EE143 Midterm Exam #2 Solutions Fall 2003
Problem 1
(a) (i )Let translational error be (xt, yt).
After subtracting the translational error, we have:
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x
+3 -xt 0 -2 -xt
y +3 - yt 0 +1 - yt
Since thermal run out/in error is antisymmetrical : [+3 -xt ] = - [-2 -xt] gives xt = +0.5 µm
Since rotational error is antisymmetrical : [+3 - yt ] = - [+1 - yt ] gives yt = +2 µm
or
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x 2.5 0 -2.5
y 1 0 -1
(ii ) thermal run out error = +2.5 µm
(iii) rotational error = +1 µm (counterclockwise)
(b) DOF = 1
2λ (R
k)2 . Therefore DOF ratio increases by (1/0.8)2 -1 = 0.56= 56%
(c ) (i) Negative resist (ii) Resist sensitivity = 50 mJ/cm2
Problem 2
(a )Geometrical shadowing is due to the original substrate topography. Self shadowing is due to the
additional topography change with the deposited film.
(b) R = constant × exp [-Ea/kT] or Ea = - k × ln R(1) - ln R(2)
1/T(1) -1/T(2)
R = 20000 Å/min at T =1073K
R = 2000 Å/min at T=973K
Therefore Ea = 2.1eV
(c )(i ) ks is proportional to exp [-Ea/kT]. Higher T gives higher ks.
(ii)
hG = DG
δ with D ∝ T3/2/P and δ= 2
3
L
ρUL
µ
Since ρ = P/kT , therefore hG increases with T when other control parameters are same.
(d) PECVD can be performed at a lower processing temperature than thermal CVD. The deposited film
properties (mechanical stress, density, etc) can also be tailored with plasma and substrate bias conditions.
Also possible to form special compound films which cannot be formed by thermally activated chemical
reactions.
100
50
Position x in µm
% Resist thickness remaining after development
100
50
Position x in µm
% Resist thickness remaining after development
100
50
Photon energy density
with 2 sec exposure in mJ/cm2
Position x in µm
Resist sensitivity
Photon
intensity
100
50
Photon energy density
with 2 sec exposure in mJ/cm2
Position x in µm
Resist sensitivity
Photon
intensity
