[솔루션] 반도체공학 1판( 저자 Neamen 1st - Introduction to Semiconductor Devices) PDF

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Solutions Manual Exercise Solutions Chapter 1Exercise ProblemsEX1. For a body-centered cubic, 2 atoms per unit cell. ThenVolume Density == 22 −a 3 o b475 10. x 8 g 3Or Volume Density = 187 10./ xa 22 tomscm 3####### ____EX1.For a (132) →→FH 111 IK HF 6 KI66 26 pqs 3,, ,,So that pq s ===623,,####### ...

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An Introduction to Semiconductor Devices Chapter 1 Solutions Manual Exercise Solutions ______________________________________________________________________________________

Chapter 1 Exercise Problems

(b) Also (113) plane ⇒

FH

EX1.1 For a body-centered cubic, 2 atoms per unit cell. Then 2

Volume Density =

3

=

ao

b

2

4 .75 x10

−8

Intercepts of plane are p = 3 , q = 3 , s = 1 _______________________________________

g

3

Or 22

Volume Density = 187 . x10 atoms / cm

Test Your Understanding Exercises

3

____________________________________ EX1.2 For a (132) →

F H

1

I F K H

1 1 , → p q s ,

IK

1 1 1 → ( 3, 3, 1) , , 1 1 3

6 6 6 , , 6 2 3

TYU1.1 Body-centered cubic = 2 atoms per unit cell. 2 Density = 3 ao

IK

So that p = 6, q = 2, s = 3

____________________________________

Then a o = 3.42 A Now

EX1.3 (a) For a body-centered cubic, (100) plane, effectively 1 atom per plane. Then Surface Density =

1 atom 2

=

ao

b

1

4.75 x10

−8

g

14

=

(a ) o

c h b ao

2

=

−8

4. 75x 10

g

2

(3.42 ) 3 4

r = 1.48 A

°

2

____________________________________ TYU1.2

d=

2

ao 2

(4.83) 2 4

°

⇒ d = 3.42 A

TYU1.3

2

Surface Density = 6.27 x 10 atoms / cm

=

____________________________________

Or 14

5 x10

or

2

2

−23

= 4 x10

22

°

4 r = ao 3 ⇒ r =

Or Surface Density = 4 .43 x10 atoms / cm (b) (110) plane, effectively 2 atoms per plane. Then Surface Density 2 atoms

2

3

⇒ ao =

Density =

2

8 3

ao

____________________________________

=

8

b5.43x10 g −8

3

Or Density = 5x 10 22 cm −3

EX1.4 (a) For p = 1 , q = 1 , s = 3 ; then [113] direction

____________________________________

1

An Introduction to Semiconductor Devices Chapter 1 Solutions Manual Problem Solutions ______________________________________________________________________________________

Chapter 1 Problem Solutions

(b) Face-centered cubic lattice d d = 4r = a 2 ⇒ a = =2 2 r 2

1.1 (a) fcc: 8 corner atoms × 1/8 = 1 atom 6 face atoms × ½ = 3 atoms Total of 4 atoms per unit cell

h = 16 2 r F 4 πr IJ 4 atoms per cell, so atom vol. = 4G H 3K Then

Ratio =

−8

g

3

⇒

Density of Ga = 2.22 x10

22

Unit cell vol. = a cm

F 4 rI = H3K

3

FG 4πr IJ H3K 3

Then

Density of As = 2.22 x10 cm _______________________________________

FG 4π r IJ H 3 K × 100% ⇒ Ratio = F 4r I H 3K 3

2

1.3

3

8 Ge atoms per unit cell 8 Density = ⇒ 8 3 5.65x10−

g

22

3

2 atoms per cell, so atom vol. = 2

−3

22

Ratio = 74%

−3

4 As atoms per unit cell, so that

b

3

16 2 r (c) Body-centered cubic lattice 4 d = 4r = a 3 ⇒ a = r 3

4 Ga atoms per unit cell 5.65 x10

FG 4πr IJ H 3 K ×100% ⇒

4

1.2

b

Ratio = 68%

(d) Diamond lattice −3

Density of Ge = 4.44 x10 cm _______________________________________

Body diagonal = d = 8r = a 3 ⇒ a =

1.4 (a) Simple cubic lattice; a = 2 r

Unit cell vol. = a 3 =

F 8r I H 3K

FG 4πr IJ H3K 3

FG 4 πr IJ H3K 3

Then

FG 4πr IJ H 3 K × 100% ⇒ Ratio = 34% Ratio = F 8r I H 3K 3

Then

8

3

3

r

3

8 atoms per cell, so atom vol. 8

1 atom per cell, so atom vol. = (1)

8r

8 3

3

Unit cell vol = a3 = (2 r) = 8 r3

FG 4 πr IJ H 3 K × 100% ⇒ Ratio =

3

3

(c) Diamond: 8 corner atoms × 1/8 = 1 atom 6 face atoms × ½ = 3 atoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell _______________________________________

4

3

3

(b) bcc: 8 corner atoms × 1/8 = 1 atom 1 enclosed atom = 1 atom Total of 2 atoms per unit cell

Density =

c

3

Unit cell vol = a = 2 2 r

3

Ratio = 52.4%

_______________________________________

2

An Introduction to Semiconductor Devices Chapter 1 Solutions Manual Problem Solutions ______________________________________________________________________________________

B-type: 1 atom per unit cell, so . x10 23 cm −3 Density (b) = 118 _______________________________________

1.5 We have 8r = a o 3 so that

r=

5.65 3 8

= 1.223 A

1.9

°

Simple cubic; a o = 2 r = 4.2 A

Then °

d = 2 r = 2.45 A _______________________________________

bcc: 4r = a o

1.6 From Problem 1.4, percent volume of fcc atoms is 74%. Therefore after coffee is ground,

fcc: 4 r = ao 2 ⇒ ao =

3

8

From 1.3d, a =

(5.43) 3

°

= 5.94 A

2 8(2.1) 3 °

1.10 (b) ° . + 1.0 ⇒ a = 2.8 A a = 18

r °

. A = 118 8 8 Center of one silicon atom to center of nearest ° neighbor = 2r ⇒ 2.36 A

(c)

(b) Number density 8 22 −3 = ⇒ Density = 5x10 cm 3 −8 5.43x10

Cl: Density (same as Na) = 2.28 x10 cm

so that r =

b

=

Na: Density =

= ρ=

N ( At. Wt.)

b5 x10 g(28.09) ⇒ 23

NA

6. 02 x 10

ρ = 2.33 grams / cm 3 _______________________________________

ρ=

1.8 °

(a) a = 2r A = 2( 1.02) = 2.04 A Now

−3

−3

4.85x 10

− 23

b 2.8x10 g −8

3

⇒

_______________________________________ 1.11

so that rB = 0.747 A ° (b) A-type; 1 atom per unit cell 1 Density = ⇒ 8 3 2.04 x10 −

. )= 8 A (a) a 3 = 2 (2.2 ) + 2 (18 so that ° a = 4.62 A

g

23

3

ρ = 2.21 gm / cm3

2rA + 2rB = a 3 ⇒ 2rB = 2.04 3 − 2.04

b

22

= 2.28 x10 cm

b2.8 x10 g −8

(d) Na: At.Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell 1 1 (22.99 )+ (35. 45) − 23 2 = 4.85 x10 = 2 6.02 x10 23 Then mass density is

22

=

12

22

g

(c) Mass density

°

a o = 9.70 A _______________________________________

3

a 3

3 4( 2.1)

= 4.85 A

or

1.7 °

4(2.1)

diamond: 8r = a o 3 ⇒ a o =

Volume = 0.74 cm _______________________________________

(a) a = 5.43 A

3 ⇒ ao =

°

−3

. x10 cm Density(A) = 118

Density of A =

1

b4.62 x10 g −8

3

3

°

⇒ 1.01x1022 cm −3

An Introduction to Semiconductor Devices Chapter 1 Solutions Manual Problem Solutions ______________________________________________________________________________________

Density of B =

1

b4.62 x10 g

22

⇒ 1.01x10 cm

−8

1.15 (a) Planes intercept at: (i) p = 1, q = ∞ , s = ∞ (ii) p = 3, q = 1, s = ∞ (iii) p = 3, q = ∞ , s = 2

−3

(b) Same as (a) (c) Same material _______________________________________ 1.12 (a) Surface density 1 1 = 2 = 8 a 2 4.62 x10 −

b

3 .31 x10

14

g

(b) Vector components: (i) p = 1, q = 1, s = 0 (ii) p = 3, q = 1, s = 1 (iii) p = 1, q = 2 , s = 3 _______________________________________

⇒

2

2

−2

cm

1.16 (a)

Same for A atoms and B atoms (b) Same as (a) (c) Same material _______________________________________

(b)

1.13

1

(a) Vol density =

a3o

F 1 , 1 , 1I ⇒ (313) H 1 3 1K F 1 , 1 , 1I ⇒(121) H 4 2 4K

_______________________________________ 1

Surface density =

2

ao

1.17

2

°

(a) d = ao = 5.25 A

(b) Same as (a) _______________________________________

(b) d =

ao

No . atoms

2

b5. 43x10 g −8

2

(110) plane: Density =

b5. 43x 10 g

2

(111) plane: Density =

1.18 ° (a) d = a o = 5.20 A

No . atoms 2

2

(b) d = 14

= 9.59 x10 cm No . atoms 2

ao

=

3

(5.20) 2 2

= 3.68 A°

(5.20 ) 3

= 3. 00 A ° 3 3 _______________________________________

c 3 2h

14

2

2

(c) d =

a1 6f ⋅ 3 + a1 2f ⋅ 3 = 7.83x10 b5. 43x 10 g c 3 2h −8

ao 2

−2

2

ao

=

°

−2

14

4

−8

d=

= 6. 78 x10 cm

ao

=

ao 3

= 3.03 A 3 _______________________________________ (c)

2

ao =

°

= 3.71 A

2

1.14

(100) plane: Density =

2

=

1.19 cm

−2

ao =

4r

=

4(2.25 )

3 3 (a) Density = 2 2 = 3 ao 5.196 x10− 8

(a) (110) = highest density (b) (100) = lowest density _______________________________________

b

4

g

. A = 5196

3

°

= 1.43 x10 22 cm −3

An Introduction to Semiconductor Devices Chapter 1 Solutions Manual Problem Solutions ______________________________________________________________________________________

(b) d =

ao

2

=

5196 2 .

2 2 (d) Surface density 2 2 = 2 = −8 ao 2 . x10 5196

b

1.23

= 3.67 A

FH θ IK = a 4 = 2 ca 3 h 4

°

We find

o

g

1 3

so that

θ

2

2

2

or 14

o

cos

= 54.74°

or

−2

= 5.24 x 10 cm _______________________________________

θ = 109.5° _______________________________________

1.20

1.24

22 −3 Density of silicon atoms = 5x10 cm and 4 valence electrons per atom, so 23

(a) Ratio =

−3

Density of valence electrons = 2 x10 cm _______________________________________

(b) Ratio =

5x10 22 2 x1015 22

−5

⇒ 8x 10 % −6

⇒ 4 x10 %

5x10 _______________________________________

1.21 Density of GaAs atoms 8 atoms 22 −3 = = 4.44 x10 cm 8 3 5.65x10 −

b

4 x10 16

1.25

b5 x10 g(30 .98 ) ⇒ (a) Fraction by weight ≈ b5 x10 g(28.06 )

g

16

22

An average of 4 valence electrons per atom, −3 23 Density of valence electrons = 1.77 x 10 cm _______________________________________

. x10 −4 % 110

(b) Fraction by weight

b10 g(10.82 ) b 5x10 g( 30.98) +b 5 x10 g( 28.06) ⇒ 18

≈

1.22 Silver: Group 1 = 1 valence electron per atom 3 At.Wt. = 107.88, Mass density = 10.50 gm / cm Mass per unit cell, 107.88 −22 = = 1.79x 10 6. 02 x 1023 Now − 22

ρ=

1. 79x 10 a 3o

FG 1.79x10 JI H 10.50 K − 22

⇒ ao =

16

22

−4

7.71 x10 % _______________________________________

1.26 1

Volume density =

d

1 /3

3

15 −3 = 2x 10 cm

So −6

d = 7.94 x10 cm = 794 A

or

°

°

We have aO = 5.43 A So

°

ao = 2.57 A Then, density of valence electrons 1 = −8 2.57 x10 or

b

22

d 794 d = ⇒ = 146 ao 5.43 ao

g

3

_______________________________________

−3

Density = 5.89x 10 cm _______________________________________

5

An Introduction to Semiconductor Devices Chapter 2 Solutions Manual Exercise Solutions ______________________________________________________________________________________

Chapter 2 Exercise Problems EX2.1 (a)

E =

hc

λ

Then

b 6.625x10 gb 3x10 g b10 gx10 − 34

=

10

p

−19

1.99 x10

LM 1 − 1 PO a 4π ∈ f ⋅ 2( =) N( n + 1) n Q −b 9.11x10 gb 1.6x 10 g = 4 π b8 .85 x10 g (2 )b1.054 x10 g L 1 − 1 OP ×M (n + 1) n Q N We want L 1 − 1 PO ≤ 0.20 E − E = −13.58 M N(n + 1) n Q

− 19

− 19

1.6x 10

= 1.24 eV

− 34

− 16

J Then

= 1.24 x 10 3 eV 1.6x 10−19 ____________________________________

2

FH pIK m

1

2

=

m

2

b 3. 68x 10 g E= 2b 5x10 g − 26

− 31

p

n +1

− 19

(n + 1)

2

2

≥

1 n2

− 0 .0147

n=5 ____________________________________

2m

2

EX2.4 = 1.35 x10 −21 J

b gb1.6x10 g = 1.6x10

ΔE = 10

−8

− 19

− 27

J

so mv 1Δv = 1.6 x10 −27 −3

= 8.46x 10 eV

E = 20 meV = 3.2 x10 2 mE =

2

We find

2

or Δv =

p=

2

2

(b) and

4

− 34

n

1

− 21

1.6 x10

2

or

and 1. 35x10

−19

2

EX2.2 (a) 6. 625x 10 −34 h p= = = 3.68 x10 −26 kg − m / s 180x 10−10 λ Now 2

2

2

2

− 12

1.99 x10 −16

mv =

2

−31

10

−8

1

4

o

b6.625x10 gb3x10 g = 1. 99x 10 E= b10 x10 g E=

− mo e

En +1 − En =

(b)

E=

= 8.68 x10 −9 m

EX2.3

E=

so

7.636 x10

− 26

____________________________________

J

Then

E=

6. 625x 10

λ = 86.8 A°

or E = 1.99 x10

=

or

−8

4

− 34

h

λ=

b

−21

2 9.11x10

J −31

b

1. 6 x10−27

9.11x 10−31

gb

2 x 105

g

−3

= 8. 78 x10 m / s or

= 0.878 cm / s ____________________________________ Δv

gb3.2 x10 g −21

or 26

p = 7.636 x10 − kg − m / s

6

An Introduction to Semiconductor Devices Chapter 2 Solutions Manual Exercise Solutions ______________________________________________________________________________________ EX2.8 From Example 2.8, we can write

EX2.5 From Example 2.5 N=

4π ( 2 m)

=

h

3/ 2

⋅

3

b

2 3

⋅E

4π 2 9.11x10

b6.625x10 × b 3.2 x 10

−31

− 34

−19

F 1 I = 4.6kT H 0.01K

3 / 2 2 eV 1 eV

g F2I g H3K g − b1.6x10 g

E − E F = kT ln

3 /2

3 /2

−19

____________________________________ 3 /2

Test Your Understanding Exercises

or

TYU2.1

N = 8 .29 x10 27 m −3 = 8 .29 x10 21 cm−3 ____________________________________

EX2.6

(a)

(a)

L− a E − E f O MN kT PQ L −( 0. 3+ 0. 0259)O = exp NM 0.0259 QP F

f F = exp

f F = 3. 43x10

f F = exp

L MN

−7

1 − f F = 4.98 x10

(b)

LM −( 0.35 + 2 (0.0259 ))OP N 0.0259 Q

1 − f F = exp

−6

−( 0.3 + 2( 0.0259)) 0.0259

or

PQO

−7

1 − f F = 183 . x10 ____________________________________

or

TYU2.2

−6

f F = 1.26 x10 ____________________________________

fF = 1 + exp

EX2.7 f F = 0.005 = exp

or

LM −0.20 O N 0.0259 PQ

= ln (200 )

or kT =

0. 20

ln( 200) We can write

= 0.03775 eV

0.03775 = (0.0259 )

kT

KI

=

1 1 + exp( 4)

f F = 0.018 = 18% . ____________________________________

F +0.20I = 1 = 200 H kT K 0.005

kT

F H

1 E − EF

or

exp

Then +0.20

F

or

or (b)

LM −aE − E f PO N kT Q LM −(0.35 + 0.0259 ) OP = exp N 0.0259 Q

1 − f F ≅ exp

FTI H300 K

which yields T = 437 K ____________________________________

7

An Introduction to Semiconductor Devices Chapter 2 Solutions Manual Problem Solutions ______________________________________________________________________________________

Chapter 2 Problem Solutions

λ=

h

− 34

=

p

2.1 E = hν =

hc

λ

2.31x10

⇒