chapter-3-solutions - Prob. 3.1 Chapter 3 Calculate Bohr...

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Unformatted text preview: Prob. 3.1 Chapter 3 Calculate Bohr radius for donor in Si (ma. = 0.26 mg). From Eq.(2-IO) with n == 1 and using err— 11.8 for Si: r ' 2 mnq r: 2.41 x10‘9m = 24.1.11 z 41:6,.60112 z l1.8(8.85x10‘12)(6.63x10‘34)2 7c(026)(9.l1x10'3‘)(1.6x10“9)2 Note that this is more than four lattice spacings a (5.435;) flab. 3,2 Plot F ermi fitnction for E;- = 1 eV. f(E)={1+e(E-E,)lkT]—I We will choose E in eV and therefore use [CT = 0.0259 Egefl gE-EpykT 0.75 -9.6525 0.99994 0.90 -3.8610 0.97939 0.95 -1 .9305 0.87330 0.98 —0.7722 0.68399 LOZ +0.7722 0.31600 1.05 +1.9305 0.12669 1.10 +3.8610 0.02061 1.25 +9.6525 0.00006 18 18 Prob. 3.3 Calculate the demity—af—statas' efi'ective mass associated with the X minimum for the given band structure. Given that near the energy minimum along [I O O], the band structure is : E = EO - _ + which can be Taylor expanded near the minima : Mela—2r:Tl—lz-zlfi‘f-l‘zfigfl awe ~A—2B)+:§a2k3 +§Bz(k: +k3) 2 The effective mass is defined as : m' = .32}. , dkz ' Along (l 0 0) direction (longitudinal direction), the efl‘ectlve mass becomes 2 . h2 7:2 m = ——-—-— z m ’ dzE Au 2 die? Along the two tansverse directions, the effective mass becomes : Finally, the density-of-states effective mass is given by: hz “320‘ 2 B 4)1/ 3 t i '2 0.0.x»: =(m,m, )1”: l9 x mmwmwmw «WNW m 19 Mb. 3.4 F or the given band structure, find the temperature at which the number of electrons in the 1‘ minimal and the X minima equal. From Eq.(3 ~ 15), we have 2 0. 35 3:33;;71? fir Cr Given that there are 6 X minima along the<100 > directions, Ngoge - (o. { ELL—‘5 "r 4 0.30 \ 0.065 fi-om Eq.(3 ~16b) we get : 30)3/ 2 Nap x @095)” L\ l 0.35 ekT Whennr=nx,weobtaih 99.5. e” =6x [ 0.30 0.065 3 '2' That is: kT == 0.08576V or, T = 988K mi). 3.5 For the given b and structure, calculate and sketch how the conductivity varies fi‘om low T to high Tandfind the ratio of the conductivities at 1000°C and 300°C. n = nr +nL z nr[l+(15 0.30 )3/2 6 k1 = constant, independent of temperature according to the problem From Eq.(3 - 15), we have : -932 2.4:.“ng k? "r Nor EVE: nercre kT EVE“ i . 3/2 45, -5 15 "“ ‘4‘ nlchLe kr .ekT =[_m‘fl "FekT =(15)3/2,,ren r 20 20 6 = qD’rflr + "£1111: Q‘D‘I‘Hr + "L /2 LEE = anflr 1+ ’1 =q——-—-——-—-—-—-—-—oh30 pf 1+ 2+(15)3’2e77 :2 .939 1+g§2-e *7 ‘ 9"“1" 50 0.30 1+(15)3’2e k7 [ 0.30 1+———5:(')1 e 1‘7 WWI“? 0.30 [1+58‘1xe *7 T<<%—, 6 =60 =qnul- E 1+1.16 J = = T»k’° °°" q""r(1+ss.1 6(1000°C) 6 (300°C) : 0.254 9.523;“? 50 111' 50 0.30 ) 21 21 If the F to L separation is assumed to be 0.35 eV instead of 0.30 6V, we get: 581 l+—-——e “' 6 = qwr 1+58.1><e'F 6(1000°C) _ 6(300°C) 0.322 i: ‘2: 22 22 my. 3.6 Find ngbr Si fi'om F ig. 3-1 7. g 111n§1=h1 r , / E 1 Inan =1“ N'ng ‘— ] 2 'FmSi, an .1108 mi: 4x10'3 n,’1=3x10” at —1—=Zx10'3 T1 1 1 Eg = ZkN’Iil/"ifl/(E‘g‘fi‘) 2(8.62x10‘5) -1n(3x10‘4/108) (4— 2)- 10—3 =1.3eV This result is only approximate, since we neglect the temperature dependencias och, Ny, , and Es. 23 23 W Show that Eq.(3—25) results from Eqs.(3-15) and (3-19) and find lhe position of zhe Fermi level relative to E; at 300K EC Emem—m _..._. 0.347eV «— —————— ——-——-—Ei EV no = Nce'(Ee“ErV"T = Ncg'erO/Wewr-E,)IkT = "f e(£,-E,)lk1‘ p0 : "‘2 me = nie(E,—E,)/k7‘ no =1016 =LSX10‘0 Xe(E;—E¢)/0‘0259 E,— ~ Ei = 0.0259 x1n(6.667 x105) = 0.347ev Prob. 3.8 Find the displacement of E; fiom the middle of Eg for Si. E; is not exactly in the middle of the gap because the density of states NC and N y differ. Equating qu.(3-21) and (3—23). Nee-(a—EJ/kr = Mia/21:1 Eg/z ~ (EC — 12,) = len(Nv/Nc)1"2 = kT1n(m;/m;)3/‘ For Si at 300K, Eg/Z — (EC ~ E1) = 0.0259 xi- 1n (0.56/1.1)= —0.013 eV Thus, E,- is about kT/2 below the center of the gap. 24 24 b. 3.9 (a) Explain why holes appear at the top of the valence band. Electron energy is plotted “up” in band diagrams such as Fig. 3-5. Thus conduction band electrons relax to the bottom of the conduction band. Holes, having positive charge, have energies which increase oppositely to that of negatively charged electrons. Thai is, hole energy would be plotted “ wn” on an electron energy diagram such as Fig.3—S. Holes therefore relax to the lowest hole energy available to them; Le. the “mp” of the valence band. (b) Explain why Si doped with 10“ cm” donors is n—type at 400K, but Ge is not. According Fig. 3-17, the intrinsic concemmtion n; at 400K is m (400K) ~ 10” cm" for Ge ~ 10” cm'3 for Si Thus at this temperature, Nd » n; for Si, N, « n,- for Ge. Prob. 3.12. For 5: with M- Na 2 4 x 10” cm", find Epanng. no =Nd — Na =4x10‘5 = niemF‘E‘mT -.— 0.0259ln(4><10‘5/1.5X101°)= 0.324eV R” = —(qno)“' = —(1.6x10“‘9 x4x10‘5r‘ = 4562.5 cm’lc 25 25 Prob. 3.11. (a) F ind the value ofnojbr minimum conductivity 6 = «nun +pup) = q(nun + upnf/n) do '8'”— : (IO-Ln "' Rang/"2) Setting this equal to zero and defining n”. as the electron concentration for minimum conductivity, we have "3. = nfup/um nm = rim/Rpm” (b) What is am? 6min = Mum/Hymn + um/un/up ) = anrxiunup (c) Calculate cm. and 0'.- for Si For Si, om = 20.6x10"")(15><10“’)(1350><4:30)”2 = 3.9x10‘5(Q- cm)" at. = mm" + up): (1.6x10'19)(15x101°)(1830) = 4.4x10“(Q-cm “ or, take the reciprocal of p; in Appendix III. 26 26 2 1‘ ar 0.1 cm long and 100 [MW2 in cross sectional area is doped with 10" cm'3 antimony. Find the current at 300K with 10V applied. Fm Fig.3-23, 1.1,. a 700 cmz/V-s §=qfinno =1.6x10“9 x700x10” =11.2(Q~cm)'l = p" ' =0.08939.~cm R pL/A = 0.0893 ><0.1/10‘6 = 8.93 x103 :2 >1: V/R =10/(s.93 x103) = 1.12 mA apcat for a length of 111m. w 8 = lOV/lO“ cm = 105 V/cm, which is in the velocity saturation regime. From 319.24, V; = 107 cm/s , = (1.6 x10“9)(10")(10”)ao") = 0.16 A (I!) How long does it take an average electron to (in)? 1 gm in pure Si at an electric field of100 V/cm? Repeatjbr 105 V/cm. From Appendix m, p, = 1350 cmzN-s ,ow ficld: vd= 9,8 = 1350 x 100 = 1.35 x 105 cm/s t= L/vd =1o4/(1.3s x 105) = 7.4 x 104° 5 = 0.74 as high field: seaming-limited velocity v, ——~ 107 cm/s (Fig. 3-24) =1o4/107 =10'“s = 10 ps 27 27 Pgoh. 3:13 A perfect ['11- Vsemiconductar is doped with column VI and 11 impurities. For the given 31", p1,, calculate the energy levels inn-calmed in the bandgap. qt . =1.5x10"9x10"3 “n a t '7" ma m2 "’n 100000—47) cm m=l.6x10"3]kg=0.176mo m; = 1.408»;0 From Eq.(3 - 8) and using the value of the ground state energy of a H atom (l Rydberg)=l3.6ev: ‘ 13.61% ED = 62"“) =14.2meV below EC 7' E A = 113.6meV above EV [3-99. 3,14 Find V” with Hall probes misaligned. Displacement of the probes by an amount 5 gives a small IR drop V5 in addition to V”. The Hall voltage reverses when .6 is reversed; however, V3; is insensitive m the direction of the magnetic field. Thus, wifhfapositive: VA}, = VH + V; negative: VXB = —-VH + V5 subtracting, VXB - VXB = 2V“ We obtain the true Hall voltage t‘mtn VH = g-(VA'B — VKB). 28 28 diheposition ofthe Fermi Ievelfor 11 electrons in an infinite 1-D potential well 1001i he probability of exciting a carrier to the first excited state. 6219112 E6 =-E-m-Z2—, where m= fieeelcctron mass, L =100A 52(0.00120) eV = 0.0432 eV mLz =0‘0588cV l 1 31-15, I + e(0.0588-0.0432)/0.0259 l+e "T At 300K, this is 0.354. 29 29 ob.6 Electron mobility (cmzN-s) 0 100 200 300 400 500 Tcmpaam(K) £mb.3.l7 i '25. E 5 Lu O 1 0 0 2 O O 3 O 0 4 O 0 5 0 0 Temperature (K) When freeze-out occurs, ionized impurity scattering disappears, and only the phonon scattering remains. In real Si, other mechanisms, including neutral impurity scattenng, contribute to mobility. 30 30 Prgp. 3.; 8 Find the hole concentration and mobility with Hall measurement on a 1743722 semiconductor bar. The voltage measured is the Hall voltage‘plus the ohmic drop. The Sign of V” changes with the magnetic field, but the ohmic voltage does not. VH, ‘ V” True, VHaI; = z 3 3 mV Thus the ohmic drop is 3.2 - 3.0 = 0.2 mV From Eq.(3 — 50) (3xlO'3AX10x10‘5Wb/cm2) = 3.125x10‘7cm‘3 1’” “ q(20><10”4cm/)(3 x 1041/) 0.2 mV p: 3"“ =0.033Q~cm= l fin Qflppo 50051:» x 20m 1 1 =—-—-—=-——-—-———-———-——-—=600cm2/V~s up qppo 1.6x10"9(o.033)(3.125x10”) ( ) rob. 3.19 Calculate the conductivity of a hypothetical semiconductor at 600K. The intrinsic conductivity is given as -5. - V'NCNve M x2000=4x10'5(52-cm)" 9 6i zqniq‘l'fl +up)=q 5 4x10'6 = 1.6x1049(10‘9)(2000)e"51F As T goes fiom 300K to 600K, Eg, NC. N, do not change, and I? E! ..__L. .. e 3”» increases to e “(273), wherer, = 300K. Therefore, E _ . —6 e 2k(2x):\!___~.f‘_§i(l___._=3.s4x10‘5 1.6x10"9><10‘9><2000 49—— = 0.11352 - cm)‘l 6 = __5 3.54X10 31 Prob. §,20 Calculate the number of electrons, holes, and n; in the unknown semiconductor with 151.- 0.25eV below Ea; Ec 0.2SeV - “26" ED Incomplete ionization : fwd): Jim = 0.1267 l+eW9 n =(1— f)Nd = 8.733x 1014cnf‘3 Ec~EP Also, I: = Nee- "T E —E M NC = "3 k7 = 8.733X1014 x 3025103259 =1.359x10‘9cm"3 = Nv _Er'Ev p = me k7 = 1.359 x10” Xe‘(“‘“°~25)’°'°259 = 7.591 x 104cm‘3 n, = np = 8.142X109cm“3 W Referring to Fig. 3.25, find the type, concentration and mobility of the majorin carrier. Given, 3, = 10‘4 Wb/cmz From the sign of VAR, we can see the majon'ty carriers are electrons. —3 -4 =_Lx§z._=__——_(_‘_1%l}3L)—__T:3stx1o”cm‘3 qt(—VAB) 1.6x10 (10 )(2x10‘) -3 p=i=kfla=~fllflfi=aoom~cm L/wt L/wt 0.5/0.01x10 1 fl 1 ”” " pqm0 (o.ooz)(1.6x10“9)(3.125x10”) "o = 10,000 cm2(V - 8Y4 32 32 ...
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This note was uploaded on 03/19/2008 for the course EE 339 taught by Professor Banjeree during the Spring '08 term at University of Texas at Austin.

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chapter-3-solutions - Prob. 3.1 Chapter 3 Calculate Bohr...

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