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。