Download PDF solution manual for microelectronic circuit design 5th edition by jaege

CHAPTER Solution Manual for Microelectronic Circuit Design 5th Edition by Jaege 2.1 Based upon Table 2.1, a resistivity of 2.82 -cm < m-cm, and aluminum is a conductor 2.2 Based upon Table 2.1, a resistivity of 10 15 -cm > 10 -cm, and silicon dioxide is an insulator 2.3 2.4 1.8 cm  2.82 x10 6  cm   144  4 4 A 5x10 cm 1x10 cm L a R   L b R    2.82x10  cm 6 A 1.8 cm 5x10 4  287  4 cm0.5x10 cm 2.5 a R  L  1.66x106  cm 1.8 cm A  b R  L  1.66 x106  cm 5x10   5x10 A 4  94.5  cm1x10 4 cm 1.8 cm 169  cm0.5x10 4 cm 2.6 2-1 ©R C Jaeger & T N Blalock 3/23/15 = BT ni 1010 ổ -E ữ ố kT ứ expỗ 31 B=1.08x10 G 1.12 ư÷ è 8.62x10 T ø Using a spreadsheet, solver, or MATLAB yields T=305.23K 31 æ =1.08x10 T expỗ- Define an M-File: function f=temp(T) f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); Then: fzero('temp',300) | ans = 305.226 K 2.7 For silicon, B = 1.08 x 1031 and EG = 1.12 eV: ni = 5.07 x10-19/cm3 1.69 x 1013/cm3 6.73 x109/cm3 For germanium, B = 2.31 x 1030 and EG = 0.66 eV: ni = 2.63 x10-4/cm3 2.93 x 1015/cm3 2.27 x1013/cm3 2.8 (a) Define an M-File: function f=temp(T) ni=1E15; f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T)); 15 ni = 10 /cm for T = 602 K 15 b N 15 =10 cm , n =10 D p= n n 15 cm 10 15 +  :n = 10 i 10 i 30 = 15 =6.18x10  15 + 10   15 =1.62 x10 / cm 14 / cm 1.62 x10 c At room temperature, ND >> ni2. \n = ND =1015 electrons / cm3 and p = ni2 = 1020 =105 holes / cm3 n 1015 2.9 T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm 2-2 ©R C Jaeger & T N Blalock 3/23/15 T = 100 K: ni = 6.03 x 10 -19 /cm 10 10 T = 450 K: ni = 3.82 x10 /cm T = 300 K and EG = 1.42 eV: ni = 2.21 x10 /cm T = 100 K: ni = 6.03 x 10 -19 /cm T = 450 K: ni = 3.82 x10 /cm 2-3 ©R C Jaeger & T N Blalock 3/23/15 2.10 cm ửổ Vử ổ ữỗ-2000 ữ = +1.40x10 v = - E = ỗ-700 V n ố n - s ứố v = + E = ỗ+250 V ố p Vử cm ữ = -5.00 x10 s ửổ ữỗ-2000 - s øè cm ø p n j s cm ø cm æ cm n = = -qnv ổ 19 -1.60 x10 17 ỗ C ửổ ữỗ 10 cm A2 ữ = 2.24x10 1.40 x10 ø cm è è s ø cm ưỉ cm A j p = qnv p = 1.60 x1019 C ổ10 3 ữỗ -5.00 x105 ÷ = -8.00 x1011 è cm øè s ø cm 2.11 j n n = qnv = ỗ ổ 19 ỗ C 1.60x10 ố j p = qnv p = 1.60x1019 C  n I = j ã Area =1.60x10 ổ ỗ ố A cm öæ cm ö =1.60x10 A =1.60 MA cm øè s ø cm cm ưỉ cm A pA = 160 102 107 ÷ =1.60x1010 2 cm øè s ø cm cm   2 4 4 10 cm 25x10 cm = 400 mA 18 ữỗ ữ 10 10 ữỗ 2.12 2.13 v = j = 2500A / cm = 2.5x10 cm Q 0.01C / cm s 2.14 æ = - E = - 1000 cm ửổ ỗ ố n ổ v p = + E = +ỗ 400 ố V ữỗ -1500 V - s ứố cm ÷ = +1.50 x106 s cm ø cm ưỉ V cm ÷ = -6.00x10 s V - s ữỗ-1500 ứố cm ứ p jn = -qnv n = -1.60 x1019 C jp = qnv p = 1.60 x10 ổ ỗ 103 ố ổ 19 C ỗ 17 10 ố ửổ ữỗ +1.50 x106 cm øè ưỉ s cm -6.00x10 ữỗ cm ứố cm ữ s ứ ữ A = -2.40 x10 10 cm ø A = -9.60x10 2 cm 2-4 ©R C Jaeger & T N Blalock 3/23/15 2.15 a  E = 5V =10, 000 5x10 cm b V V = ỗ 10 è cm V  5x10 ỉ ÷ 4 cm = 50 V cm ø 2.16 p   n i p i  i  n For intrinsic silicon,  = q  n +  n = qn  +   £105 W-ni cm 1 for an insulator 10 5 W- cm 1 10 £ = 2.497x10 = q n+  p  cm cm 19  ỉ  1.602x10 C 1800+ 700 è v - sec ø 20 ỉ -E G 5.152 x10 with ni = = BT kT ø è cm -5 B =1.08x1031 K 3cm6, k =8.62x10 eV/K and E G =1.12eV Using MATLAB as in Problem 2.6 yields T 316.6 K ỗ ữ ữ expỗ 2.17 p i i n n i + p  For intrinsic silicon,  = q  n +  n = qn   ³1000 W- cm1 for a conductor n= i n2 i =  q n + p  1.203x10 39 ³ 1000 W- cm  1  cm 1.602x10 C 120 + 60 v - sec ỉ E = BT expỗ - ữ with 19 = 3.468x1019 cm3 G cm è kT ø G =1.12eV B =1.08x1031 K 3cm6, k = 8.62x10-5eV/K and E This is a transcendental equation and must be solved numerically by iteration Using the HP solver routine or a spread sheet yields T ≥ 2579.3 K Note that this temperature is far above the melting temperature of silicon 2-5 ©R C Jaeger & T N Blalock 3/23/15 2.18 No free electrons or holes (except those corresponding to ni) 2.19 Since Ge is also from column IV, acceptors come from column III and donors come from column V (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi 2.20 (a) Gallium is from column and silicon is from column Thus silicon has an extra electron and will act as a donor impurity (b) Arsenic is from column and silicon is from column Thus silicon is deficient in one electron and will act as an acceptor impurity 2.21 (a) Germanium is from column IV and indium is from column III Thus germanium has one extra electron and will act as a donor impurity (b) Germanium is from column IV and phosphorus is from column V Thus germanium has one less electron and will act as an acceptor impurity 2.22 ỉ j E =  A = j = ỗ 5000 cm ố ữ ứ V 0.02W- cm=100 cm , a small electric field 2.23 16 æ10 N= ỗ ố atoms cm ữ0.180m2m0.5mỗ ứ ổ104 cm ử3 ữ =1800 atoms ố m ứ 2-6 âR C Jaeger & T N Blalock 3/23/15 2.24 18 (a) Since boron is an acceptor, NA = x 10 /cm Assume ND = 0, since it is not specified The material is p-type 10 18 (b) At room temperature, ni =10 / cm and NA - ND = 7x10 / cm >> 2ni 20 n 10 / cm So p = 7x10 / cm and n = = 18 =14.3 / cm p 7x10 / cm ỉ 1.12 (c) At 200K, n i2 =1.08x10 31 2003 expỗ-ỗ ữữ = 5.28x10 / cm 5 18 i 200ø è 8.62x10 n = 7.27x10 / cm i N A - N D >> 2n , so p = 7x10 18 i / cm and n = 5.28x10 = 7.54x10 10 / cm 18 7x10 2.25 17 (a) Since arsenic is a donor, ND = x 10 /cm Assume NA = 0, since it is not specified The material is n-type 2.26 18 18 (a) Arsenic is a donor, and boron is an acceptor ND = x 10 /cm , and NA = x 10 /cm Since NA > ND, the material is p-type 10 18 3 (b) At room temperature, n i =10 / cm and N A - N D = 5x10 / cm >> 2n i n 10 20 / cm 18 3 = So p = 5x10 / cm and n = 18 = 20.0 / cm p 5x10 / cm i 2.27 17 (a) Phosphorus is a donor, and boron is an acceptor ND = x 10 /cm , and NA = x 17 10 /cm Since NA > ND, the material is p-type 10 17 / cm and N - N >> 2n (b) At room temperature, n =10 = 4x10 / cm A i 17 So p = 4x10 / cm and n = n2 i p = 10 20 / cm D i 17 = 250 / cm 4x10 / cm 2-7 ©R C Jaeger & T N Blalock 3/23/15 2.28 2.29 2.30 16 ND = x 10 /cm Assume NA = 0, since it is not specified 16 N D  NA : The material is D A n 10 20 16 3 n=  16 2x10 / cm 5x10 / cm | p  n 5x10 n-ty pe | N N  5x10 / cm  2n  2x10 / cm 10 3 i i N D  N A 5x10 16 / cm | Using the equations in Fig 2.8, n 885 1    ưỉ cm2 q n  ỉ ỗ885 19 n 1.602x10 C ố V s ữỗ 16 5x10 ứố cm 0.141 cm ÷ ø 2.31 18 NA = 2.5x10 /cm Assume ND = 0, since it is not specified 18 N  N : The material is p-type | N  N / cm  2.5x10 A D A p=2.5x10 18 / cm N D  NA  2.5x10 18 / cm 1020  18  40 / cm p 2.5x10 p 1.602x10 ỉ 19 cm ưỉ 2.5x10 C ỗ58.7 ữỗ V s ứố cm è 18 10 / cm3 | Using the equations from Fig 2.8, n 187  q p   2n  2x10 i i 3 D n2 | n= cm cm and p 198 V s Vs cm cm and p  58.7 Vs Vs  42.5 m cm ÷ ø 2-8 ©R C Jaeger & T N Blalock 3/23/15 2.32 19 Indium is from column and is an acceptor NA = x 10 /cm Assume ND = 0, since it is not specified N A  ND : material is p-type | N A  ND  8x1019 / cm3  2ni  2x1010 / cm3 n 10 20 p  8x10 19 / cm | n = i  19 1.25 / cm p 8x10 cm cm 19 N D  N A  7x10 / cm | Using Fig 2.8, n  66.2 and p  46.1 Vs Vs 1 1.69 m cm     q p ỉ cm ưỉ8x10 19 p 1.602x10 19 C ỗ46.1 ữỗ ữ V s ứố cm ø è 2.33 16 16 Phosphorus is a donor: N D  4.5x10 / cm | Boron is an acceptor: N A  5.5x10 / cm N  N : The material is p-type | N  N 10 16 / cm  2n  x10 10 / cm A D A 16 n p 10 / cm | n  ND  NA  q n n D i 20 10  16 10 / cm p 10 i cm cm and p 153 10 / cm | Using Fig 2.8,  n  727 V s Vs ö  4.08  cm  cm ửổ 10 ổ 1.602x10 19 C 153 ữỗ ữ V  s øè cm ø è 17 16 ỗ 2.34 An iterative solution is required Using the equations from Fig 2.8 and trial and error: NA  p p p 1018 70.8 7.08 x 1019 x1018 61.0 1.22 x 1020 1.90 x 1018 61.6 1.17 x 1020 1.89 x 1018 61.6 1.16 x 1020 2-9 ©R C Jaeger & T N Blalock 3/23/15 2.35 An iterative solution is required Using the equations in Fig 2.8 and trial and error: p NA p p 1016 318 3.18 x 1018 x 1016 214 8.55 x 1018 7.5 x 1016 170 1.28 x 1019 7.2 x 1016 173 1.25 x 1019 2.36 Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, but the hole and electron concentrations remain unchanged See Problem 2.39 for example However, it is physically impossible to add exactly equal amounts of the two impurities 2.37 An iterative solution is required Using the equations in Fig 2.8 and trial and error: ND n n n 1015 1360 1.36 x 1018 1.5 x 1015 1340 2.01 x 1018 1.6 x 1015 1340 2.14 x 1018 1.55 x 1015 1340 2.08 x 1018 2.38 Based upon the value of its resistivity, the material is an insulator However, it is not intrinsic because it contains impurities The addition of the impurities has increased the resistivity Since ND  N A =0, n=p=ni, and   q n ni  pni   qni n  p  NA  ND 1020 / cm3 which yields p  45.9 and n  64.3 using the equations from Fig 2.8 1 æ cm ö  5.66x10 -cm  qni n  p £ 19 10 3 cm ÷ è v  sec ứ 1.602x10 64.3 45.9ỗ C10 2-10 âR C Jaeger & T N Blalock 3/23/15 2.39 (a) An iterative solution is required Using the equations in Fig 2.8 and trial and error: ND n n n 1019 108 1.08 x 1021 x 1019 67.5 4.73 x 1021 x 1021 64.3 6.43 x 1021 9.67 x 1019 64.5 6.24 x 1021 (b) An iterative solution is required using the equations in Fig 2.8 and trial and error: NA x1020 p 45.9 p p 4.96 x 1021 1.2 x1020 45.8 5.93 x 1021 1.4 x1020 45.7 6.17 x 1021 1.37 x 1020 45.7 6.26 x 1021 - 11 ©R C Jaeger & T N Blalock 3/23/15 2.40 (a) For the ohm-cm starting material: To change the resistivity to 0.25 ohm-cm: Iterative solutions are required using the equations with Fig 2.8 aand trial and error: NA p p p Ohm-cm 2.51x1016 249 6.25 x 1018 0.25 Ohm-cm 2.2 x 1017 147 2.5 x 1018 17 Additional acceptor concentration = 2.2 x 10 - 2.5 x 10 16 10 /cm (b) If donors are added: 16 = 1.95 x ND ND + NA n ND - NA n n x 1016 5.5 x 1016 864 0.5 x 1015 4.32 x 1018 x 1016 7.5 x 1016 794 2.5 x 1016 1.98 x 1019 765 3.5 x 1016 2.68 x 1019 772 3.24 x 1016 2.50 x 1019 8.5 x 1016 x 1016 5.74 x 1016 8.24 x 1016 16 So ND = 5.7 x 10 /cm must be added to change achieve a resistivity of 0.25 ohm-cm The silicon is converted to n-type material 2-12 ©R C Jaeger & T N Blalock 3/23/15 2.41 16 /cm and p = 318 cm /V-s from equations with Fig 2.8 0.509     qp p  qp NA  1.602x10 19 C3181016  Boron is an acceptor: NA = 10 Now we add donors until  = 4.5 (-cm)   q n n n D : 4.5 cm |  n   N  N n  cm -1 1  A 2.81x10  1.602x10 19 C V  cm  s 19 Using trial and error: ND ND + NA n ND - NA 7x n n x 1016 x 1016 752 1016 4x 5.26 x 1019 x 1016 x 1016 845 1016 3x 3.38 x 1019 x 1016 x 1016 885 1016 3.2 x 2.66 x 1019 4.2 x 1016 5.2 x 1016 877 1016 2.81 x 1019 2.42 Phosphorus is a donor: N D = 10   q n  q N  1.602x10 n n 19  D 16 /cm and  = 1180 cm /V-s from Fig 2.8 n 16 C 1180 10   Now we add acceptors until  = 5.0 (-cm) -1  1.89   cm : 1 3.12x1019 5 cm   qp p |  p   N  N   19 p A D p 1.602x10 C V  cm  s Using trial and error: NA ND + NA 1.10E+17 2.10E+17 3.60E+17 3.40E+17 1.00E+17 2.00E+17 3.50E+17 3.30E+17  NA - ND 9.00E+16 1.90E+17 3.40E+17 3.20E+17 p 147 116 95.6 97.4 p p 1.33E+19 2.20E+19 3.25E+19 3.12E+19 2.43 T (K) 50 75 100 150 200 250 300 350 400 VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5 - 13 ©R C Jaeger & T N Blalock 3/23/15 2.44 ổ j = -qD ỗ - dn ÷= è n dx ø  j = 1.602x10 19 dn qV  dx T n æ cm ửổ ỗ C 0.025V ố -10 ữỗ 350 V - s ứố 0.25x10 kA ÷ 18 - ø cm = -56.1 cm 2.45 2.46 2.47 At x = 0: drift jn = qnnE = 1.60 x10 19 16 ưỉ V A ữỗ10 ữỗ+25 ữ =14.0 cm V - s ứố cm ứố cm ứ ổ Cỗ350 cm è j drift p = q pE = 1.60 x10 dn = qDn diff jp  p diff jn 19  = 1.60 x10 dx dp = -qDp dx = -1.60 x10 ổ C ửổ ỗ 19 ửổ ữỗ ữ 18 150 cm ửổ1.01x10 +25 V = +606 A è V - s øè cm3 øè cm ứ cm2 cm2 ửổ 104 -1016 A ổ ữỗ Cỗ350 ì 0.025 ữỗ 4 ữ = -70.0 s øè x10 cm ø cm è ö cm2 ửổ1018 -1.01x1018 ổ 19 Cỗ150 ì 0.025 ố ữỗ s øè 4 2x10 cm A ÷ = 30.0 ø cm A jT =14.0 + 607 - 70.0 + 30.0 = +580 cm 2-14 ©R C Jaeger & T N Blalock 3/23/15 At x=1 m assuming linear distributions:   18 p 1m =1.005x10 / cm drift jn = qnnE = 1.60 x10   æ 19 diff n  = qD dx = 1.60 x10 j dp A Cỗ350 ữỗ ữ = +7.00 ữỗ+25 cm A è V - s øè cm øè ưỉ cm ø V  n V ưỉ1.005x10 cm ç è ç æ 19 15 j drift = qp pE = 1.60 x10 19 C  150 dn ưỉ 5x10 cm ỉ p 15 , n 1m = 5x10 / cm ửổ 18 ữỗ V - s ứố ữỗ +25 cm 10 ứố ửổ cm ữỗ -10 ữ = +603 cm cm ø ö 16 p 2.48 C è 350 × 0.025 s øè x10 cm ø = -70.0 cm ổ ửổ10 cm ỗ -1.01x10 18 = -1.60 x10 19 C 150 ì 0.025 ữỗ 4 dx è s øè 2x10 cm j + 7.00 + 603 - 70.0 + 30.0 = -570 A T cm j diff = -qD p A 4÷ 18 A ÷ = 30.0 cm ø NA = 2ND 2.49 2.50 - 15 ©R C Jaeger & T N Blalock 3/23/15 2.51 + An n-type ion implantation step could be used to form the n region following step (f) in Fig 2.17 A mask would be used to cover up the opening over the p-type region and leave the opening over the n-type silicon The masking layer for the implantation could just be photoresist 2.52 2-16 ©R C Jaeger & T N Blalock 3/23/15 ... x 1020 45.7 6.26 x 1021 - 11 ©R C Jaeger & T N Blalock 3/23/15 2.40 (a) For the ohm-cm starting material: To change the resistivity to 0.25 ohm-cm: Iterative solutions are required using the equations... -9.60x10 2 cm 2-4 ©R C Jaeger & T N Blalock 3/23/15 2.15 a  E = 5V =10, 000 4 5x10 cm b V V = ỗ 10 ố cm V  5x10 ỉ ÷ 4 cm = 50 V cm ø 2.16 p   n i p i  i  n For intrinsic silicon,... 2.6 yields T 316.6 K ỗ ữ ữ expỗ 2.17 p i  i  n  n i + p  For intrinsic silicon,  = q  n +  n = qn   ³1000 W- cm1 for a conductor n= i n2 i =  q n + p  1.203x10 39 ³ 1000