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ece3901_HW2_solutions

ece3901_HW2_solutions - Homework 0 Sob’hom SQT CHAPTER 3...

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Unformatted text preview: Homework 0? Sob’hom SQT CHAPTER 3 a (a) See Fig. 3-10(a). (b) . EC 13v C (d) EC Light W fi3v O (f) See Fig. 3.15(a). (g) See Fig. 3.150)). (11) See Fig. 3.15(C) (C) Ec iiv 0 130/4 0 c ( ) 13C Thermal 5 energy liv O 32 (a) Given: Ud : 103 cm/sec E : AWL : ZV/cm [JD 2 Ud/8 = 500 cmZ/V-sec (b) (i) Lattice scattering (ii) Ionized impurity scattering (See the Relationship to Scattering discussion in Subsection 3.1.3.) (C_)}1inuinsic is higher than #heavily doped / . . v o . - - - o ' Reason: In inmnStc material the scattering 15 due excluswely t0 lattice scattenng. In heavily doped materials, ionized impurity scattering is also important. The more scattering there is, the lower the mobility. (d) Given N91 and NM >> ni, we know from Eqs. (3.8) that 1 1 p — -—— ...n—type wafer l; p , — w W ...p-type wafer 2 qflnNot (#1pr A2 In most semiconductors including GaAs, ,ttn is greater than it for a given doping and system temperature. Since we are given N131 : NAg, taking tiie wafer temperatures to be the same. and with ,Lln > up, we conclude from the above equations that p(wafer 2} > p(wafer 1). Note that the conclusion here is consistent with Fig. 3.8(b). (e}-DN=(kaq)ttn : (0.0259)(l300) = 33.7 cmz/sec a 1 (a) p = ...Eq.(3.8a) QHnND l = w = 0.501 ohm-cm it from Fig. 3.5(11) (1.6 x10“19)(1248)(1016) i” p 2: 0.5 ohnhcm ...by inspection from Fig. 3.8(a) (b) Since NA = ND , n = p : ni = 1010/cm2. Moreover, the total number of scattering centers is ND + NA = 2 x 1015/cm3. Thus, from Fig. 3.5(a), ,un = 1165 cmZN-sec, ,up = 419 ch/V-sec, and i 1 p _ (10.1“ +lup)ng _ (1.6x 10-19)(1584){1010) = 3.95 X 105 ohm-cm (c) Here n 2 p : n1: 10:0/cm3. With NA = 0 and ND 2 0, one has the maximum possible carrier mobilities. From Fig. 3.5(a), unmax: 1358 cmE/V‘sec and tipmax : 461 cmZ/Vvsec. p : —-—m~1— = ——-————m1-—_—— : 3.44 x 105 ohm-cm (101“ + ppm (1.6 x10'19)(1819)(1010) Because of the lower mobilities in compensated material, p(pan b) > p(p3rt c). (d) R = pm p : RA/l = (500)(10*2)/(1) : Sohm—cm Since the bar is n-type, we conclude frorn Fig. 3.8(3) that ND 5 9 x 1014/c1n3. (e) For a sample where ND >> m, p = UCIiUnND Furthermore, since the sample is lightly doped, lattice scattering will dominate and tin will decrease with increasing T. Fig. 3.73 confirms the preceding observation. Thus, with p cc l/un, heating up the sample causes the resistivity to increase. 3V7 212 _ . The briefexplanat'ion how one arrives at a given answer, an explanation applicable to all the energy band diagrams, is given immediately below. Sketches indicating the gencrm’jbrm of the expected answers follow the explanations. (a) for all cases. The semiconductor is concluded to be in equilibrium because the Fermi level has the same energy value (it is constant) as a function of position. (b) Vvs. x has the same functional form as the "upside down" of E; (or E] or EV). The sketches that follow were consmicted taking the arbitrary reference voltage to be V = 0 at I: O. (c) 8 vs. I is determined by simplyr noting the slope ofthe energy bands as a function of posuton. (d) For electrons, PE : EC—EF and KB 2 EmEC; for holes PE : EPEV and KE : EWE. (e) The general carrier concentration variation with position can be deduced by noung EF—Ei vs. x. Under equilibrium conditions, rt : niexp[(EF—E.)//<T] and p : riiexp[(Ei—EF)/kT] if the semiconductor is nondegeneraie. (1) Since JNtdn’ft = gunng , the general variation of JNIdn'rt with position can be deduced by conceptually fonning the product of the 8 vs. x dependence sketched in part (c) and the n vs. I dependence sketched in pan (e). Under equilibrium conditions, JN = JNldrift + JNidiff = 0- Thus JNtdtrr = —JNtdrtn- Diagram (3} Drift Diff Diagram (b) 3-16 Diagram {‘c) Diagram ((1) Diagram (e) Diagram (0 346 (a) One assumes the minority carrier drift current is negligible compared to the diffusion current in deriving the equation; i.e., diffusion is taken to be the dominant mode of minority carrier transport — hence the name DIFFUSION equation. (b) The equation is only valid for minority can-iers. (c) The recombinationigeneration term appearing in the equation, namely —Anp/rn, is valid only under low-level injection conditions. M ’0 ’0 ._ __.L-«-‘- W120 lcm +31 (loge; CLO f“ EVFNA / -m I V? {H 300K .ow‘ {hfiwfi 305a) Cm Book We limo! NY '3— 116? U’fieo C) ’l’mmtoblé CW1 {mejv OOBLICJIV] From fo‘ft a) “Om/Mme, #LlfckVIESS {157/ +mw=003cwl 0““ R = 100m L. ”am R: 195- :> me £— Wt w'flmmx Rm , ft W'fnmn 0,03 R , fl; «— fl" EM. : ‘Qnom Mj/ngflofi ‘Mth ’ W tuna»; _. thnovn '2:ka flu/tax . 9‘ W R :- flmm L anal {QMRX'J M N mun H A W bemotmé QEFOVH (2616‘, a) (Nfl)nom: JOAN; Qnom: {OD/(IL 1 W1 143 U3 RMW‘: 4%}:- /j;\l‘fl : Rh WHNMMX mom am I W W .. "9'" a V M n W RWfin "‘ Rhona LNPAM‘XX ') Q‘MCV]: :26 figk ) Sun/xi {our {7/ . __ , fnom U‘hflmm ‘ V “Rwy“ r helm {MU ' (Mammy —) Rmmfl 6?,3JLRIL QR : , Mflm .\ MM Rhom fmm (NF-‘Jmm J> QMax: MH' 23160, In +EV‘IM5 Qt +013rance 5 “H41 ’0 u) XIV/V1! ‘( .‘ LID/Tl) *_ (Qmin“RHOM)xw a ‘ ~32 13? o ‘0 d ,— m 0 9 w ° 0 W 41W / (04305.5 1&Rmax*Rhom)X\Dop/O :+ “H.532, (WC/00,9 7%“ R \ ' ‘ \Qesb’Mflc: hummfl We Muck/x WOFSE W1 IC— flaériwfll‘o” ’ «Hm m mm ,pgrm Mam {cm m EOE ‘fioOIKCJL EE3901 Semiconductor Devices A2000 (0 (ii) (iii) (M (V) DOPING CARRIER I... ENERGY BAND DIAGRAM D Momfien \EHH [11:36 IE'HL‘ EC m3 hams El 25+}? 51 %E+17 ; Nondfien m3 /¢m3 cm TYPE INTRINS IC DEGEN NONDEGEN MORE ON THE OTHER SIDE!!! EE3901 Semiconductor Devices A2000 (0 (ii) (iii) (M (V) DO P ING CARR]: ER CONC l 0‘ E a ENERGY BAND DIAGRAM E E 2 (L131! NA ND H >4 [12 II: m C] %&2 fig 5:: El H D Z ’ NEW; 1391 km; flmg Ivfirimic Danae en 1E“: IEHO , /UM /(vvn3 Ei _______ Fm__EF 2E+319 5c. _ I . ¢ #_ __ _... __ v _. cm N/Q N/A E; , % {0.01'?&V:SKT 5E+17 SE”? ’15” 6; We [01113 .- 0 H5 ?eV '5 3 ' lI/CV‘n flm Eimfli ‘_Imm 1E+14 5,6 ———m—#¥~—— ...
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