ece3901_HW5_solutions

ece3901_HW5_solutions - Homework ’5 VSOlU-thA 68+ 16.1 ....

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Unformatted text preview: Homework ’5 VSOlU-thA 68+ 16.1 . ' Biasing Energy Band Block Charge 3:151 Doping Condition Diagram Diagram (3) P depletion (d) ’1 accumulation VErCCLI'OHS it = —!n (ND/hi) = _|n(lil'_5_) = -1151 kT/q 1010 £3}? = —l 1.51 (kT/q) = -(l l.51)(0.0259) 3 -0.298V (b) Using Eq.(16.i6) with NA -—) —ND, (2)01,8)(8.85X10‘14)(2){0,298) “2 (1.6x10'19)(10‘5) = 0.882pm W = WT = [2‘580 (2%) l/2 —qN D (c) Evaluating Eq.(16.12) atx = 0 yields 83. Thus, with NA —> —ND in Eq. (16.12), _ _ (1.6><10-‘9)(1()‘5)(0.882><10-4) K380 (l 1.8)(8.85xt0-14) = —1.35 x 104 V/cm (d) Substituting into Eq.(16.26) gives Ks V0 2 2(1)}: + K—xogs ...83 evaluated at gas ': 2¢F O -5 4 z — (2)(0.293) _ W 3.9 = -l.00 V Except for the dOping type, the parameters used in this problem are identical to those assumed in constructing Fig. 16.10. Since (p3 = 29151:, the IVGI calculated in part (d) should correspond to the depletion/inversion transition point in Fig. 16.10. Indeed, in the figure VT E 1V. M Inversion Depletion Flat band VG = VT Accumulation . . . “macrpfl U‘leW-B 16.9 (a) . The Fermi level inside the semiconductor is position independent. (b)... n: = (llq)lEi(bulk) — BF} = 0.3 V (C)... qbs = (l/q)[Ei(bulk) — Ei(surface)] = 451: z 0.3 V (d)... Ep(metal) — Ep(semi) : —qVG .,.Eq. (2.1) VG I (l/q)[EF(semi) — Ep(meta1)] = 0.6 V (e) Based on the delta-depletion approximation, K31 ZqNA K0 K580 where from prior parts of the problem VG = 0.6V and $5 = 0.3V. Also, (bl: : (kT/Cl) IDWA/ni) 01‘ NA : niefibF/(HYQ) : (1010)60.3/0.0259 : 1'073X 1015/0113 Thus x0 = fluZW—h—Oflm = 0,10ym 53 MM (p (1L8) (2)(l_6><10'19)(1.(}73X10]5)(0-3)]m K0 K880 3.9 (11.8)(8.85><10"4) (f) w WT (ta/2%) 16.13 (a) p-type ...For p-type devices accumulation (Cmax) occurs for negative VG and inversnon Cmin) occurs at positive V0. The exact opposite is true for n-type devices. (b) At point (2) the p-type MOS—C is far into inversion. Thus (c) At point (I) the MOS-C is clearly deep into accumulation. 16.15 (a) p-typc ...There is an inversion—layer of negative charge 7 electrons 7 shown in the bloc c arge diagram. The semiconductor must therefore be p—type. (Aiso, the depletion- region charge is negative or clearly due to acceptor ions.) (b) Inversion biased ...As noted in pan (a), there is an inversion layer with n3 > NA shown on t e iagram. (c) (d) ~AQ due to ac signal at S highifrequency ...
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ece3901_HW5_solutions - Homework ’5 VSOlU-thA 68+ 16.1 ....

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