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Unformatted text preview: CHAPTER "16 15;; Biasing EL“ PEPE; Minn (a) (b) (C) (d) (e) depletion ﬂat band depl/inv
transition accumulation inversion Energy Band
Diagam 161 Block Charge
Diagam electrons #:%:HE ‘CYCCUOHS % ‘ECCU‘OHS MOS E17 \ ionized
acceptors
electrons
M O (b) The part (a) charge diagram is in agreement with the p/qNA versus x plot in
Fig. 16.8(c). To obtain the total charge in the semiconductor at each point one adds the
separate block charges shown in part (a). p x
ionized
acceptors electrons The spike nearx = 0 in the Fig. 16.8(c) plot simply reﬂects the forming inversion layer of
electrons at the surface. By deﬁnition, at the onset of inversion Usurface = N A. Thus, at the special VT bias point p3 = —q(n3urface + N A) = —2qNA, or p/qN A = —2 at x = 0 at the
onset of inversion. (0) Since ¢F/(kT/q) 2 12, inverting Eq.(l6.8a) yields NA = nie‘l’F/(kT/Q) = 1.00><1010e12 =1.63><1015/cm3 1/2
i : 0.706pm (1.6X10‘19)(l.63><1015) ; 1/2 ~14
Wr— _ [23:0 (2%)] : [(2)(11.8)(8.85x10 )(24)(0.0259) (From Fig 16.9 one also reads WT E Glam.) The above WT is indeed consistent with the positioning of the end of the approximate
charge distribution (the dashedline distribution) in Fig. 16.8(6). 16—2 16.3 The required WT versus doping plot appropriate for GaAs and the MATLAB program script
that generated the plot are reproduced below. 1 10
10° E 2 G) E O Q E f. 3
10“
10.2 ‘ zgiiii‘ I :Q;_£_;J i‘221:::f‘ ‘ H 1014 1015 1016 1017 1018 NAorND(anN3) MATLAB program script...
%WT versus NA or ND for GaAs at 300K %Initialization
clear; Close; %Constants and parameters q=1.6e—l9: e0=8.85e~14; ni=2.25e6; KS212.85; kT=0.0259; , NB=logspace(l4,18); %NB = NA or ND %Plotting result %WT calculation loglog(NB,WT); grid
oF=kT.*1og(NB./ni); xlabel('NA or ND (cmA—B)‘);
WTzsqrt(4*KS*eO.*eF./(q.*NB)); ylabe1('WT (micrometers)');
WT:(1.0e4).*WT; %WT in micrometers text(1.0el7,2.3,'GaAs, BOOK‘) 16—3 16 4
(a) 1‘1 = —1n (ND/mi) = «(1015
1010 ) = —11.51
kT/q $1: = e11.51(kT/q) 2—(11.51)(O.0259) =0.298V (b) Using Eq.(16.l6) with NA —+ ~ND, 1/2_ [(2)01.8)(8.85X10'14)(2)(0.298) W= WT=[2KS£0 (2%)] —
(1,6x1019x1015) 1/2
2 0.882;:111
qND (0) Evaluating Eq.(16.12) at x = 0 yields 83. Thus, with NA —> wND in Eq. (16.12), 49 15 4
8 = __q_I_YP_W = (1.6X10 )(10 )(O.882X10 ) :_1.35 X 104 V/cm K380 (11.8)(8.85X10‘14) (d) Substituting into Eq.(16.26) gives VG = 2¢F + g5 083 ...85 evaluated at $8 = 2m;
0
5 4
= — (2)(0t298) _ W91
= 4.00 V ‘ Except for the doping type, the parameters used in this problem are identical to those
assumed in constructing Fig. 16.10‘ Since (pg = 2¢p, the WC} calculated in part ((1) should correspond to the depletion/inversion transition point in Fig. 16.10. Indeed, in the ﬁgure
VT 3 1V. (e) The MATLAB program script yielding a computer generated computation of (1)12, W, 83,
and VG is listed on the next page and included on the Instructor's disk as mﬁle
P_16_04.m. Note that a normalized 433, (pg/2%, is taken to be one of the input variables
Also, donor dopings must be input as a negative concentration in running the program. 164 MATLAB program script... %Autocalculation of EF, W, ES and VG %Initialization
clear; close:
format compact %Constants and parameters
q=l.6e—l9; e0=8.85e—14; ni=1.0e10: Ks=11.8; KO=3.9; F kT=0.0259; %Input variables xo=input('Please input xo in cm, x0 = ');
N=input('Please input NA or —ND in cmAB, N = ‘);
r=input(‘Please input oS/ZoF, oS/ZoF = ‘);
NB=abs(N); s=N/NB; %0F and WT calculation
®F=s*kT*lOg(NB/ni) oS=r*2*cF; WO=sqrt(2*KS*eO*QS/(q*N)); %WO in cm
W=(l.0e4)*W0 %W in micrometers %Surface Electric Field (ES) calculation
Es: (q*N*WO) / (KS*e0) %VG calculation
VG=¢S+KS*XO*ES/KO 16.5
(a) In general we can write
A130 VG = (pg +£§~x083 ...Eq.(16.26)
K0
(is : E3: Us ...Eq.(B.2)
and Eq.(B.l6) evaluated at the surface gives
A F (U 511 F)
= U .1: _l_.
85 S q LD Substituting the above 455 and 83 expressions into the general Vg—gbs relationship yields the
desired result; 16—5 V dip +3 WKSXOFUy
G q s SKOLD(SF) (b) The required VG versus Us computation is rformed by the MATLAB mﬁle listed
below. Setting 1: = 0.1;tm and ND = 1015/cm yields a plot identical to Fig» 16.10 except
the entire plot is reﬂected through the origin of coordinates. MATLAB program script...
%VG versus US Calculation %Initialization
clear; close
format compact %Universal and System Constants
q=1.60e—19; e0=8.85e—l4; kT20.0259: %Device andeaterial Constants KS=11.8; KO=3.9; ni=1.00e10; LDzsqrt((KS*eO*kT)/(2*q*ni)); s=input('Employ xo=l.Oe—5cm and ND=l.Oe15/cm3? 1Yes, Z—No..l‘);
ifs=ﬂ Net=l.0e15;
xo=1.0e—5;
else Net=input('Input the net semi doping in cm—3, ND~NA = ');
xo=input('Input the oxide thickness in cm, x0 = ');
end N=abs(Net); sign=~Net/N; UF=sign*log(N/ni) %Computation Proper US=UF~21:1:UF+21; S=US./abs(US);
Fzsqrt(exp(UF).*(exp(US)+US~1)+eXp(~UF).*(exp(US)US~1));
VszT*(US+S*(KS*xo)/(KO*LD).*F): %Plot result
plot(US,VG); grid
ifs=ﬂ
axis([—40, 10, ~4, 4]);
end
xlabel(‘US'); ylabel('VG (volts)‘) 16—6 16.6
(a) Eq.(16.28) may be viewed as a quadratic equation with \[ES as the variable. [1+——LV ]1/2— 1“
(12/27)2 } £3401] (INA ){[1+ VG :1/21\
K0 2Ks£0 , \ (b/2)2 ‘ Substituting the #33 result into Eq.(16.15) then yields 01' WE: 2 2
VH3) Wm
K680
We have indeed obtained the text result. (b) (i) 4b,: = «ﬂln(ND/ni) = — (0.0259) 111(191—5) 2: —O.298V
q 1010 x _ 4 1/2
WT _[2KS«€O (2%)]1/2:[(2)(11.8)(8.85X10 1 )(2)(0298)] : 0.882111“
(1.6X10'19)(1015) (From Fig. 16.9 with ND 2 1015/cm3 one would estimate WT E 0.9;1m.) 167 (ii) From Eq.(16.34d), L = ~—~—1— ...inv(a) —~> 00) KSxo = ——1————— = 0.255
1 + (3.9)(O.882) (11.8)(0.1) (iii) Some care must be exercised in working this part of the problem. An acceptable
approach is to proceed as in Problem 16.4, ﬁrst calculating 83 using Eq.(16.12) and
then substituting into Eq.(l6.26). In fact, the parameters are the same as in Prob. 16.4
and thus the expected answer is VT = —1.00V. Alternatively, one might consider
substituting into Eq.(16.28) directly; VG = VT when (1)5 = 2(1)}: However, Eq.(16.28)
is only valid for p—type devices and simply changing N A to ~ND will not yield the
correct VT. [For an ntypc device the "+" between the two right—hand terms in
Eq.(16.28) is replaced with a "—" Sign] Nevertheless, Eq.(16.28) can be used if we
ﬁrst act as if the doping was p—type, and then just change the sign of the result noting
the voltage symmetry between ideal n and ptype devices , VT = "—,.[(2¢F) +kK—Sxo quA (2¢F) ] (¢F> 0)
» o K880
: _ [(megg) + £1 1.8)(10'5) [(2)(1.6><10'19)(1015)(2)(O.298)]1/ 2‘
l 39 (11.8)(8.85X10‘14) f :2 —1.00 V (— expected result (iv) The parameters used in this problem are identical to those assumed in constructing
Fig. 16.13. Thus the part (ii) C/CO value should correspond to the high—frequency
inversion value on the ﬁgure and the VT calculated in part (iii) should be the
depletion/inversion transition voltage shown in the ﬁgure. This is indeed the case. 16~8 lﬁl (a)‘¢ has the same shape as the "upside down" of the bands. ¢
W (b) 8 is proportional to the slope of the bands. Also, as emphasized in a footnote on
p.581, 80x E 383 in an ideal MOS—C. (c) . Inside the semiconductor EF is position independent. (d) Noting
n : nie(EF“‘Ei)/kT we conclude
ln(n) ND 0 W (e) Since E1: = E at the Si~Si02 interface, lnlxzo E 111 = lOlo/cm3 l. (0 ND 5 ”bulk : nie {Epg—Ei(bulk)]/kT : {1010)60'29/0'0259 = 7.29 x 1014/cm3 (g) (p3 = (l/q)[Ei(bulk) ~ Ei(surface)] = —0.29 V 16w9 (h) Some care must be exercised in completing this part of the problem. Simply employing
Eq. (16.28) with N A replaced by ——ND yields an incorrect result because 85 < 0 when an
n—bulk MOS C is depletion biased. Speciﬁcally, for an n«bulk device 2qND 1/2
8 =4 _ l
s K350 ( $3)
and /
__ KSXO ZCIND ‘
VG — ~~ (‘
$8 Ko K380 $8)
Thushere
5 19 14 1/2
VG = _O‘29_(11.8)(2X10 )[(2)(1.6><10 )(729X10 )(029)]
(3.9) (11.8)(8.85X10‘l4)
01' VG = 0.78 V (i) VG = Mex +¢s
A¢0x = VG~¢S = —O.78 +0.29 = —0.49 V q Kgxg ND : (1.6x1019) (11.8)(2x105)2(7.29x1014) 0) V5 = — = 0.2045 v
__ [(3350 2 _ (3.9)2(8.85x1014) 2 C = 1 s 1 = 0.45
Co V1+Vg/V5 V1+il.787il§ii (Eqs. 16.15 and 16.341) may alternatively be used to compute C/Co.) L631 Inversion . . . e, 4
Depletion . . . C, 3
Flat band ‘ . . b, l
VG = VT a . . d, 2
Accumulation . . . a 5 16 10 16.9 (a) . The Fermi level inside the semiconductor is position independent.
(b)... ¢F = (Ilq)[Ei(bulk)  EF] = 0.3 V (c)... 453 = (1/q)[Ei(bulk) — Ei(surface)] = (1);: = 0.3 V (d)... Ep(metal) — Ep(semi) = —qVG ....Eq (2.1)
VG = (l/q)[EF(semi) — EF(metal)] = 0.6 V (e) Based on the delta—depletion approximation, VG = ¢S+EEA ZqNA 493 4: Eq.(16.28)
K0 K880 where from prior parts of the problem VG = 0.6V and (#5 = 0.3V. Also, (PF = (161761) 1n(NA/ni)
01‘ NA = mew/(15774) = (1010)e03/00259 2.1.073 x1015/cm3
Thus x0 = Jﬂ... = MALE : 0.1%,“
Es. 2(INA 4,8 {113) (2)(1.6><10‘19)(1.073X1015)(0.3)] /
K0 K580 ‘ 3.9 (11.8)(8.85X10‘14) (f) lllllﬂllll‘lul 1611 (g) : [Hype
j lowfrequency 4
; xin depletion region 3 .......................... (h) Expressions (i) and (iv) are clearly wrong because they do not apply to depletion.
Employing Eq. (16.28), we conclude VT 5 1V and VG E 0.6VT. Thus referring to Eqa (16.37), expression (ii) is close but not the correct expression. Finally, noting that at; the
speciﬁed bias point, 903 = (in: and W = [MWJUZZ WT/{f cINA
we conclude
C = ._.C_'Q_._ = 4L— : Expression (iii)
K 5x0 QKSXO 1612 (a)
Ec
‘ EF=Ei
/
// Ev
M O S
(b)
M O S M O S
holes
\
electrons
Vg>0 I Vg<0
(C)
C
Co
V
O G Justification: When VG > 0, electrons pile—up in the Si immediately adjacent to the oxide
giving rise to a low‘frequency C = Col Similarly, when VG < 0, holes pile~up in the Si
immediately adjacent to the oxide giving rise to a low~frequency C = C 0. (Actually, C E Co, but in the deltadepletion formulation the carrier layers are taken to be 8—functi0ns
at the Si surface.) Note that, within the framework of the delta»dep1etion formulation, there
is no ”depletion" or depletion—like region inside the given device. 16—13 Part (a) l Part (1)) i
No charge
anywhere S O S (1i) VG > 0 but small ionized donors \ ace layer of
electrons holes (inv layer)\ ionized
donors (iv) VG < 0 but small — (ii) answer with semiconductor regions interchanged.
(v) VG < O and large ~— (iii) answer with semiconductor regions interchanged“ (C) 16—14 1 .12 (a) are standard low— and highfrequency C—V curves that result when the semiconductor component of the MOSC is in equilibrium under dc. biasing
conditions. Curve 6 is a nonequilibrium deepdepletion characteristic. (b) In accumulation C—> Co = KomAglxo Since both devices exhibit the same capacitance
in accumulation, the two devices have the same oxide thickness. With x0 being the same,
the lower capacitance of device b in inversion indicates this device has a lower doping. (WT increases with decreasing doping, thereby giving rise to a smaller capacitance; also
see Fig. 16.14b.) 16.13 (a) ...For p—type devices accumulation (C max) occurs for negative VG and
inversmn Cmin) occurs at positive VG. The exact opposite is true for notype devices. (b) At point (2) the ptype MOSC is far into inversion. Thus (c) At point (1) the MOS—C is clearly deep into accumulation. 16~ 15 (d) From Fig. P1613, Cmax = IOOpF. However, [(08vo Cmax = C0 = x0 2 {gm __(3.9)(8.85><10‘14)(3x103) x — —————————————————— = 0.104ym
O Cmax (1040)
(c) In the deltadepletion formulation
C = “91—— inv (a) ——9 co) (1634(1)
1 + KOWT
l(5x0 Thus :QEQQQW)=W119QM :
WT K0 (C 1 (3.9) (20 1) ”Wm Employing Fig. 16.9, we conclude N A E 5 X 1014/cm3. 16.14
(a)
VG
14 3
(b) CMAX : C0 : Koé‘oAG = (3.9)(8.85><10 )(10 ) 2 345 pF “‘0 10'5 (c) (pg = ~(kT/q) ln(ND/ni) = 410259 1n(2><1015/1010) = —O.316 lIND WT = [m (_2¢F)]1r2 : [(2)01.8)(8.85X10l4)(2)(0.316) 1/2 : 6.42X10'5 cm (1.6x10l9x2x1015) 1646 CMN=_..£Q.._:__3AA____= 11.1pF
1 + 182V; 1 (3.9)(6.42X10'5)
KSxo + 5
(11.8)(10 )
((1) By deﬁnition, if VG = VT. . . qbs :2 2% = —0.632 V
(e) VT = (2%) ~st0 .2qu (—29%) (Also see Prob. 16.7h.)
K0 K580 1/2
= 1.23 V ~2(0.316) — (11 (3.9) (11.8)(835x10'14) .8)(10'5) [ma.6x1019)(2x1015)(0.632) §\\\\\\\\\\u (h) Under the speciﬁed operating conditions, the MOS—C is expected to exhibit a total deep
depletion characteristic exempliﬁed by the dashed line in the part (a) answer. 1647 15.15 (a) ...There is an inversion—layer of negative charge  electrons «— shown in the
bloc c arge diagram. The semiconductor must therefore be patype. (Also, the depletiona
region charge is negative or clearly due to acceptor ions) (b) ...As noted in part (a), there is an inversion layer with n3 > N A
shown ont e ragram. (C) (d) AQ ~AQ due to
ac signal at
high~frequency 1618 (e) operating
point (0 NOTE: Because the added depletionlayer charge is farther from the surface than the inversion layer Charge, there is NOT a oneto'one correspondence between the two
charges. Also, the charge on the metal will be slightly different than under equilibrium inversion conditions at the same VG bias. 16—19 1 6. l 6 The MATLAB program script yielding deep—depletion p—type MOSC C—V characteristics
and a sample plot (x0 = 0.2ttm, N A = 7.8X10 4[cm3) are reproduced below. Note that the
sample plot has been extended to VG = SVT (as opposed to stopping at VG = 3VT per the
directions in the ﬁrst printing of the book). If the sample C~V curve is converted to an
n~type characteristic AND translated approximately 2V along the voltage axis in the
negative direction, the sample plot becomes a very good match to the experimental totai~
deepdepletion data displayed in Fig 16.17. MATLAB program script...
%p—type Deep Depletion MOS—C C—V’Characteristics %Initialization and Input
clear; close
format compact NA=input('P1ease input the bulk doping in /cm3, NA=');
xo=input('Please input the oxide thickness in cm, xo=‘); %Constants and Parameters
e0=8.85e—l4: q=1.6e—19; k=8.617e—S; KS=11.8; KO==3.9; ni=l.0e10; T2300; kT=k*T; %Computed Constants
UF=log (NA/mi) ,
LD=sth((kT*KS*eO)/(2*q*ni)); %C~V Computation for US < UF ( or VG < VI)
US=UF—21:O.5:UF: ,
Fzsqrt(exp(UF).*(exp(US)+US—l)+exp(—UF).*(exp(US)—USl));
VGl=kT*(US+(US,/abs(US)).*(KS*xo)/(KO*LD).*F):
DENOMlzexp(UF).*(l—exp(~US))+exp(—UF).*(exp(US)—1):
W1=(US./abs(US)).*LD.*(2*F)./DENOM1:
cl=1.0./(1+(KO*W1)./(KS*xo)); %CwV Computation for US > UF (or VI < VG < SVT)
FI=sqrt(exp(UF).*(exp(—UF)+UF—l)+exp(—UF),*(exp(UF)—UF~1));
VI=kT*(UF+(KS*XO)/(KO*LD).*FI): oF=kT*UF; VT=2*QF+ (KS*xo/KO) *Sqrt( (4*q*NA*QJF) / (KS*eO) ) ;
Vdelta=(q/2)*(KS*XO“2*NA)/(KO“2*eO); VGZ=VI+O.1:O.1:5*VT; c2=1./sqrt(l+VGZ./Vdelta): 1620 %Combining and Plotting results
c=[cl, c2] ; VG: [VG1,VG2] .‘ plot (VG, c) ; grid axis ( f3*VT, 5*V’I', O, 1. 1}) xlabel ( 'VG (volts) ' ); ylabel ( 'C/CO') 16~21 (a)
Co : [<08er :3 x0 = [(08vo
x0 CO
14 3
x0 = (3.9)(8.85X10 )(4.75><10 ) = 0200”!“
82x1012
(b) From Fig. 16.17, C/C0(inv) E 0.39.
C(inv) = m, C0 .
1 + KoWetrOnV)
KSXO
 st C0 ' (11.8)(0.2) 1
W = ——0 —_1 = ~———— ————1
6mm") K0 [C(inv) l (3.9) (039 )
= 0.946pm
and
~ 5
Weff(1nv) = 9.46X 10 = 3.25 X 10.2 LD 2.91 x 103 (c) If Weffﬁnv) is equated to WT, one estimates from Fig. 16.9 that ND 5 8.5 X 1014/cm3
or Up = —ln(ND/n;) = —in(8.5><1014/1.00><1010) = —11.35. Substituting Up = 41.35 into
the expression for Weff/LD one computes Weff/LD = 3.374 X 10‘2. Weff/LD is too large
implying ND and IUFI are somewhat larger. Trying UF 2 —11.45 yields Weff/LD = 3.223 X 10’2; trying Up = —l 1.40 yields Weff/LD = 3.298 X 10‘2. Clearly
U1: is bracketed between —11.40 and 41.45. Subsequent calculations give Weff/LD = 3.268 x 10—2, 3.253 x 102, 3.238 x 102 when U}: =~11.42,~11.43, and
w11.44, respectively. The best value appears to be 1 U}: = 1143 and ND = nie‘UF z (1010)e1143 = 9.20X1014/cm3? NOTE: We actually pushed the calculation here beyond the accuracy of the data to illustrate
the procedure. 1622 ...
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