Table Of Content

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 1.1 n =BT3/2e−Eg/2kT i (a) Silicon ⎡ ⎤ −1.1 (i) n =(5.23×1015)(250)3/2exp⎢ ⎥ i ⎢2(86×10−6)(250)⎥ ⎣ ⎦ =2.067×1019exp[−25.58] n =1.61×108 cm−3 i ⎡ ⎤ −1.1 (ii) n =(5.23×1015)(350)3/2exp⎢ ⎥ i ⎢2(86×10−6)(350)⎥ ⎣ ⎦ =3.425×1019exp[−18.27] n =3.97×1011 cm−3 i (b) GaAs ⎡ ⎤ −1.4 (i) n =(2.10×1014)(250)3/2exp⎢ ⎥ i ⎢2(86×10−6)(250)⎥ ⎣ ⎦ =(8.301×1017)exp[−32.56] n =6.02×103 cm−3 i ⎡ ⎤ −1.4 (ii) n =(2.10×1014)(350)3/2exp⎢ ⎥ i ⎢2(86×10−6)(350)⎥ ⎣ ⎦ =(1.375×1018)exp[−23.26] n =1.09×108 cm−3 i ______________________________________________________________________________________ 1.2 ⎛−Eg⎞ a. n =BT3/2exp⎜ ⎟ i ⎝2kT ⎠ ⎛ −1.1 ⎞ 1012 =5.23×1015T3/2exp⎜ ⎟ ⎝2(86×10−6)(T)⎠ ⎛ 6.40×103⎞ 1.91×10−4 =T3/2exp⎜− ⎟ ⎝ T ⎠ By trial and error, T ≈368 K b. n =109 cm−3 i ⎛ ⎞ −1.1 109 =5.23×1015T3/2exp⎜ ⎟ ⎜2(86×10−6)(T)⎟ ⎝ ⎠ ⎛ 6.40×103⎞ 1.91×10−7 =T3/2exp⎜− ⎟ ⎝ T ⎠ By trial and error, T ≈268° K ______________________________________________________________________________________ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3 Silicon ⎡ ⎤ −1.1 (a) n =(5.23×1015)(100)3/2exp⎢ ⎥ i ⎢2(86×10−6)(100)⎥ ⎣ ⎦ =(5.23×1018)exp[−63.95] n =8.79×10−10 cm−3 i ⎡ ⎤ −1.1 (b) n =(5.23×1015)(300)3/2exp⎢ ⎥ i ⎢2(86×10−6)(300)⎥ ⎣ ⎦ =(2.718×1019)exp[−21.32] n =1.5×1010 cm−3 i ⎡ ⎤ −1.1 (c) n =(5.23×1015)(500)3/2exp⎢ ⎥ i ⎢2(86×10−6)(500)⎥ ⎣ ⎦ =(5.847×1019)exp[−12.79] n =1.63×1014 cm−3 i Germanium. ⎡ ⎤ −0.66 (a) n =(1.66×1015)(100)3/2exp⎢ ⎥=(1.66×1018)exp[−38.37] i ⎢2(86×10−6)(100)⎥ ⎣ ⎦ n =35.9 cm−3 i ⎡ ⎤ −0.66 (b) n =(1.66×1015)(300)3/2exp⎢ ⎥=(8.626×1018)exp[−12.79] i ⎢2(86×10−6)(300)⎥ ⎣ ⎦ n =2.40×1013 cm−3 i ⎡ ⎤ −0.66 (c) n =(1.66×1015)(500)3/2exp⎢ ⎥=(1.856×1019)exp[−7.674] i ⎢2(86×10−6)(500)⎥ ⎣ ⎦ n =8.62×1015 cm−3 i ______________________________________________________________________________________ 1.4 n2 (2.4×1013)2 (a) n-type; n =1015 cm−3; p = i = =5.76×1011 cm−3 o o n 1015 o n2 (1.5×1010)2 (b) n-type; n =1015 cm−3; p = i = =2.25×105 cm−3 o o n 1015 o ______________________________________________________________________________________ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5 n2 (1.8×106)2 (a) p-type; p =1016 cm−3; n = i = =3.24×10−4 cm−3 o o p 1016 o n2 (2.4×1013)2 (b) p-type; p =1016 cm−3; n = i = =5.76×1010 cm−3 o o p 1016 o ______________________________________________________________________________________ 1.6 (a) n-type (b) n =N =5×1016 cm−3 o d n2 (1.5×1010)2 p = i = =4.5×103 cm−3 o n 5×1016 o (c) n =N =5×1016 cm−3 o d From Problem 1.1(a)(ii) n =3.97×1011 cm−3 i (3.97×1011)2 p = =3.15×106 cm−3 o 5×1016 ______________________________________________________________________________________ 1.7 n2 (1.5×1010)2 (a) p-type; p =5×1016 cm−3; n = i = =4.5×103 cm−3 o o p 5×1016 o n2 (1.8×106)2 (b) p-type; p =5×1016 cm−3; n = i = =6.48×10−5 cm−3 o o p 5×1016 o ______________________________________________________________________________________ 1.8 (a) Add boron atoms (b) N = p =2×1017 cm−3 a o n2 (1.5×1010)2 (c) n = i = =1.125×103 cm−3 o p 2×1017 o ______________________________________________________________________________________ 1.9 (a) n =5×1015 cm−3 o n2 (1.5×1010)2 p = i = ⇒ p =4.5×104 cm−3 o n 5×1015 o o (b) n > p ⇒n-type o o (c) n ≅ N =5×1015 cm−3 o d ______________________________________________________________________________________ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a. Add Donors N =7×1015 cm−3 d b. Want p =106 cm−3 =n2/N o i d So n2 =(106)(7×1015)=7×1021 i ⎛−Eg⎞ =B2T3exp⎜ ⎟ ⎝ kT ⎠ ⎛ ⎞ 7×1021 =(5.23×1015)2T3exp⎜ −1.1 ⎟ ⎜(86×10−6)(T)⎟ ⎝ ⎠ By trial and error, T ≈324° K ______________________________________________________________________________________ 1.11 (a) I = AσΕ=(10−5)(1.5)(10)⇒I =0.15 mA AΕ Iρ (1.2×10−3)(0.4) (b) I = ⇒Ε= = ( ) =2.4 V/cm ρ A 2×10−4 ______________________________________________________________________________________ 1.12 J 120 J =σΕ⇒σ= = =6.67(Ω−cm)−1 Ε 18 σ (6.67) σ≅eμnNd ⇒Nd = eμ = (1.6×10−19)(1250)=3.33×1016 cm−3 n ______________________________________________________________________________________ 1.13 1 1 1 (a) ρ≅ eμN ⇒Nd = eμρ= (1.6×10−19)(1250)(0.65)=7.69×1015 cm−3 n d n Ε (b) J = ⇒Ε=ρJ =(0.65)(160)=104 V/cm ρ ______________________________________________________________________________________ 1.14 σ 1.5 (a) σ≅eμnNd ⇒Nd = eμ = (1.6×10−19)(1000)=9.375×1015 cm−3 n σ 0.8 (b) Na = eμ = (1.6×10−19)(400)=1.25×1016 cm−3 p ______________________________________________________________________________________ 1.15 (a) For n-type, σ≅eμN =(1.6×10−19)(8500)N n d d For 1015 ≤ N ≤1019 cm−3 ⇒1.36≤σ≤1.36×104(Ω−cm)−1 d (b) J =σE=σ(0.1)⇒0.136≤J ≤1.36×103 A/cm2 ______________________________________________________________________________________ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 D =(0.026)(1250)=32.5 cm2/s; D =(0.026)(450)=11.7 cm2/s n p J =eD dn =(1.6×10−19)(32.5)⎜⎛1016 −1012 ⎟⎞=−52 A/cm2 n n dx ⎜⎝ 0−0.001 ⎟⎠ J =−eD dp =−(1.6×10−19)(11.7)⎜⎛1012 −1016 ⎟⎞=−18.72 A/cm2 p p dx ⎜⎝ 0−0.001 ⎟⎠ Total diffusion current density J =−52−18.72=−70.7 A/cm2 ______________________________________________________________________________________ 1.17 dp J =−eD p p dx ⎛−1⎞ ⎛−x⎞ =−eD (1015)⎜ ⎟exp⎜ ⎟ p ⎜L ⎟ ⎜L ⎟ ⎝ p ⎠ ⎝ p ⎠ (1.6×10−19)(15)(1015) ⎛−x⎞ J = exp⎜ ⎟ p 10×10−4 ⎜L ⎟ ⎝ p ⎠ J =2.4 e−x/Lp p (a) x = 0 J =2.4 A/cm2 p (b) x=10 μm J =2.4 e−1 =0.883 A/cm2 p (c) x=30 μm J =2.4 e−3 =0.119 A/cm2 p ______________________________________________________________________________________ 1.18 a. N =1017 cm−3 ⇒ p =1017 cm−3 a o n2 (1.8×106)2 n = i = ⇒n =3.24×10−5 cm−3 o p 1017 o o b. n=n +δn=3.24×10−5+1015 ⇒n=1015 cm−3 o p= p +δp=1017 +1015 ⇒ p=1.01×1017 cm−3 o ______________________________________________________________________________________ ⎛N N ⎞ 1.19 V =V ln⎜ a d ⎟ bi T ⎜⎝ ni2 ⎟⎠ ⎡(5×1015)(5×1015)⎤ (a) (i) V =(0.026)ln⎢ ⎥=0.661 V bi ⎢⎣ (1.5×1010)2 ⎥⎦ ⎡(5×1017)(1015)⎤ (ii) V =(0.026)ln⎢ ⎥ =0.739 V bi ⎢⎣ (1.5×1010)2 ⎥⎦ ⎡(1018)(1018)⎤ (iii) V =(0.026)ln⎢ ⎥=0.937 V bi ⎢⎣(1.5×1010)2⎥⎦ Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎡(5×1015)(5×1015)⎤ (b) (i) V =(0.026)ln⎢ ⎥=1.13 V bi ⎢⎣ (1.8×106)2 ⎥⎦ ⎡(5×1017)(1015)⎤ (ii) V =(0.026)ln⎢ ⎥=1.21 V bi ⎢⎣ (1.8×106)2 ⎥⎦ ⎡(1018)(1018)⎤ (iii) V =(0.026)ln⎢ ⎥=1.41 V bi ⎢⎣(1.8×106)2⎥⎦ ______________________________________________________________________________________ 1.20 ⎛N N ⎞ V =V ln⎜ a d ⎟ bi T ⎜⎝ ni2 ⎟⎠ or (n2) ⎛V ⎞ (1.5×10)2 ⎛0.712⎞ N = i exp⎜ bi ⎟= exp⎜ ⎟=1.76×1016 cm−3 a Nd ⎜⎝VT ⎟⎠ 1016 ⎝0.026⎠ ______________________________________________________________________________________ 1.21 ⎛N N ⎞ ⎡ N (1016) ⎤ V =V ln⎜ a d ⎟=(0.026)ln⎢ a ⎥ bi T ⎝ ni2 ⎠ ⎢⎣(1.5×1010)2⎥⎦ For N =1015 cm−3, V =0.637 V a bi For N =1018 cm−3, V =0.817 V a bi ______________________________________________________________________________________ 1.22 ⎛ T ⎞ kT =(0.026)⎜ ⎟ ⎝300⎠ kT (T)3/2 200 0.01733 2828.4 250 0.02167 3952.8 300 0.026 5196.2 350 0.03033 6547.9 400 0.03467 8000.0 450 0.0390 9545.9 500 0.04333 11,180.3 Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ⎛ ⎞ −1.4 n =(2.1×1014)(T3/2)exp⎜ ⎟ i ⎜2(86×10−6)(T)⎟ ⎝ ⎠ ⎛N N ⎞ V =V ln⎜ a d ⎟ bi T ⎝ n2 ⎠ i T n V i bi 200 1.256 1.405 250 6.02 × 103 1.389 300 1.80 × 106 1.370 350 1.09 × 108 1.349 400 2.44 × 109 1.327 450 2.80 × 1010 1.302 500 2.00 × 1011 1.277 ______________________________________________________________________________________ 1.23 −1/2 ⎛ V ⎞ C =C ⎜1+ R ⎟ j jo⎝ V ⎠ bi ⎡(1.5×1016)(4×1015)⎤ V =(0.026)ln⎢ ⎥=0.684 V bi ⎢ (1.5×1010)2 ⎥ ⎣ ⎦ ⎛ 1 ⎞−1/2 (a) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.255 pF ⎛ 3 ⎞−1/2 (b) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.172 pF ⎛ 5 ⎞−1/2 (c) Cj =(0.4)⎜⎝1+0.684⎟⎠ =0.139 pF ______________________________________________________________________________________ 1.24 −1/2 ⎛ V ⎞ (a) C =C ⎜1+ R ⎟ j jo⎝ V ⎠ bi ⎛ 5 ⎞−1/2 For VR = 5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.00743 pF ⎛ 1.5⎞−1/2 For VR = 1.5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.0118 pF Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 0.00743+0.0118 C (avg)= =0.00962 pF j 2 v (t)=v (final)+(v (initial)−v (final))e−t/τ C C C C where τ=RC=RC (avg)=(47×103)(0.00962×10−12) j or τ=4.52×10−10 s Then v (t)=1.5=0+(5−0)e−ti/τ C 5 ⎛ 5 ⎞ =e+r1/τ⇒t =τln⎜ ⎟ 1.5 1 ⎝1.5⎠ t =5.44×10−10 s 1 (b) For V = 0 V, C = C = 0.02 pF R j jo ⎛ 3.5⎞−1/2 For VR = 3.5 V, Cj =(0.02)⎜⎝1+0.8⎟⎠ =0.00863 pF 0.02+0.00863 C (avg)= =0.0143 pF j 2 τ=RC (avg)=6.72×10−10 s j v (t)=v (final)+(v (initial)−v (final))e−t/τ C C C C 3.5=5+(0−5)e−t2/τ=5(1−e−t2/τ) so that t =8.09×10−10 s 2 ______________________________________________________________________________________ 1.25 ⎛ V ⎞−1/2 ⎡(5×1015)(1017)⎤ C =C ⎜1+ R ⎟ ; V =(0.026)ln⎢ ⎥ =0.739 V j jo⎜⎝ Vbi ⎟⎠ bi ⎢⎣ (1.5×1010)2 ⎥⎦ For V =1V, R 0.60 C = =0.391 pF j 1 1+ 0.739 For V =3 V, R 0.60 C = =0.267 pF j 3 1+ 0.739 For V =5 V, R 0.60 C = =0.215 pF j 5 1+ 0.739 1 1 (a) f = = ⇒ f =6.57 MHz o ( )( ) o 2π LC 2π 1.5×10−3 0.391×10−12 1 (b) f = ⇒ f =7.95MHz o ( )( ) o 2π 1.5×10−3 0.267×10−12 Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1 (c) f = ⇒ f =8.86 MHz o ( )( ) o 2π 1.5×10−3 0.215×10−12 ______________________________________________________________________________________ 1.26 ⎡ ⎛V ⎞ ⎤ ⎛V ⎞ a. I =I ⎢exp⎜ D ⎟−1⎥−0.90=exp⎜ D ⎟−1 S ⎣ ⎝VT ⎠ ⎦ ⎝VT ⎠ ⎛V ⎞ exp⎜ D ⎟=1−0.90=0.10 ⎝V ⎠ T V =V ln(0.10)⇒V =−0.0599 V D T D b. IF = IS ⋅⎢⎡⎣exp⎛⎜⎝VVTF ⎞⎟⎠−1⎥⎤⎦ = exp⎝⎛⎜00.0.226⎠⎞⎟−1 I I ⎡ ⎛V ⎞ ⎤ ⎛ −0.2 ⎞ R S ⎢exp⎜ R ⎟−1⎥ exp⎜ ⎟−1 ⎣ ⎝VT ⎠ ⎦ ⎝0.026⎠ 2190 = −1 I F =2190 I R ______________________________________________________________________________________ ⎡ ⎛V ⎞ ⎤ 1.27 ID =IS⎢⎢⎣exp⎜⎜⎝VTD ⎟⎟⎠−1⎥⎥⎦ ( ) ⎛ 0.3 ⎞ (a) (i) I = 10−11 exp⎜ ⎟⇒1.03μA D ⎝0.026⎠ ( ) ⎛ 0.5 ⎞ (ii) I = 10−11 exp⎜ ⎟⇒2.25 mA D ⎝0.026⎠ ( ) ⎛ 0.7 ⎞ (iii) I = 10−11 exp⎜ ⎟⇒4.93 A D ⎝0.026⎠ ( )⎡ ⎛−0.02⎞ ⎤ (iv) ID = 10−11 ⎢⎣exp⎜⎝ 0.026 ⎟⎠−1⎥⎦ =−5.37×10−12 A ( )⎡ ⎛−0.20⎞ ⎤ (v) ID = 10−11 ⎢⎣exp⎜⎝ 0.026 ⎟⎠−1⎥⎦≅−10−11 A ( ) (vi) I =−10−11 A D ( ) ⎛ 0.3 ⎞ (b) (i) I = 10−13 exp⎜ ⎟⇒0.0103μA D ⎝0.026⎠ ( ) ⎛ 0.5 ⎞ (ii) I = 10−13 exp⎜ ⎟⇒22.5μA D ⎝0.026⎠ ( ) ⎛ 0.7 ⎞ (iii) I = 10−13 exp⎜ ⎟⇒49.3 mA D ⎝0.026⎠

Microelectronics Circuit Analysis and Design, Solutions Manual PDF

665 Pages·2009·9.1 MB·English

by D. A. Neamen

#Technique #Electronics

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Language:	English
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