Chapter 18 - CHAPTER 18 I S. 2 (a) In theory the two...

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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 QI-r 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 gate-oxide lies directly beneath the MOS gates. The field—oxide is typically much thicker than the gate-oxide. (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. lS-l (a) (b) ¢MS = %(¢M’¢S) z [X"“(EF—Ec)poly-Si]“‘[X'+(Ec—EF)FB,cryslalline-Si] = $030 “‘ EF)poly-Si - (Ec - EF)FB. crystalline-Si] = —0.4 V (Note that the computational equation developed here is the same as Eq. 18.24.) (c) biased. When VG = 0 the polysilicon side of the part (a) diagram is ' lowered, yielding little band bending because of heavy doping accumulation 18—2 18,3 In general ¢Ms = qi[¢M'—x'— (EC—EFnB] where I —0.03eV ...Al (See Fig. 18.3 caption.) (DM' —- x' = -0.18eV ...n+ poly (See Fig. 18.3 caption.) A i (En—Ep)p+ poly * x'= (1'+EG)-x' = 196 = Law mp" P01}! 1 Also (EC-5F)“; = (Ea—El) + (Er-5F)“; E Eel?- + (Er-Emm 01' I E‘s/2 — (kT/q)ln(ND/ni) ...n-type crystalline Si — LEG/2 + (kT/q)ln(NA/ni) ...p-type crystalline Si The results of the p+ polycrystalline-gate computation based on the above relationships are presented in the following plot. The MA'ILAB program script used to generate the plot is also listed on the next page. Although it leads to only a minor difference, it should be mentioned that, instead of employing Ec— E15 life/2. the more accurate value of Ec—Ei = 0.57eV was used in constructing Fig. 18.3, 1.2 (EC‘EF)FB oMS (volts) 0 9 m m .0 4'. 0.2 .. . :. 10M 101$ 10‘! 10" 10“ NA or N D (tame) [8-3 MATLAB program script... %Metal-Semiconductor 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 M-S Workfunction Difference %EcEF=(Ec—EF)FB NB=logspace(14,18): EcEFn=EG/2—kT.*log(NB.lni): EcEFp=EG/2+kT . *log (:13. /ni) ,- QMSn2A-ECEFn: eMszA-EcEFp; %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 Sxo-Ax Pox = lQ—Ex ...0 .<. x' S Ax, where x": x-xo+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 = 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 10Ax. 18-6 M (3) Given pic“ = p0 = constant, x0 2 mobile _ 1 _ _ 90% AVG( ions “' “EOE! 1!)de - _ (1.6 x 10-19)(1018)(10-5)2 (2)(3.9)(8.85 x 10-14) = — 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 18-5 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 VFB-axis 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 ‘ 1.6x10-19 _ o 0.1 02 03 0.4 0.5 .zocum). 18-7 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 spread-out 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 Si-SiOz 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 D-field 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 1_&..8 (a) In an ideal version of an MOS-C, 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 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 2.9x10-3 =. 3.24X10‘8 coullem2 18-9 1.8.2 We infer from the C41 characteristics that the MOS—C is a p-bulk device. Also, we know that acceptor—like 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. 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 1-2 and 1-3), 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.) 18-10 111 (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 Donor-like Filled (H Donor-like . ; (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 18-11 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 donor-like level very close to E. is always filled and neutral for the non-constant capac— itance portion of the C—V curve. There will be no observable C-V 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. - 18-12 18,12 The two halves of the MOS-C 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 MOS-C 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 turn-on 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) Vfiu-a 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- (El-591:3 = 56/2 +q¢1= ms = (IICIH‘DM’ — x' - (Ea-EF)FB] = (llq)l¢ MLXLEG/Z] — tap Since 4512 increases with doping, ¢Mg decreases and V-r' 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 V-r'. 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 18-14 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)(3.85x10-14) (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 p-channel 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. 18-15 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 ideal-device 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, + (11-8) (106) 1’2 (3-9) (11.3)(8.85x10-14) = 1.32V Finally, substituting into the N1 expression, we obtain (3.9)(8.85x1o-14) N1 = 10n+ (1.6x10-19x10-6) (0.5 — 1.32 + 1.02) or N; = 5.31 X 101I boron ions/cm2 18-16 ...
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Chapter 18 - CHAPTER 18 I S. 2 (a) In theory the two...

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