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Unformatted text preview: CHAPTER 18 I S. 2
(a) In theory the two quantities are numerically identical. (b) The MOS—C or MOSFET 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 detect
sodium ion contamination would be T = lSO'C, VG such that 80,; < 106 V/cm, and t = 5 ,
minutes. ' (c) 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 Si—SiOz interface when
the oxidation process is abruptly terminated. (d) D11 is greatest on { 111 } Si surfaces, smallest on { 100} Si surfaces, and the ratio of
rnidgap states on the two surfaces is approximately 3:1. (e) MOS device structures exhibit both an increase in the apparent ﬁxed charge within the
oxide and an increase in the interfacial trap concentration. (0 In response to —BT stressing, the negative—bias instability causes a shifting of the C—V
curve toward negative biases. Alkali ion contamination leads to a C—V curve voltage
translation in the direction opposite to the applied bias. (g) The VT = VT' + VFB relationship was derived assuming QIr changes little over the range
of surface potentials between $3 = 0 and $3 = 2th; This becomes a poor assumption if the
device contains a large density of interfacial traps — if the device is unannealed, for
example. (11) A depletion—mode transistor is a MOSFET that is "on" or conducting when VG = 0. (i) The ﬁeld—oxide lies outside the active region in M08 devices and integrated circuits; the gateoxide lies directly beneath the MOS gates. The ﬁeld—oxide is typically much thicker
than the gateoxide. (3) Simply stated, the "body effect“ refers to the deep depletion condition that is created
beneath the gate when the back or body of a MOSFET is reverse biased relative to the
source. The body effect is utilized to adjust the threshold voltage. lSl (a) (b) ¢MS = %(¢M’¢S) z [X"“(EF—Ec)polySi]“‘[X'+(Ec—EF)FB,cryslallineSi] = $030 “‘ EF)polySi  (Ec  EF)FB. crystallineSi] = —0.4 V
(Note that the computational equation developed here is the same as Eq. 18.24.) (c) biased. When VG = 0 the polysilicon side of the part (a) diagram is '
lowered, yielding little band bending
because of heavy doping accumulation 18—2 18,3
In general ¢Ms = qi[¢M'—x'— (EC—EFnB] where
I —0.03eV ...Al (See Fig. 18.3 caption.)
(DM' — x' = 0.18eV ...n+ poly (See Fig. 18.3 caption.) A
i (En—Ep)p+ poly * x'= (1'+EG)x' = 196 = Law mp" P01}! 1
Also
(EC5F)“; = (Ea—El) + (Er5F)“; E Eel? + (ErEmm
01' I E‘s/2 — (kT/q)ln(ND/ni) ...ntype crystalline Si — LEG/2 + (kT/q)ln(NA/ni) ...ptype crystalline Si The results of the p+ polycrystallinegate computation based on the above relationships are
presented in the following plot. The MA'ILAB program script used to generate the plot is
also listed on the next page. Although it leads to only a minor difference, it should be
mentioned that, instead of employing Ec— E15 life/2. the more accurate value of Ec—Ei =
0.57eV was used in constructing Fig. 18.3, 1.2 (EC‘EF)FB oMS (volts)
0 9
m m .0
4'. 0.2 .. . :. 10M 101$ 10‘! 10" 10“ NA or N D (tame) [83 MATLAB program script... %MetalSemiconductor Workfunction Difference %Initialization
clear: close %Constants and Parameters
ni=1.0e10:
EG=1.12:
kT=0.0259:
s=menu('Specify the gate material',‘Al','n+ poly','p+ poly'):
if s==1, A=—G.O3; elseif s==2, A=—O.18: else A=EG;
end %Ca1cu1ate MS Workfunction Difference
%EcEF=(Ec—EF)FB NB=logspace(14,18):
EcEFn=EG/2—kT.*log(NB.lni):
EcEFp=EG/2+kT . *log (:13. /ni) ,
QMSn2AECEFn: eMszAEcEFp; %Plotting result
semilogx(NB,@MSn,NB,EMSp): grid
xlabel('NA or ND (cm—3)’): ylabe1('gMS (volts)') 13—4 143.43
(a) AVg(fixed charge) = — QF/Co ...Eq. (18.15) (b) We are given or mathematically 0 ...0 Sx SxoAx
Pox = lQ—Ex ...0 .<. x' S Ax, where x": xxo+Ax
m2 Substituting pox into Eq.(18.11) gives x0 Ax
AV = .—1— xpoxd = X'(XI+IO'AX)dxl and (c) AVG(pan b) = 1_ A;
AVG(part a) 3x0 Ifo = 10'7cm and xo = 10'5cm —> AVG(b)/AVg(a) = 0.997
Ifo = 107cm and x0 = 106cm —) AVG(b)/AV(;{a) = 0.967 Provided x0 >> Ax, it is essentially impossible to distinguish between charge distributed a
short distance into the oxide and charge right at the interface. For verythin Oxides the
difference becomes detectable, but not all that signiﬁcant. even when x0 is only 10Ax. 186 M (3) Given pic“ = p0 = constant, x0 2
mobile _ 1 _ _ 90%
AVG( ions “' “EOE! 1!)de  _ (1.6 x 1019)(1018)(105)2
(2)(3.9)(8.85 x 1014) = — 23.2V (b) Herc
pion = QMMXO)
where Io
QM = I pion(1)dx = We
0 Substituting pig“ = QM3(XD) into Eq.(18.l3) gives . . 2
AVGlmoblle = QM '_ x0 90% . —— _ — p = _
Ions Co K080 0x0 {(050 Clearly the AVG here is twice that in part (:1). AVG = .—46.4V 185 18.6 I
(a) If the MOS—C is ideal except for ms at 0 and Q}: at 0, then A plot of VFB versus x0 data should be a straight line with an extrapolated VFBaxis
intercept equal to ms and a slope of —QF/Kom. . .i
(b) The given VFB versus x0 data is plotted below. A least squares fit through the data
yields . VFB = —0.596—(3.02><104)xo “:0 in cm
Thus 
(0M3 = 0.596 V
14 4 
QF/q = _K080(slope)/q = = 6,52x1010/cm2
‘ 1.6x1019
_ o 0.1 02 03 0.4 0.5 .zocum). 187 1_&.l (a) . If there is no charge in the oxide, if pox = D, then em = constant and the
oxide energy bands are a linear function of position. However, if pox at 0, EM becomes a function of position and the oxide energy bands in turn exhibit curvature. A concave
curvature as pictured in Fig. P18.7 is indicative of a signiﬁcant positive charge, alkali ions,
in the oxide. ' (b) . The normal component of the D—ﬁeld, where D = 88 , must be continuous
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 D—ﬁeld
equal to the charge/cm2 along the interface. Note from Fig. P1857 that the slope of the
bands is zero and therefore 8 = (1/q)(dEcldx) = O on the oxide side of the interface. On
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 pictured situation we in fact
require Qinmrface = [(58085 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
Qinwrface. 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—Si02 interface. We conclude Q}: at 0. ' Although a conclusion has been reached, we need to address an apparent inconsistency.
In this problem and in Exercise 18.3, we have indicated that the ﬁxed charge will cause a
discontinuity in the interfacial D—ﬁeld at the SiSiOz interface. However, in deriving
Eq.(18.1 l), the D—ﬁeld was explicitly assumed to be continuous across the Si—SiOz
interface. Eq. (18.11) in turn was used to establish the AVg(ﬁxed charge) expression.
This apparent inconsistency is resolved if the mathematical development is examined
carefully. To be precise, by including Qp in p0x in the Eq. (18.1 1) derivation, we actually
took the ﬁxed charge to be slightly inside the oxide. The Dﬁeld discontinuity then occurs
at x = x0" instead of exactly at x = x0. 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. 188 1_&..8 (a) In an ideal version of an
MOSC, ﬂat band always occurs
at VG = 0, with the ideal device
exhibiting the same value of C at
flat band as the nonideal device. Because Q” = 0, 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 = 0. (b) Given. I CMAx = C0 zm 10
we conclude '
 14 3
Xe = KQeOAG = (3.9)(8.85X10 )(2.9x10 ) z smxwgcm
C0 200x10“12
Also
CMIN = _._£O_.__
l + KOWT
[(8350
making Km, c0 (11 8)(5X10’6) 200
=———— _——_ =__'—..._._ __ =_ I X '5 ____ '
WT K0 CMIN 1) (3.9) 67 1) 300 10 cm 03pm Referring to Fig. 16.9, the plot of WT versus NA or ND, we conclude a WT = 0.3pm
results when
N D E 1015/cm3 (0) Since QM =0 and Qn = 0,  ' _' QF
AVGlflaIband _ VFB ‘¢MS“’CTO’ VFB = "0311 in the statement of the problem. Also, for an ND = low/cm3 Al(n—Si) device,
we conclude from Fig. 18.3 that (ﬁlms = 0.24V. Thus 200><l '12
QF = Co(¢Ms—VFB) = £Q(¢MS—VFB) = ——Q~(0.24+0.71) AG 2.9x103
=. 3.24X10‘8 coullem2 189 1.8.2 We infer from the C41 characteristics that the MOS—C is a pbulk device. Also, we know that acceptor—like traps are negatively charged when ﬁlled with an electron and neutral when empty. For a pbulk MOSC the effect of biasing on the occupation and charge state of the
acceptorlike traps is Summarized in the following ﬁgure. EC
0 .
0
E1:
_ Ev
Aec Depl _ Inv (strong) We also note
. er
AV = — —~
G CO Thus, relative to the "after" or negligible er situation, the "before" characteristics will be
shifted positively (Qr'r is negative) and the displacement will systematically increase as one progresses from accumulation, through depletion, to inversion. The deduced "before"
characteristics are pictured below.
C Before
After 18.11! From the answer to Problem 1.5 (see Solutions Manual pages 12 and 13), we know that
there are 6.78 X 1014 atoms/cm2 and 9.59 X 1014 atoms/cm2 on the (100) and (110) surface planes, respectively. If one assumes the number of residual "dan ling bonds" is
proportional to the number of Si surface atoms, then the (I 10) surface should exhibit
the higher density of residual "danglin g bonds" or interfacrail traps. lgxperiments conﬁrm
the above conclusion.) 1810 111 (a) We note that the interfacial traps will be neutral when the MOS—C is accumulation or
lightly depletion biased, but become positively charged when the device is lips! > 143:1
depletion biased or inversion biased. Empty Donorlike
Filled (H
Donorlike . ;
(neutral) \ Clearly, there is no shift in the C—V mine when the device is accumulation and l¢5 < ¢Fl
depletion biased. However, when lqbsl > I¢FI depletion biased or inversion biased, the characteristics are translated AVG = —er/Co = constant negative value along the voltage
ans. (b) For acceptor—like interfacial traps Position Trap _
of E1: Occupation State acc, Igbsl < ¢1:i depl
lips} > ¢Fi depl, inv 1811 From the preceding one concludes, (c) If the states are very close to 15'c they retain the same charge over the non—constant
capacitance portion of the C—V characteristic. Since the states are donor—like and always
empty for all depletion biasing, one expects a positive (211 and a negative shifting for the
entire depletion part of the C—V characteristic. \ One cannot see the
place where Q 11
becomes neutral —
it is lost in the level
accumlation portion
of the curve. VG (d) A donorlike level very close to E. is always ﬁlled and neutral for the nonconstant capac—
itance portion of the C—V curve. There will be no observable CV shift due to such states. C One cannot see Q n
become positive — it is
lost in the ﬂat inversion
portion of the curve. \ Same as ideal Note: This problem points out the difﬁculty of detecting Err states that are very close to
the band edges.  1812 18,12 The two halves of the MOSC may be viewed as separate capacitors. LlJiMOS'C Before irradiation, each half will contribute precisely one—half of the observed capacitance, '
each yielding a C—V characteristic like that labeled "% before" in the ﬁgure below. After
irradiation, the C—~V characteristic of the affected half (labeled "% after" in the ﬁgure below) will be shifted toward negative voltages doe to the apparent QF. Graphically combining the
"% before" and "iv after" curves yields the total expected "after" curve. ' ' irradiate a MOSC 18.13 (a) The shift in the gd — V0 characteristic after +B’I‘ srressing is symptomatic of mobile ions
in the oxide. (b) Conceptually extrapolating the gd — VG curves into the ,Vc. axis, we conclude that the
turnon voltage has shifted negatively ~ 2V after +BT stressing. The device is now obviously "off" when VG = — 2V and V0 == — 3V. Moreover, the VG = —4V state after
stressng is equivalent to the VG =  2V state before stressing. Thus, It)
VG = —4V Z VG =—2v, —3v 18—‘13 18, 14
(at) VT will shift in the —VG direction. Since Q1: is positive, an apparent QF causes a
negative shift in the threshold voltage.  ____+__> +VG (apparent Q: at 0) VT e Vﬁn (b) The gate material affects $345. With (DM'~— x' = —0.03 eV for Al (see the Fig. 18.3
caption) and (DM'— x' = 0.63 eV for Cu (from Table 18.1), rims and hence VT will increase
by 0.66V in going from an A] to a Cu gate. ____—_> +VG (Al —>Cu gate) Vﬁua VT (c) The substrate doping affects both mg and VT'. As given by Eq(18.22), ' z £1 1/441“ . = t: N_A
Al VT 2¢p +K0 x0 K880 991: where 451: q In( Hi)
S
° (Etr’EFkB .=. EG/Z (El591:3 = 56/2 +q¢1=
ms = (IICIH‘DM’ — x'  (EaEF)FB] = (llq)l¢ MLXLEG/Z] — tap Since 4512 increases with doping, ¢Mg decreases and Vr' increases with increasing N A.
However, the increase in VT' is greater than the decrease in qus, and VT = VT' + ms
increases w1th an increase in substrate doping. . —.—l__> +VG (increased NA) Vro —> VT and (d) In general, x0 enters into the determination of both VT‘ and VFB. However, because the
MOSFET is speciﬁed to be ideal except for (ms at 0, 3:0 in this problem affects only Vr'. Inspecting the VT' expression quoted in part (b), one rapidly concludes VT', and therefore
VT, decrease with decreasing x0. —I——*——> +VG (decreased x0) VT 9— P30
(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. ———+——————> +VG (Boron implantation) VT0—> VT 1814 15.15 (a) Adding the voltage shift due to the ion implanted charge (Eq. 18.25) to the regular ﬂat
band expression (Eq. 18.20), one obtains VFB=¢MS_Q_F._Q_WM._M_~Q_E Co Co Co‘ Co
_ __ x g QMYM Q1T(0) g1
‘Wg qK0:0[q+ q + q +4] For the given device (1.6X10'19)(5X10'6) (2x10i1 VFB = —0.46— +0+0—4x10”) a 0 (3.9)(3.85x1014)
(b)
V. = 2 _...K_'_S.. __
T ¢F KO Io K380 ( ¢F = ~%ln(Nn/ni) = —0.0259 1n(1015/1010) = 4293‘,
' 19 15 U2
VT =  (2)(0.298)—(11'8) (5x106) (3'9) (11.8)(8.85x10‘l4)
= —0.80V
and .
VT = V++VFB = Vi =  0.80v (c) Enhancement mode device. For the given pchannel device there is no inversion Iayer at zero 133 an erefore no drain current when VG = 0. A MOSFET which is "off"
at zero bias is referred to as an enhancement mode device. 1815 18‘ 15
Combining Eqs.(18.21), (18.20), and (18.25), one can write _@__Qhﬂ_9ﬂﬂ_ﬂ
Co Co Co Co Since there are no interfacial traps and no mobile ions in the oxide, QM = 0 and Qn = 0.
Also C0 = K omlxo and Q] = —qNI. Thus the V1 expression simpliﬁes to VT = Vr+ ¢Ms VT = VT+ ¢MS—4K:;0(QF/q—NI) Solving the preceding equation for N] then gives [[NI = %F+%K§’:° (VT V'l“¢MS)]] Qpiq and VT are speciﬁed in the statement of the problem. However, we need to
determine ms and VT'. Because the MOSFET is an Al—SiOz—Si device, we can read (has
directly from Fig. 18.3. For N A = IOU/c1113, one ﬁnds ms = —1.02V. The idealdevice
threshold voltage can be computed using Eq.(l8.22).  4N
V =2 +53“: ‘7 A
T W K0 0 K880 45F $1: = italumni/m) = 0.0259111001711010) = 0.417v v; = (mm, + (118) (106) 1’2
(39) (11.3)(8.85x1014) = 1.32V
Finally, substituting into the N1 expression, we obtain (3.9)(8.85x1o14) N1 = 10n+
(1.6x1019x106) (0.5 — 1.32 + 1.02) or
N; = 5.31 X 101I boron ions/cm2 1816 ...
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