1
3.155J/6.152J
Microelectronic Processing
Fall Term,2003
Bob O'Handley
Martin Schmidt
Problem set 4 Solutions Out Oct,8,2003 Due Oct.15,2002
1,Plot resolution and depth of field as a function of exposure wavelength for a projection
aligner with 100nm < λ < 500nm,Assume NA = 0.26,Recalculate on the same plot for
NA =0.41,Discuss the implication of these plots for the technologist that must
manufacture transistors with 0.5 mμ features.
NANA
k
R
λλ 6.0
1
==
22
2
)(
5.0
)( NANA
k
DOF
λλ
±=±=
To resolve 0.5 mμ feature size with NA=0.26,one can choose exposure wavelength less
than 216 nm (e.g,193 nm ArF DUV) and also carefully control the resist layer’s
topography to keep the pattern in focus.
-4000
-3000
-2000
-1000
0
1000
2000
3000
4000
100 150 200 250 300 350 400 450 500
Resolution and Depth of Focus Vs Exposure wavelength
R(NA=0.26)
DOF (NA=0.26)
DOF (NA=0.26)
Re
so
l
u
t
i
on
and
D
e
p
t
h
o
f
Fo
cus
(
nm
)
Exposure wavelength (nm)
ArF 193nm
If NA = 0.41,one can choose exposure wavelength less than 340 nm (e.g,ArF 193nm or
KrF 248nm).
2
-2000
-1500
-1000
-500
0
500
1000
1500
2000
100 150 200 250 300 350 400 450 500
Resolution and Depth of Focus Vs,Exposure wavelength
R(NA=0.41)
DOF (NA=0.41)
DOF (NA=0.41)
Re
so
l
u
t
i
on
and
D
e
p
t
h
o
f
Fo
cus
(
nm
)
Exposure wavelength (nm)
ArF 193nm KrF 248nm
2) A 0.6 mμ thick layer of resist has Q
0
=40mJ/cm
2
and Q
f
= 160 mJ/cm
2
,Calculate the
resist contrast and CMTF,If the resist thickness is cut in half,Q
f
reduces to 70 mJ/cm
2
while Q
0
is unchanged,Assuming NA = 0.4,use the figure below to determine the
minimum linewidthfor an aligner with S = 1.0 using both resist thicknesses with a source
of 365 nm,The figure below plots MTF of the aligner for a set of lines and spaces,The
lines and spaces are of equal width (W),and the spatial frequency is normalized by the
Rayleigh criteria,R,In other words,a normalized spatial frequency of 0.5,corresponds to
a linewidth W equal to R (since the equivalent source spacing of the lines is 2W).
3
0.6 mμ resist:
66.1
)40/160(log
1
)/(log
1
10010
===
QQ
f
γ
6.0
40160
40160
110
110
/1
/1
0
0
=
+
=
+
=
+
=
γ
γ
QQ
QQ
CMTF
f
f
0.3 mμ resist:
11.4
)40/70(log
1
)/(log
1
10010
===
QQ
f
γ
273.0
4070
4070
110
110
/1
/1
0
0
=
+
=
+
=
+
=
γ
γ
QQ
QQ
CMTF
f
f
Normalized spatial frequency:
f=(line/unit length)/(cut off frequency)=(1/2W)/(1/R)=R/2W?W=R/2f.
nm
NA
R 557
61.0
==
λ
To resolve the feature,we need MTF≥CMTF,The minimum linewidth,i.e,the
maximum spatial frequency,can be directly read from the graph as the intersections
between MTF=CMTF and S=1.
The maximum frequencies from the graph are 0.33 and 0.63 with respect to 0.6 mμ and
0.3 mμ resists,corresponding to a minimum linewidth of 844nm and 442nm.
3) Estimate the diffraction-limited resolution for an X-Ray exposure system using
photons with an energy of 1 keV and a mask to wafer separation of 20 microns.
nmm
E
hchc
E 24.11024.1
106.110
10998.21062.6
9
193
834
=×=
××
×××
==?=
λ
λ
Minimum resolvable feature size due to the X-ray exposure system’s Fresnel diffraction
(Proximity printing) is,mg μλ 157.0)201024.1()(
2/132/1
=××=
0.6 mμ resist
CMTF
0.3 mμ resist
CMTF
Maximum frequencies
R/2W