift doc to the ion implantod charge (Eel. 18.25) to tho mgular ﬂat
hand expression (En. 18.20), one obtains K D s s s e
Formogivcndovice
veg e —o.4s—~———m—“'ﬁxmw)(5xmﬁ) {Eh(1'3” +n+o—exm"} “= 0
{s.s}(s.ssx1o14)
(b)
V'i‘ = MFHEExe 43:3 (—s’JFJ on: e Jqllnwpmi} = —ﬂﬂ2591n(1{}15ﬂﬂm} = 4.2931: VT = — (amass) — M (3.9} ‘1 IS La
exms[W
(11.3)(s.ssx1o14} I1 —U.3ﬂV
and VT = v++veg = v; = _ onus? {o} W device. For the given pchsnnsl deviee there is no inversion
lsyer at zero unis = a motors no drain current when VG = ID. A MDSFET which is "off“ at zero bias is tofurod to as an enhnnocmmt mode device: 1315 13. 15
Combining 1345.08.21], (13.210), and {13.25}, one: can write Sinoo men: an: no intorfacial traps mid no mohilo ions in the oxide. QM = ﬂ and {2n = {1.
Also Co = Kgﬁy’xn and Q1 5 —qNI. Thus tho VT oxprossion simplifies Io VT=VI+ MS— —XL[QFJ' NIi
T 11* qKoEn tr Solving the preceding equation for N1 ﬂicn givos [[NI = &+;—Kﬂ(VT—Vi¢Mslﬂ q Jro (2qu and ﬁrm: speciﬁod in tho statemont of the problem. Howovor. wo need to
dolorrnino W5 and VT'. Boonuso tho MDSFET is an Al—SiUz—Si dovioo, we can read W8
directly from Fig. 18.3. For NA = lol'l'xcmi ono ﬁnds my, = —1.ozv. The. ideaIAdovioo g ﬂimshold voltago can be computed using Eq.(13.22). VI =2 +£i ”N3
T ¢F Koxu K359 $1: $1: = kgtlnwmi} = oozsolnuol'inom) = onlw . 49 11 1H
VT=(21w.41?)+(“‘3)(10‘5J[(4)”'ﬁxm no man?) (3+9) {1 1.3}{335x1 0'14}
. = 1.32V

i Finally, substituting into the N1 expression, we: obtain
44
N1= 1o” +ml (os— 1.32 +132) (1.6XIﬂ'wjﬂﬂ'5} N1 = 5.31): 11]“ boron ionoiom1 13115 ...
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