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chap16 - CHAPTER"16 15 Biasing EL“ PEPE...

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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 16-1 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 dashed-line 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%)] — (1,6x10-19x1015) 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 m-file 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. 16-4 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 cmA-B, 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 m-file 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? 1-Yes, 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 :|1/2-1\ 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)(0-298)] : 0.882111“ (1.6X10'19)(1015) (From Fig. 16.9 with ND 2 1015/cm3 one would estimate WT E 0.9;1m.) 16-7 (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 n-typc 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 p-type 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 )(7-29X10 )(0-29)] (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 : (1.6x10-19) (11.8)(2x10-5)2(7.29x1014) 0) V5 = —- = 0.2045 v __ [(3350 2 _ (3.9)2(8.85x10-14) 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)e03/0-0259 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 16-11 (g) : [Hype j low-frequency 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 16-12 (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 delta-depletion 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 high-frequency C—V curves that result when the semiconductor component of the MOS-C 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 p-type MOS-C 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)(3x10-3) x — —————————————————— = 0.104ym O Cmax (1040) (c) In the delta-depletion 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)(8.85X10-l4)(2)(0.316) 1/2 : 6.42X10'5 cm (1.6x10-l9x2x1015) 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.6x10-19)(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 16-18 (e) operating point (0 NOTE: Because the added depletion-layer charge is farther from the surface than the inversion layer Charge, there is NOT a one-to'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 MOS-C 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~ deep-depletion 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)—US-l)); 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): 16-20 %Combining and Plotting results c=[cl, c2] ; VG: [VG1,VG2] .‘ plot (VG, c) ; grid axis ( f-3*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”!“ 82x10-12 (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) (0-39 ) = 0.946pm and ~ -5 Weff(1nv) = 9.46X 10 = 3.25 X 10.2 LD 2.91 x 10-3 (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 10-2, 3.238 x 10-2 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)e11-43 = 9.20X1014/cm3? NOTE: We actually pushed the calculation here beyond the accuracy of the data to illustrate the procedure. 16-22 ...
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