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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Chapter 1 1.1 ni = BT 3 / 2 e (a) Silicon (i) − Eg / 2 kT ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦ = 2.067 × 1019 exp [ −25.58] ni = ( 5.23 × 1015 ) ( 250 ) 3/ 2 ni = 1.61× 108 cm −3 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦ = 3.425 × 1019 exp [ −18.27 ] ni = 3.97 ×1011 cm −3 (ii) ni = ( 5.23 × 1015 ) ( 350 ) (b) GaAs (i) ni = ( 2.10 × 1014 ) ( 250 ) 3/ 2 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦ = ( 8.301× 1017 ) exp [ −32.56] ni = 6.02 × 103 cm −3 (ii) ni = ( 2.10 × 1014 ) ( 350 ) 3/ 2 ⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦ = (1.375 × 1018 ) exp [ −23.26] ni = 1.09 × 108 cm −3 ______________________________________________________________________________________ 1.2 a. ⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ⎛ ⎞ −1.1 1012 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎟ −6 ⎝ 2(86 × 10 )(T ) ⎠ ⎛ 6.40 × 103 ⎞ 1.91×10−4 = T 3 / 2 exp ⎜ − ⎟ T ⎝ ⎠ By trial and error, T ≈ 368 K b. ni = 109 cm −3 ⎛ ⎞ −1.1 ⎟ 109 = 5.23 × 1015 T 3 / 2 exp ⎜ ⎜ 2 ( 86 × 10−6 ) (T ) ⎟ ⎝ ⎠ ⎛ 6.40 × 103 ⎞ 1.91× 10−7 = T 3 / 2 exp ⎜ − ⎟ 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 (a) ni = ( 5.23 × 1015 ) (100 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦ = ( 5.23 × 1018 ) exp [ −63.95] ni = 8.79 ×10−10 cm −3 (b) ni = ( 5.23 × 1015 ) ( 300 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦ = ( 2.718 × 1019 ) exp [ −21.32] ni = 1.5 × 1010 cm −3 (c) ni = ( 5.23 × 1015 ) ( 500 ) 3/ 2 ⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦ = ( 5.847 × 1019 ) exp [ −12.79] ni = 1.63 × 1014 cm −3 Germanium. (a) ni = (1.66 × 1015 ) (100 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ = (1.66 × 1018 ) exp [ −38.37 ] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦ ni = 35.9 cm −3 (b) ni = (1.66 × 1015 ) ( 300 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ = ( 8.626 × 1018 ) exp [ −12.79] exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦ ni = 2.40 × 1013 cm −3 (c) ni = (1.66 × 1015 ) ( 500 ) 3/ 2 ⎡ ⎤ −0.66 ⎥ = (1.856 × 1019 ) exp [ −7.674] exp ⎢ −6 ⎢⎣ 2 ( 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 2 = 5.76 × 1011 cm −3 ni2 1.5 × 1010 = = 2.25 × 10 5 cm −3 no 1015 ______________________________________________________________________________________ (b) n-type; no = 1015 cm −3 ; po = Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5 (a) p-type; p o = 1016 cm −3 ; no = ( ni2 1.8 × 10 6 = po 1016 ) ( 2 = 3.24 × 10 − 4 cm −3 ) 2 ni2 2.4 × 1013 = = 5.76 × 1010 cm −3 po 1016 ______________________________________________________________________________________ (b) p-type; p o = 1016 cm −3 ; no = 1.6