Solution Manual for Semiconductor Physics and Devices 3ed (Neamen)

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Problem Solutions Chapter Problem Solutions FG 4πr IJ H3K atoms per cell, so atom vol = 1.1 (a) fcc: corner atoms × 1/8 = atom face atoms × ½ = atoms Total of atoms per unit cell Then FG 4πr IJ H K × 100% ⇒ Ratio = (b) bcc: corner atoms × 1/8 = atom enclosed atom = atom Total of atoms per unit cell Ratio = 74% 16 r (c) Body-centered cubic lattice d = 4r = a ⇒ a = r (c) Diamond: corner atoms × 1/8 = atom face atoms × ½ = atoms enclosed atoms = atoms Total of atoms per unit cell Unit cell vol = a = F rI H 3K FG 4πr IJ H3K atoms per cell, so atom vol = 1.2 (a) Ga atoms per unit cell Density = b 5.65 x10 −8 g Then FG 4πr IJ H K × 100% ⇒ Ratio = F 4r I H 3K ⇒ Density of Ga = 2.22 x10 cm 22 −3 As atoms per unit cell, so that Density of As = 2.22 x10 cm 22 −3 (d) Diamond lattice (b) Body diagonal = d = 8r = a ⇒ a = Ge atoms per unit cell ⇒ Density = −8 5.65 x10 b Ratio = 68% g Unit cell vol = a Density of Ge = 4.44 x10 cm 22 −3 F 8r I = H 3K r FG 4πr IJ H3K atoms per cell, so atom vol 1.3 (a) Simple cubic lattice; a = 2r Then Unit cell vol = a = (2 r ) = 8r 3 FG 4πr IJ H K × 100% ⇒ Ratio = F 8r I H 3K 3 FG 4πr IJ H3K atom per cell, so atom vol = (1) Then FG 4πr IJ H K × 100% ⇒ Ratio = Ratio = 34% 1.4 From Problem 1.3, percent volume of fcc atoms is 74%; Therefore after coffee is ground, Ratio = 52.4% 8r (b) Face-centered cubic lattice d =2 2r d = 4r = a ⇒ a = c Unit cell vol = a = 2 r h = 16 2r Volume = 0.74 cm 3 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Problem Solutions Then mass density is 1.5 (a) a = 5.43 A so that r = ° a From 1.3d, a = (5.43) ρ= r =ρ= N ( At Wt ) NA 6.02 x10 1.9 (a) Surface density 1 = = −8 a 4.62 x10 ° b ° 14 3.31x10 cm 23 Density(A) = 118 x10 cm −3 x10 cm Density(B) = 118 (a) Vol density = −3 Surface density = 1.7 (b) b2.8x10 g ao ao (b) Same as (a) ° (c) = 2.28 x10 cm 22 ⇒ 1.01x10 cm 22 22 1.01x10 cm g 2 −2 1.10 B-type: atom per unit cell, so −8 Same for A atoms and B atoms (b) Same as (a) (c) Same material g 12 ° (b) Same as (a) (c) Same material so that rB = 0.747 A (b) A-type; atom per unit cell Density = ⇒ −8 2.04 x10 Na: Density = −8 −8 23 rA + 2rB = a ⇒ 2rB = 2.04 − 2.04 23 b4.62 x10 g Density of B = b4.62 x10 g ⇒ (a) a = rA = 2(1.02) = 2.04 A Now a = 18 + 1.0 ⇒ a = 2.8 A ° Density of A = 1.6 b ⇒ a = 4.62 A b5x10 g(28.09) ⇒ ρ = 2.33 grams / cm (a) a = 2(2.2 ) + 2(1.8) = A so that ° 22 = −8 1.8 g (c) Mass density b2.8x10 g ° (b) Number density −3 22 = ⇒ Density = x10 cm −8 5.43 x10 b −23 ρ = 2.21 gm / cm = = 118 A 8 Center of one silicon atom to center of nearest neighbor = 2r ⇒ 2.36 A 4.85 x10 Cl: Density (same as Na) = 2.28 x10 cm 22 1.11 Sketch −3 1.12 (a) −3 (d) Na: At.Wt = 22.99 Cl: At Wt = 35.45 So, mass per unit cell 1 (22.99) + (35.45) −23 = = 4.85 x10 23 6.02 x10 (b) F , , 1I ⇒ (313) H 1K F , , I ⇒ (121) H 4K ⇒ −3 −3 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 1.13 (a) Distance between nearest (100) planes is: d = a = 5.63 A = ° d = 3.98 A (iii) ° or (iii) FI HK c h b 4.50 x10 −8 (iii) 15 x10 cm 114 b g −2 g ⇒ 2.85 x10 cm 14 g −2 3a b g 1.16 −2 d = 4r = a then 4r 4(2.25) ° a= = = 6.364 A 2 (a) atoms Volume Density = −8 6.364 x10 −2 (110) plane, surface density, atoms −2 14 ⇒ 6.99 x10 cm −8 2 4.50 x10 b g 22 x10 cm 155 Same as (a),(iii), surface density 2.85 x10 cm −3 (b) Distance between (110) planes, a 6.364 = a = = ⇒ 2 or (111) plane, surface density, 14 g (c) (111) plane, surface density, atoms −2 14 = ⇒ 7.83 x10 cm −8 5.43 x10 14 b I K b Same as (a),(i); surface density 4.94 x10 cm = F H (b) (110) plane, surface density, atoms −2 14 = ⇒ 9.59 x10 cm −8 2 5.43 x10 (b) Body-centered cubic (i) (100) plane, surface density, (ii) g 6.78 x10 cm g b b (110) plane, surface density, atom −2 14 ⇒ 3.49 x10 cm −8 2 4.50 x10 (111) plane, surface density, 1 atoms = = = a a ( x) ⋅a ⋅ 2 = (110) plane, surface density, atoms −2 14 ⇒ 6.99 x10 cm −8 2 4.50 x10 14 g b −2 1.15 (a) (100) plane of silicon – similar to a fcc, atoms ⇒ surface density = −8 5.43 x10 ° = 14 ° Simple cubic: a = 4.50 A (i) (100) plane, surface density, atom −2 14 = ⇒ 4.94 x10 cm −8 4.50 x10 (ii) ⇒ 9.88 x10 cm (111) plane, surface density, 1 3⋅ + 3⋅ = = −8 3 4.50 x10 a 1.14 (a) b b4.50x10 g −8 = (c) Distance between nearest (111) planes is: a 5.63 d= a 3= = 3 or d = 3.25 A atoms (ii) (b)Distance between nearest (110) planes is: a 5.63 d= a 2= = 2 or Chapter Problem Solutions −2 (c) Face centered cubic (i) (100) plane, surface density g ⇒ Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 4.50 A Chapter Problem Solutions ° 1.20 (c) Surface density atoms = = 2a b5x10 g(30.98) ⇒ (a) Fraction by weight ≈ b5x10 g(28.06) 16 22 b 6.364 x10 −8 g −6 x10 110 (b) Fraction by weight or 14 3.49 x10 cm b10 g(10.82) ≈ b5x10 g(30.98) + b5x10 g(28.06) ⇒ 18 −2 16 1.17 Density of silicon atoms = x10 cm valence electrons per atom, so 22 −3 23 Density of valence electrons x10 cm 7.71x10 and Volume density = −3 16 x100% ⇒ −5 x10 % 1x10 15 x10 22 ° We have a O = 5.43 A So d d 794 = ⇒ = 146 a O 5.43 aO 1.19 (b) Percentage = 15 ° 23 22 = x10 cm −6 Density of valence electrons 1.77 x10 cm x10 d d = 7.94 x10 cm = 794 A g (a) Percentage = So An average of valence electrons per atom, x10 −6 1.21 −3 1.18 Density of GaAs atoms atoms 22 −3 = = 4.44 x10 cm −8 5.65 x10 b 22 x100% ⇒ −6 x10 % −3 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Problem Solutions Chapter Problem Solutions p = 5.4 x10 2.1 Computer plot λ= 2.2 Computer plot h p = −25 6.625 x10 5.4 x10 kg − m / s −34 ⇒ −25 or λ = 12.3 A 2.3 Computer plot ° (ii) K.E = T = 100 eV = 1.6 x10 2.4 For problem 2.2; Phase = p= 2πx − ωt = constant λ λ= Then λ 2π dx dx ⋅ − ω = or = v = +ω λ dt 2π dt F I H K p For problem 2.3; Phase = 2πx λ p = 5.4 x10 mT ⇒ h p ⇒ λ = 1.23 A −17 −24 J kg − m / s ° (b) Proton: K.E = T = eV = 1.6 x10 p= + ωt = constant mT = b 1.67 x10 −27 −19 gb1.6x10 g −19 or p = 2.31x10 Then λ 2π dx dx ⋅ + ω = or = v p = −ω λ dt 2π dt F I H K λ= h p = −23 6.625 x10 2.31x10 kg − m / s −34 ⇒ −23 or λ = 0.287 A 2.5 E = hν = hc λ ⇒λ = hc b g So b6.625x10 gb3x10 g ⇒ 2.54 x10 λ= (4.90)b1.6 x10 g Gold: E = 4.90 eV = (4.90) 1.6 x10 −34 −19 For T = eV = 1.6 x10 J p= mT −5 cm b Cesium: E = 1.90 eV = (1.90) 1.6 x10 −19 g b6.625x10 gb3x10 g ⇒ 6.54 x10 λ= (1.90)b1.6 x10 g −34 λ= J −5 b gb1.6x10 g p = −22 6.625 x10 3.13x10 kg = m / s −34 −22 ⇒ λ = 0.0212 A cm ° (d) A 2000 kg traveling at 20 m/s: p = mv = (2000)(20) ⇒ or p = x10 kg − m / s (a) Electron: (i) K.E = T = eV = 1.6 x10 9.11x10 −27 or 2.6 mT = h 10 λ = 0.654 µm p= b p = 313 x10 −19 or J or or λ = 0.254 µm −19 2(183.92 ) 1.66 x10 = 10 −19 So ° (c) Tungsten Atom: At Wt = 183.92 E −31 −19 gb1.6x10 g λ= J h p = 6.625 x10 x10 −34 ⇒ or −19 λ = 1.66 x10 or −28 A ° J −19 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Problem Solutions or 2.7 E avg = kT = E = 1.822 x10 (0.0259) ⇒ Also or b b 9.11x10 −31 g(0.01727)b1.6x10 g h λ= −19 pavg = 7.1x10 = p −26 7.1x10 kg − m / s kg − m / s −34 h = λ ⇒ 1822 x10 −26 ⇒ ° −34 6.625 x10 125 x10 p = 5.3x10 or λ = 93.3 A −34 (b) p= −26 6.625 x10 λ = 364 A −26 Now 6.625 x10 = p or λ= gb2 x10 g ⇒ Now mE avg h −31 p = 1822 x10 Now = −3 J ⇒ E = 114 x10 eV p = mv = 9.11x10 E avg = 0.01727 eV pavg = −22 ⇒ −10 −26 kg − m / s Also ° p v= 2.8 = m 5.3 x10 −26 = 5.82 x10 m / s 9.11x10 −31 or hc E p = hν p = v = 5.82 x10 cm / s λp Now Now Ee = pe 2m and pe = h λe ⇒ Ee = FG h IJ 2m H λ K E= hc λp FG h IJ = 2m H λ K e e FG 10h IJ = 2m H λ K = b (a) E = hν = p or −21 hc λp hc = 100h −31 ⋅ mc = gb3x10 g mc E = 1.64 x10 −15 mv = hc λ b6.625x10 gb3x10 g = −34 1x10 −15 −10 J −15 b = 1.6 x10 −19 gV V = 12.4 x10 V = 12.4 kV ⇒ (b) p = b mE = = 6.02 x10 J = 10.3 keV −31 so 100 9.11x10 −23 −31 gb1.99 x10 g −15 kg − m / s Then b9.11x10 gb2 x10 g 2 −3 E = e ⋅ V ⇒ 1.99 x10 λ= 2.9 J ⇒ E = 9.64 x10 eV Now 100 −31 E = 1.99 x10 9.11x10 b9.11x10 gb5.82 x10 g 2.10 So (a) E = E = 1.54 x10 which yields 100h λp = mc Ep = E = mv = or Set E p = E e and λ p = 10λ e Then 10 h p = 6.625 x10 6.02 x10 −34 −23 ⇒ λ = 0.11 A ° Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Problem Solutions 2.11 (a) ∆p = h ∆x 1.054 x10 = 10 −34 ⇒ −6 ∆p = 1.054 x10 −28 (b) ∆t = kg − m / s or ∆E = 3.16 x10 −20 −h J ⇒ ∆E = 0.198 eV = 1.054 x10 12 x10 −26 ∆p = 8.78 x10 (b) ∆E = ⋅ ( ∆p)2 = m ∆E = 7.71x10 −23 −h −34 ⇒ −10 kg − m / s b8.78x10 g ⋅ −26 x10 ∆E = ( ∆p)2 ⋅ m ∆E = 7.71x10 −26 b8.78x10 g ⋅ −26 = x10 + V ( x )Ψ2 ( x , t ) = jh ∂Ψ1 ( x , t ) ∂t ∂Ψ2 ( x , t ) ⋅ ∂ −26 ∂t Ψ1 ( x , t ) + Ψ2 ( x , t ) +V ( x ) Ψ1 ( x , t ) + Ψ2 ( x , t ) ∂ Ψ1 ( x , t ) + Ψ2 ( x , t ) ∂t which is Schrodinger’s wave equation So Ψ1 ( x , t ) + Ψ2 ( x , t ) is also a solution (b) If Ψ1 ⋅ Ψ2 were a solution to Schrodinger’s wave equation, then we could write kg − m / s −h ⇒ ∂ 2m ∂x −7 J ⇒ ∆E = 4.82 x10 eV aΨ ⋅ Ψ f + V ( x)aΨ ⋅ Ψ f 2 = jh 2.14 ∆p = ∂ Ψ2 ( x , t ) + V ( x )Ψ1 ( x , t ) = jh = jh −26 2 ⋅ ⇒ J ⇒ ∆E = 4.82 x10 eV ∂x 2m ∂x −29 (a) Same as 2.12 (a), ∆p = 8.78 x10 ∂ Ψ1 ( x , t ) ⋅ −h 2.13 s ∂x 2m Adding the two equations, we obtain −4 (b) 2m and 2.12 ∆x −16 2.16 (a) If Ψ1 ( x , t ) and Ψ2 ( x , t ) are solutions to Schrodinger’s wave equation, then −28 (a) ∆p = g⇒ p = hc h b (1) 1.6 x10 −19 ∆t = 6.6 x10 F I = pc H hK λ So ∆E = c( ∆p) = b3x10 gb1.054 x10 g ⇒ E= −34 or (b) hc 1.054 x10 ∂ ∂t aΨ ⋅ Ψ f which can be written as h ∆x = 1.054 x10 10 p = mv ⇒ ∆v = = 1.054 x10 −2 ∆p m LMΨ ∂ Ψ + Ψ ∂ Ψ + ∂Ψ ⋅ ∂Ψ OP 2m N ∂x ∂x Q ∂x ∂x LM ∂Ψ + Ψ ∂Ψ OP +V ( x )Ψ ⋅ Ψ = jh Ψ N ∂t ∂t Q Dividing by Ψ ⋅ Ψ we find −h L ∂ Ψ ∂ Ψ ∂Ψ ∂Ψ O ⋅ + ⋅ + M P m N Ψ ∂x Ψ ∂x Ψ Ψ ∂x ∂x Q L ∂Ψ + ∂Ψ OP +V ( x ) = jh M N Ψ ∂t Ψ ∂x Q −34 = 1.054 x10 −h −32 ⇒ −36 2 2 m/s 2 h ∆x = 1.054 x10 10 −10 ∆p = 1.054 x10 −34 ⇒ 1 2 1 2 −24 2 2 2.15 1 (a) ∆p = 2 or ∆v = x10 −32 1500 kg − m / s 11 1 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 2.19 Since Ψ1 is a solution, then −h ⋅ ∂ Ψ1 ⋅ + V ( x ) = jh ⋅ ⋅ Note that L P= z M N F −xIO expG J P dx H a KQ a F −2 x IJdx = expG z Ha K a F −a I F −2 x IJ = expG Ha K a H K or L F −2a IJ − 1OP = − expF −1I P = −1MexpG H 2K N H 4a K Q ao 2 1 ∂Ψ2 ∂ Ψ2 2 Ψ2 ∂t + V ( x ) = jh −h ao ∂Ψ2 o z Ψ( x , t ) dx = = A z +1 −1 A o F P= z G H F −xII expG J J dx H a KK a F −2 x IJdx = expG z Ha K a F −a I F −2 x IJ = expG Ha K a H K or LM F −1I OP P = −1 exp( −1) − exp H KQ N ao o ao o +1 =1 −1 z z F P = zG H A ⋅ sin ( nπx )dx LM x − sin(2nπx)OP N 4nπ Q ao o o ao ao o +1 o or =1 o P = −1 exp( −2 ) − which yields P = 0.865 which yields A = or A=+ ,− ao F −xII expG J J dx H a KK a F −2 x IJdx = F −a I expFG −2 x IJ = expG z Ha K aH2K Ha K a or o which yields P = 0.239 (c) 2.18 Ψ( x , t ) = A sin(nπx ) exp( − jωt ) +1 o ao 2 o o A = or A = +1 , − , + j , − j o ao sin (πx )dx LM x − sin(2πx)OP ⋅ N 4π Q Ψ( x , t ) dx = = A 2 ao which yields +1 which yields P = 0.393 (b) −1 or o o 2.17 Ψ( x , t ) = A sin(πx ) exp( − jωt ) +1 o o ∂Ψ ∂Ψ ⋅ ⋅ ⋅ − V ( x) = m Ψ1 Ψ2 ∂x ∂x This equation is not necessarily valid, which means that Ψ1 Ψ2 is, in general, not a solution to Schrodinger’s wave equation o o m Ψ2 ∂x Ψ2 ∂t Subtracting these last two equations, we obtain o ao Since Ψ2 is also a solution, we may write −h 2 = jh * 2 Ψ ⋅ Ψ dx = Function has been normalized (a) Now LM ∂ Ψ + ∂Ψ ∂Ψ OP m N Ψ ∂x Ψ Ψ ∂x ∂ x Q −h z ∞ ∂Ψ1 m Ψ1 ∂x Ψ1 ∂t Subtracting these last two equations, we are left with Chapter Problem Solutions 2,+ j 2,− j 12 o o Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 2.20 (a) kx − ωt = constant Then ω dx dx = vp = + k −ω = 0⇒ dt dt k or vp = 15 x10 2.22 15 x10 hnπ E= ma b1.054 x10 g π n = 2b9.11x10 gb100 x10 g −34 2 2 −31 −22 E = 6.018 x10 n = 10 m / s −10 (J) or −3 E = 3.76 x10 n Then v p = 10 cm / s (b) (eV ) −3 k= 2π λ ⇒λ = 2π = k n = ⇒ E1 = 3.76 x10 eV 2π x10 15 −2 n = ⇒ E = 1.50 x10 eV or −2 λ = 41.9 A n = ⇒ E = 3.38 x10 eV ° Also 2.23 p= or h = λ 6.625 x10 41.9 x10 −34 −10 hnπ ⇒ (a) E = 2 ma 2 b1.054 x10 g π n = 2b9.11x10 gb12 x10 g −34 x10 p = 158 −25 kg − m / s 2 −31 Now E = hν = or hc λ E = 4.74 x10 b6.625x10 gb3x10 g = −34 41.9 x10 −17 −20 = 4.81x10 n −10 (J) So −10 J ⇒ E = 2.96 x10 eV E1 = 4.18 x10 −20 J ⇒ E1 = 0.261 eV E = 1.67 x10 −19 J ⇒ E = 1.04 eV (b) 2.21 b ψ ( x ) = A exp − j kx + ωt g E − E = hν = where or = λ= h b 9.11x10 −31 g(0.015)b1.6x10 g ω= E h = λ ∆E b6.625x10 gb3x10 g ⇒ 1.67 x10 −19 − 4.18 x10 −20 −6 or −34 λ = 1.59 µm or k = 6.27 x10 m hc ⇒λ= λ = 159 x10 m −19 1.054 x10 Now hc −34 mE k= or so 13 Chapter Problem Solutions −1 b (0.015) 1.6 x10 −19 1.054 x10 2.24 (a) For the infinite potential well g hnπ E= −34 so ω = 2.28 x10 rad / s 13 n = or 13 2 ma 2 2 ma E ⇒n = hπ 2 b gb10 g b10 g = 182 x10 b1.054 x10 g π 10 −5 −2 −34 −2 2 56 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual n = 1.35 x10 28 ∆E = hπ ma hπ = or 2 ma (2n + 1) b1.054 x10 g π (2)b1.35x10 g ∆E = 2b10 gb10 g −34 28 −5 ∆E = 1.48 x10 −30 where K = J −30 Energy in the (n+1) state is 1.48 x10 Joules larger than 10 mJ (c) Quantum effects would not be observable where K = E1 = ma b1.054 x10 gπ = 2b1.66 x10 gb10 g −34 2 −27 where K = −14 E1 = 2.06 x10 eV where K = For an electron in the same potential well: −34 −31 2 2π −14 a 3π ∂x 2.26 Schrodinger’s wave equation 2m ∂x h We know that −a V ( x ) = for a 2 + mE and x ≤ ≤x≤ −a a 2 ma 16π h ma ∂ ψ ( x , y, z) + ∂ X 2 YZ +a 2 ∂y ∂ ψ ( x , y , z) + ∂z 2 mE ∂x ∂Y + XZ ∂y ∂ Z + XY ∂z + mE h Dividing by XYZ and letting k = obtain ψ ( x) = XYZ = mE h , we ∂ X ∂Y ∂ Z ⋅ + ⋅ + ⋅ +k =0 X ∂x Y ∂y Z ∂z We may set ∂x h Solution is of the form ψ ( x ) = A cos Kx + B sin Kx 2 ψ ( x , y , z) = h Use separation of variables, so let ψ ( x , y , z ) = X ( x )Y ( y )Z ( z ) Substituting into the wave equation, we get so in this region ∂ ψ ( x) 9π h + ( E − V ( x ))ψ ( x ) = ψ ( x ) = for x ≥ 2 so E = + ma so E = 4π ∂ ψ ( x , y , z) 2 2.27 The 3-D wave equation in cartesian coordinates, for V(x,y,z) = E1 = 3.76 x10 eV 4π h or ∂ ψ ( x) 2 so E = a Fourth mode: ψ ( x ) = B sin Kx or b1.054 x10 g π E = 2b9.11x10 gb10 g π h Third mode: ψ ( x ) = A cos Kx 2.25 For a neutron and n = 1: hπ π so E1 = a ma Second mode: ψ ( x ) = B sin Kx −2 h Boundary conditions: +a −a ,x= ψ ( x ) = at x = 2 So, first mode: ψ ( x ) = A cos Kx (n + 1)2 − n 2 mE where K = (b) Chapter Problem Solutions (1) 14 2 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Exercise Solutions (b) fF = + exp F 0.30 + 0.0259 I H 0.0259 K f F = 3.43 x10 E3.6 ⇒ F 400 I = 0.03453 H 300 K kT = ( 0.0259 ) (a) −6 fF = + exp E3.5 (a) + exp FE−E I H kT K + exp F F F E3.7 kT = 0.03453 eV − fF = so −6 ⇒ 0.35 I F + exp H 0.03453K −5 (b) F 0.35 + 0.0259 I + exp H 0.0259 K − fF = or − f F = 4.98 x10 1 − f F = 3.96 x10 (b) − fF = −5 (a) F 0.35 I + exp H 0.0259 K − f F = 1.35 x10 F 0.30 + 0.03453I ⇒ H 0.03453 K f F = 6.20 x10 Then − fF = f F = 1.69 x10 fF = F FE−E I H kT K = = F E − E I + expF E − E I + exp H kT K H kT K exp ⇒ (b) 1 − fF = − F I H 0.03453K 0.30 ⇒ 0.35 + 0.03453 I F + exp H 0.03453 K − f F = 1.46 x10 −7 22 −5 −4 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Exercise Solutions Chapter Exercise Solutions Now E4.1 ni = 2.8 x10 b no = 2.8 x10 exp 19 F −0.22 I H 0.0259 K 15 ni = 5.65 x10 ni = 2.38 x10 cm 12 E F − E v = 112 − 0.22 = 0.90 eV po = 1.04 x10 exp 19 E4.4 (a) 200K F −0.90 I H 0.0259 K b ni = 4.7 x10 or po = 8.43 x10 cm 24 Then −3 Now So gb I expF −112 I gFH 400 H 0.03453K 300 K 19 1.04 x10 or or no = 5.73 x10 cm 19 17 gb −3 I expF −1.42 I gFH 200 H 0.01727 K 300 K x10 18 or −3 ni = 1.904 Then E4.2 po = 7.0 x10 exp 18 ni = 1.38 cm F −0.30 I H 0.0259 K (b) 400K po = 6.53 x10 cm 13 b −3 17 ni = 1.08 x10 eV E c − E F = 1.42 − 0.30 = 112 So no = 4.7 x10 exp 17 ni = 3.28 x10 cm −112 E4.5 (a) 200K or −3 2 18 20 Then ni = 2.16 x10 cm 10 −3 (b) 400K 19 F 400 I expF −0.66 I = b1.04 x10 gb6 x10 g H 300 K H 0.03453K i ni or I expF −0.66 I gb6x10 gFH 200 H 0.01727 K 300 K 19 ni = 4.67 x10 ni = 5.90 x10 −3 or F 200 I = 0.01727 For 200K : kT = ( 0.0259 ) H 300 K Now F 200 I expF −112 I n = b2.8 x10 gb1.04 x10 g H 300 K H 0.01727 K 19 b ni = 1.04 x10 E4.3 (a) 19 Then F I H 0.0259 K no = 0.0779 cm 18 or Now 19 18 or ni = 7.39 x10 Then ni = 7.68 x10 cm (b) I expF −1.42 I gb7 x10 gFH 400 H 0.03453K 300 K ni = 4.7 x10 or −3 29 Then −3 ni = 8.6 x10 cm 14 F 400 I = 0.03453 H 300 K For 400K : kT = ( 0.0259 ) 35 −3 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual E4.6 E Fi − E midgap = E4.11 FG m IJ Hm K po = N a − N d = x10 − x10 16 * kT ln Chapter Exercise Solutions p 15 or * n F 0.067 I = (0.0259 ) ln H 0.48 K po = 1.5 x10 cm 16 Then or no = E F − E midgap = −38.2 meV ni po −3 b1.8x10 g = 16 15 x10 or E4.7 rn F m I = (1)(131 )F I Hm K H 0.067 K = n ∈r ao −4 no = 2.16 x10 cm o * E4.12 (b) so r1 ao n= = 195.5 Nd + FN I H2K Then E4.8 11 x10 = x10 + a f Then π a f N C F1 η F = π ni = 11x10 and b2.8x10 g(0.60) 19 po + pa NV exp a By trial and error v a 28 19 T ≅ 552 K E F − E v = (0.0259 ) ln 17 = 0.130 eV or po + pa g E4.13 19 pa i 19 LM −a E − E f OP 4N N kT Q = 1.04 x10 LM −0.045 OP exp 1+ 4b10 g N 0.0259 Q 1+ 14 −3 = b5x10 g + n 28 ni2 = N C NV exp E4.9 pa LM − E OP = 11x10 N kT Q FTI = b2.8 x10 gb1.04 x10 g H 300K L −112 OP × exp M N (0.0259)aT 300f Q 19 or no = 1.9 x10 cm 14 + ni which yields For η F = , F η F = 0.60 no = d 15 −3 E4.14 = 0.179 E F − E Fi = ( 0.0259 ) ln or E4.10 Computer plot LM x10 OP N 5x10 − x10 Q 18 16 LM1.7 x10 OP N 15 x10 Q E F − E Fi = 0.421 eV 36 15 17 10 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Exercise Solutions Chapter Exercise Solutions a σ ≅ eµ n N d − N a (b) b = 1.6 x10 E5.1 no = 10 − 10 = x10 cm 15 so po = Now ni no 14 14 b15 x10 g = 10 x10 b ρ= = 2.5 x10 cm −3 b −19 g g(1350)b9 x10 g(35) σ = 4.8 ⇒ ρ = 0.208 Ω − cm σ = eµ n N d = 14 so ρ b1.6x10 gµ N or J drf = 6.80 A / cm −1 E5.6 J drf = e µ n no + µ p po Ε ≈ eµ n no Ε = 1.6 x10 16 σ = 4.8 (Ω − cm) −3 14 f g(1000)b3x10 g ⇒ −19 −19 n d = = 10 0.1 Then µ n N d = 6.25x10 E5.2 J drf ≅ eµ p po Ε Then b 120 = 1.6 x10 Using Figure 5.4a, −19 N d ≅ x10 cm 16 −3 Then g (480) po (20) µ n ≈ 695 cm / V − s so −3 po = 7.81x10 cm = N a 16 19 E5.7 (a) R = E5.3 Use Figure 5.2 (a) V I = = 2.5 kΩ (b) R = 2.5 x10 = 10 ρ = 2.08 Ω − cm (i) µ n ≅ 500 cm / V − s , (ii) ≅ 1500 cm / V − s (b) b ρ 1.2 x10 −3 −6 g⇒ (i) µ p ≅ 380 cm / V − s ,(ii) ≅ 200 cm / V − s (c) From Figure 5.4a, N a ≅ x10 cm E5.4 E5.8 FG 10 IJ expFG − x IJ H 10 K H L K 2 Use Figure 5.3 [Units of cm / V − s ] J diff = eDn −3 (a) For N I = 10 cm ; µ n ≅ 1350 , µ p ≅ 480 : 15 −3 (b) N I = 15 x10 cm ; µ n ≅ 700 , µ p ≅ 300 : 17 −4 n −4 Dn = 25 cm / s , Ln = 10 cm = µm Then −x J diff = −40 exp A / cm −3 F I H K −3 (d) N I = x10 cm ; µ n ≅ 4500 , µ p ≅ 220 17 dx (a) x = ; J diff = −40 A / cm E5.5 (a) For −3 (b) x = µm ; J diff = −14.7 A / cm N I = x10 cm ; µ n ≅ 1000 cm / V − s , 16 −3 15 = − eDn (c) N I = 11 x10 cm ; µ n ≅ 800 , µ p ≅ 310 : 17 dn 15 (c) x = ∞ ; J diff = µ p ≅ 350 cm / V − s 51 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual E5.10 At x = , E5.9 J diff = − eD p so b 20 = − 1.6 x10 dp FG x10 IJ dx H −L K Then F x10 IJ 6.4 = b1.6 x10 g(10)G H L K dx −19 J diff = − eD p g(10) (0 − 0.010) 15 p 17 Which yields or 17 15 = − eD p −19 ∆p = 1.25x10 = x10 − p p( x = 0.01) = 2.75 x10 cm dp p ∆p Then 17 Chapter Exercise Solutions −4 LP = 5x10 cm −3 52 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter Exercise Solutions Chapter Exercise Solutions E6.1 δn(t ) = x10 FG −t IJ Hτ K F −t IJ expG H µs K 14 no δn(t ) = 10 15 (a) t = ; δn = 10 cm 15 LM1 − expFG t IJ OP N H µs K Q (c) As t → ∞ , δn( ∞ ) = x10 cm δn(t ) = δn(0) exp so 14 −3 E6.5 FG − x IJ HL K L = D τ = (10)b10 g = 31.6 µm Then F − x IJ cm δn( x ) = δp( x ) = 10 expG H 31.6 µmK δn( x ) = δp( x ) = δn(0) exp p −3 −6 P (b) t = µs ; δn = 3.68 x10 cm 14 (c) t = µs ; δn = 183 x10 cm 13 −3 −3 P po 15 −3 E6.2 R= δn E6.6 n-type ⇒ Minority carrier = hole dp d (δp( x )) = − eD p J diff = − eD p dx dx τ no Then 15 (a) R = (b) R = (c) R = 10 10 −3 ⇒ R = 10 cm s 21 −6 3.68 x10 10 J diff = −3 ⇒ R = 3.68 x10 cm s 20 −1 b g(10)b10 g expF −10 I H 31.6K −b316 x10 g − 1.6 x10 14 −6 −19 15 −3 or J diff = +0.369 A / cm Hole diffusion current 1.83 x10 10 −1 13 −6 ⇒ −3 R = 1.83x10 cm s 19 −1 J diff ( electrons) = − J diff (holes) so E6.3 (a) p-type ⇒ Minority carrier = electrons F −t IJ (b) δn(t ) = δn(0) expG Hτ K Then F −t IJ δn(t ) = 10 expG H µs K J diff = −0.369 A / cm E6.7 δp = cm −3 b exp − t τ po g b4πD t g expa −1 5f ⇒ δp = 73.0 (4π )(10)b10 g expa −5 5f ⇒ δp = 14.7 (4π )(10)b5x10 g expa −15 5f ⇒ δp = 115 (4π )(10)b15x10 g expa −25 5f ⇒ δp = 0.120 (4π )(10)b25x10 g 1/ p (a) E6.4 (a) p-type ⇒ Minority carrier = electrons L F −t IJ OP (b) δn(t ) = g ′τ M1 − expG N Hτ KQ or L F −t IJ OP δn(t ) = b10 gb5x10 gM1 − expG N H µs K Q (b) no no 20 Electron diffusion current no 15 (c) −6 (d) Then 63 −6 1/ 1/ −6 −6 1/ −6 1/ Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual x = µ p Ε o t = (386)(10)t (ii) x − µ p Ε o t (a) x = 38.6 µm ; (b) x = 193 µm −2 Then g ⋅ expLM −b x − µ Ε t g OP δp = b4πD t g MN D t PQ (a) (i) x − µ Ε t = 1.093x10 − (386)(10)b10 g = 7.07 x10 L −b7.07 x10 g OP expa −1 5f δp = ⋅ exp M (4π )(10)b10 g MN 4(10)b10 g PQ or L −b7.07 x10 g OP δp = 73.0 exp M MN 4(10)b10 g PQ b exp − t τ po b = 5.08 x10 − (386)(10) 15x10 (b) x = 579 µm ; (d) x = 965µm E6.8 Chapter Exercise Solutions p −6 g = −7.1x10 −3 δp = 1.05 E6.9 Computer Plot o 1/ p E6.10 p p (a) E F − E Fi = ( 0.0259 ) ln o −2 −6 −3 −6 −6 1/ (ii) E Fi − E Fp = (0.0259 ) ln b g = −7.07 x10 L −b−7.07 x10 g OP δp = 73.0 exp M MN 4(10)b10 g PQ −6 14 10 E6.11 (a) p-type −3 E Fi − E F = (0.0259 ) ln −6 (b) (i) x − µ p Ε o t (b) E Fn − E Fi = (0.0259 ) ln b = 2.64 x10 − (386)(10) 5x10 −2 FG 6x10 − 10 IJ H 1.5x10 K −6 L −b7.1x10 g OP δp = 14.7 exp M MN 4(10)b5x10 g PQ −3 g = 7.1x10 15 15 10 E Fi − E F = 0.3294 eV δp = 20.9 −3 FG x10 IJ ⇒ H 15 x10 K 14 10 E Fn − E Fi = 0.2460 eV E Fi − E Fp = (0.0259 ) ln −6 FG 5x10 + x10 IJ H 1.5x10 K 15 14 10 E Fi − E Fp = 0.3304 eV δp = 11.4 x − µ pΕ ot E6.12 b = 1.22 x10 − (386)(10) x10 −2 Then 10 FG 5x10 IJ ⇒ H 15 x10 K −3 (ii) 14 x − µ pΕ ot −3 Then 16 E Fi − E Fp = 0.2697 eV δp = 20.9 −3.21x10 − (386)(10) 10 or FG 10 + 5x10 IJ ⇒ H 1.5x10 K E Fn − E Fi = 0.3486 eV −6 Then 10 (b) E Fn − E Fi = (0.0259 ) ln −3 16 E F − E Fi = 0.3473 eV −3 FG 10 IJ ⇒ H 15 x10 K −6 g = −7.1x10 15 R= b = 6.50 x10 − (386)(10) 15x10 −2 −6 LM −b7.1x10 g OP MN 4(10)b15x10 g PQ −3 δp = 115 exp −3 an + δnfa p + δpf − n an + δn + n f + τ a p + δp + n f δp = 11.4 (c) (i) x − µ p Ε o t −3 n-type; no = 10 cm ; po = 2.25 x10 cm −3 g = 7.1x10 τ po o o i i no Then −3 −3 R = 1.83x10 cm s 20 −6 o δp = 1.05 64 −1 o i Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 11 Exercise Solutions Chapter 11 Exercise Solutions E11.5 φ fp = 0.376 V E11.1 φ ms = (0.555 − 0.376) ⇒ φ ms = +0.179 V F 3x10 IJ = 0.376 V (a) φ = (0.0259 ) lnG H 15 x10 K R 4(11.7)b8.85x10 g(0.376) UV x =S T b1.6x10 gb3x10 g W 16 fp 10 1/ −14 dT −19 E11.6 From E11.3, φ ms = −0.981 V Cox = 16 or FG 10 IJ = 0.288 V H 15 x10 K R 4(11.7)b8.85x10 g(0.288) UV =S T b1.6x10 gb10 g W or VFB = φ ms − 15 10 −19 or Cox −7 F / cm b1.6x10 gb8x10 g = −0.981 − −19 1.73 x10 10 −7 E11.7 From E11.4, φ ms = −0.931 V b1.6x10 gb8x10 g = −0.931 − −19 FG 8x10 IJ = 0.342 V H 15 x10 K R 4(11.7)b8.85x10 g(0.342) UV =S T b1.6x10 gb8x10 g W 1.73 x10 −7 VFB = −1.01 V 10 −19 10 or 15 1/ −14 x dT Qss′ g = 1.73x10 15 VFB φ fn = (0.0259) ln 200 x10 −8 VFB = −1.06 V x dT = 0.863 µm E11.2 b (3.9) 8.85x10 −14 or 1/ −14 x dT t ox = Then x dT = 0.180 µm (b) φ fp = (0.0259 ) ln ∈ox E11.8 From E11.5, φ ms = +0.179 V 15 b1.6x10 gb8x10 g = +0.179 − −19 VFB x dT = 0.333 µm 10 1.73 x10 −7 or E11.3 FG 3x10 IJ = 0.376 V H 15 x10 K F E + φ IJ = φ′ − G χ′ + H 2e K 16 φ fp = (0.0259) ln φ ms VFB = +0.105 V E11.9 From E11.3, φ ms = −0.981 V and φ fp = 0.376 V 10 R 4(11.7)b8.85x10 g(0.376) UV = 0.18 µm x =S T b1.6x10 gb3x10 g W Now Q ′ ( max ) = b1.6 x10 gb3 x10 gb 0.18 x10 g g m −14 fp dT = 3.20 − (3.25 + 0.555 + 0.376) or −19 16 −19 φ ms = −0.981 V 16 SD or ′ ( max ) = 8.64 x10 C / cm QSD −8 E11.4 φ fp = 0.376 V From Equation [11.27b] φ ms = −(0.555 + 0.376) ⇒ φ ms = −0.931 V 163 −4 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual b gb1.6x10 g F 250x10 I − 0.981 + 2(0.376) ×G H (3.9)b8.85x10 gJK −8 ′ Cmin −19 VTN = 8.64 x10 − 10 11 Cox −8 Chapter 11 Exercise Solutions FG ∈ IJ FG x IJ H ∈ KH t K = 1+ −14 VTN = +0.281 V FG 10 IJ = 0.288 V H 15 x10 K R 4(11.7)b8.85x10 g(0.288) UV = 0.863 µm x =S T b1.6x10 gb10 g W Then Q ′ ( max) = b1.6 x10 gb10 gb 0.863 x10 g dT ′ Cmin 15 Also ′ C FB 1/ 15 −19 Cox 1+ FG ∈ IJ FG IJ F kT I FG ∈ IJ H ∈ K H t K H e K H eN K ox s s ox a SD or ′ ( max ) = 1.38 x10 C / cm QSD −8 Also b = −8 −8 F 3.9 I F I 1+ H 11.7 K H 220 x10 K b g gb3x10 g (0.0259)(11.7) 8.85x10 −14 b1.6x10 −8 gb1.6x10 g = 1.28x10 C / cm Now, from Equation [11.28] V = b −1.38 x10 − 1.28 x10 g F 220x10 I + 0.97 − 2(0.288) ×G H (3.9)b8.85x10 gJK Qss′ = x10 −19 10 IJ ⇒ K = −4 15 ox = 0.294 Cox 10 −19 s F I FG H KH E11.10 From Figure 11.15, φ ms = +0.97 V −14 dT From E11.9, x dT = 0.18 µm Then ′ Cmin = −4 3.9 0.18 x10 Cox 1+ −8 11.7 250 x10 or φ fn = (0.0259) ln ox −19 16 or ′ C FB −8 Cox TP = 0.736 −8 E11.13 −14 or Cox = VTP = +0.224 V ∈ox t ox = b (3.9) 8.85x10 −14 200 x10 −7 −3 By trial and error, let N d = x10 cm , then φ fn = 0.383 , φ ms ≅ 1.07 , ′ ( max ) = 1x10 QSD −7 and F I a f H K F 50I (650)b1.73x10 gaV = H 2K = b2.81x10 gaV − 0.4 f GS or specified ID −3 GS Then E11.12 Cox −7 VTP = −0.405 V which is between the limits ′ Cmin g⇒ Cox = 1.73 x10 F / cm Now W µ n Cox VGS − VTN ID = L E11.11 16 −8 = a ∈ox f t ox + ∈ox ∈s xdT = ∈ox t ox VGS = V ⇒ I D = 1.01 mA VGS = V ⇒ I D = 7.19 mA t ox t ox + FG ∈ IJ x H∈ K VGS = V ⇒ I D = 19 mA ox dT s or 164 − 0.4 f Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual E11.14 F I µ C aV − V f 2H LK Now F W I (650)b1.73x10 g (1.75 − 0.4) 100 x10 = H LK which yields F W I = 0.976 H LK W ID = (b) φ fp = (0.0259 ) ln n ox GS TN ∈ox Cox = t ox = b (3.9) 8.85x10 −14 220 x10 −8 g = 1.57 x10 F 60I (310)b157 x10 gaV H 2K I = 1.46 x10 aV − 0.4 f or SG ∆V = (0.333) F / cm or D ∆V = (0.333) or − 0.4 f Cox = 1.73 x10 V / cm (a) γ = b 1.6 x10 −14 1.73 x10 γ = 0.105 V (i) F W I F 310I b157 x10 g(1.25 − 0.4) H L KH K −7 ∆V = (0.105) or F I = 11.4 H LK Cox = t ox = or b 200 x10 −8 g = 1.73x10 2(0.288) + − 2( 0.288) + − F / cm Cox = 1.73 x10 V / cm −19 gm = n ox GS T −7 g(11.7)b8.85x10 gb10 g −14 1.73 x10 16 g m = 2.91 mA / V 1/ −7 or γ = 0.333 V 2(0.288) F W I µ C aV − V f H LK = (20)(400)b1.73x10 g(2.5 − 0.4) or Cox 2( 0.288) ∆V = 0.0888 V −7 −7 2e ∈s N a b 10 E11.19 Now 1.6 x10 15 ∆V = 0.052 V ∆V = (0.105) (3.9) 8.85x10 FG 10 IJ = 0.288 V H 15 x10 K (ii) which yields W ∈ox −7 1/ VSG = V ⇒ I D = 3.74 mA −14 15 or (b) φ fp = (0.0259 ) ln E11.17 (a) g(11.7)b8.85x10 gb10 g −19 VSG = 1.5 V ⇒ I D = 1.77 mA −6 2(0.347 ) E11.18 SG 200 x10 = = 2( 0.347 ) + − ∆V = 0.269 V VSG = V ⇒ I D = 0.526 mA γ = (2)(0.347 ) + − 2(0.347) ∆V = 0.156 V Then E11.16 10 (ii) −3 16 −7 −7 ID = −7 FG 10 IJ = 0.347 V H 15 x10 K (i) −7 −6 E11.15 Chapter 11 Exercise Solutions 1/ 165 1/ Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Now CM C gdT = + g m RL = + (2.91)(100) or CM C gdT E11.20 fT = = = 292 a µ n VGS − VT f 2πL (400)(2.5 − 0.4) b 2π 0.5 x10 −4 g or f T = 53.5 GHz 166 Chapter 11 Exercise Solutions Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 12 Exercise Solutions Chapter 12 Exercise Solutions E12.1 I D1 I D2 Now −14 FG V IJ H V K = expFG V = F V IJ H expG HV K F I IJ − V = V lnG HI K GS exp − VGS GS t L 2(11.7)b8.85x10 g OP ∆L = M N b1.6x10 gb2 x10 g Q Vt GS −19 IJ K × or VGS 2(0.365) + 2.5 − 0.20 = t D2 Then VGS − VGS = (0.0259 ) ln(10) ⇒ VGS − VGS = 59.64 mV 0.1867 ⇒ L L = 0.934 µm E12.4 f 2L (1000)b10 gb10 g = aV 2b10 g or I ( sat ) = 0.50 x10 aV − 0.4 f or I ( sat ) = 50aV − 0.4 f µA (b) I ( sat ) = WC v aV − V f = b10 gb10 gb5x10 gaV or I ( sat ) = x10 aV − 0.4 f or I ( sat ) = 50aV − 0.4f µA (a) I D ( sat ) = E12.2 φ fp = ( 0.0259) ln FG x10 IJ = 0.365 V H 1.5x10 K 10 −19 or GS − VT −3 GS ID = L − ∆L I D′ ID E12.3 I D′ ID = = 1 − 0.1188 GS D 2( 0.365) + 0.60 GS D ox sat −3 GS −8 T GS L f 16 ∆L = 0.1188 µm I D′ − 0.4 −4 D 1/ 2(0.365) + 2.5 − × aV −4 VDS ( sat ) = VGS − VT = − 0.4 = 0.60 V Now L 2(11.7)b8.85x10 g OP ∆L = M N b1.6x10 gb2 x10 g Q µ n CoxW −8 16 −14 2( 0.365) + 0.40 Then D1 GS 16 ∆L = 0.1867 µm t or 1/ = 0.4 f −5 D ⇒ GS D GS = 1135 E12.5 L → kL = (0.7 )(1) ⇒ L = 0.7 µm = 1.25 ⇒ ∆L I F 1− H LK ∆L L = W → kW = (0.7 )(10) ⇒ W = µm 1.25 − t ox → kt ox = (0.7 )(250) ⇒ t ox = 175 A 1.25 Na → Na x10 15 ⇒ N a = 7.14 x10 cm 0.7 k VD → kVD = (0.7 )(3) ⇒ VD = 2.1 V or ∆L = 0.20 L VDS ( sat ) = 0.80 − 0.40 = 0.40 V 183 = ° 15 −3 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual E12.6 Cox = ∈ox = t ox b (3.9) 8.85x10 250 x10 −14 −8 g = 1.38x10 or or −7 +0.4 = +0.959 + ∆V which yields ∆V = −0.559 V 15 Implant Donors for negative shift Now ∆VT Cox eDI ∆VT = ⇒ DI = Cox e so 10 1/ −14 x dT VT = VTO + ∆V FG 3x10 IJ = 0.316 V H 15 x10 K R 4(11.7)b8.85x10 g(0.316) UV =S T b1.6x10 gb3x10 g W φ fp = (0.0259) ln −19 15 −4 x dT = 0.522 x10 cm Now ∆VT = b − 1.6 x10 DI = gb3x10 gb0.522 x10 g 1.38 x10 R 0.3 LM + 2(0.522) − 1OPUV ×S T 0.8 N 0.3 QW −19 −4 15 VT = VTO + ∆V or ∆V = VT − VTO = −0.4 − 0.959 ⇒ ∆V = −1.359 V Implant donors for a negative shift Now eDI ∆V Cox ∆V = ⇒ DI = Cox e so 15 10 1/ −14 −19 15 x dT = 0.863 µm b ′ ( max) = 1.6 x10 QSD −19 DI = gb10 gb0.863x10 g −4 15 ′ ( max ) = 1.38 x10 QSD b −19 12 −8 −8 −9 10 −9 TN −8 −14 +0.35 + 2(0.288) or VTN = +0.959 V b (3.9 ) 8.85x10 −14 200 x10 −8 g = 1.73x10 −7 F / cm 1.6 x10 −19 g⇒ DI = 1.47 x10 cm gb5x10 g = 8x10 Then b1.38x10 − 8x10 gb200x10 g V = (3.9)b8.85x10 g Qss′ = 1.6 x10 b (1.359) 1.73x10 −7 or or Cox = −2 E12.8 Using the results of E12.7 VTO = +0.959 V F 10 IJ = 0.288 V = (0.0259 ) lnG H 15 x10 K R 4(11.7)b8.85x10 g(0.288) UV =S T b1.6x10 gb10 g W We find g⇒ 11 E12.7 φ ms ≅ +0.35 or 1.6 x10 −19 DI = 6.03 x10 cm ∆VT = −0.076 V x dT b (0.559) 1.73x10 −7 −7 or φ fp Chapter 12 Exercise Solutions Now 184 −2 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Chapter 15 Exercise Solutions Chapter 15 F I ⇒ I (max) = 1.33 A H 3K At the maximum power point, F 2I R 15 = 30 − H 3K Exercise Solutions I C ( max ) = E15.1 (a) Collector Region R ∈ aV + V f FG N IJ FG IJ UV x =S T e H N KH N + N K W Neglecting V compared to V ; R 2(11.7)b8.85x10 g(200) x =S T 1.6x10 F 10 IJ F I UV ×G H 10 K H 10 + 10 K W 1/ s bi R L a n d bi a which yields R L = 22.5 Ω d R (b) VCC = 15 V ⇒ I C ( max) = A −14 n VCE = VCC − I C RL We have = 15 − (2 ) RL ⇒ R L = 7.5 Ω Maximum power at the center of the load line, or at VCE = 7.5 V , I C = A Then P( max) = (1)(7.5) ⇒ P( max) = 7.5 W −19 1/ 16 14 16 14 or x n = 50.6 µm (b) Base Region RS 2(11.7)b8.85x10 g(200) T 1.6x10 F 10 IJ F ×G H 10 K H 10 + 10 C −14 xp = E15.3 VCE = VCC − I C RE ⇒ VCE = 20 − I C (0.2 ) so = 20 − I C ( max )(0.2 ) ⇒ I C ( max) = 100 mA −19 16 16 I UV KW 1/ 14 14 or x p = 0.506 µm E15.2 Maximum power at the center of the load line, or P( max) = ( 0.05)(10) ⇒ P( max) = 0.5 W (a) VCC = 30 V , VCE = 30 − I C RC Now, maximum power PT = 10 W , PT = 10 = I CVCE Maximum power at VCE = VCC = Then, maximum power at I C = 10 VCE E15.4 30 = For VDS = , I D ( max ) = = 15 V 10 15 = I D ( max ) = 1.2 A 24 20 ⇒ For I D = , VDS ( max ) = VDD ⇒ A VDS ( max) = 24 V Then, 223 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Maximum power at the center of the load line, or at I D = 0.6 A , VDS = 12 V Chapter 15 Exercise Solutions E15.6 θ dev − case = Then P( max) = (0.6)(12 ) ⇒ P( max) = 7.2 W Tj ,max − Tamb PD ,rated PD ( max) = E15.5 Power = I DVDS = (1)(12 ) = 12 W (c) Heat sink: Tsnk = Tamb + P ⋅ θ snk − amb or Tsnk = 25 + (12 )(4 ) ⇒ Tsnk = 73° C = = 200 − 25 50 = 3.5 ° C / W Tj ,max − Tamb θ dev − case + θ case − snk + θ snk − amb 200 − 25 ⇒ 3.5 + 0.5 + PD ( max ) = 29.2 W Now Tcase = Tamb + PD ( max ) θ case − snk + θ snk − amb = 25 + (29.2 )(0.5 + ) ⇒ (b) Case: Tcase = Tsnk + P ⋅ θ case − snk or Tcase = 73 + (12 )(1) ⇒ Tcase = 85° C Tcase = 98° C (a) Device: Tdev = Tcase + P ⋅ θ dev − case or Tdev = 85 + (12 )(3) ⇒ Tdev = 121° C 224 ... 0.0544 2 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual b g k = 0.1 A * −1 ⇒ 10 m For Curve A: d E −1 x10 g g b10 g b1.054 2m b −34 dk which yields m −31 For Curve... Z 2 ⋅ = − k y and ⋅ = −kz Z ∂z Y ∂y From the boundary conditions, we find 28 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual Now k y a = n y π and k z a = nz... X = A sin k x x nx π and X ( x = a ) = ⇒ k x = Chapter Problem Solutions 3h π 2 ma 2 ∆E = 15 2 Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual a=4 A (i) ∆E =