施敏-半导体器件物理英文版-第一章习题

Solutions Manual to Accompany SEMICONDUCTOR DEVICESPhysics and Technology2nd EditionS. M. SZEUMC Chair ProfessorNational Chiao Tung UniversityNational Nano Device LaboratoriesHsinchu, TaiwanJohn Wiley and Sons, IncNew York. Chicester / Weinheim / Brisband / Singapore / TorontoContentsCh.1 Introduction--------------------------------------------------------------------- 0 Ch.2 Energy Bands and Carrier Concentration-------------------------------------- 1 Ch.3 Carrier Transport Phenomena-------------------------------------------------- 7 Ch.4p-n Junction--------------------------------------------------------------------16 Ch.5 Bipolar Transistor and Related Devices----------------------------------------32 Ch.6 MOSFET and Related Devices-------------------------------------------------48 Ch.7 MESFET and Related Devices-------------------------------------------------60 Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices---------------68 Ch.9Photonic Devices-------------------------------------------------------------73 Ch.10 Crystal Growth and Epitaxy---------------------------------------------------83 Ch.11 Film Formation----------------------------------------------------------------92 Ch.12 Lithography and Etching------------------------------------------------------99 Ch.13 Impurity Doping---------------------------------------------------------------105 Ch.14 Integrated Devices-------------------------------------------------------------113CHAPTER 21. (a) From Fig. 11a, the atom at the center of the cube is surround by fourequidistant nearest neighbors that lie at the corners of a tetrahedron. Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is1/2 [(a /2)2 + (a 2/2)2]1/2 = a 3/4 = 2.35 Å.(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)for an area of a 2, therefore we have2/ a 2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm 2Similarly we have for (110) plane (Fig. 4a and Fig. 6)(2 + 2 ×1/2 + 4 ×1/4) /a 22 = 9.6 × 1015 atoms / cm 2,and for (111) plane (Fig. 4a and Fig. 6)(3 × 1/2 + 3 × 1/6) / 1/2(a 2)(a23) =2232a= 7.83 × 1014 atoms / cm 2.2. The heights at X, Y, and Z point are ,43,41and 43.3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere.4 Maximum fraction of cell filled= no. of sphere × volume of each sphere / unit cell volume = 1 × 4ð(a /2)3 / a 3 = 52 %(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere. The fcc also contains half a sphere at each of the six faces for a total of three spheres. The nearest neighbor distance is 1/2(a 2). Therefore the radius of each sphere is 1/4 (a 2).4 Maximum fraction of cell filled= (1 + 3) {4ð[(a /2) / 4 ]3 / 3} / a 3 = 74 %.(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a total of three spheres, and 4 spheres inside the cell. The diagonal distancebetween (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a isD =21222222+ + a a a = 34a The radius of the sphere is D/2 =38a4 Maximum fraction of cell filled= (1 + 3 + 4)33834a π/ a 3 = ð3/ 16 = 34 %.This is a relatively low percentage compared to other lattice structures.4. 1d = 2d = 3d = 4d = d 1d +2d +3d +4d = 01d • (1d +2d +3d +4d ) = 1d • 0 = 021d +1d •2d +1d •3d + 1d •4d = 04d 2+ d 2 cos è12 + d 2cos è13 + d 2cos è14 = d 2 +3 d 2 cos è= 04 cos è =31−è= cos -1 (31−) = 109.470 .5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest three integers having the same ratio are 6, 4, and 3. The plane is referred to as (643) plane.6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and Asare 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and four arsenic atoms per unit cell, therefore4/a 3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm 2,Density = (no. of atoms/cm 3 × atomic weight) / Avogadro constant = 2.22 × 1022(69.72 + 74.92) / 6.02 × 1023 = 5.33 g / cm 3.(b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are formed, because Sn has four valence electrons while Ga has only three. The resulting semiconductor is n -type.7. (a) The melting temperature for Si is 1412 ºC, and for SiO 2 is 1600 ºC. Therefore,SiO 2 has higher melting temperature. It is more difficult to break the Si-O bond than the Si-Si bond.(b) The seed crystal is used to initiated the growth of the ingot with the correctcrystal orientation.(c) The crystal orientation determines the semiconductor’s chemical and electricalproperties, such as the etch rate, trap density, breakage plane etc.(d) The temperating of the crusible and the pull rate.8. E g (T ) = 1.17 – 636)(4.73x1024+−T T for Si∴ E g ( 100 K) = 1.163 eV , and E g (600 K) = 1.032 eVE g (T ) = 1.519 –204)(5.405x1024+−T T for GaAs∴E g ( 100 K) = 1.501 eV, and E g (600 K) = 1.277 eV .9. The density of holes in the valence band is given by integrating the product N (E )[1-F (E )]d E from top of the valence band (V E taken to be E = 0) to the bottom of the valence band E bottom :p = ∫bottomE 0N (E )[1 – F (E )]d E (1)where 1 –F(E) = ()[]{}/kT1 e/1 1F E E −+− = []1/)(e1−−+kTE EF If E F – E >> kT then1 – F (E ) ~ exp ()[]kT E E F −− (2)Then from Appendix H and , Eqs. 1 and 2 we obtainp = 4ð[2m p / h 2]3/2∫bottomE 0E 1/2 exp [-(EF – E ) / kT ]d E (3)Let x a E / kT , and let E bottom = ∞−, Eq. 3 becomesp = 4ð(2m p / h 2)3/2(k T)3/2exp [-(E F / kT )]∫∞− 0x 1/2e x d xwhere the integral on the right is of the standard form and equals π / 2.4 p = 2[2ðm p kT / h 2]3/2 exp [-(E F / kT )]By referring to the top of the valence band as E V instead of E = 0 we have,p = 2(2ðm p kT / h 2)3/2 exp [-(E F – E V ) / kT ]orp = N V exp [-(E F –E V ) / kT ]where N V = 2 (2ðm p kT / h 2)3 .10. From Eq. 18N V = 2(2ðm p kT / h 2)3/2The effective mass of holes in Si is m p = (N V / 2) 2/3 ( h 2 / 2ðkT ) = 3236192m 101066.2××−()()()30010381210625623234−−××..π = 9.4 × 10-31 kg = 1.03 m 0.Similarly, we have for GaAsm p = 3.9 × 10-31 kg = 0.43 m 0.11. Using Eq. 19(()C VV C N NkTE E E i ln 22)(++== (E C + E V )/ 2 + (3kT / 4) ln32)6)((n p m m (1)At 77 KE i = (1.16/2) + (3 × 1.38 × 10-23T ) / (4 × 1.6 × 10-19) ln(1.0/0.62) = 0.58 + 3.29 × 10-5 T = 0.58 + 2.54 × 10-3 = 0.583 eV.At 300 KE i = (1.12/2) + (3.29 × 10-5)(300) = 0.56 + 0.009 = 0.569 eV.At 373 KE i = (1.09/2) + (3.29 × 10-5)(373) = 0.545 + 0.012 = 0.557 eV.Because the second term on the right-hand side of the Eq.1 is much smallercompared to the first term, over the above temperature range, it is reasonable to assume that E i is in the center of the forbidden gap.12. KE =()())(/)( / d d e C F top C F topCE E x kTE E C E E kT E E C E E C EeE E EE E E E −≡−−−−−−−∫∫= kT ∫∫∞−∞− 021 023d e d e x x x x x x= kTΓ Γ2325 = kTππ505051...×× = kT 23.13. (a) p = mv = 9.109 × 10-31 ×105 = 9.109 × 10-26 kg–m/sλ = p h = 2634101099106266−−××..= 7.27 × 10-9m = 72.7 Å(b) n λ=λp m m 0= 06301.× 72.7 = 1154 Å .14. From Fig. 22 when n i = 1015 cm -3, the corresponding temperature is 1000 / T = 1.8.So that T = 1000/1.8 = 555 K or 282 .15. From E c – E F = kT ln [N C / (N D – N A )]which can be rewritten as N D – N A = N C exp [–(E C – E F ) / kT ]Then N D – N A = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm -3or N D = 1.26 × 1016 + N A = 2.26 × 1016 cm -3A compensated semiconductor can be fabricated to provide a specific Fermi energy level.16. From Fig. 28a we can draw the following energy-band diagrams:17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boronimpurities are ionized. Thus p p = N A = 1015 cm -3n p = n i 2 / n A = (9.65 × 109)2 / 1015 = 9.3 × 104 cm -3.The Fermi level measured from the top of the valence band is given by:E F – E V = kT ln(N V /N D ) = 0.0259 ln (2.66 × 1019 / 1015) = 0.26 eV(b) The boron atoms compensate the arsenic atoms; we havep p = N A – N D = 3 × 1016 – 2.9 × 1016 = 1015 cm -3Since p p is the same as given in (a), the values for n p and E F are the same as in (a). However, the mobilities and resistivities for these two samples are different.18. Since N D >> n i , we can approximate n 0 = N D andp 0 = n i 2 / n 0 = 9.3 ×1019 / 1017 = 9.3 × 102 cm -3From n 0 = n i exp−kT E E i F ,we haveE F – E i = kT ln (n 0 / n i ) = 0.0259 ln (1017 / 9.65 × 109) = 0.42 eV The resulting flat band diagram is :19.Assuming complete ionization, the Fermi level measured from the intrinsic Fermi level is 0.35 eV for 1015 cm -3, 0.45 eV for 1017 cm -3, and 0.54 eV for 1019cm -3.The number of electrons that are ionized is given byn ≅ N D [1 – F (E D )] = N D / [1 + e ()T k E E F D /−− ]Using the Fermi levels given above, we obtain the number of ionized donors as n = 1015 cm -3 for N D = 1015 cm -3n = 0.93 × 1017 cm -3 for N D = 1017 cm -3n = 0.27 × 1019 cm -3 for N D = 1019 cm -3Therefore, the assumption of complete ionization is valid only for the case of 1015 cm -3.20. N D +=kT E E F D e /)(16110−−+ =135016e 110.−+ = 1451111016.+= 5.33 × 1015 cm -3The neutral donor = 1016 – 5.33 ×1015 cm -3 = 4.67 × 1015 cm -34 Theratio of +DD N N O = 335764..= 0.876 .CHAPTER 31. (a) For intrinsic Si, µn = 1450, µp = 505, and n = p = n i = 9.65×109We have 51031.3)(11×=+=+=p n i p n qn qp qn µµµµρ Ω-cm(b) Similarly for GaAs, µn = 9200, µp = 320, and n = p = n i = 2.25×106We have 81092.2)(11×=+=+=p n i p n qn qp qn µµµµρ Ω-cm.2. For lattice scattering, µn ∝ T -3/2T = 200 K, µn = 1300×2/32/3300200−− = 2388 cm 2/V-s T = 400 K, µn = 1300×2/32/3300400−− = 844 cm 2/V-s.3. Since21111µµµ+=∴500125011+=µ µ = 167 cm 2/V-s.4. (a) p = 5×1015 cm -3, n = n i 2/p = (9.65×109)2/5×1015 = 1.86×104 cm -3µp = 410 cm 2/V-s, µn = 1300 cm 2/V-s ρ =pq p q n q p p n µµµ11≈+ = 3 Ω-cm(b) p = N A – N D = 2×1016 – 1.5×1016 = 5×1015 cm -3, n = 1.86×104 cm -3µp = µp (N A + N D ) = µp (3.5×1016) = 290 cm 2/V-s,µn = µn (N A + N D ) = 1000 cm 2/V-s ρ =pq p q n q p p n µµµ11≈+ = 4.3 Ω-cm(c) p = N A (Boron) – N D + N A (Gallium) = 5×1015 cm -3, n = 1.86×104 cm -3µp = µp (N A + N D + N A ) = µp (2.05×1017) = 150 cm 2/V-s,µn = µn (N A + N D + N A ) = 520 cm 2/V-s ρ = 8.3 Ω-cm.5. Assume N D − N A >> n i , the conductivity is given byσ ≈ qn µn = q µn (N D − N A )We have that16 = (1.6×10-19)µn (N D − 1017)Since mobility is a function of the ionized impurity concentration, we can use Fig. 3 along with trial and error to determine µn and N D . For example, if we choose N D = 2×1017, then N I = N D + + N A - = 3×1017, so that µn ≈ 510 cm 2/V-s which gives σ = 8.16.Further trial and error yieldsN D ≈ 3.5×1017 cm -3andµn ≈ 400 cm 2/V-swhich givesσ ≈ 16 (Ω-cm)-1.6. )/( )( 2n n bn q p n q i p p n +=+=µµµσFrom the condition d σ/dn = 0, we obtainbn n i / =Thereforeb b b n q n b b bn qìi p i i p i m 21 )1(1)/( +=++=µρρ.7. At the limit when d >> s, CF =2ln π= 4.53. Then from Eq. 16226.053.410501011010433=×××××=××=−−−CF W I V ρ Ω-cm From Fig. 6, CF = 4.2 (d/s = 10); using the a/d = 1 curve we obtain 78.102.4105010226.0)/(43=×××=⋅⋅=−−CF W I V ρ mV.8. Hall coefficient,7.42605.0)101030(105.2106.1101049333=××××××××==−−−−W IB A V R z H H cm 3/C Since the sign of R H is positive, the carriers are holes. From Eq. 2216191046.17.426106.111×=××==−H qR p cm -3Assuming N A ≈ p , from Fig. 7 we obtain ρ = 1.1 Ω-cm The mobility µp is given by Eq. 15b 3801.11046.1106.1111619=××××==−ρµqp p cm 2/V-s.9. Since R ∝ ρ and p n qp qn µµρ+=1, hence pn p n R µµ+∝1From Einstein relation µ∝D 50//==p n p n D D µµpA n D nD N N N R .R µµµ+=15011We have N A = 50 N D .10. The electric potential φ is related to electron potential energy by the charge (− q )φ = +q1(E F − E i )The electric field for the one-dimensional situation is defined asE (x ) = −dx d φ=dx dE q i1n = n i exp−kT E E i F = N D (x )HenceE F − E i = kT lni D n )x (N dx )x (dN )x (N q kT (x) D D1 −=E.11. (a) From Eq. 31, J n = 0 anda qkT e N e a N q kT n dx dnDx ax axnn+=−==−− )(- - )(00µE(b) E (x ) = 0.0259 (104) = 259 V/cm.12. At thermal and electric equilibria, 0)()(=+=dxx dn qD x n q J nn n E µ xN N LN N N D LN N L xN N N D dx x dn x n D x L L n n L L n nn n )( ))((1)()(1)(000000−+−−=−−+−=−=µµµE000 0n ln )(D -N ND x N N LN N N V L n n L L Lµµ−=−+−=∫.13. 1116610101010=××==∆=∆−L p G p n τ cm -3151115101010≈+=∆+=∆+=n N n n n D no cm -3.cm 101010)1065.9(3-111115292≈+×=∆+=p N n p D i 14. (a)815715101021010511−−=××××=≈t th p p N νστ s 48103109−−×=×==p p p D L τ cm20101021010167=×××==−sts s th lr N S σν cm/s (b) The hole concentration at the surface is given by Eq. 67.cm 10 2010103201011010102)10(9.65 1)0(3-98481781629≈ ×+××−×+××=+−+=−−−−lr p p lr p L p no n S L S G p p τττ15. pn qp qn µµσ+=Before illuminationnon no n p p n n == ,After illumination,G n n n n p no no n τ+=∆+=Gp p p p p no no n τ+=∆+=.)( )()]()([G q p q n q p p q n n q p p n no p no n no p no n τµµµµµµσ+=+−∆++∆+=∆16. (a) diff ,dxdp qD J pp −== − 1.6×10-19×12×410121−××1015exp(-x /12)= 1.6exp(-x /12) A/cm 2(b) diff,drift ,p total n J J J −== 4.8 − 1.6exp(-x /12) A/cm 2(c) En n qn J µ=drift , Q ∴ 4.8 − 1.6exp(-x /12) = 1.6×10-19×1016×1000×E E = 3 − exp(-x /12) V/cm.17. For E = 0 we have022=∂∂+−−=∂∂xp D p p t p np p no n τat steady state, the boundary conditions are p n (x = 0) = p n (0) and p n (x = W ) =p no .Therefore[]−−+=p p no n non L W L x W p p p x p sinh sinh )0()([]−=∂∂−===p ppno n x npp L W L D p p q xp qD x J coth )0()0(0[]−=∂∂−===p p pno n Wx np p L W L D p p q xpqD W x J sinh 1)0()(.18. The portion of injection current that reaches the opposite surface by diffusion isgiven by)/cosh(1)0()(0p p p L W J W J ==α26105105050−−×=××=≡p p p D L τ cm98.0)105/10cosh(1220=×=∴−−αTherefore, 98% of the injected current can reach the opposite surface.19. In steady state, the recombination rate at the surface and in the bulk is equalsurface,surface ,bulk,bulk ,p n p n p p ττ∆=∆so that the excess minority carrier concentration at the surface∆p n , surface = 1014⋅671010−−=1013 cm -3The generation rate can be determined from the steady-state conditions in the bulkG = 6141010− = 1020 cm -3s -1From Eq. 62, we can write22=∆−+∂∆∂p p p G x p D τThe boundary conditions are ∆p (x = ∞) = 1014 cm -3 and ∆p (x = 0) = 1013 cm -3Hence ∆p (x ) = 1014(p L x e /9.01−−)where L p = 61010−⋅= 31.6 µm.20.The potential barrier heightχφφ−=m B = 4.2 − 4.0 = 0.2 volts.21. The number of electrons occupying the energy level between E and E +dE isdn = N (E )F (E )dEwhere N (E ) is the density-of-state function, and F (E ) is Fermi-Dirac distribution function. Since only electrons with an energy greater than m F q E φ+ and having a velocity component normal to the surface can escape the solid, the thermionic current density isdE e E v hm qv J kT E E x q E x F m F )(21 323)2(4−−∞+∫∫==φπwhere x v is the component of velocity normal to the surface of the metal. Since the energy-momentum relationship)(2122222z y x p p p mm P E ++==Differentiation leads to mPdP dE =By changing the momentum component to rectangular coordinates,zy x dp dp dp dP P =24πHencezmkTp y mkT p x x mkT mE p p zy x p mkTmE p p p x dp e dp e dp p e mhq dp dp dp e p mhq J zy f x x x f z y x 22 2)2( 3 p p 2/)2(322200y 2222 2−∞∞−−∞∞−−−∞∞∞−∞=∞−∞=−++−∫∫∫∫∫∫== where ).(220m F x q E m p φ+= Since 21 2=−∞∞−∫a dx eax π, the last two integrals yield (2ðmkT )21.The first integral is evaluated by setting u mkTmE p Fx =−222 .Therefore we have mkTdpp du xx = The lower limit of the first integral can be written as kT q mkT mE q E m m F m F φφ=−+22)(2 so that the first integral becomes kTq u ktq mm e mkT du e mkT φφ−−∞=∫ / Hence−==−kTq T A eT h qmk J mkTq mφπφexp 42*232.22. Equation 79 is the tunneling probability 110234193120m 1017.2)10054.1()106.1)(220)(1011.9(2)(2−−−−×=××−×=−=h E qV m n β []6121001019.3)220(241031017.2sinh(201−−−×=−××××××+=T .23. Equation 79 is the tunneling probability[]403.0)2.26(2.24)101099.9sinh(61)10(1210910=−×××××+=−−−T ()[]()912999108.72.262.24101099.9sinh 61)10(−−−−×=−×××××+=T .19234193120m 1099.9)10054.1()106.1)(2.26)(1011.9(2)(2−−−−×=××−×=−=hE qV m n β24. From Fig. 22As E = 103 V/sνd ≈ 1.3×106 cm/s (Si) and νd ≈ 8.7×106 cm/s (GaAs)t ≈ 77 ps (Si) and t ≈ 11.5 ps (GaAs)As E = 5×104 V/sνd ≈ 107 cm/s (Si) and νd ≈ 8.2×106 cm/s (GaAs)t ≈ 10 ps (Si) and t ≈ 12.2 ps (GaAs).cm/s105.9m/s 109.5 101.9300101.382 22 velocity Thermal 25.643123-0×=×=××××===−m kT m E v thth For electric field of 100 v/cm, drift velocitythn d v v <<×=×==cm/s 1035.110013505E µ For electric field of 104 V/cm.th n v ≈×=×=cm/s 1035.110135074E µ.The value is comparable to the thermal velocity, the linear relationship betweendrift velocity and the electric field is not valid.CHAPTER 41. The impurity profile is,space charge per unit area in the p -side must equal the total positive space charge per unit area in the n -side, thus we can obtain the depletion layer width in the n -side region:141410321088.0××=××n W Hence, the n -side depletion layer width is:m0671µ.W n =The total depletion layer width is 1.867 µm.We use the Poisson ’s equation for calculation of the electric field E (x).In the n -side region,()V/cm108640)100671(1031006710m 067134144×−===×−××=∴××−=⇒==+=⇒=−−.)x (.x qx .N qK ).x (K x N q )x (N q dx d n maxsn D sn D sn D s E E E E E E εεµεεIn the p -side region, the electrical field is:()()()V/cm10864010802108020m 8023242242×−===×−××=∴×××−=⇒=−=+×=⇒=−−.)x (.x a qx .a qK ).x (K ax q )x (N q dx d p maxsp s'p 'sp A s E E E EE E εεµεεThe built-in potential is:()()()V 52.0 0 0=−−=−=∫∫∫−−−−nn ppx side n x x x side p bi dx x dx x dx x V E E E.2. From ()∫−=dx x V bi E, the potential distribution can be obtainedWith zero potential in the neutral p -region as a reference, the potential in the p -side depletion region is()()()[]()(()()×−×−××−= ×−×−−=×−××−=−=−−−−−∫∫34243114243 0242108.032108.03110596.7108.032108.0312 108.02 x x x x qadx x a q dx x x V sxxs p εεEWith the condition V p (0)=V n (0), the potential in the n -region is()×−×−××−= ×+×−××−=−−−−734277342141098.010067.1211056.41098.010067.121103x x x x q x V s n εThe potential distribution isDistance p-region n-region-0.80.000-0.70.006-0.60.022-0.50.048-0.40.081-0.30.120-0.20.164-0.10.2110.2590.2594133330.10.3057885330.20.3476037330.30.3848589330.40.4175541330.50.4456893330.60.4692645330.70.4882797330.80.5027349330.90.51263013310.5179653331.0670.518988825Potentia l DistributionD i s t a nce (um)P o t e n t i a l (3. The intrinsic carriers density in Si at different temperatures can be obtained by using Fig.22 inChapter 2 :Temperature (K)Intrinsic carrier density (n i )250 1.50×1083009.65×109350 2.00×10114008.50×10124509.00×10135002.20×1014The V bi can be obtained by using Eq. 12, and the results are listed in the following table.T niVbi (V)250 1.500E+080.7773009.65E+90.717350 2.00E+110.6534008.50E+120.4884509.00E+130.3665002.20E+140.329Thus, the built-in potential is decreased as the temperature is increased.The depletion layer width and the maximum field at 300 K areV/cm. 10476.11085.89.1110715.910106.1m9715.010106.1717.01085.89.112241451519max151914×=××××××===××××××==−−−−−s D D bis W qN qN V W εµεE18162/11818141952/1max 10110755.1 10101085.89.1130106.121042 .4DDD D D A D A sRN N N N N N N N qV +=×⇒+×××××=×⇒+≈−−ε.E We can select n-type doping concentration of N D = 1.755×1016 cm -3 for the junction.5. From Eq. 12 and Eq. 35, we can obtain the 1/C 2 versus V relationship for doping concentration of1015, 1016, or 1017 cm -3, respectively.For N D =1015 cm -3,()()()V V N qåV V C B s bi j−×=×××××−×=−=−−837.010187.110108.8511.9101.60.837221161514192For N D =1016 cm -3,()()()V V N qåV V C B s bi j−×=×××××−×=−=−−896.010187.110108.8511.9101.60.896221151614192For N D =1017 cm -3,()()()V V N qåV V C s bi j −×=×××××−×=−=−−956.010187.110108.8511.9101.60.95622114171419B2When the reversed bias is applied, we summarize a table of 2j C 1/ vs V for various N D values asfollowing,V N D =1E15N D =1E16N D =1E17-4 5.741E+165.812E+155.883E+14-3.5 5.148E+165.218E+155.289E+14-3 4.555E+164.625E+154.696E+14-2.5 3.961E+164.031E+154.102E+14-2 3.368E +163.438E+153.509E+14-1.5 2.774E +162.844E+152.915E+14-1 2.181E +162.251E+152.322E+14-0.5 1.587E+161.657E+151.728E+149.935E+151.064E+151.134E+14Hence, we obtain a series of curves of 1/C 2 versus V as following,The slopes of the curves is positive proportional to the values of the doping concentration.1/C ^2 vs V-4.5-4-3.5-3-2.5-2-1.5-1-0.5Applied Volta ge1/C ^The interceptions give the built-in potential of the p-n junctions.6. The built-in potential is()V5686.01065.9106.180259.01085.89.111010ln 0259.0328ln 323919142020322=×××××××××××= =−−i s bi n q kT a q kT V εFrom Eq. 38, the junction capacitance can be obtained ()()()3/12142019-3/125686.0121085.89.1110101.612−×××××=−==−R R bi s s j V V V qa W C εεAt reverse bias of 4V, the junction capacitance is 6.866×10-9 F/cm 2.7. From Eq. 35, we can obtain()()22221jsR bi D B s bi jC q V V N N q V V C εε−=⇒−=We can select the n-type doping concentration of 3.43×1015cm -3.8. From Eq. 56,16915151571515108931065902590020exp 1002590020exp 1010101010exp exp ×=××−+ ×××=−+−=−=−−−−......n kT E E kT E E N U G it i p i t n tth n p σσυσσand()3152814192cm 10433)10850(108589111061422−−−−×=⇒×××××××=≅⇒>>.N ....C q V N V V d j s R D bi R εQ()m 266.1cm 1066.1210106.1)5.0717.0(1085.89.11225151914µε=×=××+××××=+=−−−A bi s qN V V W Thus2751619A/cm 10879.71066.121089.3106.1−−×=×××××==qGW J gen .9. From Eq. 49, and Di noN n p 2=We can obtain the hole concentration at the edge of the space charge region,()3170259.08.01629)0259.08.0(2cm 1042.2101065.9−×=×==ee N n p D i n .10. ()()()1/−=−+=kTqV s p n n p eJ x J x J J V. 017.0195.010259.00259.0=⇒−=⇒−=⇒V e e J J VVs11. The parameters aren i = 9.65×109 cm -3D n = 21 cm 2/secD p =10 cm 2/sec τp0=τn0=5×10-7 secFrom Eq. 52 and Eq. 54()()()3150259.07.0297192/cm 102.511065.910510106.1711−−−×=⇒−××××××=⇒−××=−=D DkT qV D i po p kT qV p no p n p N e N e N n D q e L p qD x J a τ()()()3160259.07.0297192/cm 10278.511065.910521106.12511−−−×=⇒−××××××=⇒−××=−=−A A kT qV A i no n kTqV npon p n N e N e N n D q eL n qD x J a τWe can select a p-n diode with the conditions of N A = 5.278×1016cm -3 and N D =5.4×1015cm -3.12. Assume ôg =ôp =ôn = 10-6 s, D n = 21 cm 2/sec, and D p = 10 cm 2/sec(a) The saturation current calculation.From Eq. 55a and p p p D L τ=, we can obtain+=+=002011n n Ap p D i np n pn p s D N D N qn L n qD L p qD J ττ()2126166182919A/cm 1087.6102110110101011065.9106.1−−−−×=+×××=And from the cross-sectional area A = 1.2×10-5 cm 2, we obtainA 10244.81087.6102.117125−−−×=×××=×=s s J A I .(b) The total current density is −=1kt qV s e J J ThusA.10244.8110244.8A 1051.41047.510244.8110244.8170259.07.0177.0511170259.07.0177.0−−−−−−−×=−×=×=×××=−×=e I e I VV。

______________________________________________________________________________________Chapter 1Problem Solutions1.1 (a)fcc: 8 corner atoms 18/1atom6 face atoms32/1atomsTotal of 4 atoms per unit cell (b)bcc: 8 corner atoms 18/1atom1 enclosed atom=1 atom Total of 2 atoms per unit cell(c)Diamond: 8 corner atoms 18/1atom6 faceatoms 32/1atoms4 enclosedatoms= 4 atomsTotal of 8 atoms per unit cell_______________________________________ 1.2 (a)Simple cubic lattice: r a 2Unit cell vol33382rra1 atom per cell, so atom vol 3413r ThenRatio%4.52%10083433rr(b)Face-centered cubic latticerd aa rd22224Unit cell vol 33321622rr a4 atoms per cell, so atom vol3443r ThenRatio%74%10021634433rr (c)Body-centered cubic latticeraa rd3434Unit cell vol 3334ra2 atoms per cell, so atom vol 3423r ThenRatio%68%1003434233r r (d)Diamond lattice Body diagonal raa rd3838Unit cell vol3338r a8 atoms per cell, so atom vol 3483r ThenRatio%34%1003834833rr _______________________________________1.3(a)oA a43.5; From Problem 1.2d,ra38Then oAa r176.18343.583Center of one silicon atom to center ofnearest neighboroAr 35.22______________________________________________________________________________________ (b)Number density22381051043.58cm 3(c)Mass density23221002.609.28105..AN W t At N 33.2grams/cm3_______________________________________1.4(a)4 Ga atoms per unit cell Number density381065.54Density of Ga atoms 221022.2cm34 As atoms per unit cell Density of As atoms 221022.2cm3(b)8 Ge atoms per unit cell Number density381065.58Density of Ge atoms221044.4cm3_______________________________________ 1.5From Figure 1.15 (a)aa d4330.0232oAd 447.265.54330.0(b)aa d7071.022oAd 995.365.57071.0_______________________________________1.674.5423232222sin a a 5.109_______________________________________ 1.7(a) Simple cubic: oAr a 9.32(b)fcc:oAr a515.524(c) bcc:oA r a 503.434(d) diamond:oAra007.9342_______________________________________ 1.8 (a)Br 2035.122035.12oBAr 4287.0(b)oAa 07.2035.12(c)A-atoms: # of atoms1818Density381007.21231013.1cm3B-atoms: # of atoms3216Density381007.23231038.3cm3_______________________________________ 1.9(a)oAr a 5.42# of atoms1818Number density38105.412210097.1cm3______________________________________________________________________________________Mass density AN W t At N ..23221002.65.12100974.1228.0gm/cm3(b)oAr a196.534# of atoms 21818Number density3810196.5222104257.1cm3Mass density23221002.65.12104257.1296.0gm/cm3_______________________________________ 1.10From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground,Volume = 0.74 cm3_______________________________________1.11(b)oAa 8.20.18.1(c)Na: Density38108.22/1221028.2cm3Cl: Density221028.2cm3(d)Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell23231085.41002.645.352199.2221Then mass density21.2108.21085.43823grams/cm3_______________________________________ 1.12(a)oAa 88.122.223Then oA a 62.4Density of A:22381001.11062.41cm3Density of B:22381001.11062.41cm3(b)Same as (a) (c)Same material_______________________________________ 1.13oAa619.438.122.22(a) For 1.12(a), A-atomsSurface density28210619.411a1410687.4cm2For 1.12(b), B-atoms: oAa 619.4Surface density14210687.41acm2For 1.12(a) and (b), Same material(b) For 1.12(a), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density______________________________________________________________________________________14210315.321a cm 2For 1.12(b), A-atoms;oAa 619.4Surface density212a1410315.3cm2B-atoms;Surface density14210315.321acm2For 1.12(a) and (b), Same material_______________________________________ 1.14 (a)Vol. Density31oaSurface Density212oa(b)Same as (a)_______________________________________ 1.15 (i)(110) plane(see Figure 1.10(b))(ii) (111) plane(see Figure 1.10(c))(iii) (220) plane,1,1,21,21Same as (110) plane and [110]direction(iv) (321) plane6,3,211,21,31Intercepts of plane at6,3,2sq p [321] direction is perpendicular to(321) plane_______________________________________1.16(a)31311,31,11(b)12141,21,41_______________________________________ 1.17Intercepts: 2, 4, 331,41,21(634) plane_______________________________________ 1.18(a)oAa d 28.5(b)oAa d734.322(c)oAa d048.333_______________________________________ 1.19(a) Simple cubic(i) (100) plane:Surface density2821073.411a141047.4cm 2(ii) (110) plane:Surface density212a141016.3cm 2(iii) (111) plane: Area of planebh21where oAa b 689.62Now2222243222a a a hSooAh793.573.426______________________________________________________________________________________Area of plane881079304.51068923.62116103755.19cm 2Surface density16103755.19613141058.2cm2(b) bcc(i) (100) plane:Surface density 1421047.41acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19613141058.2cm2(c) fcc(i) (100) plane:Surface density 1421094.82acm2(ii) (110) plane: Surface density222a141032.6cm 2(iii) (111) plane:Surface density16103755.19213613151003.1cm2_______________________________________ 1.20 (a)(100) plane: - similar to a fcc:Surface density281043.52141078.6cm 2(b)(110) plane:Surface density281043.524141059.9cm2(c)(111) plane: Surface density281043.5232141083.7cm2_______________________________________1.21oAr a703.6237.2424(a)#/cm338310703.64216818a2210328.1cm3(b)#/cm222124142a210703.62281410148.3cm2(c)oA a d74.422703.622(d)# of atoms2213613Area of plane: (see Problem 1.19)oAa b4786.92oAa h2099.826Area88102099.8104786.92121bh______________________________________________________________________________________15108909.3cm2#/cm215108909.32=141014.5cm2oAa d87.333703.633_______________________________________ 1.22Density of silicon atoms 22105cm3and4 valence electrons per atom, soDensity of valence electrons 23102cm3_______________________________________ 1.23Density of GaAs atoms22381044.41065.58cm3An average of 4 valence electrons peratom,SoDensity of valence electrons231077.1cm3_______________________________________ 1.24 (a)%10%10010510532217(b)%104%10010510262215_______________________________________ 1.25 (a)Fraction by weight7221610542.106.2810582.10102(b)Fraction by weight5221810208.206.2810598.3010_______________________________________ 1.26Volume density 1631021dcm3So610684.3dcmoAd 4.368We haveoo Aa 43.5Then85.6743.54.368oa d _______________________________________ 1.27Volume density 1531041dcm 3So61030.6dcmoAd630We have oo Aa 43.5Then11643.5630oa d _______________________________________。

半导体器件物理施敏课后答案【篇一：半导体物理物理教案(03级)】>学院、部：材料与能源学院系、所；微电子工程系授课教师：魏爱香，张海燕课程名称；半导体物理课程学时：64实验学时：8教材名称：半导体物理学2005年9-12 月授课类型：理论课授课时间：2节授课题目（教学章节或主题）：第一章半导体的电子状态1.1半导体中的晶格结构和结合性质1.2半导体中的电子状态和能带本授课单元教学目标或要求：了解半导体材料的三种典型的晶格结构和结合性质；理解半导体中的电子态, 定性分析说明能带形成的物理原因，掌握导体、半导体、绝缘体的能带结构的特点本授课单元教学内容（包括基本内容、重点、难点，以及引导学生解决重点难点的方法、例题等）：1．半导体的晶格结构：金刚石型结构；闪锌矿型结构；纤锌矿型结构2．原子的能级和晶体的能带3．半导体中电子的状态和能带（重点，难点）4．导体、半导体和绝缘体的能带（重点）研究晶体中电子状态的理论称为能带论，在前一学期的《固体物理》课程中已经比较完整地介绍了，本节把重要的内容和思想做简要的回顾。

本授课单元教学手段与方法：采用ppt课件和黑板板书相结合的方法讲授本授课单元思考题、讨论题、作业：作业题：44页1题本授课单元参考资料（含参考书、文献等，必要时可列出）1.刘恩科，朱秉升等《半导体物理学》，电子工业出版社2005?2.田敬民，张声良《半导体物理学学习辅导与典型题解》?电子工业出版社20053. 施敏著，赵鹤鸣等译，《半导体器件物理与工艺》，苏州大学出版社，20024. 方俊鑫，陆栋，《固体物理学》上海科学技术出版社5．曾谨言，《量子力学》科学出版社注：1.每单元页面大小可自行添减；2.一个授课单元为一个教案；3. “重点”、“难点”、“教学手段与方法”部分要尽量具体；4.授课类型指：理论课、讨论课、实验或实习课、练习或习题课。

授课类型：理论课授课时间：2节授课题目（教学章节或主题）：第一章半导体的电子状态1.3半导体中的电子运动——有效质量1.4本征半导体的导电机构——空穴本授课单元教学目标或要求：理解有效质量和空穴的物理意义，已知e（k）表达式，能求电子和空穴的有效质量，速度和加速度本授课单元教学内容（包括基本内容、重点、难点，以及引导学生解决重点难点的方法、例题等）：1．半导体中e（k）与k的关系（重点，难点）2．半导体中电子的平均速度3．半导体中电子的加速度4．有效质量的物理意义（重点，难点）【篇二：《半导体器件物理》理论课程教学大纲】=txt>课程编码：01222316 课程模块：专业方向课修读方式：限选开课学期：5 课程学分：2.5课程总学时：51 理论学时：36实践学时：15一、课程性质、内容与目标本课程是高等学校本科集成电路设计与集成系统、微电子技术专业必修的一门专业主干课，是研究集成电路设计和微电子技术的基础课程。

施敏-半导体器件物理英文版-第一章习题施敏-半导体器件物理英文版-第一章习题施敏半导体器件物理英文版第一章习题1. （a ）求用完全相同的硬球填满金刚石晶格常规单位元胞的最大体积分数。

（b ）求硅中（111）平面内在300K 温度下的每平方厘米的原子数。

2. 计算四面体的键角，即，四个键的任意一对键对之间的夹角。

（提示：绘出四个等长度的向量作为键。

四个向量和必须等于多少？沿这些向量之一的方向取这些向量的合成。

）3. 对于面心立方，常规的晶胞体积是a 3，求具有三个基矢：（0,0,0→a/2,0,a/2），(0,0,0→a/2,a/2,0),和(0,0,0→0,a/2,a/2)的fcc 元胞的体积。

4. （a ）推导金刚石晶格的键长d 以晶格常数a 的表达式。

（b ）在硅晶体中，如果与某平面沿三个笛卡尔坐标的截距是10.86A ，16.29A ，和21.72A ，求该平面的密勒指数。

5. 指出（a ）倒晶格的每一个矢量与正晶格的一组平面正交，以及（b ）倒晶格的单位晶胞的体积反比于正晶格单位晶胞的体积。

6. 指出具有晶格常数a 的体心立方（bcc ）的倒晶格是具有立方晶格边为4π/a 的面心立方（fcc ）晶格。

[提示：用bcc 矢量组的对称性：)(2x z y a a -+=，)(2y x z a b -+=，)(2z y x a c -+= 这里a 是常规元胞的晶格常数，而x ，y ，z 是fcc 笛卡尔坐标的单位矢量：)(2z y a a ρρρ+=，)(2x z a b ρρρ+=，)(2y x a c ρρρ+=。

] 7. 靠近导带最小值处的能量可表达为.2*2*2*22++=z z y y xx m k m k m k E η 在Si 中沿[100]有6个雪茄形状的极小值。

如果能量椭球轴的比例为5:1是常数，求纵向有效质量m*l 与横向有效质量m*t 的比值。

8. 在半导体的导带中，有一个较低的能谷在布里渊区的中心，和6个较高的能谷在沿[100] 布里渊区的边界，如果对于较低能谷的有效质量是0.1m0而对于较高能谷的有效质量是1.0m0，求较高能谷对较低能谷态密度的比值。

212k mE =）（1222k V E mE -= ）（2232k V E mE -= Chapter 11.22 Calculate the density of valence electrons in silicon.1.23 The structure of GaAs is the zincblende lattice. The lattice constant is 5.65 ︒A . Calculate thedensity of valence electrons in GaAs.1.24 (a) If 17105⨯ phosphorus atoms per 3cm are add to silicon as a substitutional impurity, determine the percentage of silicon atoms per unit volume that are displaced in the single crystal lattice.(b)Repeat part (a) for 15102⨯ boron atoms per 3cm added to silicon.1.25 (a)Assume that 16102⨯-3cm of boron atoms are distributed homogeneously throughout single crystal silicon. What is the fraction by weight of boron in the crystal? (b)If phosphorus atoms, at a concentration of 1810-3cm , are added to the material in part (a), determine the fraction by weight of phosphorus.1.26 If 16102⨯-3cm boron atoms are added to silicon as a substitutional impurity and are distributed uniformly throughout the semiconductor, determine the distance between boron atoms in terms of the silicon lattice constant.(Assume the boron atoms are distributed in a rectangular or cubic array.)1.27 Repeat Problem 1.26 for 15104⨯-3cm phosphorus atoms being added to silicon.Chapter 22.32Consider a proton in a one-dimensional infinite potential well shown in Figure2.6.(a)Derive the expression for the allowed energy states of the proton.(b)Calculate the energy state for (i)a=4︒A ,and (ii)a=0.5cm.2.33 For the step potential function shown in Figure P2.33, assume that 0V E > and that particles are incident from the +x direction traveling in the -x direction.(a) Write the wave solutions for each region.(b) Derive expressions for the transmission and reflection coefficients.2.38 An electron with energy E is incident on a rectangular potential barrier as shown in Figure2.9. The potential barrier is of width a and height E V ≥0.(a)Write the form of the wave function in each of the three regions. (b)For this geometry ,determine what coefficient in the wave function solutions is zero.(c)Derive the expression for the transmission coefficient for the electron(tunneling probability ).(d)Sketch the wave function for the electron in each region.2.39 A potential function is shown in Figure P2.39 with incident particles coming from -∞ with a total energy 2V E >.The constants k are defined asAssume a special case for which πn a 2k 2=,n=1,2,3... Derive the expression, in terms of the constants ,,,21k k and 3k , for the transmission coefficient .The transmission coefficient is defined as the ratio of the flux of particles in region Ⅲ to the incident flux in region Ⅰ.2.40 Consider the one-dimensional potential function shown in FigureP2.40.Assume the total energy of an electron is 0V E <.（a ）Write the wave solutions that apply in each region.(b)Write the set of equations that result from applying the boundary conditions.(c)Show explicitly why not , the energy levels of the electron are quantized.Chapter 33.25 Derive the density of status function for a one-dimension electron gas in GaAs(m n *=0.067m 0).Note that the kinetic energy may be written as E=(±p)2/2m n *,which means that there are two momentum states for each energy level.3.26 (a)Determine the total number (#/cm 3) of energy states in silicon between C E and at (i)T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.27 (a)Determine the total number (#/cm 3) of energy states in silicon between v E and kT E v 3- at (i) T=300K and (ii)T=400K.(b) Repeat part (a) for GaAs.3.28 (a)Plot the density of states in the conduction band of silicon over the range . (b) Repeat part (a) for the density of states in the valence band over the range v v E E eV E <<-4.0.3.29 (a)For silicon,find the ratio of the density of states in the conduction band at E=Ec+KT to the density of states in the valence band at E=Ev-KT. (b)Repeate part (a) for GaAs.Chapter 44.49 Consider silicon at T ＝300 K with donor concentrations of N d ＝1014，1015，1016，and1017，cm -3. Assume N a ＝0.（a ）Calculate the position of the Fermi energy level with respect to the conduction band for these donor concentrations.（b ）Determine the position of the Fermi energy level with respect to the intrinsic Fermi energy level for the donor concentrations given in part （a ）.4.52 Consider GaAs at T=300K with N d =0. (a)Plot the position of the Fermi energy level with respect to the intrinsic Fermi energy level as a function of the acceptor impurity concentration over the range of 1014≤N a ≤1017 cm -3. (b)Plot the position of the Fermi energy level with respect to the valence-band energy over the same acceptor impurity concentration as given in part(a).4.53 For a particular semiconductor , eV E g 50.1= ,**10n p m m =, T=300K,and 35101-⨯=cm n i .(a) Determine the position of the intrinsic Fermi energy level with respect to the center of the bandgap.(b) Impurity atoms are added so that the Fermi energy level is 0.45eV below the center of the bandgap. (i) Are acceptor or donor atoms added? (ii) What's the concentration of impurity atoms added?4.55 (a) Silicon at T=300K is doped with donor impurity atoms at a concentration of 315106-⨯=cm N d . (i) Determine E c - E F . (ii) Calculate the concentration of additional donor impurity atoms that must be added to move the Fermi energy level a distance KT closer to the conduction band edge. (b) Repeat Part (a) for GaAs if the original donor impurity concentration is 315101-⨯=cm N d .4.56 (a) Determine the position of the Fermi energy level with respect to the intrinsic Fermi level in silicon at T=300K that is doped with boron atoms at a concentration of 316102-⨯=cm N d .(b) Repeat Part (a) if the silicon is doped with phosphorus atoms at a concentration of N d =2x1016cm -3 . (c) Calculate n 0 and p 0 in parts (a) and (b).Chapter 55.8 (a) A silicon semiconductor resistor is in the shape of a rectangular bar with a cross-sectional area of 8.5×10-4cm 2, a length of 0.075 cm, and is doped with a concentration of 2×1016cm -3, boron atoms. Let T=300K. A bias of 2 volts is applied across the length of the silicon device. Calculate the current in the resistor. (b) Repeat part (a) if the length is increased by a factor of three. (c) Determine the average drift velocity of holes in the parts (a) and (b).5.18 An n-type silicon resistor has a leng th L=150μm, which W=7.5μm, and thickness T=1μm. A voltage of 2 V is applied across the length of the resistor. The donor impurity concentration varies linearly through the thickness of the resistor with N d =2×1016cm -3at the top surface and N d =2×1015cm -3at the bottom surface. Assume an average carrier mobility of n μ=750cm 2/V -s . (a)s V cm N d n⋅⨯+=-/)1051(135032/116μ)()(2min p n pn i μμμμσσ+=What is the electric field in the resistor? (b) Determine the average conductivity of the silicon. (c) Calculate the current in the resistor. (d) Determine the current density near the top surface and the current density near the bottom surface.5.22 A semiconductor material has electron and hole mobilitiesμn andμp , respectively. When the conductivity is considered as a function of the hole concentration p 0, (a) show that the minimum value of conductivity, σmin , can be written aswhere i σis the intrinsic conductivity, and (b) show that the corresponding hole concentration is2/10)/(p n n p μμ=.5.23 Consider three samples of silicon at T=300K. The n-type sample is doped with arsenic atoms to a concentration of 316105-⨯=cm N d , The p-type sample is doped with boron atoms to a concentration of 316102-⨯=cm N a .The compensated sample is doped with both the donors and acceptors described in the n-type and p-type sample. (a) Find the equilibrium electron and hole concentrations in each sample. (b) determine the majority carrier mobility in each sample. (c) calculate the conductivity of each sample. (d) and determine the electric field required in each sample to induce a drift current density of J=120A/cm 2 .5.28 (a) Assume that the electron mobility in an n-typesemiconductor is given by Where N d is the donor concentration in cm -3. Assuming complete ionization, plot the conductivity as a function of N d over the range 318151010-≤≤cm N d . (b) Compare the results of part (a) tothat if the mobility were assumed to be a constant equal to 1350 s V cm ⋅-/3. (c) If an electricfield of E=10V/cm is applied to the semiconductor. Plot the electron drift current density of parts (a) and (b).5.36 The total current in a semiconductor is constant and equal to J=-10A/cm -3. The total current is composed of a hole drift current. Assume that the hole concentration is a constant and equal to 1016 cm -3 and assume that electron concentration is given by 3/15102)(--⨯=cm e x n L x where L=15m μ. The electron diffusion coefficient is D n =27 cm 2/s and the hole mobility is s V cm p ⋅=/4202μ . Calculate (a) the electron diffusion current density for x>0, (b) the hole drift current density for, and (c) the required electric field for x>0 .Chapter 66.3 An n-type silicon sample contains a donor concentration of 31610-=cm N d . The minority carrier hole lifetime is found to be . (a) What is the lifetime of the majority carrier electrons? (b) Determine the thermal-equilibrium generation rate for electrons and holes in this material. (c) Determine the thermal-equilibrium recombination rate for electrons and holes in this material.6.5 Derive Equation (6.27) from Equations (6.18) and (6.20).6.6 Consider a one-dimensional hole flux as shown in Figure 6.4. If the generation rate of holes in this differential volume is 132010--⋅=s cm g p and the recombination rate is 1319102--⋅⨯s cm ,what must be the gradient in the particle current density to maintain a steady-state hole concentration?6.7 Repeat Problem 6.6 if the generation rate becomes zero.6.17 (a)Consider a silicon sample at T=300K doped with 1016cm -3 donor atoms. Let τp0=5x10-7s.Alight source turns on at t=0 producing excess carriers with a uniform generation rate of g'=5x1020cm -3s -1.At t=5x10-7s,the light source turns off.(i)Derive the expression(s) for the excess carrier concentration as a function of time over the range 0≤t≤∞.(ii) What is the value of the excess concentration when the light source turns off. (b) Repeat Part (a) for the case when the light source turns off at t=2x10-6s. (c) Sketch the excess minority carrier concentrations versus time for parts (a) and (b).6.18 A semiconductor is uniformly doped with 1710-3cm acceptor atoms and has the following properties :s cm D n /272=,s cm D p /122=,s n 170105-⨯=τ,and s p 17010-=τ.An external source has been turned on for 0<t producing a uniform concentration of excess carriers at a generation rate of 132110'--=s cm g .The source turns off at time t=0 and back on at time s t 6102-⨯=(a)Derive the expressions for the excess carrier concentration as a function of time for ∞≤≤t 0.(b)Determine the value of excess carrier concentration at (i)t=0 (ii)s t 6102-⨯=,and (iii)∞=t (c)Plot the excess carrier concentration as a function of time.Chapter 77.4 An abrupt silicon pn junction at zero bias dapant concentrations of N a =1017cm -3 and N d =5×1015cm -3. T=300K. ﹙a ﹚Calculate the Fermi level on each side of the junction with respect to the intrinsic Fermi level. ﹙b ﹚ Sketch the equilibrium energy band diagram for the junction and determine V bi from the diagram and the results of part ﹙a ﹚. ﹙c ﹚Calculate V bi using Equation ﹙7.10﹚, and compare the results to part ﹙b ﹚. ﹙d ﹚Determine xn, xp, and the peak electric field for this junction.7.6 A Silicon pn junction in thermal equilibrium at T=300K is doped such that E F -E Fi =0.365eV in the n region and E Fi -E F =0.330eV in the p region ﹙a ﹚Sketch the energy-band diagram for the pn junction. ﹙b ﹚Find the impurity doping concentration in each region. ﹙c ﹚Determine V bi .7.18 An ideal one-sided silicon p +n junction at T=300K is uniformly doped on both sides of the metallurgical junction. It is found that the doping relation is N a =80N d and the build-in potential barrier is V bi =0.740V . A reverse-biased voltage of VR=10V is applied. Determine ﹙a ﹚N a , N d ;﹙b ﹚x p , x n ;﹙c ﹚|E max |; and ﹙d ﹚C’j .7.19 A silicon n +p junction is biased at V R =5V. ﹙a ﹚Determine the change in built-in potential barrier if the doping concentration in the p increases by a factor of 3.﹙b ﹚Determine the ratio of junction capacitance when the acceptor doping is 3N a compared to that when the acceptor doping is N a .﹙c ﹚Why does the junction capacitance increase when the doping concentration increase ?7.20 ﹙a ﹚The peak electric field in a reverse-biased silicon pn junction is |E max |=3×105 V/cm. The doping concentrations are N d =4×1015cm -3 and N a =4×1017cm -3 .Find the magnitude of the reverse-biased voltage.﹙b ﹚Repeat part ﹙a ﹚ for N d =4×1016cm -3 and N a =4×1017cm -3﹙c ﹚Repeat part ﹙a ﹚N d =N a =4×1017cm -3Chapter 88.2 A silicon pn junction has impurity doping concentrations of N d =2×1015cm -3 and N a =8×1015cm -3. Determine the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.45 V ﹙b ﹚V a =0.55 V , and ﹙c ﹚V a =-0.55V .8.3 The doping concentrations in a G a A s pn junction are N d =1016cm -3 and N a =4×1016cm -3.Find the minority carrier concentrations at the edges of the space charge region for ﹙a ﹚V a =0.90 V ﹙b ﹚V a =1.10 V, and﹙c﹚V a =-0.95V.8.5 Consider a G a A s pn junction with doping concentrations N a=5×1016cm-3 and N d=1016cm-3. The junction cross-sectional area is A=10-3cm2and the applied forward-bias voltage is Va=1.10V. Calculate the﹙a﹚minority electron diffusion current at the space charge region, ﹙b﹚minority hole diffusion current at the edge of the space charge region. and﹙c﹚total current in the pn junction diode.8.6An n+p silicon dione at T=300K has the following parameters N d=1016cm-3, D n=25cm2/s,D p=10cm2/s , N d=1016cm-3, Τn0=Τp0=1 μs, and A=10-4cm2. Determine the diode current for ﹙a﹚a forward-bias voltage of 0.5V and ﹙b﹚a reverse-biased voltage of 0.5V.8.7 An ideal germanium pn junction diode has the following parameters: N a=4×1015cm-3, N a=2×1017cm-3, D p=48cm2/s, D n=90cm2/s, Τp0=Τn0=2×10-6s, and A=10-4cm2 . Determine the diode current for ﹙a﹚a forward-bias voltage of 0.25V and ﹙b﹚a reverse-biased voltage of 0.25V.Chapter 99.7 A Schottky diode with n-type GaAs at T = 300 K yields the 1/C’2 versus V R plot shown in Figure P9.7,where C’ is the capacitance per cm2. Determine (a)V bi (b)N d , (c) υn ,and (d)υBO.9.8 Consider a W-n-type silicon Schottky barrier at T = 300 K with N d=5×1015cm-3Use the data in Figure 9.5 to determine the barrier height. (a) determine V bi, χn, and |E max| for(i) V R = 1 V and (ii) V R=5 V . (b) Using the values of |E max| from part (a)，determine the Schottky barrier lowering parameters ΔФ and χm .9.14 A Schottky diode at T= 300 K is formed with Pt on n-type silicon with a doping concentration of N d=5×1015cm-3 . the barrier height is found to beυBn = 0.89V . Determine (a) υn ,(b) V bi , (c) J sT ,and (d) V a such at J n = 5 A/cm2 . (Neglect the barrier lowering effect)9.15(a) Consider a Schottky diode at T= 300 K that is formed with tungsten on n-type silicon .Use Figure 9.5 to determine the barrier height. Assume a doping concentration of N d = 1016cm-3and assume across-sectional area A=10-4 cm2. Determine the forward-bias voltage required to induce a current of (i) 10μA ,(ii) 100μA ,and(iii) 1mA .(b) Repeat part (a) for a temperature of T = 350 K (Neglect the barrier lowering effect)9.16 An Au-n-GaAs Schottky diode at T = 300 K has a doping concentration of N d = 1016cm-3 .(a) Using Figure 9.5, to determine the barrier height. (b) Calculate the reverse-biased saturation current J sT . (c) Determine the forward-bias voltage required to induce a current density of J n = 10A/cm2. (d )what is the change in forward-bias voltage necessary to double the current density? (Neglect the Schottky barrier lowering)Chapter 1010.10Conder a MOS device with a p-type silicon substrate with N a=2×1016cm-3.The oxide thickness t ox=15nm=150Å and the equvialent oxide charge is Q,ss=7×1010cm-2.Calculate the threshold voltage for(a)an n+ polysilicon gate.(b)a p+ polysilicon gate,and(c)an aluminum gate. 10.11 Repeat Problem 10.10 for an n-type silicon substrate with a doping of N d=3×1015 cm-310.12 A 400-Å oxideis grown on p-type silicon with N a=5×1015cm-3.The flat-band voltage is -0.9V.Calculate the surface potential at the threshold inversion point as well as the threshold voltage asssuming negligible oxide charge. Also find the maximum space charge width for this device.10.13 A MOS device with an aluminum gate is fabricated on a p-type silicon substrate. The oxide thickness t ox=22nm=220Å and the trapped oxide charge is Q,ss=4×1010cm-2.The measure threshold voltage is V T=+0.45V. Determine the p-type doping concentration.10.14 Consider a MOS device with the following paremeter :p+ polysilicon gate, n-type silicon substrate, t ox=18nm=180Å,and Q,ss=4×1010cm-2. Determine the silicon doping concentration such that the threshold voltage is in the range -0.35≦V TP≦-0.25V.。

半导体物理与器件英文版第四版课后练习题含答案Chapter 1: Crystal PropertiesMultiple Choice Questions1.Which of the following statements is correct? A. The latticestructure of a crystal can be described by three crystal axes that are normal to each other. B. For a crystal with a primitive cubic unit cell, the coordination number is 8. C. In a crystal lattice with a face-centered cubic (FCC) unit cell, each atom has only six nearest neighbors. D. The Miller indices of a crystal planeperpendicular to the x-axis and passing through point (1, 2, 3) are (1, 2, 3).Answer: A2.Which of the following statements is correct? A. The crystalstructure of diamond is face-centered cubic (FCC). B. The density of silicon is smaller than that of germanium. C. The coordination number of germanium is 4. D. The Miller indices of a crystal plane parallel to the x-axis and passing through point (1, 2, 3) are (1, 0, 0).Answer: DShort Answer Questions1.What is the difference between a lattice and a unit cell?2.Define the concept of coordination number and give anexample of a coordination number 6 crystal structure.3.Define the concept of a crystal plane and expln how Millerindices are used to describe crystal planes.Answers:1.A lattice is an infinitely repeating arrangement of pointsin space that defines the basic symmetry of a crystal, while aunit cell is the smallest repeating unit of a crystal lattice that can be used to reconstruct the entire crystal by translation.2.Coordination number is the number of nearest neighbors of anatom in a crystal lattice. An example of a coordination number 6 crystal structure is the hexagonal close-packed (HCP) structure.3.A crystal plane is an imaginary flat surface in a crystalthat can be used to define the orientation of the crystal in space.Miller indices are a set of integers that describe the orientation of a crystal plane relative to the crystal axes. The Millerindices of a plane are determined by finding the reciprocals of the intercepts of the plane with the crystal axes and thenreducing these reciprocals to the smallest set of integers that give a unique designation of the plane.。

半导体器件物理及工艺办法+施敏++答案半导体器件物理及工艺是半导体科学与工程领域的重要分支，涉及半导体器件的基本原理、结构和制造工艺等方面。

本文将介绍施敏的《半导体器件物理及工艺》一书，并给出相应的答案。

一、半导体器件物理及工艺概述半导体器件物理及工艺是研究半导体器件的基本原理、结构和制造工艺的学科。

半导体器件具有高灵敏度、高可靠性、高速度等优点，在电子、通信、自动化等领域得到广泛应用。

半导体器件物理及工艺的主要研究对象包括半导体材料、半导体器件的原理和结构、制造工艺等。

二、施敏《半导体器件物理及工艺》简介施敏的《半导体器件物理及工艺》是一本经典的教材，系统地介绍了半导体器件的基本原理、结构和制造工艺。

全书分为十章，包括半导体材料、半导体器件的基本原理、PN结二极管、双极晶体管、金属氧化物半导体场效应晶体管、光电器件、半导体集成电路等。

三、施敏《半导体器件物理及工艺》答案1.什么是半导体？请列举出三种常见的半导体材料。

答：半导体是指导电性能介于导体和绝缘体之间的材料。

常见的半导体材料包括硅、锗、砷化镓等。

2.简述PN结的形成及其基本性质。

答：PN结是由P型半导体和N型半导体相互接触形成的势垒区。

PN结的基本性质包括单向导电性、电容效应和光电效应等。

3.解释双极晶体管的工作原理。

答：双极晶体管是由P型半导体和N型半导体组成的三明治结构，通过控制基极电流来控制集电极和发射极之间的电流，实现放大作用。

4.什么是金属氧化物半导体场效应晶体管？请简述其工作原理。

答：金属氧化物半导体场效应晶体管是一种电压控制型器件，通过改变栅极电压来控制源极和漏极之间的电流，实现放大作用。

其工作原理是基于MOS结构的电容效应和隧道效应。

5.光电器件的基本原理是什么？请举例说明其应用。

答：光电器件的基本原理是光电效应，即光照射在物质表面上时，物质会吸收光能并释放电子，产生电流。

光电器件的应用包括太阳能电池、光电传感器等。

6.请简述半导体的基本制备工艺流程。

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