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1
Solutions Manual to Accompany
SEMICONDUCTOR DEVICES
Physics and Technology
2nd
Edition
S. M. SZE
UMC Chair Professor
Nat...
0
Contents
Ch.1 Introduction--------------------------------------------------------------------- 0
Ch.2 Energy Bands and ...
1
CHAPTER 2
1. (a) From Fig. 11a, the atom at the center of the cube is surround by four
equidistant nearest neighbors tha...
2
between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is
D =
2
1
222
222






+





+




 aaa
= ...
3
properties, such as the etch rate, trap density, breakage plane etc.
(d) The temperating of the crusible and the pull ra...
4
( ) ( )CVVC
NNkTEEEi ln
2
2)( ++=
= (EC+ EV)/ 2 + (3kT / 4) ln



 3
2
)6)(( np mm (1)
At 77 K
Ei = (1.16/2) + (3 ...
5
17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boron
impurities are ionized. Thus pp = NA = 10...
6
19. Assuming complete ionization, the Fermi level measured from the intrinsic
Fermi level is 0.35 eV for 1015
cm-3
, 0.4...
7
CHAPTER 3
1. (a) For intrinsic Si, µn = 1450, µp = 505, and n = p = ni = 9.65×109
We have
5
1031.3
)(
11
×=
+
=
+
=
pnip...
8
(c) p = NA (Boron) – ND + NA (Gallium) = 5×1015
cm-3
, n = 1.86×104
cm-3
µp = µp (NA + ND+ NA) = µp (2.05×1017
) = 150 c...
9
b
b
bnq
nbbbnqì
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 E...
10
pAnD
nD
NN
N
R.
R
µµ
µ
+
=
1
1
50 1
1
We have NA = 50 ND .
10. The electric potential φis related to electron potential...
11
000
0
0
n
ln
)(
D
-
N
ND
xNNLN
NN
V L
n
n
L
L
L
µµ
−=
−+
−
= ∫ .
13.
11166
10101010 =××==∆=∆
−
Lp Gpn τ cm-3
151115
101...
12
16. (a) diff,
dx
dp
qDJ pp −=
= − 1.6×10-19
×12× 4
1012
1
−
×
×1015
exp(-x/12)
= 1.6exp(-x/12) A/cm2
(b) diff,drift, pt...
13
given by
)/cosh(
1
)0(
)(
0
pp
p
LWJ
WJ
==α
26
105105050 −−
×=××=≡ ppp DL τ cm
98.0
)105/10cosh(
1
220 =
×
=∴ −−
α
Ther...
14
Differentiation leads to
m
PdP
dE =
By changing the momentum component to rectangular coordinates,
zyx dpdpdpdPP =
2
4π...
15
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 (GaA...
16
CHAPTER 4
1. The impurity profile is,
The overall space charge neutrality of the semiconductor requires that the total ...
17
( )
( ) ( )
V/cm108640
1080
2
1080
2
0m80
2
3
242
24
2
×−===



 ×−××=∴
×××−=⇒=−=
+×=⇒=
−
−
.)x(
.xa
q
x
.a
q
K)....
18
0.1 0.30578
8533
0.2 0.3476
03733
0.3 0.3848
58933
0.4 0.41755
4133
0.5 0.4456
89333
0.6 0.4692
64533
0.7 0.4882
79733
...
19
3. The intrinsic carriers density in Si at different temperatures can be obtained by using Fig.22 in
Chapter 2 :
Temper...
20
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
+
=×⇒...
21
V ND=1E15 ND=1E16 ND=1E17
-4 5.741E+
16
5.812E+
15
5.883E+
14
-3.5 5.148E+
16
5.218E+
15
5.289E+1
4
-3 4.555E+1
6
4.625...
22
The interceptions give the built-in potentialof the p-n junctions.
23
6. The built-in potential is
( )
V5686.0
1065.9106.18
0259.01085.89.111010
ln0259.0
3
2
8
ln
3
2
3919
142020
32
2
=

...
24
( )
m266.1cm1066.12
10106.1
)5.0717.0(1085.89.1122 5
1519
14
µ
ε
=×=
××
+××××
=
+
= −
−
−
A
bis
qN
VV
W
Thus
2751619
A/...
25
( ) ( )
( )
316
0259.0
7.029
7
19
2
/
cm10278.5
1
1065.9
105
21
106.125
11
−






−
−






×=
⇒




...
26
13. From








−= 1kt
qV
s eJJ
we can obtain
V78.01
10244.8
10
ln0259.01ln
0259.0 17
3
=





+


...
27
VR Js
0 1.713E-
07
0.1 1.755E-
07
0.2 1.941E-
07
0.3 2,129E-
07
0.4 2.316E-
07
0.5 2.503E-
07
0.6 2.691E-
07
0.7 2.878E...
28
When ND=1017
cm-3
, we obtain
( )
V953.0
1065.9
1010
ln0259.0 29
1719
=
×
×
=biV
15. From Eq. 39,
( )∫
∞
−=
x
nonp
n
dx...
29
( ) ( )
∫
∞ −−
−=
x
LxxkTqV
no
n
pn
dxeepq
//
1
The hole diffusion length is larger than the length of neutral region.
...
30
conditions ofW = 5 µm and NB = 101 5
cm-3
.
17. We can use following equations to determine the parameters of the diode...
31
constant. From Eq. 84, we can obtain.
V/cm1087.5)10(104110
1054
101
104
10 56
1
53
6
4
0
6
5
4
×=××=⇒=×





×
⇒=...
32
biased, the total depletion width will be increased.
22. Eg(0.3) = 1.424 + 1.247 × 0.3 = 1.789 eV
biV = ( ) −−−
∆
− q/E...
33
CHAPTER 5
1. (a) The common-base and common-emitter current gains is given by
.199
995.01
995.0
1
995.0998.0997.0
0
0
0...
34
( )
( )
.m105.364cm105.364
5.0956.0
102105
1
102
105
106.1
1005.12
12
junction)base-emittertheofhlayer widt-depletion(T...
35
( ) .465.613
102
1065.9
mc1021040scm40
17
292
37
=
×
×
==
×=⋅=== −−
D
i
no
pppp
N
n
p
DLD τ
In the collector region
( )...
36
(c) To improve γ , the emitter has to be doped much heavier than the base.
To improve Tα , we can make the base width n...
37
( )
( ) .
cosh
1
1coth
sinh
1
sinh
cosh
)1(
sinh
cosh
1
sinh
cosh
1
)1(
0
0





















...
38
[ ]
.
2
)0(
2
)
2
(
)1(
)(
0
2
0
0
n
kTqV
no
W
kTqV
no
W
kTqV
no
W
nonB
qAWp
eqAWp
W
x
xeqAp
dx
W
x
epqA
dxpxpqAQ
EB
EB...
39
( )
( )
.
cosh
1
1coth
cosh1
sinh
1






















+−


















...
40
( )
( )
.1500
102
1077.542
2 24
24
22
0
=
×
×
=≅
−
−
WLpβ
13. In the emitter region,
687
1031037401
407
354 1817
.
.
.p...
41
( ) .cm5.18624
105
1065.9 3-
15
29
=
×
×
=COp
The emitter current components are given by
A1083.526
105.0
13.4656734.30...
42
.50000
99998.01
99998.0
1 0
0
0 ≅
−
=
−
=
α
α
β
16. (a) The total number of impurities in the neutral base region is
( ...
43
Therefore,
( ) ( ) .869.11080299.0105105 533
Ω=×⋅×=×= −−−
WR BB ρ
For a voltage drop of qkT ,
.A0139.0==
B
B
qR
kT
I
Th...
44
0 5 10 15 20 25 30
1 0 0
1 5 0
2 0 0
2 5 0
IB (µA)
β0
20. Comparing the equations with Eq. 32 gives
11aIFO = , 12aIROR ...
45
99995.0
1031.9
31.9
10
1
10
105.0
1
1
1
1
23
4
=
×
⋅⋅
×
+
=
⋅⋅+
=
−−
−
no
EO
p
E
E
F
p
n
D
D
L
W
α
876.0
1031.9
10863.1...
46
.021 =+ ∞−∞ CC LL
eCeC
Hence 01 =C . In addition, for the boundary condition at 0='x ,
( )02
0
2 C
L
nCeC C
==−
( ) ( )...
47
24. Neglect the time delays of emitter and collector, the base transit time is given by
s1083.31
1052
1
2
1 12
9
−
×=
×...
48
Note that )HBT(0β increases exponentially when x increases.
27. The impurity concentration of the n1 region is -314
cm1...
49
Therefore,
25
0 cmA1025.25.4 −
×=≅ JJ
.cm4.44
1025.2
101
Area 2
5
3
=
×
×
== −
−
J
Is
28. In the 2np2n1 −− transistor, ...
50
CHAPTER 6
1.
EC
EF
Ei
EV
VG = VT
ψS
= 2ψB
METAL OXIDE n-TYPE SEMICONDUCTOR
51
2.
3.
EC
Ei
EF
EV
n+
POLYSILICON OXIDE p-TYPE SEMICONDUCTOR
n+
POLYSILICON OXIDE p-TYPE SEMICONDUCTOR
φms
VG = VFB = φφ...
52
4.
0 W x
-d
qND
QM
QP
ρS(x)
-d 0 W
x
E (x)
-d 0 W
x
ψ(x)
V Vo
ψs
(a)
(b)
(c)
53
5.
A
i
A
s
m
Nq
n
N
kT
W 2
ln
2






=
ε
= 2 1619
9
16
14
1051061
10659
105
ln026010858911
×××







...
54
9. ∫
−
−
−
−
××
×
==
6
10
0
2617
6
19
17
])10(10
2
1
[
10
106.1
)10(
1
dyyq
d
Qot
= 8×10-9
C/cm2
2
7
9
1032.2
1045.3
10...
55
DTGon
D
o
BAs
BGon
DB
o
As
DBGonD
V)VV(C)
L
Z
(
V
C
)(qN
VC
L
Z
V
C
qN
V)V(C
L
Z
I
−≅















...
56
16. The device is operated in linear region, since 1.0=DV V < 5.0)( =− TG VV V
Therefore, )(. TGon
D
D
d VVC
L
Z
V
I
g ...
57
21. 7
1045.3
49.17.0 −
×
+−=− BqF
12
19
7
107.1
106.1
1045.379.0
×=
×
××
= −
−
BF cm-2
.
22. The bandgap in degeneratel...
58
23. 42.0
1065.9
10
ln026.0 9
17
=







×
=Bψ V
o
oo
o
BAs
B
o
f
msT
C
.
.
C
....
.
C
.
.
C
qN
C
qQ
V
8
191714...
59
κ
κ
κ
ε
V
V
A
A
C
d
C ox
=′
=′
==′
2
∴scaling factor for switching energy :
1000
1111
322
==⋅⋅
κκκ
κ
A reduction of one...
60
30.
o
siA
BFBT
C
dqN
VV ++= ψ2
V.18.0
28.01.0
1063.8
103105106.1
92.002.1
F/cm1063.8
104
1085.89.3
V92.0)(ln22
V02.1
10...
61
V0.980.420.56
0
22
V42.0
1065.9
10
ln026.0
F/cm1045.3
10
10854.8
9.3
9
17
B
28
5
14
0
−=−−=
−−−=−=
=





×
=
×...
62
CHAPTER 7
1. From Eq.1, the theoretical barrier height is
eV54.001.455.4 =−=−= χφφ mBn
We can calculate Vn as
V188.0)
1...
63
(c) For V = –1 V
m0.222cm1022.2
107.4106.1
)17.0(1009.12)(2
5
1619
12
µ
ε
=×=
×××
+××
=
+
=
−
−
−
D
Rbis
qN
VV
W
V/cm10...
64
From the given relationship between C and Va, we obtain
15
2
1012.2
1
×−=






adV
C
d
(cm2
/F)2
/V
From Eq.11
...
65
27
2
2*
A/cm1081.3
0259.0
8.0
exp)300(110
exp
−
×=





 −
××=





−
=
kT
q
TAJ Bn
s
φ
The injected hole c...
66
8. From Eq.33 we obtain
D
biG
P
p
P
m V
VV
V
V
I
g
2 2
+
=
= D
bi
Psn
V
V
V
aL
Z εµ
62.4
1085.84.122
)103(107106.1
2 14...
67
14
517192
1085.84.122
)102(10106.1
2 −
−−
×××
×××
==
s
D
P
aqN
V
ε
= 0.364 V
The threshold voltage is
VT = Vbi – Vp = 0...
68
d1 = 1.68 × 10-6
cm
d1= 16.8 nm
Therefore, this thickness of the doped AlGaAs layer is 16.8 nm.
13. The pinch-off volta...
69
The 2DEG concentration is
[ ] 12
719
14
1029.1)3.1(0
10)81050(106.1
1085.83.12
×=−−×
×++××
××
= −−
−
sn cm-2
.
15. The ...
70
CHAPTER 8
1. Z0 =
C
L
nH25.1175102 2122
0 =××=⋅= −
ZCL .
2.
222
2






+





+





=
d
p
b
n
a
mc
...
71
( )
.2.27
0367.0
1.0
2.0
1exp
1.0
2.010
1.0
10
V2.01.022
01exp
2
1exp
1
2.0
2
22
2.0
322
2
2
Ω−=





=
−=


...
72
( )
( ) ( )
V8.742.576.17
100.4-3
1009.1
105.1106.1
104.4104.0104.4 4
12
1219
545
=+=
×





×
×××
−×+××=
−
...
73
= 138 V
(b) Transit time t = 7
4
10
10)312( −
×−
=
−
sv
bW
= 9×10-15
= 90 ps .
9. (a) For transit-time mode, we require...
74
( )
( )
5.6
394.2exp71
300/15000259.0
eV31.0
exp71)/exp(
=
−×=




 −
×=∆− e
CL
CU
kTE
N
N
Therefore, at Te = 3...
75
CHAPTER 9
1. Eq.9)(FromeV2.071.24/0.6)m6.0( ==µhv
α(0.6 µm) = 3×104
cm-1
The net incident power on the sample is the to...
76
58.40
296.0
1
ln
10300
11
ln
1
4
=





×
=





−
RL
cm-1
.
(b) The threshold current reduction is
( ) ( )...
77
Substituting λLn2 for m and letting dm/dλ= - ∆m/∆λ, yield


















−
∆
=∆∴
=+





∆
∆...
78
For R = 0.317
2
4
cmA1000
99.0317.0
1
ln
101002
1
10010 −
−
=











×××
+×=thJ
and so mA5105101001000 ...
79
( ) 1
C02.0
50
1 −
== o
ξ .
which is larger than that for T0 = 110 o
C. Therefore the laser with T0 = 50 o
C is worse
f...
80
Therefore, h v/q = 1.24 / λ(µm)
Thus, η= (R×1.24) / λand R = (ηλ) / 1.24.
13. The electric field in the p-layer is give...
81
14. (a) For a photodiode, only a narrow wavelength range centered at the optical signal
wavelength is important; wherea...
82
( )
( ) L
kTqV
s
L
kTqV
s
kTqV
s
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CHAPTER 10
1. C0 = 1017
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7. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility
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89
11. For the conventionally-doped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The
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94
CHAPTER 11
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x2
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95
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98
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101
CHAPTER 12
1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particles/m3
with
particle siz...
102
concern of damage to stepper lens, lower exposure speed and reduced throughput.
6. (a) A shaped beam system enables th...
103
0.6/16 × 40/60 = 0.025 µm…. (110) plane
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Sze solution

  1. 1. 1 Solutions Manual to Accompany SEMICONDUCTOR DEVICES Physics and Technology 2nd Edition S. M. SZE UMC Chair Professor National Chiao Tung University National Nano Device Laboratories Hsinchu, Taiwan John Wiley and Sons, Inc New York. Chicester / Weinheim / Brisband / Singapore / Toronto
  2. 2. 0 Contents Ch.1 Introduction--------------------------------------------------------------------- 0 Ch.2 Energy Bands and Carrier Concentration-------------------------------------- 1 Ch.3 Carrier Transport Phenomena -------------------------------------------------- 7 Ch.4 p-n Junction -------------------------------------------------------------------- 16 Ch.5 Bipolar Transistor and Related Devices---------------------------------------- 32 Ch.6 MOSFET and Related Devices------------------------------------------------- 48 Ch.7 MESFET and Related Devices------------------------------------------------- 60 Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices --------------- 68 Ch.9 Photonic Devices ------------------------------------------------------------- 73 Ch.10 Crystal Growth and Epitaxy--------------------------------------------------- 83 Ch.11 Film Formation---------------------------------------------------------------- 92 Ch.12 Lithography and Etching------------------------------------------------------ 99 Ch.13 Impurity Doping--------------------------------------------------------------- 105 Ch.14 Integrated Devices------------------------------------------------------------- 113
  3. 3. 1 CHAPTER 2 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 a2 , therefore we have 2/ a2 = 2/ (5.43 × 10-8 )2 = 6.78 × 1014 atoms / cm2 Similarly we have for (110) plane (Fig. 4a and Fig. 6) (2 + 2 ×1/2 + 4 ×1/4) / a2 2 = 9.6 × 1015 atoms / cm2 , 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 × 1014 atoms / cm2 . 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. 4 Maximum fraction of cell filled = no. of sphere × volume of each sphere / unit cell volume = 1 × 4ð(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 each of the six faces for a total of three spheres. The nearest neighbor distance is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ). 4 Maximum fraction of cell filled = (1 + 3) {4ð[(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
  4. 4. 2 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 4 Maximum fraction of cell filled = (1 + 3 + 4) 3 3 83 4            aπ / a3 = ð 3 / 16 = 34 %. This is a relatively low percentage compared to other lattice structures. 4. 1d = 2d = 3d = 4d = d 1d + 2d + 3d + 4d = 0 1d • ( 1d + 2d + 3d + 4d ) = 1d • 0 = 0 2 1d + 1d • 2d + 1d • 3d + 1d • 4d = 0 4d2 + d2 cosè12 + d2 cosè13 + d2 cosè14 = d2 +3 d2 cosè= 0 4 cosè = 3 1− è= cos-1 ( 3 1− ) = 109.470 . 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 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. (a) The melting temperature for Si is 1412 ºC, and for SiO2 is 1600 ºC. Therefore, SiO2 has higher melting temperature. It is more difficult to break the Si-O bond than the Si-Si bond. (b) The seed crystal is used to initiated the growth of the ingot with the correct crystal orientation. (c) The crystal orientation determines the semiconductor’s chemical and electrical
  5. 5. 3 properties, such as the etch rate, trap density, breakage plane etc. (d) The temperating of the crusible and the pull rate. 8. Eg (T) = 1.17 – 636)( 4.73x10 24 + − T T for Si ∴ Eg ( 100 K) = 1.163 eV , and Eg (600 K) = 1.032 eV Eg(T) = 1.519 – 204)( 5.405x10 24 + − T T for GaAs ∴Eg( 100 K) = 1.501 eV, and Eg (600 K) = 1.277 eV . 9. 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 ( VE taken to be E = 0) to the bottom of the valence band Ebottom: p = ∫ bottomE 0 N(E)[1 – F(E)]dE (1) where 1 –F(E) = ( ) [ ]{ }/kT 1 e/11 FEE − +− = [ ] 1/)( e1 −− + kTEE F If EF – E >> kT then 1 – F(E) ~ exp ( )[ ]kTEEF −− (2) Then from Appendix H and , Eqs. 1 and 2 we obtain p = 4ð[2mp / h2 ]3/2 ∫ bottomE 0 E1/2 exp [-(EF – E) / kT ]dE (3) Let x a E / kT , and let Ebottom = ∞− , Eq. 3 becomes p = 4ð(2mp / h2 )3/2 (kT)3/2 exp [-(EF / kT)] ∫ ∞− 0 x1/2 ex dx where the integral on the right is of the standard form and equals π / 2. 4 p = 2[2ðmp kT / h2 ]3/2 exp [-(EF / kT)] By referring to the top of the valence band as EV instead of E = 0 we have, p = 2(2ðmp kT / h2 )3/2 exp [-(EF – EV) / kT ] or p = NV exp [-(EF –EV) / kT ] where NV = 2 (2ðmp kT / h2 )3 . 10. From Eq. 18 NV = 2(2ðmp kT / h2 )3/2 The effective mass of holes in Si is mp = (NV / 2) 2/3 ( h2 / 2ðkT ) = 3 2 3619 2 m101066.2       ×× − ( ) ( )( )300103812 106256 23 234 − − × × . . π = 9.4 × 10-31 kg = 1.03 m0. Similarly, we have for GaAs mp = 3.9 × 10-31 kg = 0.43 m0. 11. Using Eq. 19
  6. 6. 4 ( ) ( )CVVC NNkTEEEi ln 2 2)( ++= = (EC+ EV)/ 2 + (3kT / 4) ln     3 2 )6)(( np mm (1) At 77 K Ei = (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 Ei = (1.12/2) + (3.29 × 10-5 )(300) = 0.56 + 0.009 = 0.569 eV. At 373 K Ei = (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 Ei is in the center of the forbidden gap. 12. KE = ( ) ( ) )( /)( / d de C F top C F top C EEx kTEE C E E kTEE C E E C EeEE EEEEE −≡ −− −− − −− ∫ ∫ = 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 . 13. (a) p = mv = 9.109 × 10-31 ×105 = 9.109 × 10-26 kg–m/s λ = p h = 26 34 101099 106266 − − × × . . = 7.27 × 10-9 m = 72.7 Å (b) nλ = λ pm m0 = 0630 1 . × 72.7 = 1154 Å . 14. From Fig. 22 when ni = 1015 cm-3 , the corresponding temperature is 1000 / T = 1.8. So that T = 1000/1.8 = 555 K or 282 . 15. From Ec – EF = kT ln [NC / (ND – NA)] which can be rewritten as ND – NA = NC exp [–(EC – EF) / kT ] Then ND – NA = 2.86 × 1019 exp(–0.20 / 0.0259) = 1.26 × 1016 cm-3 or ND= 1.26 × 1016 + NA = 2.26 × 1016 cm-3 A compensated semiconductor can be fabricated to provide a specific Fermi energy level. 16. From Fig. 28a we can draw the following energy-band diagrams:
  7. 7. 5 17. (a) The ionization energy for boron in Si is 0.045 eV. At 300 K, all boron impurities are ionized. Thus pp = NA = 1015 cm-3 np = ni 2 / nA = (9.65 × 109 )2 / 1015 = 9.3 × 104 cm-3 . The Fermi level measured from the top of the valence band is given by: EF – EV = kT ln(NV/ND) = 0.0259 ln (2.66 × 1019 / 1015 ) = 0.26 eV (b) The boron atoms compensate the arsenic atoms; we have pp = NA – ND = 3 × 1016 – 2.9 × 1016 = 1015 cm-3 Since pp is the same as given in (a), the values for np and EF are the same as in (a). However, the mobilities and resistivities for these two samples are different. 18. Since ND >> ni, we can approximate n0 = ND and p0 = ni 2 / n0 = 9.3 ×1019 / 1017 = 9.3 × 102 cm-3 From n0 = ni exp       − kT EE iF , we have EF – Ei = kT ln (n0 / ni) = 0.0259 ln (1017 / 9.65 × 109 ) = 0.42 eV The resulting flat band diagram is :
  8. 8. 6 19. Assuming complete ionization, the Fermi level measured from the intrinsic Fermi level is 0.35 eV for 1015 cm-3 , 0.45 eV for 1017 cm-3 , and 0.54 eV for 1019 cm-3 . The number of electrons that are ionized is given by n ≅ ND[1 – F(ED)] = ND / [1 + e ( ) TkEE FD /−− ] Using the Fermi levels given above, we obtain the number of ionized donors as n = 1015 cm-3 for ND = 1015 cm-3 n = 0.93 × 1017 cm-3 for ND = 1017 cm-3 n = 0.27 × 1019 cm-3 for ND = 1019 cm-3 Therefore, the assumption of complete ionization is valid only for the case of 1015 cm-3 . 20. ND + = kTEE FD e /)( 16 1 10 −− + = 1350 16 e1 10 .− + = 1451 1 1 1016 . + = 5.33 × 1015 cm-3 The neutral donor = 1016 – 5.33 ×1015 cm-3 = 4.67 × 1015 cm-3 4 The ratio of + D D N N O = 335 764 . . = 0.876 .
  9. 9. 7 CHAPTER 3 1. (a) For intrinsic Si, µn = 1450, µp = 505, and n = p = ni = 9.65×109 We have 5 1031.3 )( 11 ×= + = + = pnipn qnqpqn µµµµ ρ Ω-cm (b) Similarly for GaAs, µn = 9200, µp = 320, and n = p = ni = 2.25×106 We have 8 1092.2 )( 11 ×= + = + = pnipn qnqpqn µµµµ ρ Ω-cm. 2. For lattice scattering, µn ∝ T-3/2 T = 200 K, µn = 1300× 2/3 2/3 300 200 − − = 2388 cm2 /V-s T = 400 K, µn = 1300× 2/3 2/3 300 400 − − = 844 cm2 /V-s. 3. Since 21 111 µµµ += ∴ 500 1 250 11 += µ µ = 167 cm2 /V-s. 4. (a) p = 5×1015 cm-3 , n = ni 2 /p = (9.65×109 )2 /5×1015 = 1.86×104 cm-3 µp = 410 cm2 /V-s, µn = 1300 cm2 /V-s ρ= pqpqnq ppn µµµ 11 ≈ + = 3 Ω-cm (b) p = NA – ND = 2×1016 – 1.5×1016 = 5×1015 cm-3 , n = 1.86×104 cm-3 µp = µp (NA + ND) = µp (3.5×1016 ) = 290 cm2 /V-s, µn = µn (NA + ND) = 1000 cm2 /V-s ρ= pqpqnq ppn µµµ 11 ≈ + = 4.3 Ω-cm
  10. 10. 8 (c) p = NA (Boron) – ND + NA (Gallium) = 5×1015 cm-3 , n = 1.86×104 cm-3 µp = µp (NA + ND+ NA) = µp (2.05×1017 ) = 150 cm2 /V-s, µn = µn (NA + ND+ NA) = 520 cm2 /V-s ρ= 8.3 Ω-cm. 5. Assume ND − NA>> ni, the conductivity is given by σ≈ qnµn = qµn(ND − NA) We have that 16 = (1.6×10-19 )µn(ND − 1017 ) Since mobility is a function of the ionized impurity concentration, we can use 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 ND≈ 3.5×1017 cm-3 and µn ≈ 400 cm2 /V-s which gives σ≈ 16 (Ω-cm)-1 . 6. )/()( 2 nnbnqpnq ippn +=+= µµµσ From the condition dσ/dn = 0, we obtain bnn i /= Therefore
  11. 11. 9 b b bnq nbbbnqì 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. 16 226.053.41050 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 cm3 /C Since the sign of RH is positive, the carriers are holes. From Eq. 22 16 19 1046.1 7.426106.1 11 ×= ×× == − HqR p cm-3 Assuming NA ≈ p, from Fig. 7 we obtain ρ= 1.1 Ω-cm The mobility µp is given by Eq. 15b 380 1.11046.1106.1 11 1619 = ×××× == − ρ µ qp p cm2 /V-s. 9. Since R ∝ ρand pn qpqn µµ ρ + = 1 , hence pn pn R µµ + ∝ 1 From Einstein relation µ∝D 50// == pnpn DDµµ
  12. 12. 10 pAnD nD NN N R. R µµ µ + = 1 1 50 1 1 We have NA = 50 ND . 10. The electric potential φis related to electron potential energy by the charge (− q) φ= + q 1 (EF − Ei) The electric field for the one-dimensional situation is defined as E(x) = − dx dφ = dx dE q i1 n = ni exp       − kT EE iF = ND(x) Hence EF − Ei = kT ln       i D n )x(N dx )x(dN )x(Nq kT (x) D D 1       −=E . 11. (a) From Eq. 31, Jn = 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 (104 ) = 259 V/cm. 12. At thermal and electric equilibria, 0 )( )( =+= dx xdn qDxnqJ nnn Eµ 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
  13. 13. 11 000 0 0 n ln )( D - N ND xNNLN NN V L n n L L L µµ −= −+ − = ∫ . 13. 11166 10101010 =××==∆=∆ − Lp Gpn τ cm-3 151115 101010 ≈+=∆+=∆+= nNnnn Dno cm-3 .cm1010 10 )1065.9( 3-1111 15 292 ≈+ × =∆+= p N n p D i 14. (a) 8 15715 10 10210105 11 − − = ×××× =≈ tthp p Nνσ τ s 48 103109 −− ×=×== ppp DL τ cm 201010210 10167 =×××== − stssthlr NS σν cm/s (b) The hole concentration at the surface is given by Eq. 67 .cm10 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 µµσ += Before illumination nonnon ppnn == , After illumination ,Gnnnn pnonon τ+=∆+= Gpppp pnonon τ+=∆+= .)( )()]()([ Gq pqnqppqnnq ppn nopnonnopnon τµµ µµµµσ += +−∆++∆+=∆
  14. 14. 12 16. (a) diff, dx dp qDJ pp −= = − 1.6×10-19 ×12× 4 1012 1 − × ×1015 exp(-x/12) = 1.6exp(-x/12) A/cm2 (b) diff,drift, ptotaln JJJ −= = 4.8 − 1.6exp(-x/12) A/cm2 (c) Enn qnJ µ=drift,Q ∴∴ 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 = ∂ ∂ + − −= ∂ ∂ x p D pp t p n p p non τ at steady state, the boundary conditions are pn (x = 0) = pn (0) and pn (x = W) = pno. 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 WL D ppq x p qDWxJ sinh 1 )0()( . 18. The portion of injection current that reaches the opposite surface by diffusion is
  15. 15. 13 given by )/cosh( 1 )0( )( 0 pp p LWJ WJ ==α 26 105105050 −− ×=××=≡ ppp DL τ cm 98.0 )105/10cosh( 1 220 = × =∴ −− 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 ∆ = ∆ so that the excess minority carrier concentration at the surface ∆pn, surface = 1014 ⋅ 6 7 10 10 − − =1013 cm-3 The generation rate can be determined from the steady-state conditions in the bulk G = 6 14 10 10 − = 1020 cm-3 s-1 From Eq. 62, we can write 02 2 = ∆ −+ ∂ ∆∂ p p p G x p D τ The boundary conditions are ∆p(x = ∞ ) = 1014 cm-3 and ∆p(x = 0) = 1013 cm-3 Hence ∆p(x) = 1014 ( pLx e / 9.01 − − ) where Lp = 6 1010 − ⋅ = 31.6 µm. 20. The potential barrier height χφφ −= 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 xv 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 ++==
  16. 16. 14 Differentiation leads to m PdP dE = By changing the momentum component to rectangular coordinates, zyx dpdpdpdPP = 2 4π Hence z mkTp y mkTp xx mkTmEp p zyx p mkTmEppp x dpedpedppe mh q dpdpdpep mh q J zyfx x x fzyx 222)2( 3 p p 2/)2( 3 222 0 0 y 222 2 2 −∞ ∞− −∞ ∞− −−∞ ∞ ∞ −∞= ∞ −∞= −++− ∫∫∫ ∫ ∫ ∫ = = where ).(22 0 mFx qEmp φ+= Since 21 2       =− ∞ ∞−∫ a dxe ax π , the last two integrals yield (2ðmkT) 21 . The first integral is evaluated by setting u mkT mEp Fx = − 2 22 . Therefore we have mkT dpp du xx = The lower limit of the first integral can be written as kT q mkT mEqEm mFmF φφ = −+ 2 2)(2 so that the first integral becomes kTqu ktq m m emkTduemkT φ φ −− ∞ =∫ / Hence       − == − kT q TAeT h qmk J mkTq m φπ 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 − − −− ×= × ×−× = − = h EqVmn β [ ] 6 12100 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( 12109 10 =         −×× ××× += − − − T ( )[ ] ( ) 9 1299 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 − − −− ×= × ×−× = − = h EqVmn β
  17. 17. 15 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 300101.382 22 velocityThermal25. 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 Eµ For electric field of 104 V/cm. thn v≈×=×= cm/s1035.1101350 74 Eµ . The value is comparable to the thermal velocity, the linear relationship between drift velocity and the electric field is not valid.
  18. 18. 16 CHAPTER 4 1. The impurity profile is, The overall space charge neutrality of the semiconductor requires that the total negative space charge per unit area in the p-side must equal the total positive space charge per unit area in the n-side, thus we can obtain the depletion layer width in the n-side region: 14 14 103 2 1088.0 ××= ×× nW Hence, the n-side depletion layer width is: m0671 µ.Wn = The total depletion layer width is 1.867 µm. We use the Poisson’s equation for calculation of the electric field E(x). In the n-side region, ( ) 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: x (µm) (ND-NA) (cm-3 ) 3×1014 a =1019 cm-4
  19. 19. 17 ( ) ( ) ( ) V/cm108640 1080 2 1080 2 0m80 2 3 242 24 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: ( ) ( ) ( ) V52.0 0 0 =−−=−= ∫∫ ∫ − − − − nn p p x siden x x x sidepbi dxxdxxdxxV EEE . 2. From ( )∫−= dxxVbi 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 ( ) ( ) ( )[ ] ( ) ( ( ) ( )       ×−×−××−=    ×−×−−=×−××−=−= −− −−− ∫ ∫ 3424311 4243 0 0 242 108.0 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 x x s p εε E With the condition Vp(0)=Vn(0), the potential in the n-region is ( )         ×−×−××−=       ×+×−××−= −− −− 7 3 427 7 3 4214 10 9 8.0 10067.1 2 1 1056.4 10 9 8.0 10067.1 2 1 103 xx xx q xV s n ε The potential distribution is Distance p-region n-region -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 0.25941 3333
  20. 20. 18 0.1 0.30578 8533 0.2 0.3476 03733 0.3 0.3848 58933 0.4 0.41755 4133 0.5 0.4456 89333 0.6 0.4692 64533 0.7 0.4882 79733 0.8 0.50273 4933 0.9 0.5126 30133 1 0.51796 5333 1.06 7 0.51898 8825 Potential Distribution -0.100 0.000 0.100 0.200 0.300 0.400 0.500 0.600 -1 -0.5 0 0.5 1 1.5 D istance (um) Potential(V)
  21. 21. 19 3. The intrinsic carriers density in Si at different temperatures can be obtained by using Fig.22 in Chapter 2 : Temperature (K) Intrinsic carrier density (ni) 250 1.50×108 300 9.65×109 350 2.00×1011 400 8.50×1012 450 9.00×1013 500 2.20×1014 The Vbi can be obtained by using Eq. 12, and the results are listed in the following table. T ni Vbi (V) 250 1.500E+0 8 0.777 300 9.65E+9 0.717 350 2.00E+1 1 0.653 400 8.50E+1 2 0.488 450 9.00E+13 0.366 500 2.20E+1 4 0.329 Thus, the built-in potential is decreased as the temperature is increased. The depletion layer width and the maximum field at 300 K are V/cm.10476.1 1085.89.11 10715.910106.1 m9715.0 10106.1 717.01085.89.1122 4 14 51519 max 1519 14 ×= ×× ×××× == = ×× ×××× == − −− − − s D D bis WqN qN V W ε µ ε E
  22. 22. 20 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 ND = 1.755×1016 cm-3 for the junction. 5. From Eq. 12 and Eq. 35, we can obtain the 1/C2 versus V relationship for doping concentration of 1015 , 1016 , or 1017 cm-3 , respectively. For ND=1015 cm-3 , ( ) ( ) ( )V V Nqå VV C Bs bi j −×= ××××× −× = − = −− 837.010187.1 10108.8511.9101.6 0.837221 16 1514192 For ND=1016 cm-3 , ( ) ( ) ( )V V Nqå VV C Bs bi j −×= ××××× −× = − = −− 896.010187.1 10108.8511.9101.6 0.896221 15 1614192 For ND =1017 cm-3 , ( ) ( ) ( )V V Nqå VV C s bi j −×= ××××× −× = − = −− 956.010187.1 10108.8511.9101.6 0.956221 14 171419 B 2 When the reversed bias is applied, we summarize a table of 2 jC1/ vs V for various ND values as following,
  23. 23. 21 V ND=1E15 ND=1E16 ND=1E17 -4 5.741E+ 16 5.812E+ 15 5.883E+ 14 -3.5 5.148E+ 16 5.218E+ 15 5.289E+1 4 -3 4.555E+1 6 4.625E+ 15 4.696E+ 14 -2.5 3.961E+ 16 4.031E+ 15 4.102E+ 14 -2 3.368E +16 3.438E+ 15 3.509E+1 4 -1.5 2.774E +16 2.844E+ 15 2.915E+1 4 -1 2.181E +16 2.251E+ 15 2.322E+ 14 -0.5 1.587E+ 16 1.657E+ 15 1.728E+ 14 0 9.935E+1 5 1.064E+ 15 1.134E+ 14 Hence, we obtain a series of curves of 1/C2 versus V as following, The slopes of the curves is positive proportional to the values of the doping concentration. 1/C^2vs V 0 1E+16 2E+16 3E+16 4E+16 5E+16 6E+16 7E+16 -4.5 -4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 Applied Voltage 1/C^2
  24. 24. 22 The interceptions give the built-in potentialof the p-n junctions.
  25. 25. 23 6. The built-in potential is ( ) V5686.0 1065.9106.18 0259.01085.89.111010 ln0259.0 3 2 8 ln 3 2 3919 142020 32 2 =         ×××× ××××× ××=        = − − i s bi nq kTa q kT V ε From Eq. 38, the junction capacitance can be obtained ( ) ( ) ( ) 3/1 2142019- 3/1 2 5686.012 1085.89.1110101.6 12         − ××××× =         − == − RRbi ss j VVV qa W C εε At reverse bias of 4V, the junction capacitance is 6.866×10-9 F/cm2 . 7. From Eq. 35, we can obtain ( ) ( ) 2 2 221 j s Rbi D Bs bi j C q VV N Nq VV C εε − =⇒ − = We can select the n-type doping concentration of 3.43×1015 cm-3 . 8. From Eq. 56, 169 1515 1571515 1089310659 02590 020 exp10 02590 020 exp10 10101010 expexp ×=××                   − +      ××× =                   − +      − =−= −− −− .. . . . . n kT EE kT EE N UG i ti p it n tthnp σσ υσσ and ( ) 315 28 1419 2 cm10433 )10850( 108589111061 422 − − −− ×=⇒ ×× ×××× × =≅⇒>> .N . ... C q V NVV d j s R DbiR ε Q
  26. 26. 24 ( ) m266.1cm1066.12 10106.1 )5.0717.0(1085.89.1122 5 1519 14 µ ε =×= ×× +×××× = + = − − − A bis qN VV W Thus 2751619 A/cm10879.71066.121089.3106.1 −− ×=×××××== qGWJ gen . 9. From Eq. 49, and D i no N n p 2 = We can obtain the hole concentration at the edge of the space charge region, ( ) 3170259.0 8.0 16 29) 0259.0 8.0 ( 2 cm1042.2 10 1065.9 −       ×= × == ee N n p D i n . 10. ( ) ( ) ( )1 / −=−+= kTqV spnnp eJxJxJJ V.017.0 195.0 1 0259.0 0259.0 =⇒ −=⇒ −=⇒ V e e J J V V s 11. The parameters are ni= 9.65×109 cm-3 Dn= 21 cm2 /sec Dp=10 cm2 /sec τp0=τn0=5×10-7 sec From Eq. 52 and Eq. 54 ( ) ( ) ( ) 315 0259.0 7.029 7 19 2 / cm102.5 1 1065.9 105 10 106.17 11 −       − −       ×= ⇒         −× × × × ××= ⇒         −××=−= D D kT qV D i po pkTqV p nop np N e N e N nD qe L pqD xJ a τ
  27. 27. 25 ( ) ( ) ( ) 316 0259.0 7.029 7 19 2 / cm10278.5 1 1065.9 105 21 106.125 11 −       − −       ×= ⇒         −× × × × ××= ⇒         −××=−=− A A kT qV A i no nkTqV n pon pn N e N e N nD qe L nqD xJ a τ We can select a p-n diode with the conditions of NA= 5.278×1016 cm-3 and ND = 5.4×1015 cm-3 . 12. Assumeôg =ôp =ôn = 10-6 s, Dn= 21 cm2 /sec, and Dp = 10 cm2 /sec (a) The saturation current calculation. From Eq. 55a and ppp DL τ= , we can obtain         +=+= 00 200 11 n n Ap p D i n pn p np s D N D N qn L nqD L pqD J ττ ( ) 212 616618 2919 A/cm1087.6 10 21 10 1 10 10 10 1 1065.9106.1 − −− − ×=         +×××= And from the cross-sectional area A= 1.2×10-5 cm2 , we obtain A10244.81087.6102.1 17125 −−− ×=×××=×= ss JAI . (b) The total current density is         −= 1kt qV s eJJ Thus A.10244.8110244.8 A1051.41047.510244.8110244.8 170259.0 7.0 17 7.0 511170259.0 7.0 17 7.0 − − − − −−− ×=        −×= ×=×××=         −×= eI eI V V
  28. 28. 26 13. From         −= 1kt qV s eJJ we can obtain V78.01 10244.8 10 ln0259.01ln 0259.0 17 3 =      +      × ×=⇒      +      = − − V J JV s . 14. From Eq. 59, and assume Dp= 10 cm2 /sec, we can obtain ( ) ( ) 1519 14 6 919 15 29 6 19 2 101061 108589112 10 106591061 10 10659 10 10 1061 ×× +××××××× + × ×= +≅ − − − − − − . VV..... . Wqn N nD qJ Rbi g i D i p p R ττ ( ) V834.0 1065.9 1010 ln0259.0 29 1519 = × × =biV Thus RR VJ +×+×= −− 834.010872.11026.5 711
  29. 29. 27 VR Js 0 1.713E- 07 0.1 1.755E- 07 0.2 1.941E- 07 0.3 2,129E- 07 0.4 2.316E- 07 0.5 2.503E- 07 0.6 2.691E- 07 0.7 2.878E- 07 0.8 3.065E- 07 0.9 3.252E- 07 1 3.439E- 07
  30. 30. 28 When ND=1017 cm-3 , we obtain ( ) V953.0 1065.9 1010 ln0259.0 29 1719 = × × =biV 15. From Eq. 39, ( )∫ ∞ −= x nonp n dxppqQ Current Density 0.E+00 5.E-08 1.E-07 2.E-07 2.E-07 3.E-07 3.E-07 0 0.2 0.4 0.6 0.8 1 1.2 Applied Voltage CurrentDensity RR VJ +×+×= −− 956.010872.11026.5 813 Current Density 0.00E+00 5.00E-09 1.00E-08 1.50E-08 2.00E-08 2.50E-08 3.00E-08 0 0.2 0.4 0.6 0.8 1 1.2 Applied Voltage CurrentDensity
  31. 31. 29 ( ) ( ) ∫ ∞ −− −= x LxxkTqV no n pn dxeepq // 1 The hole diffusion length is larger than the length of neutral region. ( )∫ −= ' n n x x nonp dxppqQ ( ) ( ) .C/cm10784.8 1)105( 10 )1065.9( 106.1 1)( 1 23 5 0 5 1 0259.0 1 4 16 29 19 x // ' ' n − −− −− − − − − −− ×=         −        −×− × ××=         −        −−= −= ∫ eee eeeLqp dxeepq p nn p nn n pn L xx L xx kT qV pno x LxxkTqV no 16. From Fig. 26, the critical field at breakdown for a Si one-sided abrupt junction is about 2.8 ×105 V/cm. Then from Eq. 85, we obtain ( ) ( ) 1 2 22 voltagebreakdown − == B csc B N q W V EE ε ( ) ( ) 115 19 2514 10 106.12 108.21085.89.11 − − − ×× ×××× = 258= V ( ) £gm4318cm108431 101061 2581085891122 3 1519 14 .. . .. qN VV W B bis =×= ×× ×××× ≅ − = − − − ε When the n-region is reduced to 5µm, the punch-through will take place first. From Eq. 87, we can obtain ( )/2W insert29Fig.inareashaded mcE = B B V 'V       −      = mm W W W W 2 121 43.18 5 2 43.18 5 2582' =      −      ×=      −      = mm BB W W W W VV V Compared to Fig. 29, the calculated result is the same as the value under the
  32. 32. 30 conditions ofW = 5 µm and NB = 101 5 cm-3 . 17. We can use following equations to determine the parameters of the diode. kTqV D i p pkTqV r ikTqV D i p p F e N nD qe qWn e N nD qJ / 2 2// 2 2 τττ ≅+= ( ) 1 2 22 − == D csc B N q W V EE ε ⇒ ( ) 302590 7029 7 19 2 1022 10659 10 1061 − − − ×= × ×××⇒= .e N .D .Ae N nD AqAJ . . D pkT/qV D i p p F τ ( ) ( ) 1 19 214 1 2 10612 10858911 130 22 − − − − ×× ×× =⇒== D c D csc B N . .. N q W V EEE ε 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
  33. 33. 31 constant. From Eq. 84, we can obtain. V/cm1087.5)10(104110 1054 101 104 10 56 1 53 6 4 0 6 5 4 ×=××=⇒=×      × ⇒=      × − ∫ critical W dx E EE 4 587101087.5 35 =××= − BV V (b) From Fig. 26, the critical field is 5 × 105 V/cm. ( ) ( ) 1 2 22 voltagebreakdown − == B csc B N q W V EE ε ( ) ( ) 116 19 2514 102 106.12 1051085.84.12 − − − × ×× ×××× = 8.42= V. 20. 422 4 18 cm10 102 102 − − = × × =a ( ) ( ) 238 2122 21 19 1423 21 2123 10844 10 1061 108589112 3 4 2 3 4 3 2 / c / // c / / s / c B . . .. a q W V E E EE − − − − − ×= ×      × ××× =       == ε The breakdown voltage can be determined by a selected Ec. 21. To calculate the results with applied voltage of V5.0=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 V1.1 and V103.4 4− × , respectively. The depletion widths are cm103.821 5− × and cm101.274 8− × , respectively. Also, by substituting V5−=V to Eqs. 90 and 91, the electrostatic potentials are V6.6 and V1020.3 4− × , and the depletion widths are cm109.359 5− × and cm103.12 8− × , 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-
  34. 34. 32 biased, the total depletion width will be increased. 22. Eg(0.3) = 1.424 + 1.247 × 0.3 = 1.789 eV biV = ( ) −−− ∆ − q/EE q E q E VF Cg 22 2 ( ) q/EE FC 11 − = 1.789 – 0.21– 15 17 105 1074 ln × ×. q kT – 15 18 105 107 ln × × q kT = 1.273 V 1x = ( ) 2 1 21 212       + ADD biA NNqN vN εε εε = ( ) 2 1 1519 14 46114121051061 27311085846114122       +××× ××××× − − ... .... = 4.1 5 10− × cm. Since 21 xNxN AD = 21 xx =∴ m0.82cm10282 5 1 µ.xW =×==∴ − .
  35. 35. 33 CHAPTER 5 1. (a) The common-base and common-emitter current gains is given by .199 995.01 995.0 1 995.0998.0997.0 0 0 0 0 = − = − = =×== α α β γαα T (b) Since 0=BI and A1010 9− ×=CpI , then CBOI is A1010 9− × . The emitter current is ( ) ( ) .A102 10101991 1 6 9 0 − − ×= ×⋅+= += CBOCEO II β 2. For an ideal transistor, .999 1 999.0 0 0 0 0 = − = == α α β γα CBOI is known and equals to A1010 6− × . Therefore, ( ) .mA10 1010)9991( 1 6 0 = ×⋅+= += − CBOCEO II β 3. (a) The emitter-base junction is forward biased. From Chapter 3 we obtain ( ) .V956.0 1065.9 102105 ln0259.0ln 29 1718 2 =         × ×⋅× =         = i DA bi n NN q kT V The depletion-layer width in the base is
  36. 36. 34 ( ) ( ) .m105.364cm105.364 5.0956.0 102105 1 102 105 106.1 1005.12 12 junction)base-emittertheofhlayer widt-depletion(Total 2-6- 171817 18 19 12 1 µ×=×= −      ×+×      × × × ×⋅ = −      +      =       + = − − VV NNN N q NN N W bi DAD As DA A ε Similarly we obtain for the base-collector function ( ) .V0.795 1065.9 10102 ln0259.0 29 1617 =         × ⋅× =biV and ( ) .m10.2544cm104.254 5795.0 10210 1 102 10 106.1 1005.12 2-6- 171617 16 19 12 2 µ×=×= +      ×+        ×× ×⋅ = − − W Therefore the neutral base width is .m904.010.254410364.51 -22 21B µ=×−×−=−−= − WWWW (b) Using Eq. 13a ( ) .cm10543.2 102 1065.9 )0( 3-110259.05.0 17 292 ×= × × === ee N n epp kTqV D ikTqV non EBEB 4. In the emitter region ( ) .18.625 105 1065.9 cm10721.01052scm52 18 29 38 = × × = ×=⋅== −− EO EE n LD In the base region
  37. 37. 35 ( ) .465.613 102 1065.9 mc1021040scm40 17 292 37 = × × == ×=⋅=== −− D i no pppp N n p DLD τ In the collector region ( ) .109.312 10 1065.9 cm10724.1010115scm115 3 16 29 36 ×= × = ×=⋅== −− CO CC n LD The current components are given by Eqs. 20, 21, 22, and 23: ( ) .0 A103.196 10724.10 10312.9115102.0106.1 A101.0411 10721.0 625.1852102.0106.1 A101.596 A101.596 10904.0 613.46540102.0106.1 14- 3 3219 7-0259.05.0 3 219 5- 5-0259.05.0 4 219 =−= ×= × ×⋅⋅×⋅× = ×=− × ⋅⋅×⋅× = ×=≅ ×= × ⋅⋅×⋅× = − −− − −− − −− CpEpBB Cn En EpCp Ep III I eI II eI 5. (a) The emitter, collector, and base currents are given by .A101.041 A101.596 A101.606 7- 5- -5 ×=−+= ×=+= ×=+= CnBBEnB CnCpC EnEpE IIII III III (b) We can obtain the emitter efficiency and the base transport factor: 9938.0 10606.1 10596.1 5 5 = × × == − − E Ep I I γ .1 10596.1 10596.1 5 5 = × × == − − Ep Cp T I I α Hence, the common-base and common-emitter current gains are .3.160 1 9938.0 0 0 0 0 = − = == α α β γαα T
  38. 38. 36 (c) To improve γ , the emitter has to be doped much heavier than the base. To improve Tα , we can make the base width narrower. 6. We can sketch (0))( nn pxp curves by using a computer program: 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 0 . 0 0 . 2 0 . 4 0 . 6 0 . 8 1 . 0 20 1 2 5 10 W/Lp<0.1 DISTANCE x pn (x)/pn (0) In the figure, we can see when 1.0<pLW ( 05.0=pLW in this case), the minority carrier distribution approaches a straight line and can be simplified to Eq. 15. 7. Using Eq.14, EpI is given by
  39. 39. 37 ( ) ( ) . cosh 1 1coth sinh 1 sinh cosh )1( sinh cosh 1 sinh cosh 1 )1( 0 0                       +−        =                                     +                               −=                                             − +                               − − −−=         −= = = p kTqV pp nop pp pkTqV p nop x p pp no p ppkTqV nop x n pEp L W e L W L pD qA L W L W L W e L pD qA L W L x L p L W L xW L epqDA dx dp qDAI EB EB EB Similarly, we can obtain CpI : ( ) ( ) .cosh1 sinh 1 sinh cosh sinh 1 )1( sinh cosh 1 sinh cosh 1 )1(                 +−         =                                             +                       −=                                             − +                               − − −−=       −= = = p kTqV p p nop p p p kTqV p nop Wx p pp no p ppkTqV nop Wx n pCp L W e L WL pD qA L W L W L W e L pD qA L W L x L p L W L xW L epqDA dx dp qDAI EB EB EB 8. The total excess minority carrier charge can be expressed by
  40. 40. 38 [ ] . 2 )0( 2 ) 2 ( )1( )( 0 2 0 0 n kTqV no W kTqV no W kTqV no W nonB qAWp eqAWp W x xeqAp dx W x epqA dxpxpqAQ EB EB EB = = −=     −= −= ∫ ∫ From Fig. 6, the triangular area in the base region is 2 )0(nWp . By multiplying this value by q and the cross-sectional area A, we can obtain the same expression as BQ . In Problem 3, .C103.678 2 10543.210904.0102.0106.1 15- 114219 ×= ×⋅×⋅×⋅× = −−− BQ 9. In Eq. 27, ( ) . 2 2 )0(2 )0( 1 2 2 2221 B p np np kTqV C Q W D qAQp W D W pqAD aeaI EB = = ≅ +−= Therefore, the collector current is directly proportional to the minority carrier charge stored in the base. 10. The base transport factor is
  41. 41. 39 ( ) ( ) . cosh 1 1coth cosh1 sinh 1                       +−                         +−         =≅ p kTqV p p kTqV p Ep Cp T L W e L W L W e L W I I EB EB α For 1<<pLW , ( ) 1cosh ≅pLW . Thus, ( ).21 2 1 1 sech cothsinh 1 22 2 p p p pp T LW L W L W L W L W −=         −=         =         ⋅         =α 11. The common-emitter current gain is given by . 11 0 0 0 T T γα γα α α β − = − ≡ Since 1≅γ , ( ) ( )[ ] ( ) .12 211 21 1 22 22 22 0 −= −− − = − ≅ WL LW LW p p p T T α α β If 1<<pLW , then 22 0 2 WLp≅β . 12. m54.77mc10477.5103100 37 µ=×=×⋅== −− ppp DL τ Therefore, the common-emitter current gain is
  42. 42. 40 ( ) ( ) .1500 102 1077.542 2 24 24 22 0 = × × =≅ − − WLpβ 13. In the emitter region, 687 1031037401 407 354 1817 . . .pE = ×⋅×+ += − µ ...DE scm2.2668702590 =⋅= In the base region, 631186 1021069801 1252 88 1617 . . n = ×⋅×+ += − µ ....Dp scm733063118602590 =⋅= In the collector region, 82453 1051037401 407 354 1517 . . .pC = ×⋅×+ += − µ ....DC scm75118245302590 =⋅= 14. In the emitter region, cm10506.110269.2 36 −− ×=⋅== EEE DL τ ( ) .cm04.31 103 1065.9 3- 18 292 = × × == E i EO N n p In the base region, cm10544.510734.30 36 −− ×=⋅=nL ( ) .cm13.4656 102 1065.9 3- 16 29 = × × =pon In the collector region, cm10428.310754.11 36 −− ×=⋅=CL
  43. 43. 41 ( ) .cm5.18624 105 1065.9 3- 15 29 = × × =COp The emitter current components are given by A1083.526 105.0 13.4656734.3010106.1 60259.06.0 4 419 − − −− ×= × ⋅⋅⋅× = eIEn ( ) .A10609.81 10506.1 04.31269.210106.1 90259.06.0 3 419 − − −− ×=− × ⋅⋅⋅× = eIEp Hence, the emitter current is .A10839.526 6− ×=+= EpEnE III And the collector current components are given by A1083.526 105.0 13.4656734.3010106.1 60259.06.0 4 419 − − −− ×= × ⋅⋅⋅× = eICn .A10022.1 10428.3 5.18624754.1110106.1 14 3 419 − − −− ×= × ⋅⋅⋅× =CpI Therefore, the collector current is obtained by .A10268.5 4− ×=+= CpCnC III 15. The emitter efficiency can be obtained by .99998.0 10839.526 1083.526 6 6 = × × == − − E En I I γ The base transport factor is .1 1083.526 1083.526 6 6 = × × == − − En Cn T I I α Therefore, the common-base current gain is obtained by .99998.099998.010 =×== Tγαα The value is very close to unity. The common-emitter current gain is
  44. 44. 42 .50000 99998.01 99998.0 1 0 0 0 ≅ − = − = α α β 16. (a) The total number of impurities in the neutral base region is ( ) ( ) .cm10583.51103102 1 2-13103108518 0 55 ×=−×⋅×= −== −− ××−− −− ∫ e elNdxeNQ lW AO W lx AOG (b) Average impurity concentration is .cm10979.6 108 10583.5 3-17 5 13 ×= × × = − W QG 17. For -317 cm10979.6 ×=AN , scm77.7 3 =nD , and cm10787.21077.7 36 −− ×=⋅== nnn DL τ ( ) ( ) 999588.0 10787.22 108 1 2 1 23 25 2 2 = × × −=−≅ − − n T L W α .99287.0 1010 10583.5 77.7 1 1 1 1 1 419 13 = ⋅ × ⋅+ = + = − EE G n E LN Q D D γ Therefore, 99246.00 == Tγαα .6.131 1 0 0 0 = − = α α β 18. The mobility of an average impurity concentration of -317 cm10979.6 × is about 300 Vscm2 . The average base resistivity Bρ is given by ( ) .cm-0299.0 1 Ω== WQq Gn B µ ρ
  45. 45. 43 Therefore, ( ) ( ) .869.11080299.0105105 533 Ω=×⋅×=×= −−− WR BB ρ For a voltage drop of qkT , .A0139.0== B B qR kT I Therefore, .A83.10139.06.1310 =⋅== BC II β 19. From Fig. 10b and Eq. 35, we obtain ( )AµBI ( )mACI B C I I ∆ ∆ =0β 0 0.20 --- 5 0.95 150 10 2.00 210 15 3.10 220 20 4.00 180 25 4.70 140 0β is not a constant. At low BI , because of generation-recombination current, 0β increases with increasing BI . At high BI , EBV increases with BI , this in turn causes a reduction of BCV since V5==+ ECBCEB VVV . The reduction of BCV causes a widening of the neutral base region, therefore 0β decreases. The following chart shows 0β as function of BI . It is obvious that 0β is not a constant.
  46. 46. 44 0 5 10 15 20 25 30 1 0 0 1 5 0 2 0 0 2 5 0 IB (µA) β0 20. Comparing the equations with Eq. 32 gives 11aIFO = , 12aIROR =α 21aIFOF =α , and .22aI RO = Hence, no EO p E E F p n D D L Wa a ⋅⋅+ == 1 1 11 21 α . 1 1 22 12 no CO p C C R p n D D L Wa a ⋅⋅+ ==α 21. In the collector region, cm101.414102 36 −− ×=⋅== CCC DL τ ( ) .cm101.8631051065.9 3-415292 ×=××== CiCO Nnn From Problem 20, we have
  47. 47. 45 99995.0 1031.9 31.9 10 1 10 105.0 1 1 1 1 23 4 = × ⋅⋅ × + = ⋅⋅+ = −− − no EO p E E F p n D D L W α 876.0 1031.9 10863.1 10 2 10414.1 105.0 1 1 1 1 2 4 3 4 = × × ⋅⋅ × × + = ⋅⋅+ = −− − no CO p C C R p n D D L W α A1049.1 10 31.91 105.0 1031.910 105106.1 14 44 2 419 11 − −− −− ×=       ⋅ + × ×⋅ ⋅×⋅×=       +== E EOEnop FO L nD W pD qAaI .A107.1 10414.1 10863.12 105.0 1031.910 105106.1 14 3 4 4 2 419 22 − −− −− ×=       × ×⋅ + × ×⋅ ⋅×⋅×=       +== C COCnop RO L nD W pD qAaI The emitter and collector currents are ( ) A10715.1 1 4− ×= +−= ROR kTqV FOE IeII EB α ( ) .A10715.1 1 4− ×= +−= RO kTqV FOFC IeII EB α Note that these currents are almost the same (no base current) for 1<<pLW . 22. Referring Eq. 11, the field-free steady-state continuity equation in the collector region is ( ) ( ) .02 C 2 = − −      C poC C n'xn 'dx 'xnd D τ The solution is given by ( CCC DL τ= ) ( ) .21 CC L'xL'x C eCeC'xn − += Applying the boundary condition at ∞='x yields
  48. 48. 46 .021 =+ ∞−∞ CC LL eCeC Hence 01 =C . In addition, for the boundary condition at 0='x , ( )02 0 2 C L nCeC C ==− ( ) ( ) .1.0 −= kTqV COC CB enn The solution is ( ) ( ) .1 CCB L'xkTqV COC eenxn − −= The collector current can be expressed as ( ) ( ) ( ) ( ).11 11 2221 0 −−−= −      +−−=         −+        −= == kTqVkTqV kTqV C COCnopkTqVnop 'x C C Wx n pC CBEB CBEB eaea e L nD W pD qAe W pD qA 'dx dn qDA dx dp qDAI 23. Using Eq. 44, the base transit time is given by ( ) . . DW pB s101.25 102 1050 2 10- 24 2 ×= × × == − τ We can obtain the cutoff frequency : .GHz1.2721 =≅ BTf πτ From Eq. 41, the common-base cutoff frequency is given by: .GHz1.275 998.0 1027.1 9 0 = × =≅ αα Tff The common-emitter cutoff frequency is ( ) ( ) ...ff MHz2.55102751998011 9 0 =××−=−= αβ α Note that βf can be expressed by ( ) ( ) TT ffff 0 000 1 11 β ααα αβ =×−=−= .
  49. 49. 47 24. Neglect the time delays of emitter and collector, the base transit time is given by s1083.31 1052 1 2 1 12 9 − ×= ×× == ππ τ T B f . From Eq. 44, W can be expressed by BpDW τ2= . Therefore, .m0.252 cm102.52 1083.31102 5- 12 µ= ×= ×××= − W The neutral base width should be 0.252 µm. 25. gE∆ =9.8% ×1.12≅ 110 meV. oβ ~exp         ∆ kT Eg ( ) ( )C0 C100 ° ° ∴ o o β β = exp       − kk 273 meV110 373 meV110 = 0.29. 26.       − =      − = 0259.0 424.1)( expexp )BJT( )HBT( 0 0 xE kT EE gEgBgE β β where .145.0,143.0125.09.1 45.0,247.1424.1)( ≤<++= ≤+= xxx xxxEgE The plot of )BJT()HBT( 00 ββ is shown in the following graph.
  50. 50. 48 Note that )HBT(0β increases exponentially when x increases. 27. The impurity concentration of the n1 region is -314 cm10 . The avalanche breakdown voltage (for mWW > ) is larger than 1500 V ( m100 µ>mW ). For a reverse block voltage of 120 V, we can choose a width such that punch-through occurs, i.e., . 2 2 s D PT WqN V ε = Thus, .cm1096.3 2 3− ×=      ε = D PTs qN V W When switching occurs, .121 ≅+ αα That is, 6.04.01 ln5.0 0 1 =−=       = J J W Lp α .51.1 106.39 1025 5.0 6.0 ln 4 4 0 = × × =      − − J J
  51. 51. 49 Therefore, 25 0 cmA1025.25.4 − ×=≅ JJ .cm4.44 1025.2 101 Area 2 5 3 = × × == − − J Is 28. In the 2np2n1 −− transistor, the base drive current required to maintain current conduction is KII )1( 21 α−= . In addition, the base drive current available to the 2np2n1 −− transistor with a reverse gate current is gA III −= 12 α . Therefore, when use a reverse gate current, the condition to obtain turn-off of the thyristor is given by 12 II < or KgA III )1( 21 αα −<− . Using Kirchhoff’s law, we have gAK III −= . Thus, the condition for turn-on of the thyristor is Ag II 2 21 1 α αα −+ > . Note that if we define the ratio of IA to Ig as turn-off gain, then the maximum turn-off gain maxβ is 121 2 max −+ = αα α β .
  52. 52. 50 CHAPTER 6 1. EC EF Ei EV VG = VT ψS = 2ψB METAL OXIDE n-TYPE SEMICONDUCTOR
  53. 53. 51 2. 3. EC Ei EF EV n+ POLYSILICON OXIDE p-TYPE SEMICONDUCTOR n+ POLYSILICON OXIDE p-TYPE SEMICONDUCTOR φms VG = VFB = φφms EF EC Ei EV EC Ei EF EV
  54. 54. 52 4. 0 W x -d qND QM QP ρS(x) -d 0 W x E (x) -d 0 W x ψ(x) V Vo ψs (a) (b) (c)
  55. 55. 53 5. A i A s m Nq n N kT W 2 ln 2       = ε = 2 1619 9 16 14 1051061 10659 105 ln026010858911 ×××         × × ××× − − . . ... = 1.5 ×10-5 cm = 0.15 µ m. 6. msox ox Wd C )/( min εε ε + = µ150.Wm = m From Prob. 5 8 57 14 10036 1051 911 93 108 1085893 − −− − ×= ××+× ×× =∴ . . . . .. Cmin F/cm2 . 7. V420 10659 10 ln0260ln 9 17 . . . n N q kT i A B = × ==ψ s A s WqN ε =E at intrinsic Bs ψψ = 5 1719 14 1074.0 10106.1 42.01085.89.1122 − − − ×= ×× ×××× == A ss qN W ψε cm 5 14 51719 10111 10858911 10740101061 ×= ×× ×××× = − −− . .. .. sE V/cm Eo = Es oxε ε s = 1.11 ×105 × 93 911 . . = 3.38 ×105 V/cm V = Vo+ sψ = Eod + sψ =( )75 10510383 − ×××. +0.42. = 0.59 V. 8. At the onset of strong inversion, sψ =2 Bψ ⇒ VG=VT thus, VG = o mA C WqN +2 Bψ From Prob. 5, Wm = 0.15£gm, Bψ = 0.026ln         × × 9 16 10659 105 . = 0.4 V Co = 6 14 10 1085893 − − ×× .. = 3.45×10-7 F/cm2 GV∴ = 7 51619 10453 10511051061 − −− × ××××× . .. +0.8 = 0.35 + 0.8 = 1.15V.
  56. 56. 54 9. ∫ − − − − ×× × == 6 10 0 2617 6 19 17 ])10(10 2 1 [ 10 106.1 )10( 1 dyyq d Qot = 8×10-9 C/cm2 2 7 9 1032.2 1045.3 108 − − − ×= × × ==∆ o ot FB C Q V V. 10. ∫= d otot dyyy d Q 0 )( 1 ρ )(105 11 xqot δρ ××= , where    ×≠ ×=∞ = − − 7 7 105,0 105, )( y y xδ 19711 6 106.1105105 10 1 −− − ××××××=∴ otQ = 4×10-8 C/cm2 12.0 1045.3 104 7 8 = × × ==∆∴ − − o ot FB C Q V V 11. ∫ − ×××= 6 10 0 23 )105( 1 dyyqy d Qot = 2.67×10-8 C/cm2 2 7 8 1074.7 1045.3 1067.2 − − − ×= × × ==∆∴ o ot FB C Q V V. 12. m oo ot FB N d d C q C Q V ××==∆ , where Nm is the area density of Qm. 11 19 7 1047.6 106.1 1045.33.0 ×= × ×× = ×∆ =⇒ − − q CV N oFB m cm-2 . 13. Since )( TGD VVV −<< , the first term in Eq. 33 can be approximated as DBGon V)V(C L Z ψµ 2− . Performing Taylor’s expansion on the 2nd term in Eq. 33, we obtain D / B / BD / B / B / B / BD V)()(V)()()()V( 212321232323 2 2 3 22 2 3 222 ψψψψψψ =−+≅−+ Equation 33 can now be re-written as .)10(105 3 1 106.1 10 1 362319 6 −− − ×××××=
  57. 57. 55 DTGon D o BAs BGon DB o As DBGonD V)VV(C) L Z ( V C )(qN VC L Z V C qN V)V(C L Z I −≅                 +−≅         ×−−≅ µ ψε ψµ ψ ε ψµ 22 2 2 2 32 3 2 2 where o BAs BT C )(qN V ψε ψ 22 2 += . 14. When the drain and gate are connected together, DG VV = and the MOSFET is operated in saturation )( DsatD VV > . DI can be obtained by substituting DsatD VV = in Eq. 33 [ ]         −+         −      −== 2/32/3 )(2)2( 2 3 2 2 2 BBDsat o As DsatB Dsat onVVD V C qN V V C L Z I GD ψψ ε ψµ where DsatV is given by Eq. 38. Inserting the condition 0)( == LyQn into Eq. 27 yields 0222 1 =−++= GBBDsatAs o Dsat V)V(qN C V ψψε For BDsatV ϕ2<< , the above equation reduces to TB o BAs DG V C )(qN VV =+== ψ ψε 2 22 . Therefore a linear extrapolation from the low current region to 0=DI will yield the threshold voltage value. 15. 400 109.65 105 0.026lnln 9 16 .) n N ( q kT i A =      × × == − ψ V 7 161914 10453 105106110858911 − −−− × ×××××× =≡ . ... C qN K o Asε = 0.27         +−+−≅∴ 2 2 2 112 K V KVV G BGDsat ψ = ] )27.0( 10 11[)27.0(8.05 2 2 +−+− = 3.42 V 2 7 )7.05( 12 1045.380010 )( 2 − × ××× =−= − TG on Dsat VV L CZ I µ = 2.55×10-2 A.
  58. 58. 56 16. The device is operated in linear region, since 1.0=DV V < 5.0)( =− TG VV V Therefore, )(. TGon D D d VVC L Z V I g constGV −= ∂ ∂ = = µ = 5.01045.3500 25.0 5 7 ×××× − = 1.72×10-3 S. 17. Don G D m VC L Z V I g constDV µ= ∂ ∂ = = . = 1.01045.3500 25.0 5 7 ×××× − =3.45×10-4 S. 18. o BAs BFBT C qN VV ψε ψ 2 2 ++= , ) 1065.9 10 ln(026.0 9 17 × =Bψ 7 1019 1045.3 105106.1 2 − − × ××× −−−=−= B g o f FB E C Q V ms ψφ = 102.042.056.0 −=−−− V 7 191714 1045.3 106.142.0101085.89.112 84.01 − −− × ×××××× ++−=∴ TV = 49.084.01 ++− = 0.33 V ( msφ can also be obtained from Fig. 8 to be – 0.98 V). 19. 7 1045.3 33.07.0 − × += BqF 11 19 7 108 106.1 1045.337.0 ×= × ×× = − − BF cm-2 . 20. 14.042.056.0 2 −=+−=+−= B g ms E ψφ V 7 171914 1045.3 42.010106.11085.89.112 84.002.014.0 2 2 − −− × ×××××× −−−−= −−−= o BDs B o f msT C qN C Q V ψε ψφ = 49.1− V.
  59. 59. 57 21. 7 1045.3 49.17.0 − × +−=− BqF 12 19 7 107.1 106.1 1045.379.0 ×= × ×× = − − BF cm-2 . 22. The bandgap in degenerately doped Si is around 1eV due to bandgap- narrowing effect. Therefore, 86.0114.0 =+−=msφ V 49.049.084.002.086.0 −=−−−=∴ TV V.
  60. 60. 58 23. 42.0 1065.9 10 ln026.0 9 17 =        × =Bψ V o oo o BAs B o f msT C . . C .... . C . . C qN C qQ V 8 1917141119 10215 140 106142010108589112 840 101061 980 2 2 − −−− × +−= ×××××× ++ ×× −−= ++−= ψε ψφ dd Co 1314 1045.31085.89.3 −− × = ×× = 14.20 1045.3 102.15 20 13 8 > × ×× ⇒> − − d VT 5 1057.4 − ×>∴d cm = 0.457 µm. 24. A£g1.0atV5.0 == dT IV Subthreshold swing = 1 0 0 loglog − ==         − − T VDVVD V II GTG 0log7 5.0 1.0 =−− = GVDI 12log 0 −==GVDI A101 12 0 − = ×=∴ GVDI . 25. ( )22 2 BBSB o As T V C Nq V ψψ ε −+=∆ V42.0) 1065.9 10 ln(026.0 9 17 = × =Bψ 7 7 14 109.6 105 1085.89.3 − − − × × ×× =oC F/cm2 V1.0=∆ TV if we want to reduce ID at VG = 0 by one order of magnitude, since the subthreshold swing is 100 mV/decade. ( ) V.83.0 84.084.0 109.6 101085.89.4106.12 1.0 7 171419 =∴ −+ × ×××××× = − −− BS BS V V 26. Scaling factor κ =10 Switching energy = 2 )( 2 1 VAC ⋅
  61. 61. 59 κ κ κ ε V V A A C d C ox =′ =′ ==′ 2 ∴scaling factor for switching energy : 1000 1111 322 ==⋅⋅ κκκ κ A reduction of one thousand times. 27. From Fig. 24 we have         −+−= ∆ −= ∆− = + ∆−= ++−=∆ =−∆+∆∴ −+=∆+ 1 2 111 2 22 2 ' 2' 2 022 )()( 2 222 j mj jmjj jmj mmjj r W L r LL L L LL LL rWrr rWr WWrr From Eq. 17 we have .1 2 1) 2 ' ( )regionrrectangulatheinchargespaceregionaltrapezaidtheinchargespace(         −+−=− + = − =∆ jo jmA o mA o mA o T r W LC rWqN C WqN L LL C WqN C V 28. Pros: 1. Higher operation speed. 2. High device density Cons: 1. More complicated fabrication flow. 2. High manufacturing cost. 29. The maximum width of the surface depletion region for bulk MOS nm49 cm109.4 105106.1 )1065.9/105ln(026.01085.89.11 2 )/(ln 2 6 1719 91714 2 = ×= ××× ×××××× = = − − − A iAs m Nq nNkT W ε For FD-SOI, nm49=≤ msi Wd .
  62. 62. 60 30. o siA BFBT C dqN VV ++= ψ2 V.18.0 28.01.0 1063.8 103105106.1 92.002.1 F/cm1063.8 104 1085.89.3 V92.0)(ln22 V02.1 1065.9 105 ln026.0 2 12.1 )(ln 2 7 61719 27 7 14 9 17 = +−= × ××××× ++−=∴ ×= × ×× = =×= −=      × × −−=−−== − −− − − − T o i A B i Ag msFB V C n N q kT n N q kTE V ψ φ 31. oC sidqN TV A ∆ =∆ mV46 1063.8 105105106.1 7 71719 = × ××××× = − −− Thus, the range of VT is from (0.18-0.046)= 0.134 V to (0.18+0.046) = 0.226 V. 32. The planar capacitor C = 15 6 1424 1045.3 101 1086.89.3)101( − − −− ×= × ××× = d A oxε F For the trench capacitor A = 4× 7 µm2 + 1 µm2 = 29 µm2 C = 29 ×3.45 ×10-15 F = 100 × 10-15 F. 33. 11 3 14 101.3 104 5.2 105 − − − ×= × ××== d dV CI A 34. )( Toipo VVC L Z g −= µ 4 × 10-5 = A ( TV−− 5 ) VVT 7=∴ 1 × 10-5 = A( 25 +− ) 9)2(7 =−−=∆V V. 35. 0 2 C qN VV BAs BFBT ψε ψ ++=
  63. 63. 61 V0.980.420.56 0 22 V42.0 1065.9 10 ln026.0 F/cm1045.3 10 10854.8 9.3 9 17 B 28 5 14 0 −=−−= −−−=−= =      × = ×= × ×= − − − B gf msFB EQ V C ψφ ψ V4.34 4.90.420.98 1045.3 106.142.0101085.89.112 42.098.0 8 191714 = ++−= × ×××××× ++−=∴ − −− TV V.2.02 32.234.4 V2.32 1045.3 106.1105 8 1911 0 = −= −= × ××× −=−=∆ − − T f FB V C Q V
  64. 64. 62 CHAPTER 7 1. From Eq.1, the theoretical barrier height is eV54.001.455.4 =−=−= χφφ mBn We can calculate Vn as V188.0) 102 1086.2 ln(0259.0ln 16 19 = × × == D C n N N q kT V Therefore, the built-in potential is V352.0188.054.0 =−=−= nBnbi VV φ . 2. (a) From Eq.11 /V/F)(cm103.2 02 10)6.42.6()/1( 2214 142 ×−= −− ×− = dV Cd 3-16 2 cm107.4 /)/1( 12 ×=      − = dVCdq N s D ε V06.0 107.4 107.4 ln0259.0ln 16 17 =        × × == D C n N N q kT V From Fig.6, the intercept of the GaAs contact is the built-in potential Vbi, which is equal to 0.7 V. Then, the barrier height is V76.0=+= nbiBn VVφ (b) 217 A/cm105×=sJ 8* =A A/K2 -cm2 for n– type GaAs kTq s Bn eTAJ /2* φ− = eV72.0 105 )300(8 ln0259.0)ln( 7 22* =      × × == − s Bn J TA q kT φ The barrier height from capacitance is 0.04 V or 5% larger.
  65. 65. 63 (c) For V = –1 V m0.222cm1022.2 107.4106.1 )17.0(1009.12)(2 5 1619 12 µ ε =×= ××× +×× = + = − − − D Rbis qN VV W V/cm1043.1 5 ×== s D m WqN ε E 28 F/cm1022.5 − ×== W C sε . 3. The barrier height is V64.001.465.4 =−=−= χφφ mBn V177.0 103 1086.2 ln0259.0ln 16 19 =        × × ×== D C n N N q kT V The built-in potential is V463.0177.064.0 =−=−= nBnbi VV φ The depletion width is ìm142.0463.0 103106.1 1085.89.112)(2 1619 14 =× ××× ××× = − = − − D bis qN VV W ε The maximum electric field is V/cm.1054.6 )1085.8(9.11 )1042.1()103()106.1( 0)( 4 14 51619 ×= ×× ××××× ==== − − W qN x s D ε EEm 4. The unit of C needs to be changed from µF to F/cm2 , so 1/C2 = 1.74×1015 – 2.12×1015 Va (cm2 /F)2 Therefore, we obtain the built-in potential at 1/C2 = 0 V74.0 1012.2 1057.1 15 15 = × × =biV
  66. 66. 64 From the given relationship between C and Va, we obtain 15 2 1012.2 1 ×−=       adV C d (cm2 /F)2 /V From Eq.11       ××××× =       − = −− 151419 2 1012.2 1 1085.89.11106.1 2 /)/1( 12 dVCdq N s D ε -315 cm106.5 ×= V161.0 106.5 1086.2 ln0259.0ln 16 19 =        × × == D C n N N q kT V We can obtain the barrier height V.901.0161.074.0 =+=+= nbiBn VVφ 5. The built-in potential is V605.0 195.08.0 105.1 1086.2 ln0259.08.0 ln 16 19 = −=         × × −= −= D C Bnbi N N q kT V φ Then, the work function is V.81.4 01.48.0 = += += χφφ Bnm 6. The saturation current density is
  67. 67. 65 27 2 2* A/cm1081.3 0259.0 8.0 exp)300(110 exp − ×=       − ××=      − = kT q TAJ Bn s φ The injected hole current density is 211 163 29192 A/cm1019.1 105.1101 )1065.9(12106.1 − − − ×= ××× ×××× == Dp ip po NL nqD J )1( )1( currentElectron currentHole / / − − = kTqV s kTqV po eJ eJ .103 1081.3 1018.1 5 3 11 − − − ×= × × == s po J J 7. The difference between the conduction band and the Fermi level is given by V.04.0 101 107.4 ln0259.0 17 17 =        × × =nV The built-in potential barrier is then V86.004.09.0 =−=biV For a depletion mode operation, VT is negative. Therefore, From Eq.38a m.ì0.108cm1008.1 86.0 1019.2 106.1 86.0 1085.84.122 10106.1 2 086.0 5 2 12 2 14 172192 =×> > × × > ××× ×× == <−= − − − − − a a aNqa V VV s D P PT ε
  68. 68. 66 8. From Eq.33 we obtain D biG P p P m V VV V V I g 2 2 + = = D bi Psn V V V aL Z εµ 62.4 1085.84.122 )103(107106.1 2 14 2516192 = ××× ××××× == − −− s D P aqN V ε V mg∴ = 1 84.0 62.4 105.1103.0 1085.84.124500105 44 144 × ××× ××××× −− −− = 1.28×10-3 S = 1.28 mS. 9. (a) The built-in voltage is       × −=−= 17 17 10 107.4 ln025.09.0nBnbi VV φ = 0.86 V At zero bias, the width of the depletion layer is 2 D bis qN V W ε = = 1719 12 10106.1 86.01009.12 ×× ××× − − = 1.07 × 10-5 cm = 0.107 µm Since W is smaller than 0.2 µm, it is a depletion-mode device. (b) The pinch-off voltage is 14 517192 1085.84.122 )102(10106.1 2 − −− ××× ××× == s D P aqN V ε = 2.92 V and the threshold voltage is VT = Vbi – VP = 0.86 – 2.92 = –2.06 V. 10. From Eq.31b, the pinch-off voltage is
  69. 69. 67 14 517192 1085.84.122 )102(10106.1 2 − −− ××× ××× == s D P aqN V ε = 0.364 V The threshold voltage is VT = Vbi – Vp = 0.8 – 0.364 = 0.436 V and the saturation current is given by Eq. 39 ( )2 2 TG ns Dsat VV aL Z I −= µε = 42 44 144 107.4)436.00( )101()105.0(2 45001085.84.121050 − −− −− ×=− ×××× ××××× A. 11. 0.85 – 0.0259ln 2 14 1917 1085.84.122 106.1107.4 a N N D D − − ××× ×× −      × = 0 For ND = 4.7 × 1016 cm-3 a = DD NN )107.3(107.4 ln0259.085.0 32 1 17 ×       × − = 1.52 ×10-5 cm = 0.152 µm For ND = 4.7 × 1017 cm-3 a = 0.496 × 10-5 cm = 0.0496 µm. 12. From Eq.48 the pinch-off voltage is T C BnP V q E V − ∆ −= φ = 0.89 – 0.23 – (–0.5) = 0.62 V and then, 14 2 1 18192 1 1085.83.122 103106.1 2 − − ××× ×××× == ddqN V s D P ε = 0.62 V
  70. 70. 68 d1 = 1.68 × 10-6 cm d1= 16.8 nm Therefore, this thickness of the doped AlGaAs layer is 16.8 nm. 13. The pinch-off voltage is 4 718192 1 1085.83.122 )1050(10106.1 2 − −− ××× ×××× == s D P dqN V ε = 1.84 V The threshold voltage is P C BnT V q E V − ∆ −= φ = 0.89 – 0.23 – 1.84 = –1.18 V When ns = 1.25× 1012 cm-2 , we obtain [ ] 12 7 0 19 14 1025.1)18.1(0 10)850(106.1 1085.83.12 ×=−−× ×++×× ×× = −− − d ns and then d0 +58.5 = 64.3 d0 = 5.8 nm The thickness of the undoped spacer is 5.8 nm. 14. The pinch-off voltage is 14 2717192 1 1085.83.122 )1050(105106.1 2 − −− ××× ××××× == s D P dqN V ε = 1.84 V The barrier height is 79.084.125.03.1 =++−=+ ∆ += P C TBn V q E Vφ V
  71. 71. 69 The 2DEG concentration is [ ] 12 719 14 1029.1)3.1(0 10)81050(106.1 1085.83.12 ×=−−× ×++×× ×× = −− − sn cm-2 . 15. The pinch-off voltage is 4 2 1 18192 1 1085.83.122 101106.1 5.1 2 − − ××× ×××× === ddqN V s D P ε The thickness of the doped AlGaAs is 1819 14 1 10106.1 1085.83.1225.1 ×× ×××× = − − d = 4.45×10-8 cm = 44.5 nm 93.05.123.08.0 −=−−=− ∆ −= P C BnT V q E V φ V. 16. The pinch-off voltage is 2 1 2 d qN V s D P ε = = ( )27 14 1819 1035 1085.83.122 103106.1 − − − × ××× ××× = 2.7 V the threshold voltage is p C BnT V q E V − ∆ −= φ = 0.89 – 0.24 – 2.7 = –2.05 V Therefore, the two-dimensional electron gas is [ ] 12 719 14 102.3)05.2(0 10)835(106.1 1085.83.12 ×=−−× ×+×× ×× = −− − sn cm-2 .
  72. 72. 70 CHAPTER 8 1. Z0 = C L nH25.1175102 2122 0 =××=⋅= − ZCL . 2. 222 2       +      +      = d p b n a mc fr ( ) ( ) .GHz616.1116 2 m/s103 4010 2 8 22 =× × = ++= c 3. pngbi VVqEV ++≅ )/( = 1.42+0.03+0.03 = 1.48 V ( ) ( ) .F/cm1013.6 1089.1 1016.1 nm18.9cm1089.1 25.048.1 1010 1010 106.1 1016.12 2 27 6 12 6 1919 1919 19 12 − − − − − − ×= × × == =×= −      × + × ×× = +      + = W C VV NN NN q W S bi DA DAS ε ε 4. exp1exp 0       +      −      = kT qV I V V V V II PP p From Fig. 4, We note that the largest negative differential resistance occurs between Vp<V<VV. The corresponding voltage can be obtained from the condition d2 I/d2 V = 0. By neglecting the second term in Eq. 5, we obtain
  73. 73. 71 ( ) .2.27 0367.0 1.0 2.0 1exp 1.0 2.010 1.0 10 V2.01.022 01exp 2 1exp 1 2.0 2 22 2.0 322 2 2 Ω−=      = −=      −      × −= =×==∴ =      −      + − =       −      −≅ − −− V V P PP P P P PP P P P dV dI R dV dI VV V V V VI V I dV Id V V V VI V I dV dI 5. (a) RSC = SS W SS W vA W dx vA Ix I dx I εε 2 11 2 00 ==∆ ∫∫ E ( ) Ω= ××× × = −− − 137 101005.1)105(2 1012 7124 24 (b) The breakdown voltage for ND = 1015 cm-3 and W = 12 µm is 250 V (Refer to Chapter 4). The voltage due to RSC is ( ) =×××= − 13710510 43 SCIR 68.5 V The total applied voltage is then 250 + 68.5 = 318.5 V. 6. (a) The dc input power is 100V(10-1 A) = 10W. For 25% efficiency, the power dissipated as heat is 10W(1-25%) = 7.5 W. C75)C/W10(W5.7 oo T =×=∆ (b) =×=∆ C75)C/mV60( oo BV 4.5 V The breakdown voltage at room temperature is (100-4.5) = 95.5V . 7. (a) For a uniform breakdown in the avalanche region, the maximum electric field is Em = 4.4 ×105 V/cm. The total voltage at breakdown across the diode is
  74. 74. 72 ( ) ( ) ( ) V8.742.576.17 100.4-3 1009.1 105.1106.1 104.4104.0104.4 4 12 1219 545 =+= ×      × ××× −×+××= −      −+= − − − − A S mAmB xW qQ xV ε EE (b) The average field in the drift region is ( ) V/cm102.2 104.03 2.57 5 4 ×= ×− − This field is high enough to maintain velocity saturation in the drift region. (c) ( ) ( ) .GHz19 104.032 10 2 4 7 = − = − = − A S xW v f 8. (a) In the p layer E1(x) = s m xqN ε 1 −E 0 bx ≤≤ = 3 µm E2(x) = s m bqN ε 1 −E b Wx ≤≤ = 12 µm E2(x) should be larger than 105 V/cm for velocity saturation 51 10≥−∴ s m bqN ε E or Em 11515 1066.41010 − ×+=+≥ s bqN ε N1. This equation coupled the plot of Em versus N in Chapter 3 gives N1 = 7 × 1015 cm-3 for Em = 4.2 × 105 V/cm 2 EE b V m B )( 2− =∴ +E2W = 45 45 10910 2 10310)12.4( − − ××+ ×××−
  75. 75. 73 = 138 V (b) Transit time t = 7 4 10 10)312( − ×− = − sv bW = 9×10-15 = 90 ps . 9. (a) For transit-time mode, we require n0L ≥ 1012 cm-2 . 31641212 0 cm10101/10/10 −− =×=≈ Ln (b) t = L/v = 10-4 / 107 = 10-11 s = 10 ps (c) The threshold field for InP is 10.5 kV/cm; the corresponding applied voltage is ( ) V525.0101 2 105.10 4 3 =×      × = − V The current is ( ) ( ) 86.3101025.5104600106.1 431619 0 =××××××=== −− AnqJAI n Eµ A The power dissipated in the device is then W02.2== IVP . 10. (a) Referring to Chapter 2, we have 31917 2 3 0 017 2 3 2 3 2 cm1033.371107.4 07.0 2.1 107.4 2 2 − ×=××=      ×=       =      = m m m m N kTm N nL nU CL nU CU h π (b) For Te = 300 K ( ) 4 104.4 97.11exp71 0259.0 eV31.0 exp71)/exp( − ×= −×=      − ×=∆− e CL CU kTE N N (c) For Te = 1500 K
  76. 76. 74 ( ) ( ) 5.6 394.2exp71 300/15000259.0 eV31.0 exp71)/exp( = −×=      − ×=∆− e CL CU kTE N N Therefore, at Te = 300 K most electrons are in the lower valley. However, at Te = 1500 K , 87%, i.e., 6.5/(6.5+1), of the electrons are in the upper valley. 11. The energy En for infinitely deep quantum well is 2 2* 2 8 n Lm h En = ( )( ) L E L m nh L En 13 * 22 2 2 8 =−= ∆ ∆ − L L E En ∆=∆ 12 meV31 =∆∴ E , =∆ 2E 11 meV . 12. From Fig. 14 we find that the first excited energy is at 280 meV and the width is 0.8 meV. For same energy but a width of 8 meV, we use the same well thickness of 6.78 nm for GaAs, but the barrier thickness must be reduced to 1.25 nm for AlAs. The resonant-tunneling current is related to the integrated flux of electrons whose energy is in the range where the transmission coefficient is large. Therefore, the current is proportional to the width nE∆ , and sufficiently thin barriers are required to achieve a high current density.
  77. 77. 75 CHAPTER 9 1. Eq.9)(FromeV2.071.24/0.6)m6.0( ==µhv α(0.6 µm) = 3×104 cm-1 The net incident power on the sample is the total incident power minus the reflected power, or 10 mW. ( ) 31032 105110 4 −×−− ×=− W e W = 0.231 µm. The portion of each photon’s energy that is converted to heat is %4.31 07.2 42.107.2 = − = − ν ν h Eh g The amount of thermal energy dissipated to the lattice per second is 31.4%×5 = 1.57 mW. 2. For λ = 0.898 µm, the corresponding photon energy is eV38.1 24.1 == λ E From Fig. 18, we obtain 2n (Al0.3Ga0.7As) = 3.38 ( ) ( ) ( )[ ] ( ) .%15.30315.0 38.31 2117cos138.314cos14 .21172958.0 38.3 1 sin 22 21 21 2 1 == + ′−×× = + − = ′=⇒=== o c o cc nn nn Efficiency n n θ θθ 3. From Eq. 15 2960 1393 1393 2 . . . R =      + − = (a) The mirror loss
  78. 78. 76 58.40 296.0 1 ln 10300 11 ln 1 4 =      × =      − RL cm-1 . (b) The threshold current reduction is ( ) ( ) ( ) %.6.36 58.4010 90.0296.0 1 ln 103002 1 58.40 1 ln 1 1 ln 2 11 ln 1 296.0 9.0,296.0296.0 4 21 21 = +       ⋅×⋅ − =       +             +−            + ≈ = ==−= − RL RRLRL RJ RRJRJ th thth α αα 4. From Eq. 10 cc nn n n θθ sinsin 21 2 1 ⋅=⇒= From Eq. 14 ( ) ( )( )45 101sin16.3108exp1exp1 − ×⋅−⋅×−−=∆−−≅Γ cdnC θ For θc = 84o n1 = 3.58 Γ1 = 0.794 θc = 78o n2 = 3.52 Γ2 = 0.998 . 5. From Eq. 16 we have Lnm 2=λ . Differentiating the above equation with respect to λ, we obtain λλ λ d nd Lm d dm 2=+ .
  79. 79. 77 Substituting λLn2 for m and letting dm/dλ= - ∆m/∆λ, yield                   − ∆ =∆∴ =+      ∆ ∆− λ λ λ λ λλλ λ d nd n Ln m d nd L Lnm 12 2 2 2 and ( ) ( ) ( ) .nm97.0m107.9 5.2 58.3 89.0 130058.32 189.0 4 2 =µ×=             − × =∆ − λ 6. From Eq. 22             + Γ = RL g 1 ln 11 α and Eq. 15 2 1 1       + − = n n R R1 = 0.317, R2 = 0.311 ( ) ( ) 217 311.0 1 ln 10100 1 100 998.0 1 78 270 317.0 1 ln 10100 1 100 794.0 1 84 4 4 =            × += =            × += − − o o g g       × =      ⋅′ − RRL 1 ln 10100 1 99.0 1 ln 2 1 4 For R1 = 0.317 50 317.0 1 ln 10100 1 99.0317.0 1 ln 2 1 4 =′⇒      × =      ⋅′ − L L µm For R2 = 0.311 43.50 311.0 1 ln 10100 1 99.0311.0 1 ln 2 1 24 2 =′⇒      × =      ⋅′ − L L µm. 7. From Eq. 23a
  80. 80. 78 For R = 0.317 2 4 cmA1000 99.0317.0 1 ln 101002 1 10010 − − =            ××× +×=thJ and so mA5105101001000 44 =××××= −− thI For R = 0.311 β∝Γ⇒ β2 = 0.1×0.998/0.764 = 0.13 2 4 cmA766 99.0311.0 1 ln 101002 1 10066.7 − − =            ××× +×=thJ and so mA83.310510100766 44 =××××= −− thI . 8. From the equation, we have for m = 0: 0442 =± oBB LnLn λλλ m (25a) which can be solved as         +±− ±= 2 16164 22 o B LnLnLn λ λ (25b) There are several variations of ± in this solutions. Take the solution which is the only practical one, i.e., λB ≈ λo, gives λB = 1.3296 or 1.3304 µm. m196.0 4.32 33.1 µ= × ≅d . 9. The threshold current in Fig. 26b is given by Ith = I0 exp (T/110). Therefore ( ) 1 C0091.0 110 11 − ==≡ oth th dT dI I ξ . If T0 = 50 o C, the temperature coefficient becomes
  81. 81. 79 ( ) 1 C02.0 50 1 − == o ξ . which is larger than that for T0 = 110 o C. Therefore the laser with T0 = 50 o C is worse for high-temperature operation. 10. (a) ∆I = q (µn + µp) ∆nEA ( ) ( )( )( ) 313 219 3 cm10 10126.01017003600106.1 1083.2 − −− − = ××+× × = + ∆ =−=∆∴ Aq I pairsholeelectronn pn Eµµ (b) sµτ 120 6.23 1083.2 3 = × = − (c) ( ) ( ) 9 4 3 13 105.2 102.1 10 exp10exp ×=      × −=−∆=∆ − − τtntn cm-3 . 11. From Eq. 33 A55.21055.2 1010 50001063000 3 10 85.0 6 4 106 µ=×=       × ⋅×⋅ ⋅      × ⋅= − − −− q qI p and from Eq. 35 Gain 9 1010 50001063000 4 10 = × ⋅×⋅ == − − L n Eτµ . 12. From Eq. 36       ⋅=      ⋅         =        ⋅        = − q h R q h P I h P q I opt poptp νν ν η 1 The wavelength λ of light is related to its frequency ν by ν = c/λ, where c is the velocity of light in vacuum. Therefore νh /q = hc / λq and h = 6.625×10-34 J-s, c = 3.0×1010 cm/s, q = 1.6×10–19 coul, 1 eV = 1.6×10–19 J.
  82. 82. 80 Therefore, h v/q = 1.24 / λ(µm) Thus, η= (R×1.24) / λand R = (ηλ) / 1.24. 13. The electric field in the p-layer is given by ( ) s m xqN x ε 1 1 −= EE bx ≤≤0 , where Em is the maximum field. In the p layer, the field is essentially a constant given by ( ) Wxb bqN x s m ≤<−= 1 ε EE 2 The electric field required to maintain velocity saturation of holes is ~ 105 V/cm. Therefore 51 10≥− s m bqN ε E or 1 11515 1066.41010 N bqN s m − ×+=+≥ ε E . From the plot of the critical field versus doping, the corresponding Em are obtained : N1 = 7×1015 cm-3 Em = 4.2×105 V/cm The biasing voltage is given by ( ) ( ) ( ) V.138 10910 2 103102.3 2 45 45 2 2 = ×+ ××× =+ − = − − W b V m B E E E The transit time is ( ) ps.90109 10 109 11 7 4 =×= × = − ≈ − − s bW t ν
  83. 83. 81 14. (a) For a photodiode, only a narrow wavelength range centered at the optical signal wavelength is important; whereas for a solar cell, high spectral response over a broad solar wavelength range are required. (b) Photodiode are small to minimize junction capacitance, while solar cells are large-area device. (c) An important figure of merit for photodiodes is the quantum efficiency (number of electron-hole pairs generated by incident photon), whereas the main concern for solar cells is the power conversion efficiency (power delivered to the load per incident solar energy). 15. kTE p p Dn n A VCs g e D N D N NAqNI −         += ττ 11 )a( ( )( )( ) ( ) ( ) ( )1 A1028.2 1067.51043.2 105 5.2 105 1 10 3.9 107.1 1 1066.21086.2106.12 12 2.431420 12.1 719516 191919 −−= −= ×= ××= ×        ×× + × ××××= − −− − −− − kTqV sL L kTqV s kTeV eIII IeII e e V 0 0.1 0.2 0.3 0.4 0.5 0.6 0.65 0.7 (V) )1( −kT qV s eI 0 1.1×10- 7 5×10-6 2.5×10- 4 1.1×10- 2 0.55 26.8 179 1520 (mA) IL 95 95 95 95 95 94.5 68.2 -84 -1425 (mA) (b) V68.0 1028.2 1095 ln0259.0ln 12 3 =      × × =      = − − s L OC I I q kT V (c) ( ) VIeVIP L kTqV s −−= 1
  84. 84. 82 ( ) ( ) L kTqV s L kTqV s kTqV s IVeI Ie kT qV IeI dV dP −+≈ −+−== 1 10 ( ) V.V VI I e m s LkTqV 640 1 = + =∴ ( )[ ] mW52 0259.07.241ln0259.068.01095 1ln 3 = −+−×=       −      +−== − q kT kT qV q kT VIVIP n OCLmmm 0 50 100 0 0.2 0.4 0.6 0.8 V (V) I(mA) 16. From Eq. 38 and 39 ( ) L kT/qV s IeII −−= 1 and       ≅      += s L s L oc I I q kT I I q kT V ln1ln A1049323 1002585060 −−− ×=⋅== .eeII ..kTqV Ls 02585.010 10493.23 V eI ⋅×−= − and P = I V
  85. 85. 83 V I P 0 3.00 0.00 0.1 3.00 0.30 0.2 3.00 0.60 0.3 3.00 0.90 0.4 3.00 1.20 0.5 2.94 1.47 0.51 2.91 1.48 0.52 2.86 1.49 0.53 2.80 1.48 0.54 2.71 1.46 0.55 2.57 1.41 0.6 0.00 0.00 0.00 0.50 1.00 1.50 2.00 0 0.5 1 V (volts)P(watts) ∴Maximum power output = 1.49 W Fill Factor .83.0 36.0 49.1 = × ==≡ ocL m ocL mm VI P VI VI FF 17. From Fig. 40 The output powers for Rs = 0 and Rs = 5 Ω can be obtained from the area P1 (Rs = 0) = 95 mA×0.375V = 35.6 mW, P2 (Rs = 5 Ω) = 50 mA×0.18V = 9.0 mW ∴For Rs = 0 P1/ P1 = 100% For Rs = 5 Ω P2/ P1 = 9/35.6 = 25.3%. 18. The efficiencies are 14.2% (1 sun), 16.2% (10-sun), 17.8% (100-sun), and 18.5% (1000-sun). Solar cells needed under 1-sun condition
  86. 86. 84 ( ) ( ) ( ) ( ) sun.-1000forcells1300 1%2.14 10%5.18 sun-100forcells125 1%2.14 10%8.17 sun-10forcells4.11 1%2.14 10%2.16 -1-1 ionconcentrationconcentrat = × × = = × × = = × × = × × = sunPsun P in in η η
  87. 87. 85 CHAPTER 10 1. C0 = 1017 cm-3 k0(As in Si) = 0.3 CS= k0C0(1 - M/M0)k0-1 = 0.3×1017 (1- x)-0.7 = 3×1016 /(1 - l/50)0.7 x 0 0.2 0.4 0.6 0.8 0.9 l (cm) 0 10 20 30 40 45 CS (cm-3) 3×1016 3.5×1016 4.28×1016 5.68×1016 1.07×1017 1.5×1017 0 2 4 6 8 10 12 14 16 0 10 20 30 40 50 l (cm) ND(10 16 cm -3 ) 2. (a) The radius of a silicon atom can be expressed as Å175.143.5 8 3 so 8 3 =×= = r ar (b) The numbers of Si atom in its diamond structure are 8. So the density of silicon atoms is 322 33 atoms/cm100.5 )Å43.5( 88 ×=== a n (c) The density of Si is
  88. 88. 86 3 23 2223 cm/g 1002.6 10509.28 /1 1002.6/ × ×× = × = n M ρ = 2.33 g / cm3 . 3. k0 = 0.8 for boron in silicon M / M0 = 0.5 The density of Si is 2.33 g / cm3 . The acceptor concentration for ρ = 0.01 Ω–cm is 9×1018 cm-3 . The doping concentration CS is given by 1 0 00 0 )1( − −= k s M M CkC Therefore 318 2.0 18 1 0 0 0 cm108.9 )5.01(8.0 109 )1( 0 − − − ×= − × = − = k s M M k C C The amount of boron required for a 10 kg charge is 2218 102.4108.9 338.2 000,10 ×=×× boron atoms So that borong75.0 atoms/mole1002.6 atoms102.4 g/mole8.10 23 22 = × × × . 4. (a) The molecular weight of boron is 10.81. The boron concentration can be given as 318 2 233 atoms/cm1078.9 1.014.30.10 1002.6g81.10/g1041.5 fersilicon waofvolume atomsboronofnumber ×= ×× ××× = = − bn (b) The average occupied volume of everyone boron atoms in the wafer is
  89. 89. 87 3 18 cm 1078.9 11 × == bn V We assume the volume is a sphere, so the radius of the sphere ( r ) is the average distance between two boron atoms. Then cm109.2 4 3 7− ×== π V r . 5. The cross-sectional area of the seed is 2 2 cm24.0 2 55.0 =      π The maximum weight that can be supported by the seed equals the product of the critical yield strength and the seed’ s cross-sectional area: kg480g108.424.0)102( 56 =×=×× The corresponding weight of a 200-mm-diameter ingot with length l is m.56.6cm656 g480000 2 0.20 )g/cm33.2( 2 3 ==∴ =      l lπ 6. We have 1 0 00 0 1/ −       −= k s M M kCC Fractional 0 0.2 0.4 0.6 0.8 1.0 solidified 0/CsC 0.05 0.06 0.08 0.12 0.23 ∞
  90. 90. 88 7. The segregation coefficient of boron in silicon is 0.72. It is smaller than unity, so the solubility of B in Si under solid phase is smaller than that of the melt. Therefore, the excess B atoms will be thrown-off into the melt, then the concentration of B in the melt will be increased. The tail- end of the crystal is the last to solidify. Therefore, the concentration of B in the tail-end of grown crystal will be higher than that of seed-end. 8. The reason is that the solubility in the melt is proportional to the temperature, and the temperature is higher in the center part than at the perimeter. Therefore, the solubility is higher in the center part, causing a higher impurity concentration there. 9. The segregation coefficient of Ga in Si is 8 ×10-3 From Eq. 18 Lkx s ekCC / 0 )1(1/ − −−= We have cm.24 ln(1.102)250 105/1051 1081 ln 108 2 /1 1 ln 1615 3 3- 0 = =       ××− ×− × =       − − = − CC k k L x s 10. We have from Eq.18 ])/exp()1(1[0 LxekekCsC −−−= So the ratio )]/exp()1(1[0/ LxekekCsC −−−= = 1/at52.0)13.0exp()3.01(1 ==×−•−− Lx = 38.0 at x/L = 2. 0.01 0.11 0.21 0 0.2 0.4 0.6 0.8 1 Fraction Solidified Cs/Co
  91. 91. 89 11. For the conventionally-doped silicon, the resistivity varies from 120 Ω-cm to 155 Ω-cm. The corresponding doping concentration varies from 2.5×1013 to 4×1013 cm-3 . Therefore the range of breakdown voltages of p+ - n junctions is given by V11600to7250/109.2)( 106.12 )103(1005.1 )( 2 171 19 2512 1 2 =×= ×× ××× = ≅ − − − − BB B cs B NN N q V Eε V4350725011600 =−=∆ BV %307250/ 2 ±=      ∆ ∴ BV For the neutron irradiated silicon, ρ= 148 ± 1.5 Ω-cm. The doping concentration is 3×1013 (±1%). The range of breakdown voltage is .V9762to9570 %)1(103/109.2/103.1 131717 = ±××=×= BB NV V19295709762 =−=∆ BV %19570/ 2 ±=      ∆ ∴ BV . 12. We have l s CC CC M M ms lm l s = − − == b b Tatliquidofweight TatGaAsofweight Therefore, the fraction of liquid remained f can be obtained as following 65.0 3016 30 = + ≈ + = + = ls l MM M f ls l . 13. From the Fig.11, we find the vapor pressure of As is much higher than that of the Ga. Therefore, the As content will be lost when the temperature is increased. Thus the composition of liquid GaAs always becomes gallium rich.
  92. 92. 90 14.       − ×=×=−= )300/( 8.88 exp105)/eV3.2exp(105)/exp( 2222 T kTkTENn ss = K300C27at0cm1023.1 0316 =≈× −− = K1173C900atcm107.6 0312 =× − = K1473C1200atcm107.6 0314 =× − . 15. )2/exp( ` kTENNn ff −= = )300//(7.94242/1.1/8.32722 1007.7101105 TkTeVkTeV eee −−− ××=×××× = 1027.5 17− × at 27o C = 300 K =2.14×1014 at 900o C = 1173 K. 16. 37 × 4 = 148 chips In terms of litho-stepper considerations, there are 500 µm space tolerance between the mask boundary of two dice. We divide the wafer into four symmetrical parts for convenient dicing, and discard the perimeter parts of the wafer. Usually the quality of the perimeter parts is the worst due to the edge effects.
  93. 93. 91 17. M kT dvf dvvf v v av π ν 8 0 0 == ∫ ∫ ∞ ∞ Where       −      = kT M kT M f 2 exp 2 4 2 2 2/3 ν ν π ν M: Molecular mass k: Boltzmann constant = 1.38×10-23 J/k T: The absolute temperature ν: Speed of molecular So that cm/sec1068.4m/sec468 1067.129 3001038.122 4 27 23 ×== ×× ××× = − − π νav . 18. cm )Pain( 66.0 P =λ
  94. 94. 92 Pa104.4 150 66.066.0 3− ×===∴ λ P . 19. For close-packing arrange, there are 3 pie shaped sections in the equilateral triangle. Each section corresponds to 1/6 of an atom. Therefore dd Ns 2 3 2 1 6 1 3 triangletheofarea trianglein thecontainedatomsofnumber × × == = 282 )1068.4(3 2 3 2 − × = d = 214 atoms/cm1027.5 × . 20. (a) The pressure at 970°C (=1243K) is 2.9×10-1 Pa for Ga and 13 Pa for As2. The arrival rate is given by the product of the impringement rate and A/πL2 : Arrival rate = 2.64×1020             2 L A MT P π = 2.64×1020       ×      × × − 2 1 12 5 124372.69 109.2 π = 2.9×1015 Ga molecules/cm2 –s The growth rate is determined by the Ga arrival rate and is given by (2.9×1015 )×2.8/(6×1014 ) = 13.5 Å/s = 810 Å/min . (b) The pressure at 700ºC for tin is 2.66×10-6 Pa. The molecular weight is 118.69. Therefore the arrival rate is s⋅×=      ×      × × × − 210 2 6 20 cmmolecular/1028.2 12 5 97369.118 1066.2 1064.2 π If Sn atoms are fully incorporated and active in the Ga sublattice of GaAs, we have an dd
  95. 95. 93 electron concentration of .cm1074.1 2 1042.4 109.2 1028.2 3-17 22 15 10 ×=      ×       × × 21. The x value is about 0.25, which is obtained from Fig. 26. 22. The lattice constants for InAs, GaAs, Si and Ge are 6.05, 5.65,5.43, and 5.65 Å, respectively (Appendix F). Therefore, the f value for InAs-GaAs system is 066.005.6)05.665.5( −=−=f And for Ge-Si system is .39.065.5)65.543.5( −=−=f
  96. 96. 94 CHAPTER 11 1. From Eq. 11 (with ô=0) x2 +Ax = Bt From Figs. 6 and 7, we obtain B/A =1.5 µm /hr, B=0.47 µm 2 /hr, therefore A= 0.31 µm. The time required to grow 0.45µm oxide is min44hr0.720.45)0.31(0.45 0.47 1 2 ==×+=+= Ax)(x B 1 t 2 . 2. After a window is opened in the oxide for a second oxidation, the rate constants are B = 0.01 µm2 /hr, A= 0.116 µm (B/A = 6 ×10-2 µm /hr). If the initial oxide thickness is 20 nm = 0.02 µm for dry oxidation, the value ofôcan be obtained as followed: (0.02)2 + 0.166(0.02) = 0.01 (0 +ô) or ô= 0.372 hr. For an oxidation time of 20 min (=1/3 hr), the oxide thickness in the window area is x2 + 0.166x = 0.01(0.333+0.372) = 0.007 or x = 0.0350 µm = 35 nm (gate oxide). For the field oxide with an original thickness 0.45 µm, the effectiveôis given by ô= .hr72.27)45.0166.045.0( 01.0 1 )( 1 22 =×+=+ Axx B x2 + 0.166x = 0.01(0.333+27.72) = 0.28053 or x = 0.4530 µm (an increase of 0.003µm only for the field oxide). 3. x2 + Ax = B )( τ+t )( 4 ) 2 ( 2 2 τ+=−+ tB AA x
  97. 97. 95       ++=+ )( 4 ) 2 ( 2 2 τt B A B A x when t >> τ , t >> B A 4 2 , then, x2 = Bt similarly, when t >> τ , t >> B A 4 2 , then, x = )( τ+t A B 4. At 980 (=1253K) and 1 atm, B = 8.5×10-3 µm2 /hr, B/A = 4×10-2 µm /hr (from Figs. 6 and 7). Since A a2D/k , B/A = kC0/C1, C0 = 5.2×1016 molecules/cm3 and C1 = 2.2×1022 cm-3 , the diffusion coefficient is given by .s/cm104.79 hr/m101.79 hr/m 102.5 102.2 2 105.8 222 29- 23 2 16 223 0 1 0 1 ×= ×= × ×× =         =        ⋅== − µ µ C CB C C A BAAk D 5. (a) For SiNxHy 2.1 1 N Si == x 4 x = 0.83 atomic % 20 83.01 100 H = ++ = y y 4 y = 0.46 The empirical formula is SiN0.83H0.46. (b) ñ= 5× 1028 e-33.3×1.2 = 2× 1011 Ù-cm As the Si/N ratio increases, the resistivity decreases exponentially.
  98. 98. 96 t A 3 C 01 OTa 52 εε = AAA 020302ONO ttt C 1 εεεεεε ++= ( )t A 32 032 ONO 2 C εε εεε + = ( )t A t A 32 2032101 23 εε εεεεε + = 6. Set Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6, area = A then ( ) ( ) 37.5 6.79.33 6.729.325 3 2 C C 32 321 ONO OTa 52 = ×× ×+ = + = εε εεε . 7. Set BST thickness = 3t, ε1 = 500, area = A1 SiO2 thickness = t, ε2 = 3.9, area = A2 Si3N4 thickness = t, ε3 = 7.6, area = A2 then .0093.0 A A 2 1 = 8. Let Ta2O5 thickness = 3t, ε1 = 25 SiO2 thickness = t, ε2 = 3.9 Si3N4 thickness = t, ε3 = 7.6 area = A then d A t A 0201 3 εεεε = .468.0 3 1 2 t t d == ε ε
  99. 99. 97 9. The deposition rate can be expressed as r = r0 exp (-Ea/kT) where Ea = 0.6 eV for silane-oxygen reaction. Therefore for T1 = 698 K             −== 11 6.0exp2 )( )( 211 2 kTkTTr Tr ln 2 =             − 300 698 300 0259.0 6.0 2T 4 T2 =1030 K= 757 . 10. We can use energy-enhanced CVD methods such as using a focused energy source or UV lamp. Another method is to use boron doped P-glass which will reflow at temperatures less than 900 . 11. Moderately low temperatures are usually used for polysilicon deposition, and silane decomposition occurs at lower temperatures than that for chloride reactions. In addition, silane is used for better coverage over amorphous materials such SiO2. 12. There are two reasons. One is to minimize the thermal budget of the wafer, reducing dopant diffusion and material degradation. In addition, fewer gas phase reactions occur at lower temperatures, resulting in smoother and better adhering films. Another reason is that the polysilicon will have small grains. The finer grains are easier to mask and etch to give smooth and uniform edges. However, for temperatures less than 575 ºC the deposition rate is too low. 13. The flat-band voltage shift is FBV∆ = 0.5 V ~ 0 C Qot 28 8 14 0 F/cm109.6 10500 1085.89.3 −− − − ×= × ×× == d C oxε . 4 Number of fixed oxide charge is
  100. 100. 98 2-11 19 8 0 cm101.2 106.1 109.65.05.0 ×= × ×× = − − q C To remove these charges, a 450 heat treatment in hydrogen for about 30 minutes is required. 14. 20/0.25 = 80 sqs. Therefore, the resistance of the metal line is 5×50 = 400 Ω . 15. For TiSi2 30 × 2.37 = 71.1nm For CoSi2 30 × 3.56 = 106.8nm. 16. For TiSi2: Advantage: low resistivity It can reduce native-oxide layers TiSi2 on the gate electrode is more resistant to high-field- induced hot-electron degradation. Disadvantage: bridging effect occurs. Larger Si consumption during formation of TiSi2 Less thermal stability For CoSi2: Advantage: low resistivity High temperature stability No bridging effect A selective chemical etch exits Low shear forces Disadvantage: not a good candidate for polycides
  101. 101. 99 Ω×= ××× ××== −− − 3 44 6 102.3 103.01028.0 1 1067.2 A L R ρ F. . ... S TL d A C 13 4 64414 1092 10360 1010110301085893 − − −−− ×= × ××××××× === εε ns42.0101.2102 133 =×××= − RC Ω×= ××× ××== − 3 44 6 102.3 103.01028.0 1 1067.2 A L R ρ F107.8 1036.0 31103.01085.89.3 13 4 414 − −− ×= × ×××××× === S TL d A C εε . 17. (a) ns93.0109.2103.2 155 =×××= − RC (b) Ω×= ××× ××== −− − 3 44 6 102 103.01028.0 1 107.1 L A R ρ F101.2 1036.0 1103.01085.88.2 d A 13 4 414 − − −− ×= × ××××× === S TL C εε (c) We can decrease the RC delay by 55%. Ratio = 093 42.0 = 0.45. 18. (a) RC = 3.2 ×103 ×8.7 × 10-13 = 2.8 ns. (b) Ω×= ××× ××== − 3 44 6 102 103.01028.0 1 107.1 A L R ρ F103.6 1036.0 31103.01085.88.2 d 13 4 414 − −− ×= × ×××××× === S TLA C εε ns.5.2107.8102.3 ns5.2107.8102 133 133 =×××= =×××= − − RC RC 19. (a) The aluminum runner can be considered as two segments connected in series: 20% (or 0.4 mm) of the length is half thickness (0.5 µm) and the remaining 1.6 mm is full thickness (1µm). The total resistance is       ×× + × ×=      += −−−− − )105.0(10 04.0 1010 16.0 103 4444 6 2 2 1 1 AA R ll ρ = 72 Ù.
  102. 102. 100 The limiting current I is given by the maximum allowed current density times cross-sectional area of the thinner conductor sections: I = 5×105 A/cm2 × (10-4 ×0.5×10-4 ) = 2.5×10-3 A = 2.5 mA. The voltage drop across the whole conductor is then A105.272 3− ××Ω== RIV = 0.18V. 20. = h: height , W : width , t : thickness, assume that the resistivities of the cladding layer and TiN are much larger than CuA and ρρ l 5.0)1.05.0( 7.2 ×− = × ×= ll Wh R AlAl ρ )25.0()25.0( 7.1 ttWh R CuCu −×− = × ×= ll ρ When CuAl RR = Then 2 )25.0( 7.1 5.04.0 7.2 t− = × ⇒ t = 0.073 µm = 73 nm . 40 nm 60 nm AlCu 0.5 µm 0.5µm
  103. 103. 101 CHAPTER 12 1. With reference to Fig. 2 for class 100 clean room we have a total of 3500 particles/m3 with particle sizes ≥0.5 µm 3500 100 21 × = 735 particles/m2 with particle sizes ≥ 1.0 µm 3500 100 5.4 × = 157 particles/m2 with particle sizes ≥ 2.0 µm Therefore, (a) 3500-735 = 2765 particles/m3 between 0.5 and 1 µm (b) 735-157 = 578 particles/m3 between 1 and 2 µm (c) 157 particles/m3 above 2 µm. 2. AD n eY 1 9 1 − = Π= A = 50 mm2 = 0.5 cm2 %1.302.1)5.01(1)5.025.0(4)5.01.0(4 ==××= −×−×−×− eeeeY . 3. The available exposure energy in an hour is 0.3 mW2 /cm2 × 3600 s =1080 mJ/cm2 For positive resist, the throughput is wafers/hr7 140 1080 = For negative resist, the throughput is rwafers/h012 9 1080 = . 4. (a) The resolution of a projection system is given by 178.0 65.0 m£g193.0 6.01 =×== NA klm λ µm       == 222 )65.0( m193.0 5.0 )( µλ NA kDOF = 0.228 µm (b) We can increase NA to improve the resolution. We can adopt resolution enhancement techniques (RET) such as optical proximity correction (OPC) and phase-shifting Masks (PSM). We can also develop new resists that provide lower k1 and higher k2 for better resolution and depth of focus. (c) PSM technique changes k1 to improve resolution. 5. (a) Using resists with high γ value can result in a more vertical profile but throughput decreases. (b) Conventional resists can not be used in deep UV lithography process because these resists have high absorption and require high dose to be exposed in deep UV. This raises the
  104. 104. 102 concern of damage to stepper lens, lower exposure speed and reduced throughput. 6. (a) A shaped beam system enables the size and shape of the beam to be varied, thereby minimizing the number of flashes required for exposing a given area to be patterned. Therefore, a shaped beam can save time and increase throughput compared to a Gaussian beam. (b) We can make alignment marks on wafers using e-beam and etch the exposed marks. We can then use them to do alignment with e-beam radiation and obtain the signal from these marks for wafer alignment. X-ray lithography is a proximity printing lithography. Its accuracy requirement is very high, therefore alignment is difficult. (c) X-ray lithography using synchrotron radiation has a high exposure flux so X-ray has better throughput than e-beam. 7. (a) To avoid the mask damage problem associated with shadow printing, projection printing exposure tools have been developed to project an image from the mask. With a 1:1 projection printing system is much more difficult to produce defect-free masks than it is with a 5:1 reduction step-and-repeat system. (b) It is not possible. The main reason is that X-rays cannot be focused by an optical lens. When it is through the reticle. So we can not build a step-and-scan X-ray lithography system. 8. As shown in the figure, the profile for each case is a segment of a circle with origin at the initial mask-film edge. As overetching proceeds the radius of curvature increases so that the profile tends to a vertical line. 9. (a) 20 sec 0.6 × 20/60 = 0.2 µm…..(100) plane 0.6/16 × 20/60 = 0.0125 µm……..(110) plane 0.6/100 × 20/60 = 0.002 µm…….(111) plane 22.12.025.120 =×−=−= lWWb µm (b) 40 sec 0.6 × 40/60 = 0.4 µm….(100)plane
  105. 105. 103 0.6/16 × 40/60 = 0.025 µm…. (110) plane 0.6/100 × 40/60 = 0.004 µm…..(111) plane 4.025.120 ×−=−= lWWb = 0.93 µm (c) 60 sec 0.6 ×1 = 0.6 µm….(100)plane 0.6/16 ×1 = 0.0375 µm…. (110) plane 0.6/100 ×1= 0.006 µm…..(111) plane =×−=−= 6.025.120 lWWb 0.65 µm. 10. Using the data in Prob. 9, the etched pattern profiles on <100>-Si are shown in below. (a) 20 sec l = 0.012 µm, 5.10 == bWW µm (b) 40 sec l = 0.025 µm, 5.10 == bWW µm (c) 60 sec l = 0.0375 µm 5.10 == bWW µm. 11. If we protect the IC chip areas (e.g. with Si3N4 layer) and etch the wafer from the top, the width of the bottom surface is 18846252100021 =×+=+= lWW µm The fraction of surface area that is lost is 22 1 2 /)( WWW − × 100%=(18842 -10002 ) /18842 × 100% = 71.8 % In terms of the wafer area, we have lost 71.8 % × 2 )2/15(π =127 cm2 Another method is to define masking areas on the backside and etch from the back. The width of each square mask centered with respect of IC chip is given by 6252100021 ×−=−= lWW = 116 µm Using this method, the fraction of the top surface area that is lost can be negligibly small. 12. 1 Pa = 7.52 m Torr
  106. 106. 104 PV = nRT 7.52 /760 × 10-3 = n/V ×0.082 × 273 n/V = 4.42 × 10-7 mole/liter = 4.42 × 10-7 × 6.02 × 1023 /1000 =2.7 ×1014 cm-3 mean–free–path P/105 3− ×=λ cm = 5× 10-3 ×1000/ 7.52 = 0.6649 cm = 6649 µm 150Pa = 1128 m Torr PV = nRT
  107. 107. 105 1128/ 760 × 10-3 = n/V × 0.082 × 273 n/V = 6.63 × 10-5 mole/liter = 6.63 ×10-5 ×6.02×1023 /1000 = 4 × 1016 cm-3 mean-free-path P/105 3− ×=λ cm = 5× 10-3 ×1000/1128 = 0.0044 cm = 44 µm. 13. Si Etch Rate (nm/min) = 2.86 × 10-13 × RT E F a eTn − ×× 2 1 = 2.86 × 10-13 ×3×1015 × 298987.1 1048.2 2 1 3 )298( × ×− ×e = 224.7 nm/min. 14. SiO2 Etch Rate (nm/min) = 0.614× 10-13 ×3×1015 × 298987.1 1076.3 2 1 3 )298( × ×− ×e = 5.6 nm/min Etch selectivity of SiO2 over Si = 025.0 7.224 6.5 = Or etch rate (SiO2)/etch rate (Si) = 025.0 86.2 614.0 298987.1 )48.276.3( =× × +− e . 15. A three–step process is required for polysilicon gate etching. Step 1 is a nonselective etch process that is used to remove any native oxide on the polysilicon surface. Step 2 is a high polysilicon etch rate process which etches polysilicon with an anisotropic etch profile. Step 3 is a highly selective polysilicon to oxide process which usually has a low polysilicon etch rate. 16. If the etch rate can be controlled to within 10 %, the polysilicon may be etched 10 % longer or for an equivalent thickness of 40 nm. The selectivity is therefore 40 nm/1 nm = 40. 17. Assuming a 30% overetching, and that the selectivity of Al over the photoresist maintains 3. The minimum photoresist thickness required is
  108. 108. 106 (1+ 30%) × 1 µm/3 = 0.433 µm = 433.3 nm. 18. e e m qB =ω 31 19 9 101.9 106.1 1045.22 − − × ×× =×× B π B = 8.75 × 10-2 (tesla) = 875 (gauss). 19. Traditional RIE generates low-density plasma (109 cm-3 ) with high ion energy. ECR and ICP generate high-density plasma (1011 to 1012 cm-3 ) with low ion energy. Advantages of ECR and ICP are low etch damage, low microloading, low aspect-ratio dependent etching effect, and simple chemistry. However, ECR and ICP systems are more complicated than traditional RIE systems. 20. The corrosion reaction requires the presence of moisture to proceed. Therefore, the first line of defense in controlling corrosion is controlling humidity. Low humidity is essential,. especially if copper containing alloys are being etched. Second is to remove as much chlorine as possible from the wafers before the wafers are exposed to air. Finally, gases such as CF4 and SF6 can be used for fluorine/chlorine exchange reactions and polymeric encapsulation. Thus, Al-Cl bonds are replaced by Al-F bonds. Whereas Al-Cl bonds will react with ambient moisture and start the corrosion process , Al-F bonds are very stable and do not react. Furthermore, fluorine will not catalyze any corrosion reactions.

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