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CHAPTER 2
2.1 Based upon Table 2.1, a resistivity of 2.6 μΩ-cm < 1 mΩ-cm, and aluminum is a conductor.
2.2
Based upon Table 2.1, a resistivity of 1015
Ω-cm > 105 Ω-cm, and silicon dioxide is an insulator.
2.3
I max = 107 A
cm2
⎛
⎝ ⎜
⎠⎟ 5μ m( )1μ m( ) 10−8cm2
μ m2
⎝ ⎜
⎠⎟ = 500 mA
2.4
⎟ ⎠
⎞⎜⎝
⎛ −= − T x
E BT n G
i 5
3
1062.8exp
For silicon, B = 1.08 x 1031
and EG = 1.12 eV:
ni = 2.01 x10-10
/cm3
6.73 x109 /cm
38.36 x 10
13 /cm
3.
For germanium, B = 2.31 x 1030
and EG = 0.66 eV:
ni = 35.9/cm3
2.27 x1013 /cm
38.04 x 10
15 /cm
3.
2.5 Define an M-File:
function f=temp(T)ni=1E14;f=ni^2-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
ni = 1014 /cm
3for T = 506 K ni = 10
16 /cm
3for T = 739 K
2.6
ni = BT 3 exp − E G8.62 x10−5
T
⎛
⎝ ⎜ ⎞
⎠⎟with B = 1.27x1029 K −3cm−6
T = 300 K and EG
= 1.42 eV: ni= 2.21 x10
6 /cm
3
T = 100 K: ni = 6.03 x 10-19
/cm3
T = 500 K: ni = 2.79 x1011 /cm
3
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2.7
vn = −μ n E = −700cm2
V − s
⎝ ⎜
⎠⎟ 2500
V
cm
⎛
⎝ ⎜
⎠⎟ = −1.75 x106 cm
s
v p = +μ p E = +250
cm2
V − s
⎛
⎝ ⎜
⎞
⎠⎟ 2500
V
cm
⎛
⎝ ⎜
⎞
⎠⎟ = +6.25 x105 cm
s
jn = −qnvn = −1.60 x10−19C ( )1017 1
cm3
⎛
⎝ ⎜
⎞
⎠⎟ −1.75 x106 cm
s
⎛
⎝ ⎜
⎞
⎠⎟ = 2.80 x104 A
cm2
j p = qnv
p = 1.60 x10−19C ( )103 1
cm3
⎛
⎝ ⎜
⎞
⎠⎟ 6.25 x105 cm
s
⎛
⎝ ⎜
⎞
⎠⎟ =1.00 x10−10 A
cm2
2.8
ni
2 = BT 3 exp −
E G
kT
⎝ ⎜
⎠⎟ B = 1.08 x1031
1010( )2
=1.08 x1031T 3 exp −1.12
8.62 x10−5T
⎛
⎝ ⎜
⎞
⎠⎟
Using a spreadsheet, solver, or MATLAB yields T = 305.22K
Define an M-File:function f=temp(T)f=1e20-1.08e31*T^3*exp(-1.12/(8.62e-5*T));
Then: fzero('temp',300) | ans = 305.226 K
2.9
scm
cmC cm A
Q jv 5
2
2
10 / 01.0 / 1000 −=−==
2.10
22
67
34104
sec104.0
cm
MA
cm
A x
cm
cm
C Qv j ==⎟
⎠
⎞⎜⎝
⎛ ⎟ ⎠
⎞⎜⎝
⎛ ==
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2.11
vn = −μ n E = −1000cm
2
V − s
⎝ ⎜
⎠⎟ −2000
V
cm
⎛
⎝ ⎜
⎞
⎠⎟ = +2.00 x106 cm
s
v p = +μ p E = +400cm
2
V − s
⎛
⎝ ⎜
⎞
⎠⎟ −2000
V
cm
⎛
⎝ ⎜
⎞
⎠⎟ = −8.00 x105 cm
s
jn = −qnvn = −1.60 x10−19C ( )103 1
cm3
⎛
⎝ ⎜
⎞
⎠⎟ +2.00 x106 cm
s
⎛
⎝ ⎜
⎞
⎠⎟ = −3.20 x10−10 A
cm2
j p = qnv p = 1.60 x10−19C ( )1017 1
cm3
⎛
⎝ ⎜
⎞
⎠⎟ −8.00 x105 cm
s
⎛
⎝ ⎜
⎞
⎠⎟ = −1.28 x104 A
cm2
2.12
( ) ( ) ( ) V100101010 b 50001010
5 45
4=⎟
⎠ ⎞
⎜⎝ ⎛ === −
− cm xcm
V V
cm
V
cm x
V E a
2.13
j p = qpv p = 1.60 x10−19C ( )1019
cm3
⎛
⎝ ⎜
⎞
⎠⎟ 107 cm
s
⎛
⎝ ⎜
⎞
⎠⎟ =1.60 x107 A
cm2
i p = j p A = 1.60 x107 A
cm2
⎛
⎝ ⎜
⎞
⎠⎟ 1 x10−4
cm( )25 x10−4cm( )= 4.00 A
2.14
For intrinsic silicon, σ = q μ nni + μ pni( )= qni μ n + μ p( )
σ ≥1000 Ω − cm( )−1
for a conductor
ni ≥σ
q μ n + μ p( )=
1000 Ω − cm( )−1
1.602 x10−19C 100 + 50( ) cm2
v − sec
=4.16 x1019
cm3
n i
2 =1.73 x10
39
cm6= BT 3 exp −
E G
kT
⎛
⎝ ⎜
⎞
⎠⎟ with
B = 1.08 x1031 K −3cm−6 , k = 8.62x10-5eV/K and EG =1.12eV
This is a transcendental equation and must be solved numerically by iteration. Using the HP
solver routine or a spread sheet yields T = 2701 K. Note that this temperature is far above themelting temperature of silicon.
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2.15
For intrinsic silicon, σ = q μ nni + μ pni( )= qni μ n + μ p( )
σ ≤10−5 Ω − cm( )−1
for an insulator
ni ≥ σ q μ n + μ p( )=
10−5 Ω − cm
( )
−1
1.602 x10−19C ( )2000 + 750( ) cm2
v − sec
⎛
⎝ ⎜
⎞
⎠⎟
= 2.270 x10
10
cm3
n i
2 =5.152 x1020
cm6= BT 3 exp −
E G
kT
⎛
⎝ ⎜
⎞
⎠⎟ with
B =1.08 x1031 K −3cm−6, k = 8.62x10-5eV/K and EG =1.12eV
Using MATLAB as in Problem 2.5 yields T = 316.6 K.
2.16
Si Si Si
SiB
Si Si Si
P
Donor electron
fills acceptorvacancy
No free electrons or holes (except those corresponding to ni).
2.17
(a) Gallium is from column 3 and silicon is from column 4. Thus silicon has an extra electron
and will act as a donor impurity.(b) Arsenic is from column 5 and silicon is from column 4. Thus silicon is deficient in oneelectron and will act as an acceptor impurity.
2.18 Since Ge is from column IV, acceptors come from column III and donors come from columnV. (a) Acceptors: B, Al, Ga, In, Tl (b) Donors: N, P, As, Sb, Bi
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2.19 (a) Germanium is from column IV and indium is from column III. Thus germanium has oneextra electron and will act as a donor impurity.(b) Germanium is from column IV and phosphorus is from column V. Thus germanium hasone less electron and will act as an acceptor impurity.
2.20
( ) field.electricsmalla,20002.0100002 cm
V cm
cm
A j
j E =−Ω⎟
⎠
⎞⎜⎝
⎛ === ρ σ
2.21
jn
drift = qnμ n E = qnvn = 1.602 x10−19( )1016( ) C
cm3
⎝ ⎜
⎠⎟ 107 cm
s
⎝ ⎜
⎠⎟ =16000
A
cm2
2.22
N = 1015
atomscm3
⎛
⎝ ⎜
⎞
⎠⎟ 1μ m( )10μ m( )0.5μ m( ) 10
−4
cmμ m
⎛
⎝ ⎜
⎞
⎠⎟
3
= 5,000 atoms
2.23
N A > N D: N A − N D =1015 −1014 = 9 x1014 /cm3
If we assume N A − N D >> 2ni =1014 / cm3 :
p = N A − N D = 9 x1014 /cm3 | n =ni
2
p=
251026
9 x1014= 2.78 x1012 /cm3
If we use Eq. 2.12: p =
9 x1014 ± 9 x1014( )2
+ 4 5 x1013( )2
2 = 9.03 x1014
and n = 2.77 x1012 /cm
3. The answers are essentially the same.
2.24
35
16
222314
3113161616
10502104
10104
102210410105
/cm x. x p
n| n /cm x N N p
/cm xn /cm x x N : N N N
i D A
i D A D A
====−=
=>>=−=−>
2.25
N D > N A: N D − N A = 3 x1017
− 2 x1017
=1 x1017
/cm3
2ni = 2 x1017 /cm3; Need to use Eq. (2.11)
n =1017 ± 1017( )
2
+ 4 1017( )2
2=1.62 x1017
/cm3
p =ni
2
n=
1034
1.62 x1017= 6.18 x1016
/cm3
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2.26
N D − N A = −2.5 x1018 / cm3
U sing Eq. 2.11: n =−2.5 x1018 ± −2.5 x1018( )
2
+ 4 1010( )2
2
Evaluating this with a calculator yields n = 0, and n =ni
2
p= ∞.
No, the result is incorrect because of loss of significant digits
within the calculator. It does not have enough digits.
2.27
(a) Since boron is an acceptor, NA = 6 x 1018 /cm3. Assume ND = 0, since it is not specified.The material is p-type.
3
318
6202318
i
318310
7.16106
10 and 106So
2n>> / 106 and10re,temperaturoomAt
/cm /cm x
/cm
p
nn /cm x p
cm x N N /cmn
i
D Ai
=====−=
(b)
At 200K, ni
2 =1.08 x1031
200( )3
exp −1.12
8.62 x10−5 200( )⎛
⎝ ⎜⎜
⎞
⎠⎟⎟ = 5.28 x10
9 /cm6
ni = 7.27 x104 /cm3 N A − N D >> 2ni , so p = 6 x1018 /cm3 and n =5.28 x109
6 x1018= 8.80 x10−10 /cm3
2.28 (a) Since arsenic is a donor, ND = 3 x 1017 /cm3. Assume NA = 0, since it is not specified. Thematerial is n-type.
3
317
6202317
i
317310
i
333103
10 and 103So
2n>> / 103 and / 10nre,temperaturoomAt
/cm /cm x
/cm
n
n p /cm xn
cm x N N cm
i
A D
====
=−=
(b) At 250K, ni
2 =1.08 x1031 250( )3
exp −1.12
8.62 x10−5 250( )⎛
⎝ ⎜⎜
⎠⎟⎟ = 4.53 x1015 /cm6
ni = 6.73 x10
7
/cm
3
N D − N A >> 2ni , so n = 3 x10
17
/ cm
3
and n =
4.53x1015
3x1017 = 0.0151/ cm
3
2.29
(a) Arsenic is a donor, and boron is an acceptor. ND = 2 x 1018 /cm3, and NA = 8 x 1018 /cm3.
Since NA > ND, the material is p-type.
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3
318
6202318
i
318310
i
716106
10 and 106So
2n>> / 106 and / 10nre,temperaturoomAt(b)
/cm. /cm x
/cm
p
nn /cm x p
cm x N N cm
i
D A
====
=−=
2.30
(a) Phosphorus is a donor, and boron is an acceptor. ND = 2 x 1017 /cm3, and NA = 5 x 1017 /cm3.
Since NA > ND, the material is p-type.
3
317
6202317
i
317310
333103
10 and 103So
2n>> / 103 and10re,temperaturoomAt(b)
/cm /cm x
/cm
p
nn /cm x p
cm x N N /cmn
i
D Ai
====
=−=
2.31
ND = 4 x 1016 /cm3. Assume NA = 0, since it is not specified.
N D > N A : material is n - type | N D − N A = 4 x1016 / cm3 >> 2ni = 2 x1010 / cm
3
n = 4 x1016 / cm3 | p =
n i
2
n=
1020
4 x1016= 2.5 x103 / cm
3
N D + N A = 4 x1016 / cm3 | Using Fig. 2.13, μ n =1030
cm2
V − sand μ p = 310
cm2
V − s
ρ =1
qμ nn=
1
1.602 x10−19C ( )1030cm
2
V − s
⎛
⎝ ⎜
⎞
⎠⎟
4 x1016
cm3
⎛
⎝ ⎜
⎞
⎠⎟
= 0.152 Ω − cm
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2.32
NA = 1018 /cm
3. Assume ND = 0, since it is not specified.
N A > N D : material is p - type | N A − N D =1018 / cm3 >> 2ni = 2 x1010 / cm3
p = 1 018 / cm3 | n =n i
2
p
=10
20
1018
=100/ cm3
N D + N
A =1018 / cm3 | Using Fig. 2.13, μ n = 375cm2
V − sand μ p =100
cm2
V − s
ρ =1
qμ p p=
1
1.602 x10−19C 100
cm2
V − s
⎛
⎝ ⎜
⎞
⎠⎟
1018
cm3
⎛
⎝ ⎜
⎞
⎠⎟
= 0.0624 Ω − cm
2.33
Indium is from column 3 and is an acceptor. NA = 7 x 1019 /cm
3. Assume ND = 0, since it is not
specified.
N A > N
D : material is p - type | N A − N
D = 7 x1019 /cm3 >> 2ni = 2 x1010 /cm3
p = 7 x1019 /cm3 | n=
ni
2
p=
1020
7 x1019
=1.43 /cm3
N D + N A = 7 x1019 / cm3 | Using Fig. 2.13, μ n =120cm2
V − sand μ p = 60
cm2
V − s
ρ =1
qμ p p=
1
1.602 x10−19C 60cm2
V − s
⎛
⎝ ⎜
⎞
⎠⎟
7 x1019
cm3
⎛
⎝ ⎜
⎞
⎠⎟
= 1.49 mΩ − cm
2.34
Phosphorus is a donor : N D = 5.5 x1016 / cm3 | Boron is an acceptor : N A = 4.5 x1016 / cm3
N D > N A : material is n - type | N D − N A =1016 / cm3 >> 2ni = 2 x1010 / cm3
n =1016 /cm3 | p =ni
2
p=
1020
1016
=104 /cm3
N D + N A =1017 / cm3 | Using Fig. 2.13, μ n = 800cm2
V − sand μ p = 230
cm2
V − s
ρ =1
qμ nn=
1
1.602 x10−19C 800cm2
V − s⎛ ⎝ ⎜ ⎞
⎠⎟
1016
cm3
⎛ ⎝ ⎜ ⎞
⎠⎟
= 0.781 Ω − cm
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2.35
ρ =1
qμ p p| μ p p =
1
1.602 x10−19C ( )0.054Ω − cm( )=
1.16 x1020
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
NA μp μp p
1018 96.7 9.67 x 1020
1.1 x1018 93.7 1.03 x 1020
1.2 x 1017 91.0 1.09 x 1020
1.3 x 1019 88.7 1.15 x 1020
2.36
ρ =1
qμ p p
| μ p p =1
1.602 x10−19
C ( )0.75Ω − cm( )
=8.32 x1018
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
NA μp μp p
1016 406 4.06 x
1018
2 x 1016 363 7.26 x
1018
3 x 1016 333 1.00 x
1019
2.4 x 1016 350 8.40 x 1018
2.37
Based upon the value of its resistivity, the material is an insulator. However, it is not intrinsicbecause it contains impurities. Addition of the impurities has increased the resistivity.
2.38
ρ =1
qμ nn| μ nn ≈ μ n N D =
1
1.602 x10−19C
( )2Ω − cm
( )
=3.12 x1018
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
ND μn μnn
1015 1350 1.35 x 1018
2 x 1015 1330 2.67 x 1018
2.5 x 1015 1330 3.32 x 1018
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2.3 x 1015 1330 3.06 x 1018
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2.39 (a)
ρ =1
qμ nn| μ nn ≈ μ n N D =
1
1.602 x10−19C ( )0.001Ω − cm( )=
6.24 x1021
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
ND μn μnn
1019 116 1.16 x 1021
7 x 1019 96.1 6.73 x 1021
6.5 x 1019 96.4 6.3 x 1021
(b)
ρ =1
qμ p p| μ p p ≈ μ p N A =
1
1.602 x10−19C ( )0.001Ω − cm( )
=6.24 x10
21
V − cm − s
An iterative solution is required using the equations in Fig. 2.8:
NA μp μp p
1.3 x 1020 49.3 6.4 x 1021
2.40
Yes, by adding equal amounts of donor and acceptor impurities the mobilities are reduced, butthe hole and electron concentrations remain unchanged. See Problem 2.37 for example.However, it is physically impossible to add exactly equal amounts of the two impurities.
2.41
(a) For the 1 ohm-cm starting material:
ρ =1
qμ p p| μ p p ≈ μ p N A =
1
1.602 x10−19C ( )1Ω − cm( )
=6.25 x1018
V − cm − s
An iterative solution is required. Using the equations in Fig. 2.8:
NA μp μp p
1016 406 4.1 x 1018
1.5 x 1016
383 5.7 x 1018
1.7 x 1016 374 6.4 x 1019
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To change the resistivity to 0.25 ohm-cm:
ρ =1
qμ p p| μ p p ≈ μ p N A =
1
1.602 x10−19C ( )0.25Ω − cm( )
=2.5 x1019
V − cm − s
NA μp μp p
6 x 1016 276 1.7 x 1019
8 x 1016 233 2.3 x 1019
1.1 x 1017 225 2.5 x 1019
Additional acceptor concentration = 1.1x1017
- 1.7x1016
= 9.3 x 1016 /cm
3
(b) If donors are added:
ND ND + NA μn ND - NA μnn
2 x 1016 3.7 x 1016 1060 3 x 1015 3.2 x 1018
1 x 1017 1.2 x 1017 757 8.3 x 1016 6.3 x 1019
8 x 1016 9.7 x 1016 811 6.3 x 1016 5.1 x 1019
4.1 x 1016 5.8 x 1016 950 2.4 x 1016 2.3 x 1019
So ND = 4.1 x 1016 /cm
3must be added to change achieve a resistivity of 0.25 ohm-cm. The
silicon is converted to n-type material.
2.42
Phosphorus is a donor: ND = 1016 /cm3 and μn = 1250 cm2 /V-s from Fig. 2.8.
σ = qμ nn ≈ qμ n N D = 1.602 x10−19C ( )1250( )1016
( )= 2.00Ω − cm
Now we add acceptors until σ = 5.0 (Ω-cm) -1:
σ = qμ p p | μ p p ≈ μ p N A − N D( )=5 Ω − cm( )
−1
1.602 x10−19C =
3.12 x1019
V − cm − s
NA ND + NA μp NA - ND μp p
1 x 1017 1.1 x 1017 250 9 x 1016 2.3 x 1019
2 x 1017 2.1 x 1017 176 1.9 x 1017 3.3 x 1019
1.8 x 1017
1.9 x 1017
183 1.7 x 1016
3.1 x 1019
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2.43
Boron is an acceptor: NA = 1016 /cm3 and μp = 405 cm2 /V-s from Fig. 2.8.
σ = qμ p p ≈ qμ p N A = 1.602 x10−19C ( )405( )1016( )=
0.649
Ω − cm
Now we add donors until σ = 5.5 (Ω-cm) -1:
σ = qμ nn | μ nn ≈ μ n N D − N A( )=5.5 Ω − cm( )
−1
1.602 x10−19C =
3.43 x1019
V − cm − s
ND ND + NA μn ND - NA μp p
8 x 1016 9 x 1016 832 7 x 1016 5.8 x 1019
6 x 1016 7 x 1016 901 5 x 1016 4.5 x 1019
4.5 x 1016 5.5 x 1016 964 3.5 x 1016 3.4 x 1019
2.44
V T =kT
q=
1.38 x10−23T
1.602 x10−19= 8.62 x10−5T
T (K) 50 75 100 150 200 250 300 350 400
VT (mV) 4.31 6.46 8.61 12.9 17.2 21.5 25.8 30.1 34.5
2.45
j = −qDn −dn
dx
⎛
⎝ ⎜
⎞
⎠⎟ = qV T μ n
dn
dx
j = 1.602 x10−19C ( )0.025V ( ) 350
cm2
V − s
⎛
⎝ ⎜
⎞
⎠⎟
1018 − 0
0 −10−4
⎛
⎝ ⎜
⎞
⎠⎟
1
cm4
= −14.0kA
cm2
2.46
j = −qD p
dp
dx= −1.602 x10−19C ( )15
cm2
s
⎝ ⎜
⎠⎟ −
1019 / cm3
2 x10−4cm
⎝ ⎜
⎠⎟exp −
x
2 x10−4cm
⎛
⎝ ⎜ ⎟
j =1.20 x105 exp −5000 x
cm
⎛
⎝ ⎜
⎞
⎠⎟
A
cm2
I 0( )= j 0( ) A = 1.20 x105 A
cm2
⎛
⎝ ⎜
⎞
⎠⎟ 10μ m2( )10−8cm2
μ m2
⎛
⎝ ⎜
⎞
⎠⎟ =12.0 mA
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2.47
j p = qμ p pE − qD p
dp
dx= qμ p p E −V T
1
p
dp
dx
⎝ ⎜
⎠⎟ = 0 → E = V T
1
p
dp
dx
E ≈ V T
1
N A
dN A
dx
= 0.025−1022 exp −104 x( )
1014 +1018 exp −104 x( ) E 0( )= −0.025
1022
1014 +1018= −250
V
cm
E 5 x10−4cm( )= −0.025
1022 exp −5( )1014 +1018 exp −5( )
= −246V
cm
2.48
jn
drift = qμ nnE = 1.60 x10−19C ( )350cm2
V − s
⎝ ⎜
⎠⎟
1016
cm3
⎝ ⎜
⎠⎟ −20
V
cm
⎛
⎝ ⎜
⎠⎟ = −11.2
A
cm2
j pdrift = qμ p pE = 1.60 x10−19C ( )150
cm2
V − s
⎛
⎝ ⎜
⎞
⎠⎟
1.01 x1018
cm3
⎛
⎝ ⎜
⎞
⎠⎟ −20
V
cm
⎛
⎝ ⎜
⎞
⎠⎟ = −484
A
cm2
jn
diff = qDn
dn
dx= 1.60 x10−19C ( )350 ⋅ 0.025
cm2
s
⎛
⎝ ⎜
⎞
⎠⎟
104 −1016
2 x10−4cm4
⎛
⎝ ⎜
⎞
⎠⎟ = −70.0
A
cm2
j pdiff = −qD p
dp
dx= −1.60 x10−19
C ( )150 ⋅ 0.025cm
2
s
⎛
⎝ ⎜
⎞
⎠⎟
1018 −1.01 x1018
2 x10−4cm
4
⎛
⎝ ⎜
⎞
⎠⎟ = 30.0
A
cm2
jT = −11.2 − 484 − 70.0 + 30.0 = −535 A
cm2
2.49 NA = 2ND
EC
EA
NA
Holes
EV
NA
ND
NA
ND
NA
ND
ED
2.50
λ =hc
E =
6.626 x10−34 J − s( )3 x108 m / s( )1.12eV ( )1.602 x10−19
J / eV ( )=1.108 μ m
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2.51
p-type silicon
n-type silicon
Si02
Al - Anode Al - Cathode
2.52 An n-type ion implantation step could be used to form the n+ region following step (f) in Fig.2.17. A mask would be used to cover up the opening over the p-type region and leave theopening over the n-type silicon. The masking layer for the implantation could just bephotoresist.
n-type silicon
p-type silicon
Si02
Photoresist
Structure after exposure anddevelopment of photoresist layer
Mask
n-type silicon
p-type silicon
Structure following ionimplantation of n-type impurity
n+
Ion implantation
Side view
Top View
Mask for ion implantation
2.53
(a) N = 81
8
⎝ ⎜
⎠⎟ + 6
1
2
⎝ ⎜
⎠⎟ + 4 1()= 8 atoms
(b) V = l3 = 0.543 x10−9
m( )3
= 0.543 x10−7cm( )
3
=1.60 x10−22cm
3
(c) D =
8 atoms
1.60 x1022cm3 = 5.00 x10
22 atoms
cm3
(d ) m = 2.33g
cm3
⎛
⎝ ⎜
⎞
⎠⎟1.60 x1022
cm3 = 3.73 x10−22
g
(e) From Table 2.2, silicon has a mass of 28.086 protons.
mp =3.73 x10−22
g
28.082 8( ) protons=1.66 x10−24 g
proton
Yes, near the actual proton rest mass.
34