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Unformatted text preview: CHAPTER 18 18.1
(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 = 150°C, VG such that 80x < 106 V/cm, and t = 5
minutes. (0) 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—Si02 interface when
the oxidation process is abruptly terminated. ((1) 01'1” is greatest on {111} Si surfaces, smallest on {100} Si surfaces, and the ratio of
midgap 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. (f) 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 QIT changes little over the range
of surface potentials between $3 = 0 and (1)5 = 2%. This becomes a poor assumption if the
device contains a large density of interfacial traps — if the device is unannealed, for
example. (h) A depletion—mode transistor is a MOSFET that is "on" or conducting when VG = 0. (i) The ﬁeldoxide lies outside the active region in MOS devices and integrated circuits; the
gate—oxide lies directly beneath the MOS gates. The ﬁeldoxide is typically much thicker
than the gate~oxide. 6) 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. 18—1 (b) ¢MS = %((I)M—(DS) = [X'—(EF—Ee)poiySi]~[x'+(Ee~EF)FB,crysxanineSi] = Ell—[(Ec — EF)poly—Si — (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 /7. 18»2 I —0.03eV ...Al (See Fig. 18.3 caption.)
‘DM' * X' = \ —«O.l8eV ...n+ poly (See Fig. 18.3 caption.) (EoEF)p+ poly  X'= (X'+EG)—X' = EG = 1‘126V uP‘” Poly
Also (EcEF)FB = (Ec~Ei) + (EiEF)FB 5 56/2 + (Er—Elﬁn;
or
I Eg/Z  (kT/q)ln(ND/ni) ...n~type crystalline Si (E —EF) =
0 FB lEG/z + (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 MATLAB 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— E E Eng, the more accurate value of Ec—Ei =
0.57eV was used in constructing Fig. 18.3. 1.2 oa mﬁw€4 0.6 . 5 3  9M5 (volts)  18 10 NA or ND (m6) 183 h¢ATLAB1nognunscﬁpL“ %Metal—Semiconductor Workfunction Difference %Initialization
clear; close %Constants and Parameters
ni=l.0e10;
EG=1.12;
kT=0.0259:
s=menu('Specify the gate material','A1','n+ poly','p+ poly”):
if s==l,
=—0.03: elseif s==2, A;—0.18; else A=EG:
end %Ca1culate M—S Workfunction Difference
%EcEF=(Ec—EF)FB NB=logspace(14,18);
EcEFn=EG/2—kT.*log(NB./ni);
ECEFp=EG/2+kT.*log(NB./ni);
aMSn=A~EcEFn; aMSp=A~EcEFp; %Plottfng result
semilogx(NB,6MSn,NB,zMSp); grid
xlabel(‘NA or ND (cm3)'); ylabe1('gMS (volts)') 18—4 18.4
(21) Given Pion = p0 = constant, x0 2 mobile _ 1 .. W0 AVG( ions ) —— ~K0£0 xpoa'x — _2K0€0
0 _ (1.6 x 1019)(1018)(105)2
(2)(3.9)(8.85 x 1014) = — 23.2V (b) Here
pion : QM5(xo) where x0
QM = I pion(X)dx = 100160
0 Substituting pion = QMsOCo) into Eq.(18.13) gives 2
b'l QM x
mm) 2 _ =  0 ~ We Co K050,
Clearly the AVG here is twice that in part (3). AVG = —46.4V 18—5 iii
(a) AV(;(ﬁxed charge) = — QF/Co ' .“Eq. (18.15) (b) We are given or mathematically 0 ...0 S x S xoAx
pox = 2
&)c' ...O S x‘ S Ax, where x": x—xO+Ax
sz Substituting pox into Eq.(18.1 1) gives xo , Ax
AV : _E—1——f xpoxdx = ~J—(Q) I x'(x'+xoAx)dx’ and (c) AVg(part b) 2 1 Ax
AVG(pan a) 3x0 Ifo = 10‘7cm and x0 = 10‘5cm —) AVG(b)/AV(;(3) = 0.997
If Ax = 10‘7cm and x0 = 10‘6cm ——> 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 very thin oxides the
difference becomes detectable, but not all that signiﬁcant, even when x0 is only 1013):. 18—6 1§.§ ,
(a) If the MOSC is ideal except for MS at 0 and Q}: at”: 0, then VFB = «ms—4%— = ms—KEZOQF
0 A plot of V123 versus x0 data should be a straight line with an extrapolated VFBaxis
intercept equal to MS and a slope of —Qp/K0£0. (b) The given VFB versus x0 data is plotted below. A least squares ﬁt through the data
yields VFB = —0.596— (3.02x104)xo .40 in cm
Thus (PMS = —0.596 V (3.9)(8.85x1014)(3.02><104)
1.6X10‘19 QF/CI =,Koso(810pe)/q = = 6.52x10101cm2 0 0.1 0.2 0.3 04 0.5
x0 (um) 18—7 l 8 . Z
(a)  If there is no charge in the oxide, if pox = 0, then an = constant and the oxide energy bands are a linear function of position. However, if p0x :3 0, gm 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 Dfield, where D = 88 , must be continuous
if there is no plane of charge at an interface between two dissimilar materials (see Subm
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. P18.7 that the slope of the
bands is zero and therefore 8 = (l/q)(dEc/dx) = 0 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 Qimerface = K 38083 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 spread—out volume charge, making alkali ions an unlikely source of
Qinterface. 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 Q1: ¢ 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 Si—SiOz interface. However, in deriving
Eq.(18.l l), the Dwﬁeld was explicitly assumed to be continuous across the Si~Si02
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 Q: in pox in the Eq. (18.11) 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. 18—8 m (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 non—ideal device. Because er = 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 CMAX = C0 =1£__.__ we conclude
14 3
x0 = KOgOAG : (3.9)(8.85X10 )(2.9X10 ) = 5.00X10’6 cm
Co 200x1012
Also
Co
CMIN = H
1 + KOWT
KSXO
ak‘n 6
m 1g WT=§S_X_Q(_§Q__ )ZWPQQ )z3‘00x1050m=0,3‘um
K0 CMlN (3.9) 67 Referring to Fig 16.9, the plot of WT versus N A or M), we conclude a WT = ()3,th
results when ND E ION/cm3 (c) Since QM = 0 and QIT = O,  _ QF
AVGlflatband ” VFB _ WS“E; VFB :_ —0.71 in the statement of the problem. Also, for an ND 2 low/cm3 AIM—Si) device,
we conclude from Fig. 18.3 that ¢Mg = —O.24V. Thus 200x1012
QF = Co(¢Ms~ VFB) = thswm) = ———————~(~0.24+0.71>
AG 2.9x103 2 3.24X10‘3 coul/cm2 189 L812 We infer from the C—V characteristics that the MOSC is a p—bulk device. Also, we know
that acceptordike traps are negatively charged when ﬁlled with an electron and neutral when
empty. For a p—bulk MOS—C the effect of biasing on the occupation and charge state of the
acceptor—like traps is summarized in the following ﬁgure. Ace Depl We also note
AVG = .. Q: Co Thus, relative to the "after" or negligible er situation, the "before" characteristics will be
shifted positively (er 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 18.19 From the answer to Problem 15 (see Solutions Manual pages 1—2 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 should exhibit
the higher density of residual "dangling bonds" or interfac1 traps. xperiments confirm
the above conclusion.) 1840 l .11 (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 141231 > Mpg!
depletion biased or inversion biased. Empty
Donorlike
Filled (+)
Donorlike
(neutral) \ Clearly, there is no shift in the C~V curve when the device is accumulation and Mg! < l¢pl
depletion biased. However, when ups! > I¢Fl depletion biased or inversion biased, the
characteristics are translated AVG = —Q[r/Co = constant negative value along the voltage
ax15. (b) For acceptorlike interfacial traps . ‘ Position Trap _ t
. Bias ‘ of E1: Occupation , State ‘
_; twwd We ,. ., , we, W’s' WP” depl 7 7 "PS! >'¢F,. (1691’ 1‘” AVG 1811 From the preceding one concludes, (c) If thestates are very close to EC they retainthe 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 (hp and a negative shifting for the
entire depletion pan of the C—V characteristic. \ One cannot see the
place where Q If
becomes neutral ——
it is lost in the level _ ’accumlation portion ideal of the curve. VG (d) A donorlike level very closeto E, is always ﬁlled and neutral for the nonaconstant capacu
itance portion of the C—V curve. There will be no observable C—V shift due to such states.
C V One cannot see Q 1T ,
become positive ——— it is
lost in the ﬂat inversion
portion of the curve. l ~ Same as ideal No te: This problem points out the difﬁculty of detecting EIT states that are very close to
the band edges. 1812 1 8. 12
The two halves of the MOSC may be viewed as separate capacitors. irradiate’lk I
g MOS—C F. é MOSC Before irradiation, each half will contribute precisely onehalf 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 "é after" in the ﬁgure below)
will be shifted toward negative voltages due to the apparent QF. Graphically combining the
"g before" and '% after" curves yields the total expected "after" curve. 18.13 (a) The shift in the gd — VG characteristic after +BT stressing is symptomatic of mobile ions
in the oxide. (b) Conceptually extrapolating the gd — VG curves into the VG axis, we conclude that the
tumon voltage has shifted negatively ~ 2V after +BT stressing. The device is now obviously "off" when VG = — 2V and VG = — 3V. Moreover, the VG = —4V state after
stressing is equivalent to the VG = — 2V state before stressing. Thus, 1813 1 8. 14
(a) VT will shift in the VG direction. Since Q: is positive, an apparent Q1: causes a
negative shift in the threshold voltage. ——.———f—___> +VG (apparent QF ¢ 0) VT(— VTO (b) The gate material affects ¢Mg With (DML x: ———0 03 eV for Al (see the Fig 18 3
caption) and <I>M'— x: —.0 63 eV for Cu (from Table 18.1), ¢M3 and hence VT will increase
by 0. 66V 1n going from an Al to a Cu gate. ____1___T_> +VG (Al ——)Cu gate) VTO ——> VT (c) The substrate doping affects both MS and VT'. As given by Eq.(l8.22), KS / 4qNA kT (Na)
V = 2 + x where = —ln ——
T ¢F K o K380 ¢F ¢F q ”i (Ea—E19123 E EG/2~(E1—EF)FB = EG/2+4¢F
¢Ms= (1/q)[<I>M X —(EcEF)FB]=(1/C1)[<I>M~X—EG/2] ¢F Since (pp increases with doping, ¢MS decreases and VT' increases with increasing N A However, the increase in VT' is greater than the decrease 1n ¢Mg, and VT: VT + ¢Ms
increases with an increase in substrate doping. ‘_—_._*____}_—p +VG (meieased NA) V10 —> VT Also 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, x0 in this problem affects only VT".
Inspecting the VT' expression quoted in part (b), one rapidly concludes VT‘, and therefore
VT, decrease with decreasing x0. ———{—_—+————> +VG (decreased x0) VT <~—~ V‘m
(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. m__+___.l._> +VG (Boron implantation) VTOw) VT 1814 18,15 ‘
(a) Adding the voltage shift due to the ion implanted charge (Eqr 18.25) to the regular ﬂat
band expression (Eq. 18.20), one obtains _ V = __ __ _ ___._ _ ____ _ __
FB ¢Ms Co C0 Co Co
2 x0 QF+QMYM+QH(0)+QJ]
(ms qK 0i 4 q q q
For the given device —19 —6
VFB = —— 0.46—W(2X1011+ o + 0  4x10“) 9. 0
(3.9)(8.85><1014) (b)
. _ Ks W
v  2 ~— *
4,1: I: _l%1n(ND/ni) = 410259 ln(’1015/1010) : —0.298V
. 19 15 1/2
W = _ (2)(0¢298)~g—1—°§) (5X106)[(4)(1.6X10 )(10 )(0.298)
(3.9) (11.8)(8.85><1014)
= —0.80V
and
VT = Vf+VFB 2 v3r = — 0.80V (c) Enhancement mode device. For the given p—channel device there is no inversion—
layer at zero 1as an t erefore no drain current when VG = 0. A MOSFET which is "off"
at zero bias is referred to as an enhancement mode device 18—15 18. 1g
Combining Eqs.(18.21), (18.20), and (18.25), one can write Since there are no interfacial traps and no mobile ions in the oxide, QM = O and Q11" = 06
Also Co = Koeo/xo and Q; = —qNI. Thus the VT expression simpliﬁes to VT = Vi+ ¢Ms~q x0 (QF/q—NI)
K060 : Solving the preceding equation for N1 then gives . K I
HN1= %F~+% 2:0(VT~VT¢Ms)ﬂ (212/51 and VT are speciﬁed in the statement of the problem. However, we need to
determine ¢MS and VT'. Because the MOSFET is an Al—SiOg—Si device, we can read W3
directly from Fig. 18.3. For N A = 1017/cm3, one ﬁnds ¢MS = —1.02V. The idealndevice
threshold voltage can be computed using eq(18.22). Vyi‘_:,2¢F+£§x0 _4CINA KO K380 (1’1? 914: = %1n(NA/ni) = 0.0259 moon/1010) = 0.417v (4)(1.6x1019x1017x0417) ”2 (11.8)(8.85X10‘14) , _ (11.8) ,6
VT~ (2)(0.417)+———(3‘9) (10 )[ = 1.32V
Finally, substituting into the N1 expression, we obtain (3.9)(8.85x1014) N1 = 1011 +
(1.6X10‘19)(10“6) (0.5 — 1.32 + 1.02) or
N; : 531 X 1011 boron ions/cm2 l8~l6 ...
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