EE143 Midterm #2 Solutions Sp 2003
Problem 1
(a) E-Beam lithography has to address and expose the pixels in a serial fashion. Optical lithography can expose all
pixels simultaneously by flood exposure ( a parallel process). The later has a higher throughput.
(b)
(i) AZ1350J is a positive resist .Kodak 747 is a negative resist.
(ii)
)45/90(10log
1 = 3.32
)7/12(10log
1= 4.27 (higher resist contrast)
(iii) Resist Kodak 747 (better sensitivity) only needs 12mJ/cm2 to fully exposure while AZ13750J requires
90 mj/cm2 to fully expose.
(c )Technology factors: k1 = 0.58 and k2 = 0.54
Lmin = kl x 248nm /0.5 =
! "
DOF = k2 x 248nm/(0.5)2 =
# $ ! "
(d) Steppers maintained at constant temperature to minimize thermal run-in/out overlay errors due to different
thermal expansion coefficients of mask and substrate.
(e) To reduce difference of standing wave's I max and I mjn , one can use a resist with absorption dyes or by placing an
antireflection coating on the reflecting interface.
Problem 2
(a) CT(Sid4) = 0.02x4.8 xl018= 9.6 xlO16 molecules/cm3
Since 1 molecule of SiCl4 gives 1 Si atom and the growth rate is assumed to mass-transfer limited,
%
22
16
2222 105
106.962.2
105105
11
1
x
xx
x
Ch
x
C
hk
dt
dy TGT
Gs
=≈⋅
+
== = 5.03x10-6cm/sec = 503 Å/sec
(b) (i) We will use the linear regime of the curve where surface-reaction mechanism determines the growth rate R
R = constant x exp[-Ea/kT] or Ea =-1000 k )2(/1000)1(/1000
)2(ln)1(ln
TT
RR
−
−
×
& ' ( ) * + * + , '- & ' ) * + * + , '
Therefore Ea = -1000 x 8.617 xl0-5
101
93
.
0
)1.0ln()2.0ln(
−
−
×= 1.5eV.
(ii) The SiH4 concentration CT is lower due to the lower partial pressure of SiH4. Therefore R will lower for both
mass transfer and surface reaction limited regions
The ks term is not affected by the dilution effect.
The hG term ( =
δ
/D) will have no change in D (
P
T2/3
∝) because Ptolal is same. The
δ
term =
µ
ρ
UL
L
3
2will
have minor changes in density and viscosity due to He dilution but will be second-order.
