# n = P kT = 10 Pa 1 N/ m 2 / Pa 1.38 x10 J / K 300 K )1 N ```M = 32, T = 300 K, P = 10-4 Pa.
Φ = (2.63 x 1020)(10-4)/ 32 • 300
λ=
= 2.68 x 1020 molecules/cm2-sec
1.38 x10 −23 J / K )(N − m / J )(10 2 cm / m )(300 K )
kT
(
2 =
2 = 7190 c m
2 πpd
2 π (10 −4 N / m 2 )(10 −4 m 2 / cm 2 )(3.6 x10 −8 cm )
or 71.9 m.
(10-4 Pa)(0.0075 torr/Pa) = 7.5 x 10-7 torr
10 −8 Pa (1 N / m 2 / Pa )
P
n =
=
kT 1.38 x10 −23 J / K (300 K )(1 N - m/J
n = 2.42 x10 molecules / cm
6
)
= 2.42 x10 12 / m 3
3
v = (Ng/N) ks hg/(ks+hg). ks = 2 x 106 exp (-1.9)(1.6 x 10-19)/(1.38 x 10-23)(1473) = 0.64
cm/sec. hg = 1 cm/sec, Ng = 3 x 1016 atoms/cm3. v = (0.64 x 1/1.64)(3 x 1016/5
x 1022) = 2.3 x 10-7 cm/sec or 0.14 &micro;m/min.
(b) At T = 1498 K, ks becomes 0.82 cm/sec, and v increases to 0.16 &micro;m/min.
The change is 0.02 &micro;m/min, a 14% increase.
(c) Setting ks = 1 yields T = 1520 K or 1247 oC.
(d) From the SiH4 curve, ks = 0.2 &micro;m/min at 1000/T = 0.93 and 0.01 &micro;m/min at
1000/T = 1.1. EA = -1000 k (∆ ln ks/∆(1000/T)) = -1000 x 8.617 x 10-5 x
(ln(.2)-ln(.01))/(.93-1.1) = 1.52 eV.
From Figs. 5.3 (a) and (b), 60 keV through 0.25 &micro;m SiO2 yields Rp = 0.19 &micro;m and ∆Rp =
0.09 &micro;m. For the Gaussian implant, N(x) = Np exp[-(x-Rp)2/2∆Rp2] with NP =
Q/∆Rp 2 π = 4.43 x 1018/cm3.
(a) N(0.25 &micro;m) = 4.43 x 1018exp [-(0.25-0.19)2/2(0.09)2] = 3.5 x 1018/cm3.
∞
(b) Q Si =
∫ 4.43 x10
18
0.25 &micro;m
QSi
 (x − 0.19 &micro;m )2 
 dx = (0 .751
exp −
 2 (0 .09 &micro;m )2 
)
2πN p ∆R p
= 7.5 x 1013/cm2
(Some one may not know how to do integral here(it should be checked from
website or other reference), if you can come out the integral fomula, this number
can be ~ 2 π Np∆Rp( there should be no order difference). If the peak position is
&lt;&lt;0.25/2&micro;m which can result in order difference, pls don’t use ~ 2 π Np∆Rp)
(c) 3 x 1015 = 4.43 x 1018exp [-(xj –0.19)2/2(0.09)2],
0.53 &micro;m， Junction depth 0.53-0.25=0.28&micro;m.
and xj = 0.34 + 0.19 =
(1015/cm2)(π)(20/2)2 cm2 = 3.14 x 1017 total boron atoms. 10-5 A corresponds to (10-5
Coul/sec)/(1.6 x 1019 Coul/charge) or 6.25 x 1013 charges/sec. 3.14 x 1017
atoms/6.25 x 1013 atoms/sec = 5030 sec or approximately 84 minutes.
(a) ρc = RA = (0.5 Ω)(10-8cm2) = 5 x 10-9 Ω-cm2.
(b) R = ρc/A = ρc /10-10cm2 = 50 Ω.
to do much better.
This value is large, but it is difficult
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