Solution manual microelectronics; circuit analysis desing 3rd edition

www.elsolucionario.net www.elsolucionario.net www.elsolucionario.net Chapter Exercise Problems EX1.1 ⎛ − Eg ⎞ ni = BT / exp ⎜ ⎟ ⎝ 2kT ⎠ GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 ) 3/ 3/ ⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ ⎛ ⎞ −0.66 ⎟ or ni = 2.40 × 1013 cm −3 exp ⎜ ⎜ ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠ EX1.2 (a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons, n (1.5 × 10 no = i = 1017 po ) 10 (b) majority carrier: electrons, no = × 1015 cm −3 minority carrier: holes, n (1.5 × 10 ) = 4.5 × 104 cm −3 po = i = × 1015 no 10 EX1.3 For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields E = 17.9 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⎠ (a) At x = (1.6 ×10 ) (10 ) (10 ) = 16 A / cm = −19 Jp 16 10−3 −3 (b) At x = 10 cm ⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm ⎝ 10 ⎠ EX1.5 ⎡N N Vbi = VT ln ⎢ a d ⎣ ni ⎡ (1016 )(1017 ) ⎤ ⎤ ⎢ ⎥ or Vbi = 1.23 V = 0.026 ln ( ) ⎥ ⎥ ⎢ ⎦ ⎣ (1.8 × 10 ) ⎦ EX1.6 ⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and −1/ www.elsolucionario.net = 2.25 × 103 cm −3 www.elsolucionario.net ⎡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 ⎡ ⎛ v ⎞ ⎤ so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎦ ⎣ ⎡ 10−3 ⎤ Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥ ⎣10 ⎦ or vD ≅ ( 0.026 ) ln (1010 ) which yields vD = 0.599 V 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 ≅ 0.864 mA and VD ≅ 0.535 V EX1.9 (a) ID = (b) ID = Then R = (c) VPS − Vγ R VPS − Vγ R − 0.7 ⇒ I D = 1.08 mA VPS − Vγ ⇒R= ID = − 0.7 = 6.79 kΩ 1.075 www.elsolucionario.net ⎡ ⎛v ⎞ ⎤ iD = I S ⎢exp ⎜ D ⎟ − 1⎥ ⎝ VT ⎠ ⎥⎦ ⎣⎢ www.elsolucionario.net ID(mA) Diode curve 1.25 1.08 Load lines (b) (a) 0.7 VD(v) EX1.10 PSpice analysis Quiescent diode current I DQ = VPS − Vγ = 10 − 0.7 = 0.465 mA 20 R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 kΩ I DQ 0.465 Then id = vI 0.2sin ω t (V ) = ⋅ or id = 9.97sin ω t ( μ A) rd + R 0.0559 + 20 ( kΩ ) 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 ⎠ or 1.2 × 10−3 IS = ⇒ I S = 1.07 × 10−10 A 0.4221 ⎛ ⎞ exp ⎜ ⎟ ⎝ 0.026 ⎠ EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA Also I = 10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R Test Your Understanding Exercises TYU1.1 (a) T = 400K ⎛ − Eg ⎞ Si: ni = BT / exp ⎜ ⎟ ⎝ 2kT ⎠ ni = ( 5.23 × 1015 ) ( 400 ) or ni = 4.76 × 1012 cm −3 3/ ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 400 ) ⎥⎦ www.elsolucionario.net EX1.11 www.elsolucionario.net Ge: ni = (1.66 × 1015 ) ( 400 ) 3/ ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 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 ) ⎥⎦ Ge: ni = (1.66 × 1015 ) ( 250 ) 3/ ⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢⎣ ( 86 × 10 ) ( 250 ) ⎥⎦ or ni = 1.42 × 1012 cm −3 GaAs: ni = ( 2.10 × 1014 ) ( 250 ) 3/ ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ ( 86 × 10−6 ) ( 250 ) ⎥⎦ or ni = 6.02 × 103 cm −3 TYU1.2 (a) n = × 1016 cm −3 , p