EE 212 FALL 19-20
Homework Assignment #3
Solutions
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Problem 1 (10 points)
A new cleaning procedure has been proposed which is based on H2SO3 saturated
with H2SO4 as an oxidant. This has been suggested as a replacement for the H 2O2
oxidizing solution used in the RCA clean. Suppose a Si wafer, contaminated with
trace amounts of Au, Fe and Cu is cleaned in the new H2SO3/H2SO4 solution. Will
this clean the wafer effectively? Why or why not? Explain.
Answer: No. Based on the table of lecture notes (Slide 8, Chpt.4), the standard oxidation
potential for H2SO4/H2SO3 is smaller than that of Fe3+/Fe, but larger than that of
Cu2+/Cu and Au3+/Au. This indicates that the solution could only remove Fe impurities
from the wafer by oxidizing them to Fe3+, but cannot remove the Au and Cu impurities
by oxidizing them to Au3+ and Cu2+, respectively.
Problem 2 (10 points)
One projection system uses an ArF laser source to produce a 193 nm wavelength
and an exposure lens with an NA of 0.9.
a) Estimate the height variation (surface roughness) tolerance on the wafer surface
of this projection system.
Answer:
The height variation tolerance is determined by the depth of focus:
𝜆
193 nm
𝐷𝑂𝐹 = 𝑘2
= 0.3
= 71.5 nm
2
(𝑁𝐴)
0.92
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This DOF can be an estimation of height variation tolerance of the wafers under this
system.
b) To improve the system DOF, an immersion lithography system is used with
water which has a refractive index of 1.44. Calculate the improvement of DOF if
this system is being used to create lines with a pitch of 200 nm.
For both a) and b), assume value of k1=0.75 and k2=0.3.
Answer:
1 − √1 − (𝜆/𝑝)2
DOF(immersion)
=
=
DOF(dry)
𝑛 − √𝑛2 − (𝜆/𝑝)2
193 nm 2
1 − √1 − (200 nm)
193 nm 2
2
√
1.44 − 1.44 − (200 nm)
= 1.99
Problem 3 (15 points)
The numerical aperture for a given lithography system is NA=0.5. To ensure the
quality of exposure, we want the depth of focus to cover the whole thickness of the
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photoresist. Also the minimum resolution required is 500nm. Assume k1=0.75 and
k2=0.5.
The photoresist is uniformly coated on the wafer with thickness of 0.7um.
a) If the focal plane of the projection system is positioned at bottom of the resist
(the boundary of photoresist and the wafer), calculate the wavelength range that
is suitable for this exposure process. Comment whether you could successfully
design this wavelength based on your result.
Answer: Based on equation for depth of focus and resolution on lecture slides, in this
condition:
𝜆
𝐷𝑂𝐹 = 𝑘2
≥ 700𝑛𝑚
(𝑁𝐴)2
𝜆
𝑅 = 𝑘1
≤ 500𝑛𝑚
𝑁𝐴
We can calculate the wavelength range:
350nm ≤ 𝜆 ≤ 333𝑛𝑚
There is no overlap, thus in this case we cannot satisfy the needs for both DOF and
resolution.
b) If the focal plane of the projection system is positioned in the middle height of
the photoresist, calculate the wavelength range that is suitable for this exposure
process.
Answer: Again, in this case
𝜆
700𝑛𝑚
𝐷𝑂𝐹 = 𝑘2
≥
= 350𝑛𝑚
2
(𝑁𝐴)
2
𝜆
𝑅 = 𝑘1
≤ 500𝑛𝑚
𝑁𝐴
Thus we can calculate the wavelength range:
175nm ≤ 𝜆 ≤ 333𝑛𝑚
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Problem 4 (10 points)
A student plans to make a grating pattern on her wafer. After trying an exposure
on an ordinary lithography system, she realizes this system cannot achieve the
resolution needed for the grating, which is less than 100 nm. So she plans to try Ebeam lithography instead. An experienced user tells her that 600nm thick E-beam
resist, with a dose of 700 uC/cm2 should work well for the desired resolution. The
pattern she plans to write is a 1cm2 area. The E-beam system has a current of
8000pA. Please help her estimate how long the E-beam exposure should be.
(Hint: For ordinary photolithography, the unit of dose in photoresist is Joule/area.
But for E-beam lithography, we normally use Coulomb/area as the corresponding
dose unit. (1 coul=1A*sec)
Answer: The time needed is just the total Coulomb divided by the current that the
system provides:
t=
𝑡𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑜𝑢𝑙𝑜𝑚𝑏 700uC/cm2 × 1cm2
=
= 24.3h
𝑐𝑢𝑟𝑟𝑒𝑛𝑡
8000pA
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It will take more than a day to finish the 1cm2 pattern using E-beam lithography!!
Problem 5 (15 points)
An X-ray exposure system uses photons with an energy of 1 keV. If the separation
between the mask and wafer is 10 µm (proximity printing system), estimate the
diffraction limited resolution that is achievable by this system.
Answer: The wavelength of 1 keV x-ray is given by:
ℎ𝑐
E = hν =
𝜆
−15
ℎ𝑐 (4.14 × 10 eV∙s)(3 × 1010 cm/s)
λ=
=
= 1.24nm
𝐸
103 eV
The X-ray system operates in proximity printing mode, so that the theoretical resolution
can be calculated by:
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𝑅 = √𝜆𝑔 = √(1.24 × 10−3 um)(5um) = 78.7nm
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Problem 6 (10 points)
An EE212 student goes to work in the semiconductor industry and they realize
that there are major issues with getting complex EUV lithography systems to work
in manufacturing. Since the industry could be in serious trouble if EUV
lithography systems can’t be made to work, they have the interesting idea of using
an EUV source with a wavelength of 13.5nm in a proximity printing system, rather
than a projection printing system. This would imply a much simpler lithography
tool and should provide high resolution. Is this a good idea or not? Why?
Answer: No, it would not work. The achievable resolution in a proximity system with
this wavelength light source would be on the order of√𝜆𝑔, where g is the separation of
the mask from the wafer. It would be very difficult to make g much less than a few µm
and even if g were 1 µm, W ≈ 100+ nm so not very good resolution would be achievable.
The other big issue would be the masks. Proximity printing systems require masks that
are transparent in regions where the photons need to expose the resist. There are
essentially no materials that are transparent at EUV wavelengths.
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