semiconductor devices physics and technology

1
Solutions Manual to Accompany

SEMICONDUCTOR DEVICES
Physics and Technology
3
rd
Edition




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

0
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




1
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
+ ( a2 /2)
2
]

1/2
= a3 /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
2/ a
2
= 2/ (5.43 × 10
-8
)
2
= 6.78 × 10
14
atoms / cm
2

Similarly we have for (110) plane (Fig. 4a and Fig. 6)
(2 + 2 ×1/2 + 4 ×1/4) / a2
2
= 9.6 × 10
15
atoms / cm
2
,
and for (111) plane (Fig. 4a and Fig. 6)
(3 × 1/2 + 3 × 1/6) / 1/2( a2 )( a
⎟
⎠

⎞
⎜
⎝
⎛
2
3
) =
2
2
3
2
a
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
= 7.83 × 10
14
atoms / cm
2
.

2. The heights at X, Y, and Z point are
,4
3
,4

1
and
4
3
.
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
/ a
3
= 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
each of the six faces for a total of three spheres. The nearest neighbor

2
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} / a
3
= 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
1
222
222
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
+
⎟
⎠
⎞
⎜
⎝
⎛
aaa
=
3
4
a

The radius of the sphere is D/2 =
3
8
a

∴ Maximum fraction of cell filled
= (1 + 3 + 4)
3
3
83
4
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎠
⎞
⎜
⎝
⎛
a
π
/ a
3
= π 3 / 16 = 34 %.
This is a relatively low percentage compared to other lattice structures.

4.
1
d =
2
d =
3
d =
4
d = d

1
d +
2
d +
3
d +
4
d = 0

1
d • (
1
d +
2
d +
3
d +
4

d ) =
1
d • 0 = 0

2
1
d
+
1
d •
2
d +
1
d •
3
d +
1
d •
4
d = 0
∴d
2
+ d
2
cosθ
12
+ d
2
cosθ
13

+ d
2
cosθ
14
= d
2
+3 d
2
cosθ= 0
∴ cosθ =
3
1
−

θ= cos
-1
(
3
1
−
) = 109.47
0
.
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.
6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As
are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and

3

four arsenic atoms per unit cell, therefore
4/a
3
= 4/ (5.65 × 10
-8
)
3
= 2.22 × 10
22
Ga or As atoms/cm
2
,
Density = (no. of atoms/cm
3
× atomic weight) / Avogadro constant
= 2.22 × 10
22
(69.72 + 74.92) / 6.02 × 10
23
= 5.33 g / cm
3
.
(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

24
+
−
T
T
for Si

∴ E
g
( 100 K) = 1.163 eV , and E
g
(600 K) = 1.032 eV
E
g
(T) = 1.519 –
204)(
5.405x10
24
+
−
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 (
V
E taken to be E = 0) to the
bottom of the valence band E
bottom
:
p =
∫
bottom
E
0
N(E)[1 – F(E)]dE (1)
where 1 –F(E) =
(
)
[
]
{
}
/kT
1
e/1 1
F
EE −
+
−
=
[
]

1
/)(
e1
−
−
+
kTEE
F

If E
F
– E >> kT then
1 – F(E) ~ exp
(
)
[
]
kTEE
F
−
−
(2)
Then from Appendix H and , Eqs. 1 and 2 we obtain
p = 4π[2m
p
/ h
2
]
3/2
∫

bottom
E
0
E
1/2
exp [-(E
F
– E) / kT ]dE (3)
Let x ≣ E / kT , and let E
bottom
=
∞
−
, Eq. 3 becomes
p = 4π(2m
p
/ h
2
)
3/2
(kT)
3/2
exp [-(E
F
/ kT)]
∫
∞−
0
x
1/2

e
x
dx
where the integral on the right is of the standard form and equals
π
/ 2.
∴ p = 2[2πm
p
kT / h
2
]
3/2
exp [-(E
F
/ kT)]
By referring to the top of the valence band as E
V
instead of E = 0 we have,
p = 2(2πm
p
kT / h
2
)
3/2
exp [-(E
F
– E
V
) / kT ]

4
or p = N
V
exp [-(E
F
–E
V
) / kT ]
where N
V
= 2 (2πm
p
kT / h
2
)
3
.

9. From Eq. 18
N
V
= 2(2πm
p
kT / h
2
)
3/2

The effective mass of holes in Si is
m

p
= (N
V
/ 2)
2/3
( h
2
/ 2πkT )
=
3
2
3619
2
m101066.2
⎟
⎠
⎞
⎜
⎝
⎛
××
−
(
)
()
()
300103812
106256
23
2

34
−
−
×
×
.
.
π

= 9.4 × 10
-31
kg = 1.03 m
0
.
Similarly, we have for GaAs
m
p
= 3.9 × 10
-31
kg = 0.43 m
0
.
10. Using Eq. 19
(
)
(
)
CVVC
NN
kT

EEE
i
ln
2
2)( ++=
= (E
C
+ E
V
)/ 2 + (3kT / 4) ln
⎥
⎦
⎤
⎢
⎣
⎡
3
2
)6)((
np
mm
(1)
At 77 K
E
i
= (1.16/2) + (3 × 1.38 × 10
-23
T) / (4 × 1.6 × 10
-19
) ln(1.0/0.62)

= 0.58 + 3.29 × 10
-5
T = 0.58 + 2.54 × 10
-3
= 0.583 eV.
At 300 K
E
i
= (1.12/2) + (3.29 × 10
-5
)(300) = 0.56 + 0.009 = 0.569 eV.
At 373 K
E
i
= (1.09/2) + (3.29 × 10
-5
)(373) = 0.545 + 0.012 = 0.557 eV.
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 E
i
is in the center of the forbidden gap.
11. KE =
()
()
)(
/)(


/

d
de
C
F
top
C
F
top
C
EEx
kTEE
C
E
E
kTEE
C
E
E
C
EeEE
EEEEE
−≡
−−
−−
−
−−
∫
∫

5
= kT
∫
∫
∞
−
∞
−

0
2
1

0
2
3
de
de
xx
xx
x
x
= kT
⎟
⎠
⎞
⎜
⎝

⎛
Γ
⎟
⎠
⎞
⎜
⎝
⎛
Γ
2
3
2
5
= kT
π
π
50
5051
.
××

=
kT
2
3
.
12. (a) p = mv = 9.109 × 10
-31
×10
5

= 9.109 × 10
-26
kg–m/s

λ
=
p
h
=
26
34
101099
106266
−
−
×
×
.
.
= 7.27 × 10
-9
m = 72.7 Å
(b)
n
λ
=
λ
p
m
m

0
=
0630
1
.
× 72.7 = 1154 Å .
13. From Fig. 22 when n
i
= 10
15
cm
-3
, the corresponding temperature is 1000 / T = 1.8.
So that T = 1000/1.8 = 555 K or 282
℃.
14. From E
c
– E
F
= kT ln [N
C
/ (N
D
– N
A
)]
which can be rewritten as N
D
– N
A

= N
C
exp [–(E
C
– E
F
) / kT ]
Then N
D
– N
A
= 2.86 × 10
19
exp(–0.20 / 0.0259) = 1.26 × 10
16
cm
-3

or N
D
= 1.26 × 10
16
+ N
A
= 2.26 × 10
16
cm
-3

A compensated semiconductor can be fabricated to provide a specific Fermi

energy level.

15. From Fig. 28a we can draw the following energy-band diagrams:
AT
77
K
Ec(0.59eV)

EF(0.53)
- · - · - · - · - · -
·-·-
·
-·
-·
Ej(O
)
AT
300
K
;·I
Ev(- 0.59)
Ec(0.56
eV
)
EF
(0.38)
Ej
(0 )

-

-
Ev(-
O.
SS)
AT
600K
Ec(0.50eV)
-·-
·-

- ·- ·- ·- ·-
·-·
- ·
EF:
Ej(O)
-


Ev(-0
.
50)
16.
(a)
Th
e ionizat
io
n energy for boron in Si is 0.045 eV At 300 K, all boron
impurities are ionized. Thus
pp

=
NA
=
10
15
cm-
3
np
=
11
? I
nA
= (9.65 x
10
9
i I
10
15
= 9.3 x
10
4
cm-
3
.
The Fe1mi level measured from the top
of
the valence band is given b
y:
Ep-
Ev

= k
Tln
(NvfND) = 0.0259 ln (2.66 x
10
19
I 10
15
)
= 0.26 eV
(b) The boron atoms compensate the arsenic atoins; we have
PP
=
NA
-
ND=
3 x
10
16
- 2.9 x 10
16
=
10
15
cm-
3
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 resi
st

ivities for these two samples are
differen
t.
17.
Since
ND
>> n
;,
we can approximate
no
=
ND
and
Po
= n? I no= 9.3 x10
19
I
10
17
= 9.3 x
10
2
c
m-
3
From
no
= n;exp F i ,
(
E

-E)
kT
we have
Ep
-
E;
=
kT
ln
(n
0
1
n;)
=0
.0
259ln(10
11
19.
65
x 10
9
)=
0
.42eV
6
Th
e resu
lt
ing flat band diagram is :

0.42eV
1.
12eV


Ej
'
t-
~,

.
Ev
-:'
18. Fr
om
Eq. 28
n = l
/2[Nn
-
NA
+~(ND
- NA)
2
+4n/
J
=
112[
2.5 x10
13
+~(2.5x10

13
)
2
+4(2.5x10
13
/ J
= 4.04x10
13
20. Assuming complete ionization, the F
en
ni level measm
ed
from the
in
trinsic
Fen
ni
level is 0.35 eV for 10
15
cm-
3
,
0.45
eV
for 10
17
cm-
3
,
and 0.54

eV
for 10
19
The number
of
electrons that are ionized is given by
Using the
Fe
nni
levels given above, we
ob
tain the number
of
ionized donors as
n = 10
15
cm-
3
for
ND
= 10
15
cm
-
3
n = 0.93 x 10
17
cm-
3
for

ND
= 10
17
cm-
3
n = 0.27 x 10
19
cm
-
3
Therefore, the ass
um
pt
ion of complete ionization is valid on
ly
for the case
of
10
15
c
m-
3
.
7

8
21. N
D
+
=

kTEE
FD
e
/)(
16
1
10
−−
+
=
1350
16
e1
10
.−
+

=
1451
1
1
10
16
.
+
= 5.33 × 10
15
cm
-3
The neutral donor = 10

16
– 5.33 ×10
15
cm
-3
= 4.67 × 10
15
cm
-3

∴

The

ratio of
+
D
D
N
N
O
=
335
764
.
.
= 0.876 .


9

CHAPTER 2
1. (a) For intrinsic Si,
μ
n
= 1450,
μ
p
= 505, and n = p = n
i
= 9.65×10
9

We have
5
1031.3
)(
11
×=
+
=
+
=
pnipn
qnqpqn
μμμμ
ρ
Ω-cm
(b)
Similarly for GaAs,
μ

n
= 9200,
μ
p
= 320, and n = p = n
i
= 2.25×10
6

We have
8
1092.2
)(
11
×=
+
=
+
=
pnipn
qnqpqn
μμμμ
ρ
Ω-cm.
2. For lattice scattering,
μ
n
∝ T
-3/2

T = 200 K,
μ
n
= 1300×
2/3
2/3
300
200
−
−
= 2388 cm
2
/V-s
T = 400 K,
μ
n
= 1300×
2/3
2/3
300
400
−
−
= 844 cm
2
/V-s.
3. Since
21
111
μμμ

+=

∴
500
1
250
11
+=
μ
μ = 167 cm
2
/V-s.
4. (a)
p = 5×10
15
cm
-3
, n = n
i
2
/p = (9.65×10
9
)
2
/5×10
15
= 1.86×10
4
cm
-3

μ
p
= 410 cm
2
/V-s,
μ
n
= 1300 cm
2
/V-s
ρ
=
pqpqnq
ppn
μμμ
1

1
≈
+
= 3 Ω-cm
(b)
p = N
A
– N
D
= 2×10
16
– 1.5×10

16
= 5×10
15
cm
-3
, n = 1.86×10
4
cm
-3

μ
p
=
μ
p
(N
A
+ N
D
) =
μ
p
(3.5×10
16
) = 290 cm
2
/V-s,
μ
n
=

μ
n
(N
A
+ N
D
) = 1000 cm
2
/V-s
ρ
=
pqpqnq
ppn
μμμ
1

1
≈
+
= 4.3 Ω-cm
(c)
p = N
A
(Boron) – N
D
+ N
A
(Gallium) = 5×10
15
cm

-3
, n = 1.86×10
4

cm
-3

μ
p
=
μ
p
(N
A
+ N
D
+ N
A
) =
μ
p
(2.05×10
17
) = 150 cm
2
/V-s,

10
μ
n

=
μ
n
(N
A
+ N
D
+ N
A
) = 520 cm
2
/V-s
ρ
= 8.3 Ω-cm.
5. Assume
N
D
− N
A
>> n
i
, the conductivity is given by
σ
≈ qn
μ
n
= q
μ
n
(N

D
− N
A
)
We have that
16 = (1.6×10
-19
)
μ
n
(N
D
− 10
17
)
Since mobility is a function of the ionized impurity concentration, we can use
Fig. 3 along with trial and error to determine
μ
n
and N
D .
For example, if we
choose
N
D
= 2×10
17
, then N
I
= N

D
+
+ N
A
-
= 3×10
17
, so that
μ
n
≈ 510 cm
2
/V-s
which gives
σ
= 8.16.
Further trial and error yields
N
D
≈ 3.5×10
17
cm
-3

and
μ
n
≈ 400 cm
2
/V-s

which gives
σ
≈ 16 (Ω-cm)
-1
.
6.
)/( )(
2
nnbnqpnq
ippn
+=+=
μμμσ

From the condition d
σ
/dn = 0, we obtain
bnn
i
/ =

Therefore
b
b
bnq
nbbbnqμ
ip
iip
i
m
2

1

)1(
1
)/(
1

+
=
+
+
=
μ
ρ
ρ
.
7. At the limit when d >> s,
CF =
2ln
π
= 4.53. Then from Eq. 16

11
226.053.41050
101
1010
4
3
3
=×××

×
×
=××=
−
−
−
CFW
I
V
ρ
Ω-cm
From Fig. 6,
CF = 4.2 (d/s = 10); using the a/d = 1 curve we obtain
78.10
2.41050
10226.0
)/(
4
3
=
×
×
×
=⋅⋅=
−
−
CFWIV
ρ
mV.
8. Hall coefficient,

7.426
05.0)101030(105.2
106.11010
493
33
=
×××××
×××
==
−−
−−
WIB
AV
R
z
H
H
cm
3
/C
Since the sign of
R
H
is positive, the carriers are holes. From Eq. 22
16
19
1046.1
7.426106.1
11
×=

××
==
−
H
qR
p
cm
-3

Assuming
N
A
≈ p, from Fig. 7 we obtain
ρ
= 1.1 Ω-cm
The mobility
μ
p
is given by Eq. 15b
380
1.11046.1106.1
11
1619
=
××××
==
−
ρ
μ
qp

p
cm
2
/V-s.
9. Since
R ∝
ρ
and
pn
qpqn
μμ
ρ
+
=
1
, hence
pn
pn
R
μμ
+
∝
1

From Einstein relation
μ
∝
D

50//

=
=
pnpn
DD
μ
μ

pAnD
nD
NN
N
R.
R
μμ
μ
+
=
1
1
50
1
1

We have N
A
= 50 N
D
.
10. The electric potential
φ

is related to electron potential energy by the charge (− q)
φ
= +
q
1
(E
F
− E
i
)
The electric field for the one-dimensional situation is defined as
E(x) = −
dx
d
φ
=
dx
dE
q
i
1


12
n = n
i
exp
⎟
⎠
⎞

⎜
⎝
⎛
−
kT
EE
iF
= N
D
(x)
Hence
E
F
− E
i
= kT ln
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
i
D
n
)x(N

dx

)x(dN
)x(Nq
kT
(x)
D
D
1
⎟
⎠
⎞
⎜
⎝
⎛
−=E
.
11.
(a) From Eq. 31, J
n
= 0 and
a
q
kT
eN
eaN
q
kT
n
dx
dn
D

x
ax
ax
n
n
+=
−
==
−
−

)(
- - )(
0
0
μ
E
(b)
E (x) = 0.0259 (10
4
) = 259 V/cm.
12. At thermal and electric equilibria,
0
)(
)( =+=
dx
xdn
qDxnqJ
nnn
E

μ


xNNLN
NND
L
NN
LxNNN
D
dx
xdn
xn
D
x
L
L
n
n
L
Ln
n
n
n
)(

))((
1)(
)(
1
)(

00
0
0
00
−+
−
−=
−
−+
−=−=
μ
μμ
E


000
0

0
n
ln
)(
D
-
N
N
D
xNNLN
NN
V

L
n
n
L
L
L
μμ
−=
−+
−
=
∫
.

13.
11166
10101010 =××==Δ=Δ
−
Lp
Gpn
τ
cm
-3

151115
101010 ≈+=Δ+=Δ+= nNnnn
Dno
cm
-3
.cm 1010

10
)1065.9(
3-1111
15
29
2
≈+
×
=Δ+= p
N
n
p
D
i

14.
(a)
8
15715
10
10210105
11
−
−
=
××××
=≈
tthp
p
N

νσ
τ
s
48
103109
−−
×=×==
ppp
DL
τ
cm
201010210
10167
=×××==
−
stssthlr
NS
σν
cm/s

13
(b) The hole concentration at the surface is given by Eq. 67
.c
m
10
2010103
2010
11010
102
)10(9.65

1)0(
3-9
84
8
178
16
29
≈
⎟
⎠
⎞
⎜
⎝
⎛
×+×
×
−×+
×
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+

−+=
−−
−
−
lrpp
lrp
Lpnon
SL
S
Gpp
τ
τ
τ

15.
pn
qpqn
μ
μ
σ
+=
Before illumination
nonnon
ppnn
=
=
,
After illumination
,
Gnnnn

pnonon
τ
+
=
Δ
+
=

Gpppp
pnonon
τ
+
=
Δ
+
=

. )(
)()]()([
Gq
pqnqppqnnq
ppn
nopnonnopnon
τμμ
μ
μ
μ
μ
σ
+=

+
−
Δ
+
+
Δ
+=Δ


16. (a)
diff ,
dx
dp
qDJ
pp
−=
=
− 1.6×10
-19
×12×
4
1012
1
−
×
×10
15
exp(-x/12)
= 1.6exp(-
x/12) A/cm

2
(b)
diff ,drift , ptotaln
JJJ −
=

= 4.8
− 1.6exp(-x/12) A/cm
2
(c) E
nn
qnJ
μ
=
drift ,
∵
∴
4.8 − 1.6exp(-x/12) = 1.6×10
-19
×10
16
×1000×E
E = 3 − exp(-x/12) V/cm.
17. For E = 0 we have
0
2
2
=
∂
∂

+
−
−=
∂
∂
x
p
D
pp
t
p
n
p
p
non
τ


14
at steady state, the boundary conditions are p
n
(x = 0) = p
n
(0) and p
n
(x = W) =
p
no
.
Therefore

[]
⎥
⎥
⎥
⎥
⎥
⎦
⎤
⎢
⎢
⎢
⎢
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝

⎛
−
−+=
p
p
nonnon
L
W
L
xW
pppxp
sinh
sinh
)0()(

[]
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−=
∂
∂
−==
=
pp

p
non
x
n
pp
L
W
L
D
ppq
x
p
qDxJ coth)0()0(
0

[]
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−=
∂
∂
−==
=
p

p
p
non
Wx
n
pp
L
W
L
D
ppq
x
p
qDWxJ
sinh
1
)0()(
.
18. The portion of injection current that reaches the opposite surface by diffusion
is given by
)/cosh(
1
)0(
)(
0
pp
p
LWJ
WJ
==

α

26
105105050
−−
×=××=≡
ppp
DL
τ
cm
98.0
)105/10cosh(
1

22
0
=
×
=
∴
−−
α

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
n
pp
ττ
Δ
=
Δ

so that the excess minority carrier concentration at the surface
Δ
p
n
,
surface
= 10
14
⋅
6
7
10
10
−
−
=10
13
cm
-3

The generation rate can be determined from the steady-state conditions in the

bulk
G =
6
14
10
10
−
= 10
20
cm
-3
s
-1


15
From Eq. 62, we can write
0
2
2
=
Δ
−+
∂
Δ∂
p
p
p
G
x

p
D
τ

The boundary conditions are
Δ
p(x =
∞
) = 10
14
cm
-3
and
Δ
p(x = 0) = 10
13
cm
-3

Hence
Δ
p(x) = 10
14
(
p
Lx
e
/
9.01
−

− )
where
L
p
=
6
1010
−
⋅ = 31.6 μm.
20. The potential barrier height
χ
φ
φ
−
=
mB
= 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
mF
qE
φ
+ and having a velocity component normal to the surface can escape
the solid, the thermionic current density is
dEeEv
h
m

qvJ
kTEE
x
qE
x
F
mF
)(
2
1


3
2
3

)2(4
−−
∞
+
∫∫
==
φ
π

where
x
v is the component of velocity normal to the surface of the metal.
Since the energy-momentum relationship
)(

2
1
2
222
2
zyx
ppp
mm
P
E
++==
Differentiation leads to
m
PdP
dE
=

By changing the momentum component to rectangular coordinates,
zyx
dpdpdpdPP =
2
4
π

Hence
z
mkTp
y
mkTp
xx

mkTmEp
p
zyx
p
mkTmEppp
x
dpedpedppe
mh
q
dpdpdpep
mh
q
J
z
yfx
x
x
fzyx
2
2


2)2(


3



p

p
2/)2(
3
2
22
0
0y
222
2

2
−
∞
∞−
−
∞
∞−
−−
∞
∞∞
−∞=
∞
−∞=
−++−
∫∫∫
∫∫ ∫
=
=

where ).(2
2
0 mFx
qEmp
φ
+=

16
Since
21


2
⎟
⎠
⎞
⎜
⎝
⎛
=
−
∞
∞−
∫
a
dxe
ax
π
, the last two integrals yield (2πmkT)
21

.
The first integral is evaluated by setting u
mkT
mEp
Fx
=
−
2
2
2
.
Therefore we have
mk
T
dpp
du
xx
=
The lower limit of the first integral can be written as

kT
q
mkT
mEqEm
mFmF
φ
φ
=
−+
2

2)(2

so that the first integral becomes
kTq
u
ktq
m
m
emkTduemkT
φ
φ
−
−
∞
=
∫


/

Hence
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−

==
−
kT
q
TAeT
h
qmk
J
m
kTq
m
φ
π
φ
exp
4
2*2
3
2
.
22.
Equation 79 is the tunneling probability

110
234
1931
2
0
m1017.2
)10054.1(

)106.1)(220)(1011.9(2
)(2
−
−
−−
×=
×
×−×
=
−
=

EqVm
n
β


[]
6
1
2
100
1019.3
)220(24
1031017.2sinh(20
1
−
−
−
×=

⎭
⎪
⎬
⎫
⎩
⎪
⎨
⎧
−××
××××
+=T
.
23.
Equation 79 is the tunneling probability

[]
403.0
)2.26(2.24
)101099.9sinh(6
1)10(
1
2
109
10
=
⎪
⎭
⎬
⎫
⎪

⎩
⎨
⎧
−××
×××
+=
−
−
−
T


()
[]
()
9
1
2
99
9
108.7
2.262.24
101099.9sinh6
1)10(
−
−
−
−
×=
⎭

⎪
⎬
⎫
⎩
⎪
⎨
⎧
−××
×××
+=T
.
19
234
1931
2
0
m1099.9
)10054.1(
)106.1)(2.26)(1011.9(2
)(2
−
−
−−
×=
×
×−×
=
−
=


EqVm
n
β

17
24. From Fig. 22
As
E = 10
3
V/s
ν
d
≈ 1.3×10
6
cm/s (Si) and
ν
d
≈ 8.7×10
6
cm/s (GaAs)
t ≈ 77 ps (Si) and t ≈ 11.5 ps (GaAs)
As
E = 5×10
4
V/s
ν
d
≈ 10
7
cm/s (Si) and

ν
d
≈ 8.2×10
6
cm/s (GaAs)
t ≈ 10 ps (Si) and t ≈ 12.2 ps (GaAs).

cm/s105.9m/s109.5
101.9
300101.382

2
2
velocity Thermal 25.
64
31
23-
00
×=×=
×
×××
=
==
−
m
kT
m
E
v
th

th

For electric field of 100 V/cm, drift velocity
thnd
vv <<×=×== cm/s1035.11001350
5
E
μ

For electric field of 10
4
V/cm.

thn
v≈×=×= cm/s1035.1101350
74
E
μ
.
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.l
m)
The overall space charge neutrality
of

the semiconductor requires that the total negative
space charge
per
unit area
in
the p-side
mu
st equal the total positive space charge
per
unit ar
ea
in
the n-side, t
hu
s
we
can obtain the depletion layer w
id
th
in
the n-side region:
0.8
X 8 X 10
14
=
W
x3x
10
14
n

2
Hence,
then-side
depletion layer w
id
th is:
The total depletion la
yer
wi
dth is 1.867
J.l
m.
We
use the Poisson
's
equation for calculation
of
the electric field E(x) .
In
the n-side region,
18

19

()
V/cm108640
)100671(103
1006710m0671
3
414

4
×−===
×−××=∴
××−=⇒==
+=⇒=
−
−
.)x(
.x
q
x
.N
q
K).x(
KxN
q
)x(N
q
dx
d
nmax
s
n
D
s
n
D
s
nD
s

EE
E
E
E
E
ε
ε
μ
εε

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

()
()
()
V/cm108640
1080
2
1080
2
0m80
2
3
2
42
2
4
2
×−===
⎦

⎤
⎣
⎡
×−××=∴
×××−=⇒=−=
+×=⇒=
−
−
.)x(
.xa
q
x
.a
q
K).x(
Kax
q
)x(N
q
dx
d
pmax
s
p
s
'
p
'
s
pA

s
EE
E
E
E
E
ε
ε
μ
εε

The built-in potential is:

() () ()
V 52.0

0


0

=−−=−=
∫∫∫
−
−−
−
nn
pp
x
siden

x
xx
sidepbi
dxxdxxdxxV EEE .

2. From
()
∫
−= dxxV
bi
E , the potential distribution can be obtained
With zero potential in the neutral p-region as a reference, the potential in the p-side
depletion region is
() ()
()
[
]
()()
()()
⎥
⎦
⎤
⎢
⎣
⎡
×−×−××−=
⎢
⎣
⎡
×−×−−=×−××−=−=

−−
−−−
∫∫
3
4
2
4311
3
4
2
43

0

0
2
42
108.0
3
2
108.0
3
1
10596.7
108.0
3
2
108.0
3
1

2
108.0
2

xx
xx
qa
dxxa
q
dxxxV
s
xx
s
p
εε
E
With the condition V
p
(0)=V
n
(0), the potential in the n-region is
7 ( 1 2 4 0.8
3
7)
= - 4.56x
l0
x
2
x
-1.

067 xl
0-
x

9
- xl
o-
T
he
potential distributi
on
is
Distance
p-r
egwn
n-r
egwn
-
0.8
0.
000
-
0.7
0.
006
-
0.6
0.
022
-

0.5
0.
048
-
0.4
0.
081
-
0.3
0.
120
-
0.2
0.164
-
0.1
0.
211
0
0.
259
0.
259413333
0.1
0.305788533
0.2
0.347603733
0.3
0.384858933
0.4

0.417554133
0.5
0.445689333
0.6
0.469264533
0.7
0.488279733
0.8
0.502734933
0.9
0.512630133
1
0.517965333
1.
067
0.518988825
20
Pot
ential
Distr
ib
ution
v.~v

-
v.Jvv
~
/
v vv
/

V•J'
/
.L
/
.~vv
v
V•'VV
___./
v.vvv
-
.(.5
05
I 5
-v
.
.vv
Distance
(urn)
21
3.
Th
e int1insic
can
iers density in Si at different temperatures can be obtained by using
Fig
.
22
in
Chapter 2 :
Temperature (K) Inuinsic catTier density

(n;)
250
1.50x10
8
300
9.65x10
9
350
2.00x10
11
400 8.50xl0
12
450
9.00xl0
13
500
2.20xl0
14
The
Vbi
ca
n be obta
in
ed by using Eq.
12
, and the results
ru·e
listed in the following table.
T ni
Vbi

(V)
250
1.500E+08
0.777
300
9.65E+9
0.717
350
2.00E+11
0.653
400
8.50E+12
0.488
450
9.00E+l3
0.366
500
2.20E+14
0.329
Thus, the built-in potential is decreased as the temperan1re is increased.
The depletion layer width and the maxi
mu
m field at 300 K are
W = 2ssVbi =
qND
2 X 11.9 X 8.
85
X 10-
14
~

0.717 = 0.
9715
1.6x
10-
19
x 1
0b
flill
~
=
qN
DW
= 1.6 x
10
-
19
x 10
15
x 9.715 x
10
-
5
= 1.
476
x
104
V/cm.
&
5
11.9 X 8.85 X

10
-
14
22

23
18
16
2/1
18
18
14
19
5
2/1
max
10
1
10755.1
10
10
1085.89.11
30106.12
104
2
.4
D
D
D
D

DA
DA
s
R
N
N
N
N
NN
NNqV
+
=×⇒
⎥
⎦
⎤
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+××
×××
=×⇒
⎥

⎦
⎤
⎢
⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
+
≈
−
−
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
,
()
(
)
()
V
V
Nqε
VV
C
Bs
bi

j
−×=
×××××
−×
=
−
=
−−
837.010187.1
10108.8511.9101.6
0.8372
2
1
16
1514192

For N
D
=10
16
cm
-3
,
()
(
)
()
V
V
Nqε

VV
C
Bs
bi
j
−×=
×××××
−×
=
−
=
−−
896.010187.1
10108.8511.9101.6
0.8962
2
1
15
1614192

For N
D
=10
17
cm
-3
,
()
(
)

()
V
V
Nqε
VV
C
s
bi
j
−×=
×××××
−×
=
−
=
−−
956.010187.1
10108.8511.9101.6
0.9562
2
1
14
171419
B
2

When the reversed bias is applied, we summarize a table of
2
j
C1/ vs V for various N

D
values as
following,