TEST BANK FOR Microelectronics Circuit Analysis and Design 4th Edition By Donald Neamen

,Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
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Chapter 1
1.1
− Eg / 2 kT
ni = BT e
(a) Silicon
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 250 )
3/ 2
(i) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦
= 2.067 × 1019 exp [ −25.58]
ni = 1.61× 108 cm −3
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 350 )
3/ 2
(ii) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= 3.425 × 1019 exp [ −18.27 ]
ni = 3.97 ×1011 cm −3

(b) GaAs
⎡ −1.4 ⎤
ni = ( 2.10 × 1014 ) ( 250 )
3/ 2
(i) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦
−6



= ( 8.301× 1017 ) exp [ −32.56]
ni = 6.02 × 103 cm −3
⎡ −1.4 ⎤
ni = ( 2.10 × 1014 ) ( 350 )
3/ 2
(ii) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10−6 ) ( 350 ) ⎥⎦
= (1.375 × 1018 ) exp [ −23.26]
ni = 1.09 × 108 cm −3
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1.2
⎛ − Eg ⎞
a. 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 ⎜ ⎟
⎜ 2 ( 86 × 10−6 ) (T ) ⎟
⎝ ⎠
⎛ 6.40 × 103 ⎞
1.91× 10−7 = T exp ⎜ − ⎟
⎝ T ⎠
By trial and error, T ≈ 268° K
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,Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1
By D. A. Neamen Problem Solutions
______________________________________________________________________________________

1.3
Silicon
⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) (100 )
3/ 2
(a) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10−6 ) (100 ) ⎥⎦
= ( 5.23 × 1018 ) exp [ −63.95]
ni = 8.79 ×10−10 cm −3


⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 300 )
3/ 2
(b) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦
−6



= ( 2.718 × 1019 ) exp [ −21.32]
ni = 1.5 × 1010 cm −3


⎡ −1.1 ⎤
ni = ( 5.23 × 1015 ) ( 500 )
3/ 2
(c) exp ⎢ ⎥
⎢⎣ 2 ( 86 × 10−6 ) ( 500 ) ⎥⎦
= ( 5.847 × 1019 ) exp [ −12.79]
ni = 1.63 × 1014 cm −3

Germanium.
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) (100 ) ⎥ = (1.66 × 1018 ) exp [ −38.37 ]
3/ 2
(a) exp ⎢
⎢⎣ 2 ( 86 × 10 ) (100 ) ⎥⎦
−6


ni = 35.9 cm −3
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) ( 300 ) ⎥ = ( 8.626 × 1018 ) exp [ −12.79]
3/ 2
(b) exp ⎢
⎢⎣ 2 ( 86 × 10 ) ( 300 ) ⎥⎦
−6


ni = 2.40 × 1013 cm −3
⎡ −0.66 ⎤
ni = (1.66 × 1015 ) ( 500 ) ⎥ = (1.856 × 1019 ) exp [ −7.674]
3/ 2
(c) exp ⎢
⎢⎣ 2 ( 86 × 10 ) ( 500 ) ⎥⎦
−6


ni = 8.62 ×1015 cm −3
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1.4

(a) n-type; no = 10 15 −3 n2
cm ; po = i =
2.4 × 1013 ( ) 2

= 5.76 × 1011 cm −3
no 1015
ni2
(b) n-type; no = 1015 cm −3 ; po =
=
1.5 × 1010 (
= 2.25 × 10 5 cm −3
)2


no 1015
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, 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 )
2

= 3.24 × 10 − 4 cm −3
po 1016

ni2
(b) p-type; p o = 1016 cm −3 ; no =
=
2.4 × 1013 (
= 5.76 × 1010 cm −3
)
2


po 1016
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1.6
(a) n-type
(b) no = N d = 5 × 1016 cm −3

ni2 (1.5 × 10 )
10 2

po = = = 4.5 × 103 cm −3
no 5 × 1016
(c) no = N d = 5 × 1016 cm −3
From Problem 1.1(a)(ii) ni = 3.97 × 1011 cm −3

po =
( 3.97 × 10 )
11 2

= 3.15 × 106 cm −3
5 × 1016
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1.7

(a) p-type; p o = 5 × 1016 cm −3 ; no =
ni2
=
(
1.5 × 1010 ) 2

= 4.5 × 10 3 cm −3
po 5 × 1016
ni2
=
1.8 × 10 6
(b) p-type; p o = 5 × 1016 cm −3 ; no =
(
= 6.48 × 10 −5 cm −3
)
2


po 5 × 1016
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1.8
(a) Add boron atoms
(b) N a = po = 2 × 1017 cm −3

(c) no =
ni2
=
(
1.5 × 1010 )
2

= 1.125 × 10 3 cm −3
po 2 × 1017
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1.9
(a) no = 5 × 1015 cm −3

n 2 (1.5 × 10 )
10 2

po = i = ⇒ po = 4.5 × 104 cm −3
no 5 × 1015
(b) n o > p o ⇒ n-type
(c) no ≅ N d = 5 × 1015 cm −3
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