半导体器件物理施敏答案

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。

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。

半导体器件物理施敏答案【篇一：施敏院士北京交通大学讲学】t>——《半导体器件物理》施敏 s.m.sze,男，美国籍，1936年出生。

台湾交通大学电子工程学系毫微米元件实验室教授，美国工程院院士，台湾中研院院士，中国工程院外籍院士，三次获诺贝尔奖提名。

学历：美国史坦福大学电机系博士（1963），美国华盛顿大学电机系硕士（1960），台湾大学电机系学士（1957）。

经历：美国贝尔实验室研究（1963-1989），交通大学电子工程系教授（1990-），交通大学电子与资讯研究中心主任（1990-1996），国科会国家毫微米元件实验室主任（1998-），中山学术奖（1969），ieee j.j.ebers奖（1993），美国国家工程院院士（1995）, 中国工程院外籍院士 (1998)。

现崩溃电压与能隙的关系，建立了微电子元件最高电场的指标等。

施敏院士在微电子科学技术方面的著作举世闻名，对半导体元件的发展和人才培养方面作出了重要贡献。

他的三本专著已在我国翻译出版，其中《physics of semiconductor devices》已翻译成六国文字，发行量逾百万册；他的著作广泛用作教科书与参考书。

由于他在微电子器件及在人才培养方面的杰出成就,1991年他得到了ieee 电子器件的最高荣誉奖（ebers奖），称他在电子元件领域做出了基础性及前瞻性贡献。

施敏院士多次来国内讲学，参加我国微电子器件研讨会；他对台湾微电子产业的发展，曾提出过有份量的建议。

主要论著：1. physics of semiconductor devices, 812 pages, wiley interscience, new york, 1969.2. physics of semiconductor devices, 2nd ed., 868 pages, wiley interscience, new york,1981.3. semiconductor devices: physics and technology, 523 pages, wiley, new york, 1985.4. semiconductor devices: physics and technology, 2nd ed., 564 pages, wiley, new york,2002.5. fundamentals of semiconductor fabrication, with g. may,305 pages, wiley, new york,20036. semiconductor devices: pioneering papers, 1003 pages, world scientific, singapore,1991.7. semiconductor sensors, 550 pages, wiley interscience, new york, 1994.8. ulsi technology, with c.y. chang,726 pages, mcgraw hill, new york, 1996.9. modern semiconductor device physics, 555 pages, wiley interscience, new york, 1998. 10. ulsi devices, with c.y. chang, 729 pages, wiley interscience, new york, 2000.课程内容及参考书：施敏教授此次来北京交通大学讲学的主要内容为《physics ofsemiconductor device》中的一、四、六章内容，具体内容如下：chapter 1: physics and properties of semiconductors1.1 introduction 1.2 crystal structure1.3 energy bands and energy gap1.4 carrier concentration at thermal equilibrium 1.5 carrier-transport phenomena1.6 phonon, optical, and thermal properties 1.7 heterojunctions and nanostructures 1.8 basic equations and exampleschapter 4: metal-insulator-semiconductor capacitors4.1 introduction4.2 ideal mis capacitor 4.3 silicon mos capacitorchapter 6: mosfets6.1 introduction6.2 basic device characteristics6.3 nonuniform doping and buried-channel device 6.4 device scaling and short-channel effects 6.5 mosfet structures 6.6 circuit applications6.7 nonvolatile memory devices 6.8 single-electron transistor iedm，iscc, symp. vlsi tech.等学术会议和期刊上的关于器件方面的最新文章教材：? s.m.sze, kwok k.ng《physics of semiconductordevice》,third edition参考书：? 半导体器件物理(第3版)(国外名校最新教材精选)(physics of semiconductordevices) 作者:(美国)(s.m.sze)施敏 (美国)(kwok k.ng)伍国珏译者:耿莉张瑞智施敏老师半导体器件物理课程时间安排半导体器件物理课程为期三周，每周六学时，上课时间和安排见课程表：北京交通大学联系人：李修函手机：138******** 邮件：lixiuhan@案2013~2014学年第一学期院系名称：电子信息工程学院课程名称：微电子器件基础教学时数： 48授课班级： 111092a,111092b主讲教师：徐荣辉三江学院教案编写规范教案是教师在钻研教材、了解学生、设计教学法等前期工作的基础上，经过周密策划而编制的关于课程教学活动的具体实施方案。

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2—1．P N +结空间电荷区边界分别为p x -和n x ，利用2T V Vi np n e =导出)(n n x p 表达式。

给出N 区空穴为小注入和大注入两种情况下的)(n n x p 表达式.解:在n x x =处 ()()⎪⎪⎩⎪⎪⎨⎧⎪⎭⎫ ⎝⎛-=⎪⎭⎫⎝⎛-=KT E E n x n KT E E n x p i Fn in n FP i i nn exp exp()()VT V i Fp Fn i n n n n e n KT E E n x n x p 22exp =⎪⎪⎭⎫ ⎝⎛-= 而()()()000n n n n nn n n n n n n p x p p p n x n n n p x =+∆≈∆=+∆=+ （n n n p ∆=∆）()()TTV V in n n V V in n n en p n p e n n n p 2020=∆+⇒=∆+2001TV V n i n n n p n p e n n ⎛⎫⇒+=⎪⎝⎭ T V V 22n n0n i p +n p -n e =0n p =(此为一般结果） 小注入：（0n n n p <<∆）T T V V n V V n i n e p e n n p 002== ()002n n i p n n =大注入: 0n n n p >>∆ 且 n n p p ∆= 所以 TV V inen p 22=或 TV Vi n en p 2=2-2．热平衡时净电子电流或净空穴电流为零,用此方法推导方程20lni ad T p n n N N V =-=ψψψ。

解:净电子电流为()n nn nI qA D n xμε∂=+∂ 处于热平衡时，I n ＝0 ，又因为 d dxψε=-所以nn d n n D dx x ψμ∂=∂,又因为n T nDV μ=（爱因斯坦关系） 所以dn nV d T=ψ， 从作积分，则2002ln ln ln ln ln i a d n p T n T po T d T T a in N NV n V n V N V V N n ψψψ=-=-=-=2-3．根据修正欧姆定律和空穴扩散电流公式证明,在外加正向偏压V 作用下,PN 结N 侧空穴扩散区准费米能级的改变量为qV E FP =∆。

半导体器件物理复习（施敏）第⼀章1、费⽶能级和准费⽶能级费⽶能级：不是⼀个真正的能级，是衡量能级被电⼦占据的⼏率的⼤⼩的⼀个标准，具有决定整个系统能量以及载流⼦分布的重要作⽤。

准费⽶能级：是在⾮平衡状态下的费⽶能级，对于⾮平衡半导体，导带和价带间的电⼦跃迁失去了热平衡，不存在统⼀费⽶能级。

就导带和价带中的电⼦讲，各⾃基本上处于平衡态，之间处于不平衡状态，分布函数对各⾃仍然是适应的，引⼊导带和价带费⽶能级，为局部费⽶能级，称为“准费⽶能级”。

2、简并半导体和⾮简并半导体简并半导体：费⽶能级接近导带底(或价带顶)，甚⾄会进⼊导带(或价带)，不能⽤玻尔兹曼分布，只能⽤费⽶分布⾮简并半导体：半导体中掺⼊⼀定量的杂质时，使费⽶能级位于导带和价带之间3、空间电荷效应当注⼊到空间电荷区中的载流⼦浓度⼤于平衡载流⼦浓度和掺杂浓度时，则注⼊的载流⼦决定整个空间电荷和电场分布，这就是空间电荷效应。

在轻掺杂半导体中，电离杂质浓度⼩，更容易出现空间电荷效应，发⽣在耗尽区外。

4、异质结指的是两种不同的半导体材料组成的结。

5、量⼦阱和多量⼦阱量⼦阱：由两个异质结或三层材料形成，中间有最低的E C和最⾼的E V，对电⼦和空⽳都形成势阱，可在⼆维系统中限制电⼦和空⽳当量⼦阱由厚势垒层彼此隔开时，它们之间没有联系，这种系统叫做多量⼦阱6、超晶格如果势垒层很薄，相邻阱之间的耦合很强，原来分⽴的能级扩展成能带(微带)，能带的宽度和位置与势阱的深度、宽度及势垒的厚度有关，这种结构称为超晶格。

7、量⼦阱与超晶格的不同点a.跨越势垒空间的能级是连续的b.分⽴的能级展宽为微带另⼀种形成量⼦阱和超晶格的⽅法是区域掺杂变化第⼆章1、空间电荷区的形成机制当这两块半导体结合形成p-n结时，由于存在载流⼦浓度差，导致了空⽳从p区到n 区，电⼦从n区到p区的扩散运动。

对于p 区，空⽳离开后，留下了不可动的带负电的电离受主，这些电离受主，没有正电荷与之保持电中性，所以在p-n结附近p 区⼀侧出现了⼀个负电荷区。