Solutions
Physics 610/Chemistry 678—Semiconductor Processing and Characterization
Quiz II—July 28, 2014
Instructions: Answer all questions thoroughly giving clear explanations of your reasoning, defining all
symbols, labeling all diagrams, and expressing your results using the appropriate units. Textbooks and
calculators are not allowed on this exam. You will receive a substantial percentage of points on a
problem if you can solve for a given variable symbolically, so wait to substitute in number values once
you have a symbolic solution. When a numerical quantity is computed, you will be expected to be within
an order of magnitude. Write your name at the top of each page, number your pages, and staple them
together when finished.
The quiz is worth a total of 100 points. Please note the point value for each problem, as they are not
equally weighted
Part I: Short-answer questions on basic principles. (5 points each)
1. Name and describe two different diffusion mechanisms of dopants in crystalline silicon. What
are the approximate activation energies for the mechanisms you named above? How does the
diffusion rate depend on temperature? Would it be a good idea to use gold as a source/drain
contact in a MOSFET? Why or why not?




Vacancy diffusion occurs when a dopant occupies a vacancy in the crystal lattice, and to
migrate further it must then occupy another nearby vacancy. The activation energy of
vacancy diffusion includes the energy needed to create vacancies and also for the dopant
to move into the vacancy; this energy is typically of order 𝐸𝑎 > 3 𝑒𝑉.
Interstitial diffusion occurs when a dopant occupies and moves in the space between the
crystal lattice. The activation energy includes that required for motion, typically of order
𝐸𝑎 < 2 𝑒𝑉.
𝐸
The temperature-dependence of the diffusion rate is exponential, i.e. exp − 𝑘 𝑎𝑇.
𝑏
𝐴𝑢 has a very high diffusion coefficient in Si and the concentration-dependence goes as
𝐷(𝐶) ∝ 𝐶 −2 . These two properties conspire to make 𝐴𝑢 diffuse quickly and thoroughly
in 𝑆𝑖. At higher temperatures, these effects can be detrimental to the functionality of a
MOSFET.
2. What is the approximate energy of formation for a vacancy defect? How much thermal energy is
there at room temperature? Do defects exist in crystals such as Silicon at room temperature? If
so, what stabilizes their formation? Does the light used for i-line lithography have enough energy
to efficiently create vacancies?



The energy of formation of vacancies is 3 𝑒𝑉.
𝑒𝑉
𝑘𝑏 𝑇 = 8.61 × 10−5 𝐾 ⋅ 300 𝐾 = 0.026 𝑒𝑉
Thermal energy at room temperature is small compared to vacancy formation energy, so
thermal excitation will not contribute much to vacancy formation. However, the increase
in entropy due to vacancies helps make their formation spontaneous, even at room
temperature.

ℎ𝑐
𝑒𝑉
The energy of 𝑖 − 𝑙𝑖𝑛𝑒 photon is 𝐸 = 𝜆 = 1240 𝑛𝑚 /365 𝑛𝑚 = 3.4 𝑒𝑉, so these
photons will create vacancies if absorbed.
3. What are the basic components of a photoresist? How do positive and negative photoresists
differ? Briefly describe the photochemical transformation processes for negative and positive
resists.




Resin matrix, photo-active compound (PAC), and solvent.
Positive resists are removed upon exposure to radiation and development; negative resists
remain upon exposure to radiation and development.
Positive Resists: The PAC acts as a dissolution inhibitor of resin in base before exposure.
After exposure, the PAC is transformed to an acid which is soluble in basic developer.
The photochemical transformation leverages the instability of a nitrogen subgroup.
Negative Resists: The PAC is soluble in polar solvents before exposure. After exposure,
the PAC crosslinks via numerous pathways yielding the PAC and matrix insoluble in
polar solvents.
4. What is the mean-free path of air molecules at room temperature for room pressure, rough pump
pressure, and the pressure used for performing metallization via evaporation.
The mean free path is given by the expression
𝜆=
𝑘𝑏 𝑇
√2𝜋𝑟 2 𝑃
≈
5
[𝑐𝑚]
𝑃(𝑚𝑡𝑜𝑟𝑟)
𝜆 (cm)
Pressure (mtorr)
6.6 × 10−6
760 × 103
5 × 10−3
103
5 × 103
10−3
5. What does it mean if a cleanroom is “class 100”? For a class 100 cleanroom, how many particles
will there be per 𝑚3 ?
Class 100 means there are 100 particles of size greater than 500 𝑛𝑚 per 𝑓𝑡 3 . Converting to
100
𝑓𝑡 3
meters, this is 𝑓𝑡 3 ⋅ (3.28 𝑚 ) = 3529 𝑚−3 ≈ 3500 𝑚−3 .
Part II:
1. You have a mask with a square aperture of area 4 𝜇𝑚2 that you would like to pattern onto an iline resist of thickness 1 𝜇𝑚 that has been spin-coated onto a Si wafer. (35 points)
a. First you use contact lithography. What will be the width of the feature that you generate
in the resist? What is the smallest feature width you could generate with this approach?
(4 points)
In the case of contact lithography with 𝜆 = 365 𝑛𝑚 and a resist thickness of 1 𝜇𝑚, the
diffraction image will be in the Fresnel regime, so the width of the feature will be
approximately 2 𝜇𝑚.
The smallest feature we can produce using this technique is given by √𝜆𝑔 =
√365 ⋅ 1000 𝑛𝑚2 = 604 𝑛𝑚.
b. Now you use a projection printer which positions the mask some distance 𝑥 from a lens
of diameter 10 𝑐𝑚 and focal length 𝑓 = 5 𝑐𝑚. Sketch the setup showing the mask, lens,
wafer, the light emanated from the aperture, and the focused light rays; make sure to
include relevant lengths in your sketch. (4 points)
c. What value of 𝑥 should you use to achieve the best resolution, and what is this
resolution? What is the reduction ratio of this setup? How flat must the substrate be to
achieve this resolution? (7 points)
The best resolution is achieved when 𝑥 causes the lens to be completely filled, i.e. 𝑤 = 𝑑,
because this will maximize the NA of the optical system. The diffraction width due to a
single-slit is given by 𝑤 = 2𝑥𝜆/𝑏, so 𝑥 = 𝑑 ⋅ 𝑏/2𝜆 = 10 𝑐𝑚 ∗ 2 × 10−4 𝑐𝑚/(2 × 365 ×
10−7 𝑐𝑚) = 27.4 cm.
Assuming maximum NA, the best resolution (diffraction-limited) is given by 𝑙𝑚 =
𝑘1 𝜆/𝑁𝐴 = 𝑘1 𝜆𝑓/𝐷. With 𝑘1 = 1, we have 𝑙𝑚 = 183 𝑛𝑚.
If we completely fill the lens with the diffraction image of our aperture, then the
reduction ratio will be 𝑏/𝑙𝑚 = 2 𝜇𝑚/0.183 𝜇𝑚 ≈ 11.
The flatness of the wafer must be of order of the 𝐷𝑂𝐹 = 𝑘2 𝜆/𝑁𝐴2 ≈ 365 𝑛𝑚/4 =
91.3 𝑛𝑚.
d. To characterize the resist, you vary the exposure dose and then measure the remaining
film thickness using an ellipsometer. The data is below.
Dose (𝑚𝐽 ⋅ 𝑐𝑚−2 )
1
5
10
20
30
45
65
90
125
150
Thickness (𝜇𝑚)
0.98
0.97
0.94
0.96
0.93
0.83
0.61
0.42
0.15
0
Plot the response curve for the resist. Is the resist a positive or negative resist? (5 points)
This is a positive resist, since the resist is removed by exposure to light.
e. Calculate the contrast and the critical modulation transfer function for the resist. (5
points)
𝐷
1
The contrast is 𝛾 = 1/log( 𝐷100 ) = log 3 = 2.1. The 𝑀𝑇𝐹 = (𝐷100 − 𝐷0 )/(𝐷100 + 𝐷0 ) =
150−50
150+50
f.
0
= 0.5.
Based on the optics of your mask aligner you know the light intensity profile on the
photoresist layer is given by the Gaussian function below. Calculate the ideal exposure
time. Explain your reasoning. (5 points)
From the graph, the intensity at ± 1𝜇𝑚 is about 2.5 𝑚𝐽 𝑐𝑚−2 𝑠 −1. With 𝐷100 from
above, the exposure time is 𝑡 = 150/2.5 = 60 𝑠. This is the required time, since only
the resist that is ± 1𝜇𝑚 from the center will be 100% photochemically altered and
removed after development.
g. Draw the quantitatively accurate resist profile (thickness versus distance) that you
predict after exposure for the ideal time followed by development. (5 points)
After 60 𝑠 of exposure and with 𝐷0 as the dose required to begin photochemical
transformation, etching will begin to occur where the intensity is 𝐷0 /𝑡 =
50/60 𝑚𝐽 𝑐𝑚−2 𝑠 −1 ≈ 0.8 𝑚𝐽 𝑐𝑚−2 𝑠 −1 , which is reached close to ±2 𝜇𝑚 from the
center of the image.
2. You have an n-type wafer with 5 x 1016 As atoms/cm3 and you decide to begin to fabricate a
MOSFET. After growing your oxide layer, you pattern and etch the wafer to open a window in the
oxide to make your MOSFET. You decide to use a boron oxide containing spun-on glass to dope
your wafer with boron. (20 points)
a. What is the surface concentration of boron in the Si at 1000°C? Describe in one sentence
how you determined this number. (2 points)
Assume solid solubility limit; from solubility curve, the Boron concentration at the surface
of Silicon at 1000 ℃ is 𝐶𝑠 = 1020 𝑐𝑚−3.
b. How long should you heat the wafer with borofilm on top at 1000°C to diffuse in a total
dose of 3 x 1014 boron atoms/cm2? (5 points)
The total dose at a given time, 𝑄(𝑡), (i.e. the total number of dopant per unit area) is
obtained by integrating the concentration over the entire thickness of the sample:
∞
𝑄(𝑡) = ∫ 𝐶(𝑥, 𝑡) 𝑑𝑥
0
For constant surface concentration boundary conditions, 𝐶(𝑥, 𝑡) = 𝐶𝑠 erfc(𝑥/2√𝐷𝑡) .
Using the table of error function algebra,
2
𝑄(𝑡) =
𝐶𝑠 √𝐷𝑡
√𝜋
We could have also obtained this result from the diffusion length √𝐷𝑡, since this length
gives one a measure of how far into the crystal the dopant has diffused. Assuming a
constant dopant concentration of 𝐶𝑠 , we obtain 𝑄(𝑡) ≈ 𝐶𝑠 √𝐷𝑡 . With 𝐷 = 2 ×
10−14 𝑐𝑚2 𝑠 −1 and solving for time, we obtain
𝑡𝑑𝑜𝑠𝑒 =
(3 × 1014 )2
𝜋𝑄 2
=
3.14
⋅
= 353 𝑠
4 ⋅ 2 × 10−14 (1020 )2
4𝐷𝐶𝑠2
c. What is the junction depth after heating the wafer at 1000°C for your calculated time? (5
points)
The junction depth, 𝑥𝑗 , is given by the depth in the crystal where the dopant concentration
is equal to the background doping concentration, in this case 𝐶(𝑥𝑗 , 𝑡𝑑𝑜𝑠𝑒 ) = 5 × 1016 As
atoms/cm3. Then solving for 𝑥𝑗 ,
𝑥𝑗 = 2√𝐷𝑡𝑑𝑜𝑠𝑒 erf𝑐 −1 (
1016
𝐶(𝑥𝑗 , 𝑡𝑑𝑜𝑠𝑒 )
)
𝐶𝑠
The 𝑒𝑟𝑓𝑐 −1 (5 × 1020 ) = 𝑒𝑟𝑓𝑐 −1 (5 × 10−4 ) = 2.8 from the reference data graph for
erfc(𝑥). So,
𝑥𝑗 = 2√4 × 10−14 × 353 × 2.8 = 210 𝑛𝑚
Note that 𝑥𝑗 is roughly the diffusion length.
d. What assumption regarding the diffusion coefficient did you make to perform the
calculations in (b)? Answer in one sentence. (2 points)
We assume that it is independent of the concentration.
e. If you had not made any assumptions how would this change your result? Answer in one
sentence. (2 points)
The diffusion coefficient increases with concentration and results in box-like, sharpened
profiles.
f.
How would changing the temperature affect the junction depth (2 points)?
Increasing the temperature would increase the diffusion coefficient of the dopant, which
would increase the junction depth for a given amount of time. Decreasing the temperature
would have the opposite effect.
g. The diffusion window you defined was 10−2 𝜇𝑚2 in area. What is the approximate lateral
size of the diffusion profile after you achieve the required doping dose? (2 points)
The lateral diffusion distance is about 3/4 ⋅ √𝐷𝑡𝑑𝑜𝑠𝑒 = 28.2 𝑛𝑚. Assuming a square
doping window with width 100 𝑛𝑚 , the width of the boron doped region would be
156 𝑛𝑚.
3. After completing the above processing, you decide to do a drive-in diffusion of the deposited
boron at 1100°C. (20 points)
a. What is the junction depth and surface concentration of boron in the Si after a 3 hr drivein? (8 points)
In this constant-total-dopant dose problem, we’ll assume that the dopant is concentrated
very near the surface, which is a good approximation given that the diffusion length from
problem 2 was 38 nm. Then the doping profile is given by
𝐶(𝑥, 𝑡 ) =
𝑆
√𝜋𝐷𝑡
exp (−
𝑥2
)
4𝐷𝑡
where 𝑆 = 3 × 1014 𝑐𝑚−2 from above. Also,
𝐷 = 𝐷(1100℃) = 2 × 10−13 𝑐𝑚2 𝑠 −1
𝑠
𝑡𝑑𝑟𝑖𝑣𝑒 = 3 ℎ𝑟 ⋅ 3600 ℎ𝑟 = 10800 𝑠
1
𝐶(𝑥𝑗 , 𝑡𝑑𝑜𝑠𝑒 ) = 5 × 1016 𝑐𝑚−3
from Problem 2. Then, 𝑥𝑗 is given by
𝑆
1/2
𝑥𝑗 = (4Dt drive ⋅ ln (
))
𝐶(𝑥𝑗 , 𝑡𝑑𝑜𝑠𝑒 )√𝜋𝐷𝑡𝑑𝑟𝑖𝑣𝑒
𝑥𝑗 = 1.92 𝜇𝑚
The surface concentration is
𝑆
√𝜋𝐷𝑡𝑑𝑟𝑖𝑣𝑒
= 3.64 × 1018 𝐵/𝑐𝑚3
Note that 𝑥𝑗 is roughly the diffusion length here as well!
b. Graph the B impurity concentration versus the distance from the surface that you have
before drive in diffusion. Quantitatively label the junction depth and 𝐶𝑠 . (5 points)
See below
c. Graph the B impurity concentration versus the distance from the surface that you have
after drive in diffusion. Quantitatively label the junction depth and 𝐶𝑠 . (5 points)
d. Explain any differences in the graphs using your understanding of the basic physical
processes occurring. (2 points)
When the dopant is initially introduce into the crystal, the temperature is lower so that the
diffusion is slower leading to a smaller 𝑥𝑗 . During the higher temperature drive in
process, the diffusion is faster and the dopant gets smeared out into the crystal while the
surface concentration gets depleted.
Reference Data and Formulas:
𝑁𝑎 = 6.02 × 1023
𝑘𝑏 = 8.61 × 10−5 𝑒𝑉/𝐾
ℎ ⋅ 𝑐 = 1240 𝑒𝑉 ⋅ 𝑛𝑚
Density of Si: 2.33 g cm-3
Density of SiO2: 2.21 g / cm-3
Density of GaAs: 5.32 g cm-3
The following data is for Silicon:
𝑥
𝐶(𝑥, 𝑡) = 𝐶𝑠 erfc (
)
2√𝐷𝑡
𝐶(𝑥, 𝑡) =
𝑆
√𝜋𝐷𝑡
exp (
−𝑥 2
)
4𝐷𝑡