Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions Chapter Exercise Solutions EX1.1 ⎛ − Eg ⎞ ni = BT / exp ⎜ ⎟ ⎝ 2kT ⎠ GaAs: ni = ( 2.1× 1014 ) ( 300 ) ⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ − ⎜ ( 86 × 10 ) ( 300 ) ⎟ ⎝ ⎠ 3/ ⎛ ⎞ − 0.66 ⎟ or ni = 2.40 × 1013 cm −3 exp⎜⎜ −6 ⎟ ( ) 86 10 300 × ⎝ ⎠ ( ) Ge: n i = 1.66 × 1015 (300) 3/ ( ) EX1.2 (a) (i) n o = N d = 2×1016 cm ( ) n i2 1.5 × 1010 = no × 1016 po = (ii) p o = N a = 1015 cm ( ) no = N d = × 1016 cm ( ni2 1.8 × 10 = no × 1016 po = (ii) = 1.125 × 10 cm −3 −3 n2 1.5 × 10 10 no = i = po 1015 (b) (i) p o = N a = 1015 cm ( −3 ) = 2.25 × 10 cm −3 −3 = 1.62 × 10 − cm −3 −3 ) n i2 1.8 × 10 = = 3.24 × 10 − cm −3 po 10 15 no = EX1.3 (a) For n-type; ρ= (b) J = ρ 1 = = 0.046 ohm-cm −19 eμ n N d 1.6 × 10 (6800) × 1016 ( ) ( ) ⋅ Ε ⇒ Ε = ρ J = (0.046)(175) = 8.04 V/cm EX1.4 Diffusion current density due to holes: dp J p = −eD p dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎝ p⎠ ⎝ p⎠ Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions (a) At x = (1.6 ×10 ) (10) (10 ) = 16 A / cm = −19 Jp (b) At x = 10−3 16 10−3 cm ⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm ⎝ 10 ⎠ EX1.5 ( )( ) ( ) ⎡ (10 )(10 ) ⎤ = (0.026 ) ln ⎢ ⎥ = 0.374 V ⎢⎣ (2.4 × 10 ) ⎥⎦ ⎡ 10 16 10 17 ⎤ (a) Vbi = (0.026 ) ln ⎢ ⎥ = 1.23 V ⎣⎢ 1.8 × 10 ⎦⎥ 16 (b) Vbi 17 13 EX1.6 −1/ ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and ⎡N N ⎤ Vbi = VT ln ⎢ a d ⎥ ⎣ ni ⎦ ⎡ (1017 )(1016 ) ⎤ ⎥ = 0.757 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ ⎞ ⎛ Then 0.8 = C jo ⎜ + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF −1/ = C jo ( 7.61) −1/ EX1.7 ⎛I ⎞ (a) V D = VT ln⎜⎜ D ⎟⎟ ⎝ IS ⎠ ⎛ 50 × 10 −6 (i) V D = (0.026) ln⎜⎜ −14 ⎝ × 10 ⎞ ⎟ = 0.563 V ⎟ ⎠ ⎛ 10 −3 (ii) V D = (0.026 ) ln⎜⎜ −14 ⎝ × 10 ⎞ ⎟ = 0.641 V ⎟ ⎠ ⎛ 50 × 10 −6 (b) (i) V D = (0.026) ln⎜⎜ −12 ⎝ × 10 ⎞ ⎟ = 0.443 V ⎟ ⎠ ⎛ 10 −3 ⎞ ⎟ = 0.521 V (ii) V D = (0.026 ) ln⎜⎜ −12 ⎟ ⎝ × 10 ⎠ Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions EX1.8 ⎛V ⎞ VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ ( − VD ) so = I D ( × 103 ) + VD ⇒ I D = ×103 and ⎛ V ⎞ I D = (10 −12 ) exp ⎜ D ⎟ ⎝ 0.026 ⎠ By trial and error, we find I D ≅ 0.866 mA and V D ≅ 0.535 V EX1.9 V PS − Vγ − 0.7 ⇒R= = 6.08 k Ω R 1.20 − (b) I D = = 0.9429 mA PD = I DV D = (0.9429 )(0.7 ) = 0.66 mW ID = (a) EX1.10 PSpice Analysis EX1.11 − 0.7 = 0.365 mA 20 V 0.026 ⇒ 71.2 Ω rd = T = I D 0.365 (a) I D = 0.25 sin ω t ⇒ 12.5 sin ω t ( μ A) 20 + 0.0712 − (b) I D = = 0.73 mA 10 0.026 rd = ⇒ 35.6 Ω 0.73 0.25 sin ω t id = ⇒ 24.9 sin ω t ( μ A) 10 + 0.0356 id = EX1.12 ⎛I ⎞ ⎛ 1.2 × 10−3 ⎞ or VD = 0.6871 V For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ × 10 ⎠ ⎝ IS ⎠ The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions or 1.2 × 10−3 ⇒ I S = 1.07 × 10−10 A ⎛ 0.4221 ⎞ exp ⎜ ⎟ ⎝ 0.026 ⎠ IS = EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA 10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R Also I = Test Your Understanding Solutions TYU1.1 (a) T = 400K ⎛ − Eg ⎞ Si: ni = BT / exp ⎜ ⎟ ⎝ 2kT ⎠ ni = ( 5.23 × 1015 ) ( 400 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 400 ) ⎥⎦ or ni = 4.76 × 1012 cm −3 Ge: ni = (1.66 × 1015 ) ( 400 ) 3/ ⎡ ⎤ −0.66 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 400 ) ⎥⎦ or ni = 9.06 × 1014 cm −3 GaAs: ni = ( 2.1× 1014 ) ( 400 ) 3/ ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 400 ) ⎥⎦ or ni = 2.44 × 109 cm −3 (b) T = 250 K Si: ni = ( 5.23 × 1015 ) ( 250 ) 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 250 ) ⎥⎦ or ni = 1.61× 108 cm −3 Ge: ni = (1.66 × 1015 ) ( 250 ) 3/ ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 250 ) ⎥⎦ or ni = 1.42 × 1012 cm −3 Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions GaAs: ni = ( 2.10 × 1014 ) ( 250 ) 3/ ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 250 ) ⎥⎦ or ni = 6.02 × 103 cm −3 TYU1.2 ( ) ( ) (a) σ = eμ p N a = 1.6 × 10 −19 (480) × 10 15 = 0.154 (ohm-cm) ρ= σ = = 6.51 Ω -cm 0.1536 ( ) ( ) (b) σ = eμ n N d = 1.6 × 10 −19 (1350) × 1017 = 43.2 (ohm-cm) ρ= σ = −1 −1 = 0.0231 Ω -cm 43.2 TYU1.3 (a) J = σ Ε = (0.154)(4) = 0.616 A/cm (b) J = σ Ε = (43.2)(4) = 172.8 A/cm TYU1.4 (a) J n = eDn ⎛ 1015 − 1016 ⎞ dn Δn so J n = 1.6 × 10−19 ( 35 ) ⎜ = eDn −4 ⎟ dx Δx ⎝ − 2.5 × 10 ⎠ ( ) or J n = 202 A / cm (b) J p = −eD p ⎛ 1014 − × 1015 ⎞ dp Δp so J p = − 1.6 × 10−19 (12.5 ) ⎜ = −eD p −4 ⎟ dx Δx ⎝ − × 10 ⎠ ( ) or J p = −24.5 A / cm2 TYU1.5 (a) no = N d = × 1015 cm −3 10 ni2 (1.5 × 10 ) po = = = 2.81× 10 cm −3 no × 1015 (b) n = no + δ n = × 1015 + 0.1× 1015 or n = 8.1×1015 cm−3 p = po + δ p = 2.81 × 10 + 1014 or p ≅ 1014 cm −3 Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions TYU1.6 ( )( ( ) ⎡ 10 15 × 10 16 ⎤ ⎞ ⎟ = (0.026 ) ln ⎢ ⎥ = 0.679 V ⎟ ⎢⎣ 1.5 × 10 10 ⎥⎦ ⎠ ⎡ 10 15 × 10 16 ⎤ (b) Vbi = (0.026 ) ln ⎢ ⎥ = 1.15 V ⎢⎣ 1.8 × 10 ⎥⎦ ⎡ 10 15 × 10 16 ⎤ (c) Vbi = (0.026 ) ln ⎢ ⎥ = 0.296 V ⎢⎣ 2.4 × 10 13 ⎥⎦ ⎛N N (a) Vbi = VT ln⎜⎜ a d ⎝ ni ( )( ( ( )( ( ) ) ) ) ) TYU1.7 ⎛V (a) (i) I D = I S exp⎜⎜ D ⎝ VT ( ) ( ) (ii) I D = 10 −16 ⎞ ⎛ 0.55 ⎞ ⎟⎟ = 10 −16 exp⎜ ⎟ ⇒ 0.154 μ A ⎝ 0.026 ⎠ ⎠ ⎛ 0.65 ⎞ exp⎜ ⎟ ⇒ 7.20 μ A ⎝ 0.026 ⎠ ( ) ⎛ 0.75 ⎞ (ii) I D = 10 −16 exp⎜ ⎟ ⇒ 0.337 mA ⎝ 0.026 ⎠ (b) (i) I D = −10 −16 A (ii) I D = −10 −16 A TYU1.8 ΔT = 100C so ΔVD ≅ × 100 = 200 mV Then VD = 0.650 − 0.20 = 0.450 V TYU1.9 Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions TYU1.10 (a) I D = (b) (c) − 0.7 = 0.325 mA − 0.7 ID = = 1.075 mA ID = ID = (d) (e) I D = TYU1.11 P = I DVD ⇒ 1.05 = I D ( 0.7 ) so I D = 1.5 mA VPS − Vγ 10 − 0.7 = ⇒ R = 6.2 kΩ 1.5 ID Now R = TYU1.12 ID 0.8 = = 30.8 mS VT 0.026 gd = TYU1.13 rd = VT 0.026 = = 2.6 k Ω I D 0.010 0.026 ⇒ 260 Ω 0.10 0.026 rd = ⇒ 26 Ω rd = TYU1.14 rd = VT 0.026 0.026 ⇒ 50 = ⇒ ID = 50 ID ID or I D = 0.52 mA TYU1.15 For the pn junction diode, − 0.7 ID = = 0.825 mA 4 − 0.3 = 0.925 mA For the Schottky diode, I D = Full file at https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition Chapter By D A Neamen Exercise Solutions TYU1.16 Vz = Vzo + I z rz ⇒ Vzo = Vz − I z rz so Vzo = 5.20 − (10−3 ) ( 20 ) = 5.18 V Then Vz = 5.18 + (10 × 10−3 ) ( 20 ) ⇒ Vz = 5.38 V TYU1.17 P = I Z VZ ⇒ I Z = P 6.5 = = 1.81 mA V Z 3.6 V PS = I Z R + V Z = (1.81)(4 ) + 3.6 = 10.8 V Full file at https://TestbankDirect.eu/ .. .Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition. .. https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition. .. https://TestbankDirect.eu/ Solution Manual for Microelectronics Circuit Analysis and Design 4th Edition by Neamen Full file at https://TestbankDirect.eu/ Microelectronics: Circuit Analysis and Design, 4th edition