semiconductor devices physics and technology

b For the 100 plane, there are two atoms one central atom and 4 corner atoms each contributing 1/4 of an atom for a total of two atoms as shown in Fig.. a For the simple cubic, a unit ce

Solutions Manual to Accompany

SEMICONDUCTOR DEVICES Physics and Technology

S M SZE Etron Chair Professor College of Electrical and Computer Engineering

National Chaio Tung University

Hsinchu, Taiwan

M K LEE Department of Electrical Engineering National Sun Yat-sen University Kaohsiung, Taiwan

John Wiley and Sons, Inc New York Chicester / Weinheim / Brisband / Singapore / Toronto

Contents

Ch.0 Introduction - 0

Ch.1 Energy Bands and Carrier Concentration in Thermal Equilibrium - 1

Ch.2 Carrier Transport Phenomena - 9

Ch.3 p-n Junction -18

Ch.4 Bipolar Transistor and Related Devices -35

Ch.5 MOS Capacitor and MOSFET -52

Ch.6 Advanced MOSFET and Related Devices -62

Ch.7 MESFET and Related Devices -68

Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices -76

Ch.9 Light Emitting Diodes and Lasers -81

Ch.10 Photodetectors and Solar Cells -88

Ch.11 Crystal Growth and Epitaxy -96

Ch.12 Film Formation -105

Ch.13 Lithography and Etching -112

Ch.14 Impurity Doping -118

Ch.15 Integrated Devices -126

CHAPTER 1

1 (a) From Fig 11a, the atom at the center of the cube is surround by four

equidistant nearest neighbors that lie at the corners of a tetrahedron

Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is

1/2 [(a/2)2 + ( a /2)2]1/2 = a /4 = 2.35 Å

(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms

each contributing 1/4 of an atom for a total of two atoms as shown in Fig 4a)

for an area of a 2, therefore we have

3) =

2 2 3

2 The heights at X, Y, and Z point are 34, 14,and 43

3 (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere

∴ Maximum fraction of cell filled

= no of sphere × volume of each sphere / unit cell volume

= 1 × 4π(a/2)3 / a3 = 52 %

(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the

eight corners for a total of one sphere The fcc also contains half a sphere at

distance is 1/2(a 2) Therefore the radius of each sphere is 1/4 (a 2)

∴ Maximum fraction of cell filled

= (1 + 3) {4π[(a/2) / 4 ]3 / 3} / a3 = 74 %

(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight

corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a

total of three spheres, and 4 spheres inside the cell The diagonal distance

between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig 9a is

D =

2

2 2

5 Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4 The smallest

three integers having the same ratio are 6, 4, and 3 The plane is referred to as

(643) plane

four arsenic atoms per unit cell, therefore

4/a3 = 4/ (5.65 × 10-8)3 = 2.22 × 1022 Ga or As atoms/cm2, Density = (no of atoms/cm3 × atomic weight) / Avogadro constant = 2.22 × 1022(69.72 + 74.92) / 6.02 × 1023 = 5.33 g / cm3 (b) If GaAs is doped with Sn and Sn atoms displace Ga atoms, donors are

formed, because Sn has four valence electrons while Ga has only three The

resulting semiconductor is n-type

7 E g (T) = 1.17 –

636)(

4.73x10 4 2+

5.405x10 4 2+

−

T

T

for GaAs

∴E g ( 100 K) = 1.501 eV, and E g (600 K) = 1.277 eV

8 The density of holes in the valence band is given by integrating the product

N(E)[1-F(E)]dE from top of the valence band ( E taken to be E = 0) to the V

bottom of the valence band Ebottom:

p = ∫E bottom

0 N(E)[1 – F(E)]dE (1)

where 1 –F(E) = { [ ( ) /kT] }

1 e/1

or p = N V exp [-(EF –EV) / kT ]

where NV = 2 (2πmp kT / h2)3

9 From Eq 18

N V = 2(2πmp kT / h2)3/2The effective mass of holes in Si is

m p = (NV / 2) 2/3 ( h2 / 2πkT )

=

3 3 6 192

m101066

10 625 6

23

2 34

Because the second term on the right-hand side of the Eq.1 is much smaller

compared to the first term, over the above temperature range, it is reasonable to

assume that Ei is in the center of the forbidden gap

) ( /

(

/d

de

C F

top C

F top

C

E E x kT

E E C E

E

kT E E C E

E e

E E

E E

E E E

x x

x x

x

x = kT

5

= kT

π

π5 0

5 0 5 1

10 626 6

Since pP is the same as given in (a), the values for np and Ep are the same as

in (a) However, the mobilities and resistivities for these two samples are

different

17 Since ND >> n;, we can approximate no= ND and

Po = n ? I no= 9.3 x1 19 I 1017 = 9.3 x 102 cm-3

- E )

20 Assuming complete ionizatio , the Fenni level measmed from the intrinsic

Fenni level is 0.35 eV for 1015 cm-3 0.45 eV for 1017 cm-3 and 0.54 eV for 1019

The number of electrons that are ionized is given by

Using the Fenni levels given above, we obtain the number of ionized donors as

21 ND+ = E E kT

F D

e ( /

161

10

.

− + =

145 1

1 1

NO = 33 5

76 4

.

= 0.876

CHAPTER 2

1 (a) For intrinsic Si, μn = 1450, μp = 505, and n = p = ni = 9.65×109

)(

1

+

=+

=

p n i p

1

+

=+

=

p n i p

3 Since

2 1

111

μμ

∴

500

1250

11

Fig 3 along with trial and error to determine μn and ND For example, if we choose ND = 2×1017, then NI = ND + + NA - = 3×1017, so that μn ≈ 510 cm2/V-s which gives σ = 8.16

Further trial and error yields

N D≈ 3.5×1017 cm-3and

μn ≈ 400 cm2/V-s which gives

σ ≈ 16 (Ω-cm)-1

6 σ =q(μn n+μp p )=qμp(bn+n i2 /n)

From the condition dσ/dn = 0, we obtain

b n

n = i/Therefore

b b b

n q

n b b bn qμ

i p

i i

p i

m

21)

1(1

)/

(

1

+

=+

+

=

μρ

ρ

π

V

From Fig 6, CF = 4.2 (d/s = 10); using the a/d = 1 curve we obtain

78.102.41050

10226.0)

106.11010

4 9 3

3 3

A V

6.1

Assuming N A ≈ p, from Fig 7 we obtain ρ = 1.1 Ω-cm

The mobility μp is given by Eq 15b

3801.11046.1106.1

11

/ p = n p =

μ

p A n D

n D N N

N R

R

μμ

μ+

1

) x ( N

dx

) x ( dN ) x ( N q

kT e

N

e a N q

kT n

dx

dn D

ax n

)(

n q

J n μn E n

x N N LN

N N D

L

N N L x N N N

D dx

x dn x n

D x

L

L n

n

L L

n

n n

n

)(

))(

(

1)

()(

1)

(

0 0

0

0 0

0

−+

E

0 0

0

0 0

)(

D-

N

N D x N N LN

N N

n

n L

L L

15 11

15 10 10

= Δ +

= Δ +

.cm 101010

)1065.9

15

2 9 2

≈+

×

=Δ+

10210105

N

νσ

τ

(b) The hole concentration at the surface is given by Eq 67

.cm 10

2010103

20101

101010

2

)10(9.65

1)

0(

3 - 9

8 4

8 17

8 16

2 9

×

×

−

×+

lr p L

p no n

S L

S G

p p

τ

ττ

15 σ =qnμn +qpμp

Before illumination

no n no

n = , =After illumination

,

G n

n n

n n = no +Δ = no+τp

G p

p p

p n = no+Δ = no+τp

)(

)(

)]

()

([

G q

p q n q p p

q n n

q

p p n

no p no n no

p no

n

τμμ

μμ

μμ

Δ+

=Δ

16 (a) ,diff

dx

dp qD

J p =− p

= − 1.6×10-19×12× 4

10 12

∴ 4.8 − 1.6exp(-x/12) = 1.6×10-19×1016×1000×E

E = 3 − exp(-x/12) V/cm

17 For E = 0 we have

02

2

=

∂

∂+

p p

no n

τ

at steady state, the boundary conditions are pn (x = 0) = pn (0) and pn (x = W) =

=

p

p no

n no n

L W L

x W p

p p x p

sinh

sinh )

0 ( )

x

n p

W L

D p p

q x

p qD x

p no n

W x

n p p

L W L

D p p

q x

p qD W

x J

sinh

1 )

0 ( )

18 The portion of injection current that reaches the opposite surface by diffusion

is given by

)/cosh(

1)

0(

)(0

p p

p

L W J

W J

Therefore, 98% of the injected current can reach the opposite surface

19 In steady state, the recombination rate at the surface and in the bulk is equal

surface ,

surface , bulk

,

bulk ,

p

n p

p

ττ

Δ

= Δ

so that the excess minority carrier concentration at the surface

From Eq 62, we can write

0 2

2

= Δ

− +

∂

Δ

∂

p p

p G x

p D

where Lp = 10 ⋅ 10−6 = 31.6 μm

20 The potential barrier height

χφ

φB = m− = 4.2 − 4.0 = 0.2 volts

21 The number of electrons occupying the energy level between E and E+dE is

dn = N(E)F(E)dE

where N(E) is the density-of-state function, and F(E) is Fermi-Dirac

distribution function Since only electrons with an energy greater than

E v h

m qv

x q

E

m F

) ( 2 3

2)2(

∞ +

=

φ

π

where v is the component of velocity normal to the surface of the metal x

Since the energy-momentum relationship

)(

2

12

2 2 2 2

z y

p m m

x dp dp dp

mkT p x

x mkT mE p p

z y x p

mkT mE p p p x

dp e

dp e

dp p e

mh q

dp dp dp e

p mh

q J

z y

f x x

x

f z y x

2 2

2 ) 2 ( 3

2 / 2 (

3

2 2

2 0

0 y

2 2 2

2

Since

2 1

, the last two integrals yield (2πmkT) 2

The first integral is evaluated by setting u

mkT

mE

p x − F =2

mE q

E

m F + φm − F = φm

2

2)(

2

so that the first integral becomes u q kT

kt q

m m

e mkT du e

A e

T h

19 31

2

)10054.1(

)106.1)(

220)(

1011.9(2)(

1019.3)

220(24

1031017.2sinh(

)101099.9sinh(

61)10

(

1 2 10 9

2.262.24

101099.9sinh61)

=

1 9 2

34

19 31

2

)10054.1(

)106.1)(

2.26)(

1011.9(2)(

24 From Fig 22

As E = 103 V/s

νd ≈ 1.3×106 cm/s (Si) and νd ≈ 8.7×106 cm/s (GaAs)

t ≈ 77 ps (Si) and t ≈ 11.5 ps (GaAs)

As E = 5×104 V/s

νd ≈ 107 cm/s (Si) and νd ≈ 8.2×106 cm/s (GaAs)

t ≈ 10 ps (Si) and t ≈ 12.2 ps (GaAs)

cm/s105.9m/s109.5

101.9

30010

1.382

22

velocity Thermal

25

6 4

31

23 -

0 0

v =μ E=1350×100=1.35×105cm/s<<

For electric field of 104 V/cm

μnE=1350×104 =1.35×107cm/s≈v th

The value is comparable to the thermal velocity, the linear relationship between

drift velocity and the electric field is not valid

CHAPTER3

1 The impmity profile is,

x (J.lm)

The overall space charge neutrality of the semicond ctor requires that the total negative

space charge per unit area in the p-side must equal the total positive space charge per

0.8 X 8 X 1014

2

Hence, then-side depletion layer width is:

The total depletion layer width is 1.867 J.lm

We use the Poisson's equation for calculation of the electric field E ( x )

In the n-side region,

( )

V/cm 10 86 4 0

) 10 067 1 ( 10 3

10 067 1 0

m 067 1

3

4 14

x q

x

N

q K )

x

(

K x N q ) x ( N

q dx

d

n max

s n

D s n

D s n D

s

E E

E

E

E E

ε

εμ

εε

In the p-side region, the electrical field is:

V/cm 10 86 4 0

10 8 0 2

10 8 0 2

0 m 8 0

2

3

2 4 2

2 4 2

x a q x

a

q K

) x

(

K ax

q ) x ( N

q dx

d

p max

s p

s

' p

' s

p A

s

E E

E

E

E E

ε

εμ

εε

The built-in potential is:

x

side n x

2 From V bi = −∫ E( )x dx, the potential distribution can be obtained

With zero potential in the neutral p-region as a reference, the potential in the p-side

4 3

11

3 4 2

4 3

2 4 2

108.03

210

8.03

110596

7

108.03

210

8.03

1210

8.02

x x

x x

qa dx

x a

q dx

x x

Pot ential Distr ib ution

3 Th e in t1ins i c can i ers density in Si at differe n t temperatures can be o btained b y us in g Fig 22 in

T hus , the bu il t - i n potent i a l is decreased as t he temperan 1 re is in creased

T he dep l e t on laye r w i dt h and the max i mu m fie l d at 300 K are

W = 2ssVbi =

qND

2 X 11 9 X 8 85 X 1 -14 ~ 0 7 1 7 = 0.

9715 1.6x 1 -19 x 1 0b flill

18 16

2 / 1 18

18 14

19 5

2 / 1 max

101

10755

85.89.11

3010

6.12104

D

D D

A

D A s

R

N N

N

N N

N

N N qV

ε

.E

We can select n-type doping concentration of N D = 1.755×10 16 cm-3 for the junction

5 From Eq 12 and Eq 35, we can obtain the 1/C 2 versus V relationship for doping concentration

of 10 15 , 10 16 , or 10 17 cm -3 , respectively

For N D=10 15 cm -3 ,

N qε

V V

bi j

8.8511.910

1.6

0.8372

2

15 14 19

2

For ND=10 16 cm -3 ,

N qε

V V

bi j

8.8511.910

1.6

0.8962

2

16 14 19

2

For ND =1017 cm-3,

N qε

V V

bi j

8.8511.910

1.6

0.9562

2

17 14 19

B 2

When the reversed bias is applied, we summarize a table of 1/ C 2 j vs V for various N D values as following,

v No= lEIS No=1E16 No=1E17

-4 5.741E+16 5.812E+l5 5.883E+l4

-3.5 5.148E+16 5.218E+l5 5.289E+l4

-3 4.555E+16 4.625E+15 4.696E+l4

-2.5 3.961E+16 4.031E+l5 4.102E+l4

-2 3.368E+l6 3.438E+l5 3.509E+l4

-1.5 2.774E+16 2.844E+15 2.915E+l4

-1 2.181E+l6 2.251E+l5 2.322E+14

-0.5 1.587E+l6 1.657E+l5 1.728E+l4

0 9.935E+l5 1.064E+15 1.134E+l4

Hence , we o bt ain a seiies of curves of 1 / C versus Vas fo ll owing,

1/C"2 VS v

<'-'T>V

- ~ VuT>V

6 The built-in potential is

0259.01085.89.111010ln0259.03

28

ln3

2

3 9 19

14 20

20 3

2 2

n q

kT a q

19 - 3

/ 1 2

5686.012

1085.89.1110

101.6

bi

s s

j

V V

V

qa W

At reverse bias of 4V, the junction capacitance is 6.866×10-9 F/cm2

7 From Eq 35, we can obtain

2

2 2

1

j s

R bi D B

s

bi

j

C q

V V N

N q

V V

15 7 15 15

10 89 3 10 65 9 02 0 02

0

10 10 10 10

exp exp

.

n kT

E E kT

E E

N U

t i p i t n

t th n p

σσ

υσσ

( )

3 15

2 8 14

19 2

cm 10 43 3

) 10 85 0 ( 10 85 8 9 11 10 6 1

4 2 2

.

C q

V N

V

V

d

j s

R D bi

∵

and

m266.1cm1066.1210

106.1

)5.0717.0(1085.89.112

15 19

V V W

Thus

2 7 5

16

19 3.89 10 12.66 10 7.879 10 A/cm10

6

n p

2 9 )

0259 0

8 ( 2

cm1042.210

1065

195

0

1

0259 0

0259 0

J

V

V s

11 The parameters are

n i= 9.65×109 cm-3 D n= 21 cm2/sec

D p=10 cm2/sec τp0=τn0=5×10-7 sec

From Eq 52 and Eq 54

( ) ( )

3 15

0259 0 7 2

9 7

19

2 /

cm102

5

110

65.9105

1010

i po

p kT

qV p

no p n

p

N

e N

e N

n D q e

L

p qD x

0259 0 7 2

9 7

19

2 /

cm10278

5

110

65.9105

2110

i no

n kT

qV n

po n p

n

N

e N

e N

n D q e

L

n qD x

12 Assumeτg =τp =τn = 10-6 s, D n= 21 cm2/sec, and D p = 10 cm2/sec

(a) The saturation current calculation

From Eq 55a and L p = D pτp , we can obtain

=

0 0

2 0

n

n A p

p D

i n

p n p

n p s

D N

D N

qn L

n qD L

p qD

J

ττ

2 12

6 16 6

18

2 9 19

A/cm10

110

1010

110

65.9106

Thus

A

10244.8110

244.8

A1051.41047.510244.8110

244.8

17 0259

0

7 17 7

5 11

17 0259

0

7 17 7

e I

V V

we can obtain

V78.0110

244.8

10ln

0259.01

ln0259

14 6

9 19

15

2 9 6

19 2

10 10 6 1

10 85 8 9 11 2 10

10 65 9 10 6 1 10

10 65 9 10

10 10

.

.

.

W qn N

n D q

J

R bi g

i D

i p

9

15 19

junction is about 2.8 ×105 V/cm Then from Eq 85, we obtain

2 2 voltage

1010

6.12

108.21085.89

10 6 1

258 10

85 8 9 11 2

15 19

14

.

.

qN

V V W

When the n-region is reduced to 5μm, the punch-through will take place first

From Eq 87, we can obtain

( W )/2

insert 29 Fig.

in area shaded

m c E

W

2

43.18

5243.18

52582

B B

W

W W

W V

i p

p kT

qV r

i kT

qV D

i p

p

N

n D q e

qWn e

N

n D q

2 2

/ /

2 2

( ) 00259 3

7 2 9 7

19

2

10 2 2 10

65 9 10 10

A e

N

n D

Aq

D

p kT

/ qV D

i p

2

10 6 1 2

10 85 8 9 11 130 2

.

N

q

W

Let Ec= 4×105 V/cm, we can obtain ND = 4.05×1015 cm-3

The mobility of minority carrier hole is about 500 at ND = 4.05×1015

∴Dp= 0.0259×500=12.95 cm2/s

Thus, the cross-sectional area A is 8.6×10-5 cm2

18 As the temperature increases, the total reverse current also increases That is,

the total electron current increases The impact ionization takes place when the

electron gains enough energy from the electrical field to create an

electron-hole pair When the temperature increases, total number of electron

increases resulting in easy to lose their energy by collision with other electron

before breaking the lattice bonds This need higher breakdown voltage

19 (a) The i-layer is easy to deplete, and assume the field in the depletion region is

constant From Eq 84, we can obtain

V/cm10

87.5)10(1041

10105

410110

4

6 4

0

6 5

( ) 2 ( )1

2 2 voltage

10210

6.12

1051085.84

2 1 22 2 1 19

14 2

3

2 1 2 1 2

3

10 84

4

10 10

6 1

10 85 8 9 11 2 3

4

2 3

4 3 2

/ c

/ /

/ c

/

/ s

/ c B

.

.

.

a q

W V

E E

E E

The breakdown voltage can be determined by a selected Ec

21 To calculate the results with applied voltage of V = 0 5 V , we can use a similar calculation in Example 10 with (1.6-0.5) replacing 1.6 for the voltage The obtained

electrostatic potentials are 1.1 and V 3.4 × 10 − 4 V , respectively The depletion widths are cm 3.821 × 10 − 5 and 1.274 × 10 − 8 cm , respectively

Also, by substituting V = − 5 V to Eqs 90 and 91, the electrostatic potentials are 6.6 V and V 20.3 × 10 − 4 , and the depletion widths are 9.359 × 10 − 5 cm and 3.12 × 10 − 8 cm , respectively

The total depletion width will be reduced when the heterojunction is forward-biased from

the thermal equilibrium condition On the other hand, when the heterojunction is

reverse-biased, the total depletion width will be increased

22 E g(0.3) = 1.424 + 1.247 × 0.3 = 1.789 eV

V = bi −Δ −(E −E )/ q−

q

E q

E

V F C g

2 2

= 1.789 – 0.21– 1517

10 5

10 7 4 ln

×

×

q

kT

10 5

10 7 ln

1

2 1 2

bi A

N N

qN

v N

εε

εε

=

2 15

19

14 46 11 4 12 10 5 10 6 1

273 1 10 85 8 46 11 4 12 2

.

.

CHAPTER 4

1 (a) The common-base and common-emitter current gains is given by

199

995 0 1

995 0 1

995 0 998 0 997 0

0

0 0 0

αβ

10 10 199 1 1

6

9 0

999 0

0

0 0

αβ

γα

CBO

I is known and equals to 10 × 10 − 6 A Therefore,

mA 10

10 10 ) 999 1 (

1

6 0

=

×

⋅ +

9

17 18

n

N N q

kT V

The depletion-layer width in the base is

( )

m 10 5.364 cm

10 5.364

5 0 956 0 10 2 10 5

1 10

2

10 5 10

6 1

10 05 1 2

1 2

junction) base

emitter the of h layer widt -

-depletion (Total

2 - 6

-17 18

17

18 19

12 1

=

−

−

V V N N N

N q

N N

N W

bi D A D

A s

D A A

ε

Similarly we obtain for the base-collector function

(9 65 10 ) 0.795V .

10 10 2 ln 0259

9

16 17

and

m 10 254 4 cm 10 4.254

5 795 0 10 2 10

1 10

2

10 10

6 1

10 05 1 2

2 - 6

-17 16

17

16 19

12 2

2 1

W (b) Using Eq 13a

10 2

10 65 9 )

0

17

2 9 2

p

D

i kT qV no

4 In the emitter region

( ) 18.625 .

105

1065.9

cm10721.01052

scm52

18

2 9

3 8

n

L D

In the base region

( ) 465.613 .

10 2

10 65 9

m c 10 2 10 40

s cm 40

17

2 9 2

3 7

p p p p

N

n p

D L

cm 10 724 10 10 115

s cm 115

3 16

2 9

3 6

n

L D

The current components are given by Eqs 20, 21, 22, and 23:

A103.19610

724.10

10312.911510

2.0106.1

A101.0411

10721.0

625.1852102.0106.1

A101.596

A101.59610

904.0

613.46540102.0106.1

14 - 3

3 2

19

7 - 0259

0 0 3

2 19

5 -

5 - 0259

0 0 4

2 19

Ep Cp Ep

I I I

I

e I

I I

e I

5 (a) The emitter, collector, and base currents are given by

A 10 1.041

A 10 1.596

A 10 1.606

7 -

5 - -5

×

=

− +

=

×

= +

=

×

= +

=

Cn BB En B

Cn Cp C

En Ep E

I I I I

I I I

I I I

(b) We can obtain the emitter efficiency and the base transport factor:

9938 0 10 606 1

10 596 1

I

γ

10 596 1

10 596 1

I

αHence, the common-base and common-emitter current gains are

160 1

9938 0 0 0

αβγα

(c) To improve γ , the emitter has to be doped much heavier than the base

To improve αT, we can make the base width narrower

6 We can sketch p n(x) p n(0) curves by using a computer program:

0.0 0.2 0.4 0.6 0.8 1.0

20

1 2

5 10