ELEN 4944 HW1 Solutions:
1. (a) The surface structures are shown below
(b) (100) plane [ a 5.43A ]
1
4
Number of atoms in this cell: 4 1
Density of atoms in this surface:
1
3.4 1014 atoms / cm 2
2
a
(110) plane
1
4
Number of atoms in this cell: 4 1
Densi
ELEN 4944 HW7 Solutions:
1. Use Eq. B.10 to solve for the time needed, according to:
t (sec) [1/ D][ x j /( 2erfc 1cfw_1016 / Cs )]2 where x j 2m 2 104 cm
a) At 900C, D( B) 1.27 1015 cm 2 / sec (using Eq. B.18 p. 539, with T=1173K),
And Cs 1.05 10 20 cm 3
ELEN 4944 HW7
Note the first three problems are from the text
.
1. Prob. 11.8 Hint: see appendix B, Eq. B.10
2. Prob. 19.10
3. Prob. 20.13
4. After exposing a layer of positive resist (n=1.7) to UV radiation, it is
found that the resist layer is about 1.1
ELEN 4944 HW6 Solutions:
1. a) Solid solubility of Boron at 950C is found from Fig. 9-12, p. 152: ~
1.5 10 20 B atoms / cm 3
Diffusion coefficient D of: boron @ 950C is found using Eq. B-18 p.539 and Table B-1,
p.538. (Note that 950C=1223K)
D( B,950C ) 10
ELEN 4944 HW8 Solutions:
1. Parameters values:
W1=0.4um, L=0.6mm, h=0.15um, S=0.1um, tm=0.3um
a) Wire to wire:
A sio2 Lt m 4 8.85 10 12 F / m 0.6mm 0.3m
C
63.72 fF
d
S
0.1m
Wire to substrate:
A sio2 LW1 4 8.85 10 12 F / m 0.6mm 0.4m
C
56.64 fF
d
h
0.15m
ELEN 4944 HW 8
1. [10 points]
(a) A 0.4 m metal line is 0.6mm long. It is atop 0.15 m of silicon dioxide and there is
an identical line parallel to it with a line-to-line spacing of 0.1 m. The space
between lines is filled with silicon dioxide. Using Al a
ELEN 4944 HW5 Solutions:
1. To calculate the dose (ions/cm2) we use Eq. 12-1 on p. 190
( I t ) /( qi A)
Where qi 1.6 1019 C; I 30mA; t 300 sec
The area being implanted corresponds to that of 30 100-mm = 10-cm wafers, or:
A 30(r 2 ) 30( 25) 2355cm2
Thus (
ELEN 4944 HW3 Solutions:
1. C 1 pF / cm 2 J 1mA / cm 2 V 1000V
t
CV 10 9 C / cm 2
3
10 6 sec 1 sec
2
J
10 A / cm
2. The atmospheric pressures are different at different elevations. Since APCVD
processes do not involve explicit pressure control systems,
ELEN 4944 HW2 Solutions:
1. (a) pn ni2 n p 1.45 1010 cm 3
(1.45 1010 ) 2
(b) As doped n 4 10 p
525 cm 3
17
4 10
17
(c) B doped p 2 1017 n
(1.45 1010 ) 2
1050 cm 3
17
2 10
(d) We will have degenerate Silicon means that we have such a high level of
dopi