Donald A. Neamen - Semiconductor physics and devices basic principles [solutions manual]-Mc Graw-Hill (2003 ) PDF

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Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 1 Problem Solutions

Chapter 1 Problem Solutions

GHF 4π3r JKI 3

4 atoms per cell, so atom vol. = 4

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

Then

FG 4πr IJ H 3 K × 100% ⇒ 3

4 Ratio =

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

Ratio = 74%

3

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

3

Unit cell vol. = a =

F 4 rI H3K

3

GHF 4π3r JKI 3

2 atoms per cell, so atom vol. = 2

1.2 (a) 4 Ga atoms per unit cell Density =

Then

4

b

5.65 x10

−8

g

3

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

⇒

3

2

Density of Ga = 2.22 x1022 cm

−3

4 As atoms per unit cell, so that 22 −3 Density of As = 2.22 x10 cm

(d) Diamond lattice

(b)

Body diagonal = d = 8r = a 3 ⇒ a =

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

b

Ratio = 68%

3

g

Unit cell vol. = a 3 =

22

− Density of Ge = 4.44 x10 cm

3

F 8r I H 3K

8

r

3 3

FG 4πr JI H 3K 3

8 atoms per cell, so atom vol. 8 1.3 (a) Simple cubic lattice; a = 2 r

Then

Unit cell vol = a = ( 2r) = 8r 3

3

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

3

8

FG 4πr IJ H3K 3

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

3

Then

FG 4πr IJ H 3 K × 100% ⇒ Ratio =

Ratio = 34%

3

1.4 From Problem 1.3, percent volume of fcc atoms is 74%; Therefore after coffee is ground, 3 Volume = 0.74 cm

Ratio = 52.4% 3 8r (b) Face-centered cubic lattice d d = 4r = a 2 ⇒ a = =2 2r 2

c

3 Unit cell vol = a = 2 2 r

h

3

3 = 16 2 r

3

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 1 Problem Solutions

Then mass density is 1.5 (a) a = 5.43 A

°

8

From 1.3d, a =

ρ= r

3 a 3

neighbor = 2r ⇒ 2.36 A

= ρ=

N ( At. Wt. )

a = 4.62 A

1.9 (a) Surface density 1 1 = 2 = a 2 4.62 x10 −8

°

b

2 rA + 2 rB = a 3 ⇒ 2 rB = 2.04 3 − 2.04 °

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

14

23

. x 10 cm Density(B) = 118

1.10 (a) Vol density =

−3

1 3

ao Surface density =

1.7 (b)

1 2

ao

b2.8 x10 g −8

°

22

3

2

(b) Same as (a)

(c) Na: Density =

g

2

⇒ 2

−2

−3

B-type: 1 atom per unit cell, so

12

−3 22 1.01x10 cm

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

Density(A) = 118 . x 10 cm

a = 18 . + 1.0 ⇒ a = 2.8 A

−3 22 ⇒ 1.01x10 cm

3.31x10 cm

g

23

3

(b) Same as (a) (c) Same material

1.6

b

−8

−8

3

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

1

b 4.62x10 g 1 Density of B = b 4.62x 10 g ⇒

6. 02 x1023

ρ = 2.33 grams / cm

3

°

Density of A =

b 5x10 g (28.09) ⇒

NA

⇒

(a) a 3 = 2 (2.2 ) + 2(1.8) = 8 A° so that

22

=

3

−8

1.8

°

g

(c) Mass density

b2.8x10 g

°

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

b

23

ρ = 2.21 gm / cm

( 5. 43) 3

= = 118 . A 8 8 Center of one silicon atom to center of nearest so that r =

4.85x 10−

= 2.28 x10 cm

1.11 Sketch

−3

1.12 (a)

−

Cl: Density (same as Na) = 2.28 x1022 cm 3 (d) Na: At.Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell 1 1 ( 22.99) + (35 .45) −23 2 = 2 = 4.85x10 23 6.02 x10

(b)

4

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

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 1.13 (a) Distance between nearest (100) planes is: d = a = 5.63 A

=

°

d = 3.98 A

(iii)

or

(iii)

F KI H c h b

3 4.50 x10 −8

1 3a

2

g

2

14

cm −2

14

(iii)

g

. x1015 cm−2 114

g

−2

b

g

b

g

d = 4r = a 2 then 4r 4(2.25) a= = = 6.364 A ° 2 2 (a) 4 atoms Volume Density = −8 6 .364 x10

−2

b

(110) plane, surface density, 2 atoms −2 14 ⇒ 6 .99 x10 cm −8 2 2 4.50 x10

b

b

1.16 ⇒ 2.85 x10

Same as (a),(i); surface density 4.94x 10 cm

=

KI

(b) (110) plane, surface density, 4 atoms −2 14 = ⇒ 9.59 x10 cm −8 2 2 5.43x 10 (c) (111) plane, surface density, 4 atoms −2 14 = ⇒ 7.83 x10 cm −8 2 3 5.43x 10

(b) Body-centered cubic (i) (100) plane, surface density, (ii)

F H

6.78x 10 cm

g

1

g

b

(110) plane, surface density, 1 atom −2 14 ⇒ 3.49 x10 cm −8 2 2 4.50x 10

(111) plane, surface density, 1 1 atoms 3 6 2 = = = 1 1 a 3 a 2 ( x) ⋅a 2 ⋅ 2 2 2

=

b

14

g

b

(110) plane, surface density, 2 atoms −2 14 ⇒ 6.99 x10 cm 2 −8 2 4.50x 10

1.15 (a) (100) plane of silicon – similar to a fcc, 2 atoms surface density = ⇒ −8 2 5.43 x10

°

=

−2 14 ⇒ 9.88 x10 cm

°

Simple cubic: a = 4.50 A (i) (100) plane, surface density, 1 atom −2 14 = ⇒ 4.94 x10 cm −8 2 4.50 x10 (ii)

2

(111) plane, surface density, 1 1 3⋅ + 3 ⋅ 4 6 2 = = −8 2 3 2 3 4.50x 10 a 2

°

1.14 (a)

b

b4.50 x10 g −8

=

(c) Distance between nearest (111) planes is: 5.63 1 a = d = a 3 = 3 3 3 or d = 3.25 A

2 atoms

(ii)

(b)Distance between nearest (110) planes is: 5.63 1 a = d = a 2 = 2 2 2 or

Chapter 1 Problem Solutions

g

. x10 22 cm− 3 155 (b) Distance between (110) planes, a 6 .364 1 = a 2 = = ⇒ 2 2 2 or

(111) plane, surface density, 14

−2

Same as (a),(iii), surface density 2.85 x10 cm (c) Face centered cubic (i) (100) plane, surface density

5

g

3

⇒

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 4.50 A

Chapter 1 Problem Solutions

°

1.20

(c) Surface density 2 atoms = = 2 2a

b5 x10 g(30.98 ) ⇒ b5 x10 g(28.06) 16

(a) Fraction by weight ≈ 2

b

2 6. 364 x10

−8

g

2

−

. x10 6 110 (b) Fraction by weight

or 14

3.49 x10 cm

22

b10 g(10. 82) ≈ b5x10 g( 30.98) + b5x10 g( 28.06) ⇒ 18

−2

16

1.17

22

−

22

6 7.71x 10

−3

Density of silicon atoms = 5x 10 cm and 4 valence electrons per atom, so −3 23 Density of valence electrons 2x 10 cm

1.21 Volume density =

1 d

1.18 Density of GaAs atoms 8 atoms −3 22 = = 4.44x 10 cm −8 3 5.65x 10

b

−6 ° d = 7.94x 10 cm = 794 A °

We have aO = 5.43 A So d d 794 = 146 = ⇒ aO a O 5.43

An average of 4 valence electrons per atom, 23

Density of valence electrons 1. 77x 10 cm

−3

1.19 16

2 x10 5x 10

22

1x10

15

5x10

22

x100% ⇒

−5

4 x10 % (b) Percentage =

= 2 x10 cm

So

g

(a) Percentage =

15

3

x100 % ⇒

−

6 2 x10 %

6

−3

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 2 Problem Solutions

Chapter 2 Problem Solutions

p = 5.4 x10

2.1 Computer plot

λ=

2.2 Computer plot

h

−25

kg − m / s

− 34

=

p

6 .625 x10

− 25

⇒

5. 4 x 10

or

λ = 12.3 A

2.3 Computer plot

°

(ii) K.E. = T = 100 eV = 1.6x 10 2.4 For problem 2.2; Phase =

p=

2 πx

− ωt = constant

λ

λ=

For problem 2.3; Phase =

2 πx

λ

h

p = 5.4 x10

− 24

J

kg − m / s

⇒ λ = 1.23 A°

p

Then dx 2 π dx λ = v = +ω ⋅ − ω = 0 or dt 2π λ dt

FH KI

p

2mT ⇒

−17

− (b) Proton: K.E. = T = 1 eV = 1. 6 x10 19 J

p=

2 mT =

b

gb

2 1.67 x10 −27 1.6 x10 −19

or

+ ωt = constant

p = 2.31 x10 −23 kg − m/ s

Then dx 2 π dx λ ⋅ + ω = 0 or = vp = −ω dt 2π λ dt

F I H K

λ=

h p

=

6 .625 x10−34 2.31 x10

− 23

⇒

or

λ = 0.287 A

2.5 E = hν =

hc

λ

⇒λ =

hc E

b

. x10 Gold: E = 4 .90 eV = ( 4 .90) 16

So

− 19

g

(c) Tungsten Atom: At. Wt. = 183.92 − 19 For T = 1 eV = 1.6x 10 J J

p=

b 6.625x 10 gb 3x 10 g ⇒ 2.54x10 λ= (4.90)b1.6x10 g −34

2mT

b

2(183. 92) 1. 66x 10

=

10

−5

− 19

cm

b

Cesium: E = 1. 90 eV = (1. 90) 1. 6 x10 So

− 19

g

b 6.625x 10 gb 3x 10 g ⇒ 6.54x10 λ= (1.90)b 1.6x10 g −34

λ= J

h p

−5

6 .625 x10 3.13x10

−22

⇒

λ = 0.0212 A ° cm

(d) A 2000 kg traveling at 20 m/s: p = mv = ( 2000)( 20) ⇒ or

λ = 0.654 µm

p = 4 x 104 kg − m / s

2.6 (a) Electron: (i) K.E. = T = 1 eV = 1 .6 x10

b

− 34

=

or

10

or

2 mT =

gb1. 6x 10 g

. x10 −22 kg = m / s p = 313

− 19

p=

−27

or

or

λ = 0.254 µm

°

gb

2 9.11x10 −31 1.6 x10 −19

−19

λ=

J

g

h p

− 34

=

6 .625 x10 4

⇒

4 x10

or

λ = 1.66 x10

or

9

−28

A

°

−19

g

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 2 Problem Solutions

or

2.7 3

Eavg =

kT =

2

3

. x10 −3 eV E = 1.822 x10 −22 J ⇒ E = 114

(0.0259 ) ⇒

2

Also − 31 4 p = mv = 9.11x 10 2 x10 ⇒

b

or E avg = 0.01727 eV

p = 1822 x10 .

Now p avg =

b

g

b

2 9.11 x10 −31 (0 .01727 ) 1 .6 x10 −19

g

=

p

6 .625 x10 x10 1822 .

λ = 364 A pavg = 7.1 x10

−26

kg − m / s

− 34

=

p

=

λ

⇒

− 26

7.1x 10

⇒

°

− 34

h

p=

6. 625x 10

− 26

(b)

Now

λ=

kg − m / s

− 34

h

λ=

or

h

g

Now

2 mE avg

=

gb

−26

6 .625 x10 125x 10

⇒

−10

p = 5.3 x10 −26 kg − m / s

or

Also °

λ = 93.3 A

p

v= 2.8

=

m

5.3 x10 −26

4

9.11x 10− 31

= 5.82x 10 m / s

or

Ep = hν p =

hc

6

v = 5.82 x 10 cm / s

λp

Now

Now 2

Ee =

pe

2m

and pe =

h

λe

⇒ Ee =

FG h JI 2m H λ K 1

E=

2

hc

λp

FG h IJ = 2mH λ K

2

1

e

e

F 10hIJ = G 2mH λ K

=

b

2

b 9.11x 10 gb5.82x10 g 2 1

−31

2.10

2

1

(a) E = h ν =

p

hc

λ

b 6.625x10 gb 3x10 g 1x10 −15

8

− 10

J

Now hc

λp

hc

=

2

⋅ 2 mc =

100h −31

E = e ⋅ V ⇒ 1.99 x10

2 mc 100

gb3 x10 g 8

⇒

(b) p =

b

2mE =

2 9.11x10

= 6. 02x 10

J = 10. 3 keV

b9.11x10 gb2 x10 g 2 −31

−31

gb1.99 x10 g

kg − m / s

Then

2.9 1

g

3

λ= mv2 =

b

− 19 = 1.6 x10 V

V = 12.4 x 10 V = 12.4 kV

2

− 23

− 15

E = 1. 64 x 10

− 15

so

So

2

E = 9.64 x10 eV

− 34

=

E = 1. 99 x10

100

1

2

or

2 9.11 x10

(a) E =

4

−3

E = 1.54 x10− 21 J ⇒

which yields 100h λp = 2mc Ep = E =

2

mv =

or

Set E p = E e and λ p = 10λ e Then

1

4

2

10

h p

=

6. 625x 10

−34

− 23

6. 02x 10

⇒

° λ = 0.11 A

−15

Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual

Chapter 2 Problem Solutions

2.11 h

(a) ∆p =

∆x

− 34

=

1. 054x 10

(b) ∆t =

⇒

−6

10

∆p = 1.054 x10 −28 kg − m / s

= hc

−20

2.12 1 .054 x10

(a) ∆p =

(b) ∆E =

1

⋅

(∆p )

2

=

m

2

∆E = 7.71 x10

1

5 x10

2

2

(∆p )2

⋅

m

∆E = 7.71x10

−26

=

1 2

b 8.78 x10 g ⋅ −26

5x 10

−26

kg − m / s

∂t

−26

2

∂

⋅

2

Ψ1( x , t) + Ψ2( x , t)

2

equation, then we could write

2

−h

⇒

2

2

∂

2m ∂x 2

aΨ ⋅ Ψ f +V (x )aΨ ⋅ Ψ f 1

2

1

2

−7

J ⇒ ∆E = 4.82 x10 eV

= jh

2.14 ∆p =

∂Ψ1(x , t )

∂ Ψ1 ( x, t ) + Ψ2 ( x, t ) ∂t which is Schrodinger’s wave equation. So Ψ1 ( x , t ) + Ψ2 ( x , t ) is also a solution. (b) If Ψ1 ⋅ Ψ2 were a solution to Schrodinger’s wave

2.13

1

+V (x )Ψ1 (x , t ) = j h

= jh

J ⇒ ∆E = 4.82 x10 −4 eV

(b)

∂x

2

+V ( x ) Ψ1 ( x, t ) + Ψ2 ( x, t )

− 29

(a) Same as 2.12 (a), ∆p = 8.78 x10

∂ Ψ1 ( x, t ) 2

⋅

2m ∂x

b8 .78x10 g ⇒

2 −23

−h

−26

⋅

s

∂Ψ2 ( x, t ) − h 2 ∂ 2Ψ2 (x , t ) + V ( x )Ψ2 ( x, t ) = jh ⋅ 2 ∂x 2m ∂t Adding the two equations, we obtain

−34

= ⇒ − 10 12 x10 ∆x ∆p = 8. 78 x10 −26 kg − m / s

∆E =

2

−h

J ⇒ ∆E = 0.198 eV

2m and h

−16

Schrodinger’s wave equation, then

or ∆E = 3.16 x10

g⇒

2.16 (a) If Ψ1 ( x , t ) and Ψ2 ( x , t ) are solutions to

− 28

8

b

(1) 1.6 x10 −19

∆t = 6.6 x10

F pI = pc H hK λ So ∆E = c( ∆p) = b3 x10 gb1.054 x10 g ⇒ hc

−34

or

(b)

E=

1.054 x10

∂ ∂t

aΨ ⋅Ψ f 1

2

which can be written as

h ∆x

=

1. 054x 10−34 10 −2

p = mv ⇒ ∆v =

∆p

=

m

= 1.054 x10 1.054 x10 −32 1500

LMΨ ∂ Ψ + Ψ ∂ Ψ + 2 ∂Ψ ⋅ ∂Ψ PO 2m N ∂x ∂x Q ∂x ∂x LM ∂Ψ + Ψ ∂Ψ PO +V ( x)Ψ ⋅ Ψ = j h Ψ N ∂ t ∂t Q Dividing by Ψ ⋅ Ψ we find 1 ∂ Ψ 1 ∂Ψ ∂Ψ O −h L 1 ∂ Ψ + ⋅ ⋅ + M P Ψ Ψ ∂ x ∂x Q 2 m N Ψ ∂x Ψ ∂x L 1 ∂Ψ + 1 ∂Ψ PO +V (x ) = jh M NΨ ∂t Ψ ∂x Q −h

− 32

⇒

−36

1

2

1

2

2

1

2

1

m/s

2

2

2
...