Microelectronics circuit analysis and design donald a neamen 4th edition solutions

www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions Chapter 1.1 ni = BT / e (a) Silicon − Eg / kT ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 250 ) ⎥⎦ = 2.067 × 1019 exp [ −25.58] ni = 1.61× 108 cm −3 (i) ni = ( 5.23 × 1015 ) ( 250 ) (ii) ni = ( 5.23 × 1015 ) ( 350 ) (b) GaAs (i) ni = ( 2.10 × 1014 ) ( 250 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 350 ) ⎥⎦ = 3.425 × 1019 exp [ −18.27 ] ni = 3.97 ×1011 cm −3 3/ 3/ ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 250 ) ⎥⎦ = ( 8.301× 1017 ) exp [ −32.56] ni = 6.02 × 103 cm −3 (ii) ni = ( 2.10 × 1014 ) ( 350 ) 3/ ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 350 ) ⎥⎦ = (1.375 × 1018 ) exp [ −23.26] ni = 1.09 × 108 cm −3 1.2 a ⎛ − Eg ⎞ ni = BT / exp ⎜ ⎟ ⎝ 2kT ⎠ ⎛ ⎞ −1.1 1012 = 5.23 × 1015 T / exp ⎜ ⎟ −6 ⎝ 2(86 × 10 )(T ) ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−4 = T / exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 368 K b ni = 109 cm −3 ⎛ ⎞ −1.1 ⎟ 109 = 5.23 × 1015 T / exp ⎜ ⎜ ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−7 = T / exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 268° K Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 1.3 Silicon (a) ni = ( 5.23 × 1015 ) (100 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) (100 ) ⎥⎦ = ( 5.23 × 1018 ) exp [ −63.95] ni = 8.79 ×10−10 cm −3 (b) ni = ( 5.23 × 1015 ) ( 300 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 300 ) ⎥⎦ = ( 2.718 × 1019 ) exp [ −21.32] ni = 1.5 × 1010 cm −3 (c) ni = ( 5.23 × 1015 ) ( 500 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 500 ) ⎥⎦ = ( 5.847 × 1019 ) exp [ −12.79] ni = 1.63 × 1014 cm −3 Germanium (a) ni = (1.66 × 1015 ) (100 ) 3/ ⎡ ⎤ −0.66 ⎥ = (1.66 × 1018 ) exp [ −38.37 ] exp ⎢ −6 ⎢⎣ ( 86 × 10 ) (100 ) ⎥⎦ ni = 35.9 cm −3 (b) ni = (1.66 × 1015 ) ( 300 ) 3/ ⎡ ⎤ −0.66 ⎥ = ( 8.626 × 1018 ) exp [ −12.79] exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 300 ) ⎥⎦ ni = 2.40 × 1013 cm −3 (c) ni = (1.66 × 1015 ) ( 500 ) 3/ ⎡ ⎤ −0.66 ⎥ = (1.856 × 1019 ) exp [ −7.674] exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 500 ) ⎥⎦ ni = 8.62 ×1015 cm −3 1.4 (a) n-type; no = 10 15 ( ) ( ) n2 2.4 × 1013 cm ; po = i = no 1015 −3 2 = 5.76 × 1011 cm −3 ni2 1.5 × 1010 = = 2.25 × 10 cm −3 no 1015 (b) n-type; no = 1015 cm −3 ; po = www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 1.5 (a) p-type; p o = 1016 cm −3 ; no = ( ni2 1.8 × 10 = po 1016 ) ( = 3.24 × 10 − cm −3 ) ni2 2.4 × 1013 = = 5.76 × 1010 cm −3 po 1016 (b) p-type; p o = 1016 cm −3 ; no = 1.6 (a) (b) n-type no = N d = × 1016 cm −3 10 ni2 (1.5 × 10 ) po = = = 4.5 × 103 cm −3 no × 1016 (c) no = N d = × 1016 cm −3 From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3 ( 3.97 × 10 ) = 11 = 3.15 × 106 cm −3 × 1016 po 1.7 (a) p-type; p o = × 1016 cm −3 ; no = ( ) ( ) ni2 1.5 × 1010 = po × 1016 = 4.5 × 10 cm −3 ni2 1.8 × 10 = = 6.48 × 10 −5 cm −3 po × 1016 (b) p-type; p o = × 1016 cm −3 ; no = 1.8 (a) Add boron atoms (b) N a = po = × 1017 cm −3 ( ) ni2 1.5 × 1010 = = 1.125 × 10 cm −3 po × 1017 (c) no = 1.9 (a) no = × 1015 cm −3 10 n (1.5 × 10 ) po = i = ⇒ po = 4.5 × 104 cm −3 no × 1015 (b) n o > p o ⇒ n-type (c) no ≅ N d = × 1015 cm −3 Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 1.10 a b Add Donors N d = × 1015 cm −3 Want po = 106 cm −3 = ni2 / N d So ni2 = (106 )( × 1015 ) = × 10 21 ⎛ − Eg ⎞ = B 2T exp ⎜ ⎟ ⎝ kT ⎠ ⎛ ⎞ −1.1 ⎟ × 1021 = ( 5.23 × 1015 ) T exp ⎜ − ⎜ ( 86 × 10 ) (T ) ⎟ ⎝ ⎠ By trial and error, T ≈ 324° K 1.11 (a) I = Aσ Ε = 10 −5 (1.5)(10) ⇒ I = 0.15 mA ( ) ( ) Iρ 1.2 × 10 −3 (0.4) = 2.4 V/cm = A ρ × 10 − (b) I = AΕ ⇒Ε= ( ) 1.12 J 120 −1 = = 6.67 (Ω − cm) Ε 18 σ (6.67) σ ≅ eμ n N d ⇒ N d = = = 3.33 × 1016 cm −3 eμ n 1.6 × 10 −19 (1250) J =σΕ ⇒σ = ( ) 1.13 1 ⇒ Nd = = = 7.69 × 1015 cm −3 eμ n N d eμ n ρ 1.6 × 10 −19 (1250)(0.65) Ε (b) J = ⇒ Ε = ρ J = (0.65)(160 ) = 104 V/cm ρ (a) ρ ≅ ( ) 1.14 σ 1.5 = = 9.375 × 1015 cm −3 eμ n 1.6 × 10 −19 (1000) σ 0.8 = = 1.25 × 1016 cm −3 (b) N a = −19 eμ p 1.6 × 10 (400) (a) σ ≅ eμ n N d ⇒ N d = ( ( ) ) 1.15 (a) For n-type, σ ≅ eμ n N d = (1.6 × 10 −19 ) ( 8500 ) N d For 1015 ≤ N d ≤ 1019 cm −3 ⇒ 1.36 ≤ σ ≤ 1.36 × 104 ( Ω − cm ) −1 (b) J = σ E = σ ( 0.1) ⇒ 0.136 ≤ J ≤ 1.36 ×103 A / cm2 Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 1.16 Dn = (0.026)(1250) = 32.5 cm /s; D p = (0.026 )(450 ) = 11.7 cm /s J n = eDn ⎛ 1016 − 1012 dn = 1.6 × 10 −19 (32.5)⎜⎜ dx ⎝ − 0.001 ( J p = −eD p ) ⎞ ⎟⎟ = −52 A/cm ⎠ ⎛ 1012 − 1016 dp = − 1.6 × 10 −19 (11.7 )⎜⎜ dx ⎝ − 0.001 ( ) ⎞ ⎟⎟ = −18.72 A/cm ⎠ Total diffusion current density J = −52 − 18.72 = −70.7 A/cm 1.17 J p = −eD p dp dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (10 15 ) ⎜ ⎟ exp ⎜ ⎟ ⎜ Lp ⎟ ⎜ Lp ⎟ ⎝ ⎠ ⎝ ⎠ (1.6 ×10 ) (15) (10 ) exp ⎛ − x ⎞ −19 Jp = 15 ⎜⎜ ⎟⎟ ⎝ Lp ⎠ 10 × 10 −4 J p = 2.4 e − x / Lp J p = 2.4 A/cm2 (a) x=0 (b) x = 10 μ m J p = 2.4 e−1 = 0.883 A/cm x = 30 μ m J p = 2.4 e−3 = 0.119 A/cm (c) 1.18 a N a = 1017 cm −3 ⇒ po = 1017 cm −3 n (1.8 × 10 no = i = 1017 po b ) ⇒ no = 3.24 × 10−5 cm −3 n = no + δ n = 3.24 × 10−5 + 1015 ⇒ n = 1015 cm −3 p = po + δ p = 1017 + 1015 ⇒ p = 1.01× 1017 cm −3 ⎛N N 1.19 Vbi = VT ln⎜⎜ a d ⎝ ni (a) (i) ⎞ ⎟ ⎟ ⎠ ( )( ⎡ × 1015 × 1015 Vbi = (0.026 ) ln ⎢ ⎢⎣ 1.5 × 1010 ( ( ) )⎤⎥ = 0.661 V ⎥⎦ )( ) ( ) ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 0.937 V ⎣⎢ (1.5 × 10 ) ⎦⎥ (ii) ⎡ × 1017 1015 ⎤ Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 ⎥⎦ ⎣⎢ 1.5 × 10 (iii) Vbi 18 18 10 www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions ( )( ⎡ × 1015 × 1015 Vbi = (0.026 ) ln ⎢ ⎢⎣ 1.8 × 10 (b) (i) ( ( ) )⎤⎥ = 1.13 V ⎥⎦ )( )⎤⎥ = 1.21 V ( ) ⎥⎦ ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 1.41 V ⎢⎣ (1.8 × 10 ) ⎥⎦ (ii) ⎡ × 1017 1015 Vbi = (0.026 ) ln ⎢ ⎢⎣ 1.8 × 10 (iii) Vbi 18 18 1.20 ⎛N N Vbi = VT ln⎜⎜ a d ⎝ ni or ⎞ ⎟ ⎟ ⎠ (n ) exp⎛⎜ V (1.5 × 10) exp⎛ 0.712 ⎞ = 1.76 × 1016 cm −3 bi ⎞ ⎜ ⎟ ⎜ V ⎟⎟ = 1016 Nd ⎝ 0.026 ⎠ ⎝ T ⎠ Na = 1.21 i ⎡ N a (1016 ) ⎤ ⎛N N ⎞ ⎥ Vbi = VT ln ⎜ a d ⎟ = ( 0.026 ) ln ⎢ 10 ⎢⎣ (1.5 × 10 ) ⎥⎦ ⎝ ni ⎠ For N a = 1015 cm −3 , Vbi = 0.637 V For N a = 1018 cm −3 , Vbi = 0.817 V 1.22 ⎛ T ⎞ kT = (0.026) ⎜ ⎟ ⎝ 300 ⎠ 200 250 300 350 400 450 500 kT 0.01733 0.02167 0.026 0.03033 0.03467 0.0390 0.04333 (T)3/2 2828.4 3952.8 5196.2 6547.9 8000.0 9545.9 11,180.3 Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions ⎛ ⎞ −1.4 ⎟ ni = ( 2.1× 1014 )(T / ) exp ⎜ ⎜ ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛N N ⎞ Vbi = VT ln ⎜ a d ⎟ ⎝ ni ⎠ T ni 200 1.256 250 6.02 × 103 300 1.80 × 106 350 1.09 × 108 400 2.44 × 109 450 2.80 × 1010 500 2.00 × 1011 Vbi 1.405 1.389 1.370 1.349 1.327 1.302 1.277 1.23 ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ −1/ ⎡ (1.5 × 10 16 )( × 10 15 ) ⎤ ⎥ = 0.684 V Vbi = ( 0.026 ) ln ⎢ ⎢⎣ (1.5 ×10 10 ) ⎥⎦ ⎞ ⎛ C j = ( 0.4 ) ⎜ + ⎟ 0.684 ⎝ ⎠ −1/ (a) ⎞ ⎛ C j = ( 0.4 ) ⎜ + ⎟ ⎝ 0.684 ⎠ −1/ (b) = 0.255 pF = 0.172 pF −1/ ⎞ ⎛ = 0.139 pF C j = ( 0.4 ) ⎜ + ⎟ ⎝ 0.684 ⎠ (c) 1.24 (a) ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ −1 / ⎞ ⎛ For VR = V, C j = (0.02) ⎜ + ⎟ ⎝ ⎠ −1 / ⎛ ⎞ For VR = 1.5 V, C j = (0.02) ⎜1 + ⎟ ⎝ ⎠ = 0.00743 pF −1 / = 0.0118 pF Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 0.00743 + 0.0118 = 0.00962 pF vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ C j (avg ) = where τ = RC = RC j (avg ) = (47 × 103 )(0.00962 × 10−12 ) or τ = 4.52 ×10−10 s Then vC ( t ) = 1.5 = + ( − ) e−ti / τ + r /τ ⎛ ⎞ = e ⇒ t1 = τ ln ⎜ ⎟ 1.5 ⎝ 1.5 ⎠ −10 t1 = 5.44 × 10 s (b) For VR = V, Cj = Cjo = 0.02 pF −1/ ⎛ 3.5 ⎞ For VR = 3.5 V, C j = ( 0.02 ) ⎜ + = 0.00863 pF ⎟ ⎝ 0.8 ⎠ 0.02 + 0.00863 C j (avg ) = = 0.0143 pF τ = RC j ( avg ) = 6.72 ×10−10 s vC ( t ) = vC ( final ) + ( vC ( initial ) − vC ( final ) ) e − t / τ ( 3.5 = + (0 − 5)e − t2 /τ = − e − t2 /τ so that t2 = 8.09 × 10 −10 ) s 1.25 ⎛ V C j = C jo ⎜⎜1 + R ⎝ Vbi ⎞ ⎟⎟ ⎠ −1 / ( )( ) ) ⎡ × 1015 1017 ⎤ ; Vbi = (0.026 ) ln ⎢ ⎥ = 0.739 V 10 ⎥⎦ ⎣⎢ 1.5 × 10 ( For V R = V, Cj = 0.60 1+ 0.739 = 0.391 pF For VR = V, Cj = 0.60 1+ 0.739 = 0.267 pF For V R = V, 0.60 Cj = (a) fo = (b) f o = 2π LC ( 1+ 0.739 = ( = 0.215 pF 2π 1.5 × 10 2π 1.5 × 10 −3 )(0.391× 10 ) −12 −3 )(0.267 × 10 ) −12 ⇒ f o = 6.57 MHz ⇒ f o = 7.95 MHz Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.17 a b logic = V logic = −0.4 V v01 = A OR B v02 = C OR D v03 = v01 OR v02 or v03 = ( A OR B ) AND (C OR D) 17.18 a For CLOCK = high, I DC flows through the left side of the circuit If D is high, I DC flows through the left R resistor pulling Q low If D is low I DC flows through the right R resistor pulling Q low For CLOCK = low, I DC flows through the right side of the circuit maintaining Q and Q in their previous state b P = ( I DC + 0.5I DC + 0.1I DC + 0.1 I DC )( 3) P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 μ W 17.19 (a) (i) For υ I = 0.1 V υ1 = 0.8 V 2.5 − 0.8 = 0.1417 mA 12 i = i = , υ O = V i1 = (ii) For υ I = 2.5 V υ1 = 0.7 + 0.8 = 1.5 V 2.5 − 1.5 = 0.0833 mA 12 υ O = 0.1 V i1 = i = 2.5 − 0.1 = 0.20 mA 12 (b) (i) υ1 = 0.7 + 0.7 = 1.4 V i3 = υ I = υ1 − 0.7 = 0.7 V (ii) υ1 = 0.7 + 0.8 = 1.5 V υ I = υ1 − 0.7 = 0.8 V www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.20 (a) vI = ⇒ V1 = 0.7 V 3.3 − 0.7 = 0.433 mA iB = iC = vo = 3.3 V i1 = (b) vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V 3.3 − 1.5 = 0.3 mA i1 = 0.8 = 0.04 mA iR = 20 iB = 0.3 − 0.04 = 0.26 mA 3.3 − 0.1 iC = = 0.8 mA vo = 0.1 V 17.21 i For i1 = ii vX = vY = 0.1 V ⇒ v′ = 0.8 V − 0.8 ⇒ i1 = 0.525 mA i3 = i4 = For v X = vY = V, v ′ = 0.8 + 0.7 + 0.7 ⇒⇒ v′ = 2.2 V − 2.2 ⇒ i1 = 0.35 mA 0.8 i4 = i1 − ⇒ i4 = 0.297 mA 15 − 0.1 i3 = ⇒ i3 = 2.04 mA 2.4 i1 = 17.22 (i) υ X = υ Y = 0.1 V υ ′ = V 3.3 − 0.8 i1 = = 0.3125 mA i3 = i = (ii) υ X = υ Y = 3.3 V υ ′ = + + = V 3 − 2 i1 = = 0.1375 mA 8 i = 0.1375 − = 0.08417 mA 15 3.3 − 0.1 i3 = = 1.333 mA Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.23 a For v X = vY = V , both Q1 and Q2 driven into saturation v1 = 0.8 + 0.7 + 0.8 ⇒ v1 = 2.3 V − 2.3 ⇒ i1 = iB1 = 0.675 mA − (0.8 + 0.7 + 0.1) i2 = ⇒ i2 = 1.7 mA i4 = iB1 + i2 ⇒ i4 = 2.375 mA i1 = 0.8 ⇒ i5 = 0.08 mA 10 = i4 − i5 ⇒ iB = 2.295 mA i5 = iB i3 = − 0.1 ⇒ i3 = 1.225 mA v0 = 0.1V − (0.1 + 0.7) = 1.05 mA iC (max) = β iB = NiL′ + i3 iL′ = b (20)(2.295) = N (1.05) + 1.225 So N = 42 17.24 (a) υ1 = 0.8 + 0.7 + 0.8 = 2.3 V i1 = υ C1 3.3 − 2.3 = 0.25 mA = i B1 = 0.8 + 0.7 + 0.1 = 1.6 V 3.3 − 1.6 = 0.85 mA i = i B1 + i = 0.25 + 0.85 = 1.10 mA i2 = = 0.08 mA 10 = i − i = 1.10 − 0.08 = 1.02 mA i5 = iB2 3.3 − 0.1 = 0.80 mA (b) iCo (max ) = β i B = i + N i L′ i3 = 3.3 − (0.1 + 0.7 ) = 0.625 mA (20)(1.02) = 0.8 + N (0.625) ⇒ N = 31.36 ⇒ N = 31 i L′ = Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.25 DX and DY off, Q1 forward active mode v1 = 0.8 + 0.7 + 0.7 = 2.2 V = i1 R1 + i2 R2 + v1 and i1 = (1 + β )i2 2[ So β = 25 Assume − 2.2 = i (1 + β ) R1 + R2 ] − 2.2 ⇒ i2 = 0.0589 mA (26)(1.75) + i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA i2 = i3 = β i2 ⇒ i3 = 1.47 mA 0.8 = 0.0589 + 1.47 − 0.16 ⇒ iBo = 1.37 mA Qo in saturation iBo = i2 + i3 − iCo = − 0.1 ⇒ iCo = 0.817 mA 17.26 (a) vI = V, Q1 forward active − 0.7 iB = = 0.717 mA iC = (25)(0.71667) = 17.9 mA iE = (26)(0.71667) = 18.6 mA (b) VI = 0.8 V iB = − (0.8 + 0.7) = 0.583 mA Because of the relative doping levels of the Emitter and collector, and because of the difference in B-C and B-E areas, we have −iC ≈ iB = 0.583 mA and iE = small value (c) vI = 3.6 Q1 inverse active − (0.8 + 0.7) iB = = 0.583 mA iE = − β R iE = −(0.5)(0.583) = −0.292 mA iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.27 (a) (i) For υ I = 0.1 V, υ1 = 0.1 + 0.8 = 0.9 V, and υ O = 2.5 V 2.5 − 0.9 = 0.1333 mA 12 i = i3 = i1 = (ii) For υ I = 2.5 V, υ1 = 0.8 + 0.7 = 1.5 V, and υ O = 0.1 V 2.5 − 1.5 = 0.0833 mA 12 i = i1 (1 + 0.1) = (0.0833)(1.1) = 0.09167 mA i1 = 2.5 − 0.1 = 0.20 mA 12 (b) (i) υ1 = 0.7 + 0.7 = 1.4 V i3 = υ I = 1.4 − 0.8 = 0.6 V (ii) υ1 = 0.8 + 0.7 = 1.5 V υ I = 1.5 − 0.8 = 0.7 V 17.28 a ii v X = vY = 0.1 V, so Q1 in saturation − (0.1 + 0.8) i1 = ⇒ i1 = 0.683 mA ⇒ iB = i2 = i4 = iB = i3 = i v X = vY = V, so Q1 in inverse active mode Assume Q2 and Q3 in saturation − (0.8 + 0.8 + 0.7) ⇒ i1 = iB = 0.45 mA − (0.8 + 0.1) i2 = ⇒ i2 = 2.05 mA 0.8 i4 = ⇒ i4 = 0.533 mA 1.5 i1 = iB = ( iB + i2 ) − i4 = 0.45 + 2.05 − 0.533 or iB = 1.97 mA − 0.1 ⇒ i3 = 2.23 mA 2.2 For Q3 : i3 = b i3 2.23 = = 1.13 < β iB 1.97 For Q2 : i2 2.05 = = 4.56 < β iB 0.45 I /I < β , then each transistor is in saturation Since ( C B ) Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.29 (a) (i) υ X = 0.1 V, υ Y = 3.3 V υ ′ = + = V 3.3 − 0.8 ii = = 0.156 mA 16 i3 = i = (ii) υ X = υ Y = 3.3 V υ ′ = + + = V 3 − 2 i1 = = 0.06875 mA 16 i = 0.06875 − = 0.02875 mA 20 3.3 − 0.1 i3 = = 0.5333 mA (b) i C1 (max ) = β i = i + N i L′ 3.3 − (0.1 + 0.7 ) = 0.15625 mA 16 (50)(0.02875) = 0.5333 + N (0.15625) ⇒ N = 5.8 ⇒ N = i L′ = (c) iC1 (max ) = β i = (50)(0.02875) = 1.44 mA < mA, ⇒ N = 17.30 For v X = vY = V, Q, in inverse active mode a − ( 0.8 + 0.8 + 0.7 ) iB1 = iB = 0.45 mA = iB1 + β R iB1 = 0.45(1 + [ 0.1]) = 0.54 mA iC = − ( 0.8 + 0.1) iB = ( iB + iC ) − = 2.05 mA 0.8 = 0.54 + 2.05 − 0.533 1.5 or iB3 = 2.06 mA Now iL′ = − (0.1 + 0.8) = 0.683 mA Then iC (max) = β F iB = NiL′ or (20)(2.06) = N (0.683) ⇒ N = 60 ′ b From above, for v0 high, I L = (0.1)(0.45) = 0.045 mA Now ⎛ − 4.9 ⎞ (21)(0.1) I L′ (max) = (1 + β F ) ⎜ ⎟ = ⎝ R2 ⎠ = 1.05 mA So I L (max) = NI L′ or 1.05 = N (0.045) ⇒ N = 23 Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.31 (a) (i) Vin = 0.1 V − (0.1 + 0.8) = 1.025 mA = i Bo = , Vout = V i RB = i RCP (ii) Vin = V − (0.7 + 0.8 + 0.7 ) = 0.70 mA = 0.7 + 0.1 = 0.8 V i RB = V out − 0.8 = 4.2 mA = (1.1)(0.7 ) = 0.77 mA i RCP = i BS i Co = β i Bo , iCS = 4.2 − β i Bo , i ES = 0.77 + (4.2 − β i Bo ) 0.7 = 0.77 + 4.2 − β i Bo − 0.7 i Bo = i ES − i Bo (1 + β )i Bo = 4.27 ⇒ i Bo = 4.27 = 0.0837 mA 51 (b) (i) Vin = 0.1 V, V out = High, i L′ = 5β R i RB = 5(0.1)(0.7 ) = 0.35 mA V out = − (0.35)(1) = 4.65 V P = (5 − 0.1)(1.025) + (0.35)(5 − 4.65) = 5.145 mW (ii) i L = 5(1.025) = 5.125 mA P = (0.77 + 4.2 )(5) + (5.125)(0.1) = 25.4 mW 17.32 v X = vY = vZ = 0.1 V a iB1 = − (0.1 + 0.8) ⇒ iB1 = 1.05 mA 3.9 Then iC1 = iB = iC = iB = iC = v X = vY = vZ = V b iB1 = − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.692 mA 3.9 Then iC1 = iB = iB1 (1 + 3β R ) = (0.692)(1 + 3[0.5]) ⇒ iC1 = iB = 1.73 mA − (0.1 + 0.8) ⇒ iC = 2.05 mA 0.8 = iB + iC − = 1.73 + 2.05 − 1.0 0.8 ⇒ iB = 2.78 mA iC = iB www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions − 0.1 = 2.04 mA 2.4 − (0.1 + 0.8) iL′ = = 1.05 mA 3.9 iC = iR + 5iL′ = 2.04 + (5)(1.05) ⇒ iC = 7.29 mA iR = 17.33 = 0.098 μ A 51 = − (0.000098)(2) ≅ V (a) (i) I L′ = μ A, i B = V B5 υ O = V = 0.098 mA 51 = − (0.098)(2) = 4.804 V (ii) I L′ = mA, i B = V B5 υ O = 4.804 − 1.4 = 3.404 V (iii) Q in saturation − V B − (V E + 0.8) = 2 − VC − (V E + 0.1) IC = = 0.13 0.13 − (V E + 0.7 ) − (V E + 0.1) I B + I C = I E = I L = 25 = + 0.13 25 = 2.10 − V E (0.5) + 37.69 − V E (7.692) ⇒ V E = 1.81 V IB = υ O = 1.81 − 0.7 = 1.11 V (b) V B = 0.7 + 0.8 = 1.5 V I B4 = − 1.5 = 1.75 mA VC = 0.7 + 0.1 = 0.8 V, I C = − 0.8 = 32.31 mA 0.13 I L = 1.75 + 32.31 = 34.06 mA Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.34 a v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode 2.8 − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.25 mA iB1 = iB = iB1 (1 + 3β R ) = 0.25(1 + [0.3]) ⇒ iB = 0.475 mA vC = 0.8 + 0.1 = 0.9 V 0.9 − (0.7 + 0.1) 0.1 = (1 + β F )(0.5) (101)(0.5) iB = iR = 0.00198 mA (Negligible) − 0.9 = = 4.56 mA 0.9 ⇒ iC = 4.56 mA 0.8 = 0.475 + 4.56 − 0.8 = 4.235 mA iB = iB + iC − ⇒ iB v X = vY = vZ = 0.1 V b iB1 = − (0.1 + 0.8) ⇒ iB1 = 2.05 mA From part (a), iL′ = β R ⋅ iB1 = (0.3)(0.25) = 0.075 mA Then iB = 5iL′ 5(0.075) = ⇒ iB = 0.00371 mA 1+ βF 101 17.35 v X = vY = vZ = 0.1 V a iB1 = − (0.1 + 0.8) + iB RB1 where iB = (2 − 0.7) − (0.9) 0.4 = RB ⇒ iB = 0.4 mA Then iB1 = 1.1 + 0.4 ⇒ iB1 = 1.5 mA iB = = iC Q3 in saturation iC = 5iL′ For v0 high, vB′ = 0.8 + 0.7 = 1.5 V ⇒ Q3′ off − 1.5 = 0.5 mA iL′ = β R iB′ = (0.2)(0.5) = 0.1 mA iB′ = Then iC = 0.5 mA Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions v X = vY = vZ = V b From part (a), ⇒ iB1 = 0.5 mA iB = = iC iB = iB1 (1 + 3β R ) = (0.5)(1 + [0.2]) iB = 0.8 mA iC = 5iL′ , ′ and from part (a), iL = 1.5 mA So iC = 7.5 mA 17.36 IB + ID = (a) IC − I D = 5.8 − 0.7 = 0.51 mA 10 − (0.7 − 0.3) = 4.6 mA Now I D = 0.51 − I B = 0.51 − IC β = 0.51 − IC 50 Then So I ⎞ ⎞ ⎛ ⎛ I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜1 + ⎟ − 0.51 = 4.6 50 50 ⎝ ⎠ ⎝ ⎠ I C = 5.01 mA IC 5.01 ⇒ I B = 0.1002 mA 50 I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA IB = β = VCE = 0.4 V (b) I D = 0,VCE = VCE ( sat ) = 0.1 V 5.8 − 0.8 ⇒ I B = 0.5 mA 10 − 0.1 IC = ⇒ I C = 4.9 mA IB = 17.37 (a) (i) υ I = , υ1 = 0.3 V − = 1.2 mA i B = i C = , υ O = 1.5 V i1 = (ii) υ I = 1.5 V, υ1 = 0.7 + 0.3 = 1.0 V 1.5 − 1.0 = 0.5 mA i B = 0.5 − = 0.465 mA 20 i1 = www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 1.5 − 0.4 = 0.9167 mA, υ O = 0.4 V 1.2 (b) (i) υ1 = 0.7 + 0.3 = 1.0 V, υ I = 0.7 V iC = i B = iC = (ii) υ1 = 1.0 V, υ I = 0.7 V 1.5 − 0.4 = 0.9167 mA 1.2 i 0.9167 iB = C = = 0.03667 mA 25 β (c) iCo (max ) = β i B = iC + N i L′ iC = 1.5 − (0.4 + 0.3) 0.4 − = 0.78 mA 20 (25)(0.465) = (0.9167) + N (0.78) ⇒ N = 13.7 ⇒ N = 13 i L′ = 17.38 a v X = vY = 0.4 V vB1 = 0.4 + 0.7 ⇒ vB1 = 1.1 V − 1.1 ⇒ iB1 = 1.39 mA 2.8 = 0.4 + 0.4 ⇒ vB = 0.8 V iB1 = vB iB = iC = iB = iC = iB = iC = iB = iC = ( No load ) = iB R2 + VBE + (1 + β )iB R4 − 0.7 iB = ⇒ iB = 0.0394 mA 0.76 + (31)(3.5) iC = β F iB ⇒ iC = 1.18 mA vB = − (0.0394)(0.76) ⇒ vB = 4.97 V b v X = vY = 3.6 V vB1 = 0.7 + 0.7 + 0.3 ⇒ vB1 = 1.7 V vB = 1.4 V vB = 0.7 V vC = 1.1 V − 1.7 ⇒ iB1 = 1.1786 mA 2.8 = iB1 (1 + β R ) = 1.18(1 + [0.1]) iB = 1.41 mA iB1 = iB 1.1 − 0.7 ⇒ iB = 0.00369 mA (31)(3.5) − 1.1 = = 5.13 mA ⇒ iC ≈ 5.13 mA 0.76 ≈ iB + iC iB = 6.54 mA iB = iR iB Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.39 (a) For the load, i RB1 = 2.5 − (0.4 + 0.3) = (0.2 ) ⇒ R B1 = 18 k Ω R B1 For υ X = υ Y = υ Z = logic 2.5 − (0.7 + 0.8) = 0.05556 mA 18 2.5 − (0.7 + 0.1) = R C1 i RB1 = i RC1 i B = 0.1 = 0.05556 + 1.7 0.7 − ⇒ RC1 = 1.63 k Ω RC1 0.7 (b) υ X = 0.4 V, υ B1 = 0.7 V, υ B = υ O ≅ 2.5 − 0.7 = 1.8 V All i B = , All iC = (c) υ B1 = 1.5 V, υ B = 0.7 V 2.5 − 1.5 = 0.0556 mA 18 2.5 − (0.7 + 0.1) = = 1.043 mA 1.63 = 0.10 mA i B1 = i C1 iB2 4[2.5 − (0.4 + 0.3)] = 0.40 mA 18 υ O = V (d) iC (max ) = β i B = N i L′ i L′ = 0.1 mA iC = (20)(0.1) = N (0.1) ⇒ N = 20 17.40 a For v X = vY = 3.6 V − 2.1 = 0.29 mA 10 − 1.8 = 0.32 mA vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 = 10 1.4 iB = iB1 + iC1 − = 0.29 + 0.32 − 0.0933 15 vB1 = 3(0.7) = 2.1 ⇒ iB1 = So iB = 0.517 mA vC = 0.7 + 0.4 = 1.1 V iC = − 1.1 = 0.951 mA 4.1 iB = iB + iC − or iB = 1.293 mA 0.7 = 0.517 + 0.951 − 0.175 ′ For v0 = 0.4 V, vB1 = 0.4 + 0.7 = 1.1 V Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions Then 1.1 − 0.7 = 0.00086 mA (31)(15) − 1.1 − 0.00086 or iL′ ≈ 0.39 mA iL′ = 10 iC (max) = β iΒ = NiL′ iB′ = So (30)(1.293) = N (0.39) ⇒ N = 99 b P = (0.29 + 0.32 + 0.951)(5) + (99)(0.39)(0.4) P = 7.805 + 15.444 or P = 23.2 mW (Assumming 99 load circuits which is unreasonably large.) 17.41 a Assume no load For v X = logic = 0.4 V iE1 = − (0.4 + 0.7) = 0.0975 mA 40 Essentially all of this current goes to ground from VCC P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW b iR1 = − (3)(0.7) = 0.0725 mA 40 − (0.7 + 0.7 + 0.4) = 0.064 mA 50 − (0.7 + 0.4) iR = = 0.26 mA 15 P = (0.0725 + 0.064 + 0.26)(5) iR = P = 1.98 mW c For v0 = 0, vC = 0.7 + 0.4 = 1.1 V iR = − 1.1 ⇒ iR = 78 mA ≈ iSC 0.050 17.42 − (0.7 + 0.3) = 1.0 mA 2.4 − 0.7 (b) i E = = 0.85 mA − RC = = 0.706 k Ω 0.85 (c) (i) P = (1.0 )(3) = 3.0 mW (a) i E = (ii) P = (0.85)(3) = 2.55 mW www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions 17.43 (a) v I = v O = 2.5 V; A transient situation vDS ( M N ) = 2.5 − 0.7 = 1.8 V vGS ( M N ) = 2.5 − 0.7 = 1.8 V ⇒ M N in saturation vSD ( M P ) = − (2.5 + 0.7) = 1.8 V vSG ( M P )5 − 2.5 = 2.5 V ⇒ M P in saturation iDN = K n (vGSN − VTN )2 = (0.1)(1.8 − 0.8) ⇒ iDN = 0.1 mA iDP = K P (vSGP + VTP )2 = (0.1)(2.5 − 0.8) ⇒ iDP = 0.289 mA iC1 = β iDP = (50)(0.289) ⇒ iC1 = 14.45 mA iC = β iDN = (50)(0.1) ⇒ iC = mA Difference between iE1 and iDN + iC is a load current (b) Assume iC1 = 14.45 mA is a constant VC = i ⋅t (V )(C ) iC1dt = C1 ⇒ t = C ∫ C C iC1 (5)(15 × 10−12 ) ⇒ t = 5.19 ns 14.45 × 10−3 (5)(15 × 10−12 ) t= ⇒ t = 260 ns 0.289 × 10−3 t= (c) 17.44 Microelectronics: Circuit Analysis and Design, 4th edition Chapter 17 By D A Neamen Problem Solutions (a) Assume R1 = R2 = 10 kΩ; β = 50 Then iR1 = iR = 0.7 = 0.07 mA 10 NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 ) 2 iDN = 0.10 mA Then iB = 0.03 ⇒ iC = (50)(0.03) = 1.5 mA iE1 = 1.53 mA ⇒ iB1 = 0.03 mA ⇒ iC1 = 1.5 mA i So DP = 0.10 mA Now, M P biased in non-saturation region vSGP = 2.5 V ⎤⎦ iDP = 0.10 = 0.10 ⎡⎣ 2(2.5 − 0.8)vSD − vSD 0.10 vSD − 0.34 vSD + 0.10 = vSD = 0.34 ± (0.34) − 4(0.10)(0.10) 2(0.10) Or vSD = 0.325 V Then vo = − 0.325 − 0.7 vo = 3.975 V v= (b) i ⋅t idt = C∫ C Cv (15 × 10−12 )(5) = i 1.53 × 10−3 t = 49 ns t= (c) Cv (15 × 10−12 )(5) = i 0.1× 10−3 t = 0.75 μ s t= ... ( max ) − VZ ( nom ) Ri = 12 − = 1.333 A For I L = 0.2 A ⇒ I Z = 1.133 A For I L = A ⇒ I Z = 0.333 A Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions. .. ) 25 + 2Vr + 3Vr = 11VA (a) Vr = 0.6 V 11VA = 25 + ( 0.6 ) = 28 ⇒ VA = 2.545 V Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions ... www.elsolucionario.org Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Problem Solutions 2.53 a R1 = kΩ, R2 = 10 kΩ ⇒ V0 = D1 and