Physics 610/Chemistry 678—Semiconductor Processing and Characterization
Quiz I—July 18, 2014
Part I: Short-answer questions on basic principles. (5 points each)
1. Briefly describe the CZ method and the Siemens process.
The CZ method grows single crystal ingots by melting an ultrapure raw material (e.g. Si and GaAs) and
then extracting and freezing the melt onto a single-crystal seed template, which is pulled out and away
from the melt.
The Siemens process forms electronic grade polycrystalline silicon (EGS) from silicate raw materials,
such as quartzite. The first steps are to react SiO2 with carbon to form 98% pure Si, which is then reacted
with gaseous HCl to form trichlorosilane. The trichlorosilane can be purified to a high degree by
fractional distillation. Finally, ultrapure trichlorosilane is reduced with hydrogen to form EGS.
2. Describe the Bridgman technique. What are challenges of growing GaAs?
The Bridgman technique uses a two zone furnace and a closed-vessel configuration to grow GaAs. One
zone heats As to create an overpressure of As in the vessel and the other zone melts GaAs and then slowly
cools it to form GaAs crystals. The challenges of growing GaAs are the high vapor pressure of As and
the complex nature of binary compounds.
3. A CZ grown crystal is doped with boron. Why is the boron concentration larger at the tail end of
the crystal than at the seed end? In a CZ growth, oxygen diffuses into the molten silicon from the
silica crucible used to contain it; will the concentration of oxygen in the crystal be larger at the
tail end or the seed end? Explain.
The boron concentration will be larger at the tail end because its equilibrium segregation coefficient, 𝑘0 ,
is less then unity, causing the growing crystal to expel boron into the melt and thereby increasing the
concentration of boron in the melt. Because the melt becomes increasingly boron-rich, the amount of
boron incorporated into the crystal increases as it grows. For oxygen, 𝑘0 > 1 and so the opposite is true
and the concentration of oxygen in the crystal decreases as the crystal grows.
4. Describe the Float-zone process for growing single-crystal Si. How does the concentration of a
dopant change in the liquefied zone during growth, assuming that the polycrystalline starting
material has a uniform dopant concentration. Explain how the Float-zone method can be used to
purify ingots of Si.
In the float-zone process, a segment or zone of a polycrystalline silicon rod is melted; this zone then
moves along the length of the rod and crystallizes silicon at its trailing edge. A single-crystal seed is used
at the base of the first zone to create a single-crystal ingot. The dopant concentration in the melted zone
can either increase or decrease during growth depending on its 𝑘0 value. If 𝑘0 < 1, then the zone gets
increasingly dopant-rich; the opposite is true for 𝑘0 > 1. The float-zone method can clean wafers if 𝑘0 <
1 because the melt zone will absorb more and more impurities as it moves along the rod length. Once the
zone reaches the end of the rod, it is allowed to cool and all the accumulated impurities will remain at this
end of the rod. This part of the ingot is then diced off and the float-zone technique is repeated to remove
further impurities with every cycle.
5. Explain how the float-zone and CZ methods can be used to produce uniform doping profiles in
ingots of Si.
In the CZ method, equilibrium doping concentrations are more appropriately determined by the effective
segregation coefficient, 𝑘𝑒 , which can approach unity if the crystal growth rate and the stagnation layer
depth are made large. One way of using the float-zone process to produce uniform doping profiles is to
first purify the ingot by float-zone cycling, and then use neutron irradiation to produce n-type Si by 𝛾 ⋅ 𝛽
decay fractional transmutation.
6. Draw the n-type and p-type wafers for the <111> and <100> surface orientations.
Part II:
1. You need a 300 mm diameter silicon ingot containing 5 x 1016 As atoms/cm3 and decide to use
the Czochralski technique to grow it. You decide to: (1) grow at equilibrium growth conditions,
(2) make the mass of the initial melt twice the mass of the final silicon ingot, (3) cut the wafers
you need out of the middle of the final single crystal ingot, and (4) grow the final ingot a total of
1 meter in length.
a. What concentration of As atoms should be in the melt? (5 points)
𝑘0 =
𝐶𝑠
𝐶𝑙
= 3 × 10−1 for As
𝑀 𝑘0 −1
The concentration of dopant in the solid phase of silicon is given by 𝐶𝑠 = 𝑘0 𝐶0 (1 − 𝑀 )
𝑀0 = 2𝑀𝑖𝑛𝑔𝑜𝑡
0
.
We want a concentration of 𝐶𝑠 = 5 × 1016 atoms of As/𝑐𝑚3 halfway through the final grown ingot, i.e.
𝑀
when
= 1/4. The thee initial concentration of As atoms in the melt should be,
𝑀0
𝐶𝑠
𝐶0 =
𝑀 𝑘0 −1
𝑘0 (1 − )
𝑀0
5 × 1016
𝐶0 =
𝑐𝑚−3
0.3(1 − 0.25)0.3−1
𝐶0 = 1.36 × 1017 𝐴𝑠 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚3
b. Calculate the mass of the silicon charge (M0). (5 points)
Final mass of ingot = 𝑀𝑖𝑛𝑔𝑜𝑡 = 𝑉𝑜𝑙𝑢𝑚𝑒 × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 𝐴 ⋅ ℎ ⋅ 𝜌𝑆𝑖 = (𝜋𝑟 2 ⋅ ℎ)𝜌, and 𝑀0 = 2𝑀𝑖𝑛𝑔𝑜𝑡 .
With 𝑟 = 15 𝑐𝑚, ℎ = 100 𝑐𝑚, and 𝜌 = 2.33 𝑔/𝑐𝑚3 ,
𝑀0 = 329.4 𝑘𝑔
c. Calculate how many grams of As should be added to the silicon charge? (5 points)
𝑚𝑎𝑠𝑠𝐴𝑠 = 𝐶0 ⋅ 2𝐴 ⋅ ℎ ⋅ 𝑀𝑊𝐴𝑟 /𝑁𝐴
Where 𝑀𝑊𝐴𝑟 = 74.92 𝑔/𝑚𝑜𝑙 is the molecular weight of arsenic, and 𝑁𝐴 = 6.022 × 1023 𝑎𝑡𝑜𝑚𝑠/𝑚𝑜𝑙𝑒
is Avogadro’s number. We have,
𝑚𝑎𝑠𝑠𝐴𝑟 = 2.40 𝑔
d. Graph the expected change in concentration of As atoms in the crystal as a function of the
length of the crystal grown. Points at 0, ½ and 1 in units of total crystal length are
sufficient. (5 points)
Here 𝐶1/2 = 5 × 1016 𝑎𝑡𝑜𝑚𝑠/𝑐𝑚3
e. You measure the As concentration in the crystal grown, and the measured values are
systematically larger than what you calculated. In one sentence, why? (5 points)
There will be a stagnant layer of dopant near the liquid/solid interface of the growing crystal. This forms
because dopants do not have time diffuse away from the interface. Therefore, we should be using 𝑘𝑒
instead of 𝑘0 in the formula for 𝐶𝑠 (𝑀).
2. You have a n-type wafer with 5 x 1016 As atoms/cm3 you decide to begin to fabricate a MOSFET.
Your first step is to grow a 0.9 micron thick oxide on the (100) Si wafer. You want to grow fast
as fast as possible so you use the maximum furnace temperature of 1200 °C.
a. How long does it take to grow the first 300 nm? (5 points)
The relationship between oxide thickness and growth time is
𝑡(𝑥) =
𝑥 2 + 𝐴𝑥
−𝜏
𝐵
The rate growth coefficients defined by 𝐴 and 𝐵, and 𝜏 are given in the reference data:
A 0.9 𝜇𝑚 oxide is more efficiently grown using a wet growth, but we’ll show the results for both wet and
dry processes.
𝑡𝑤𝑒𝑡 (300𝑛𝑚) = 0.146 ℎ𝑟 = 8.75 𝑚𝑖𝑛
𝑡𝑑𝑟𝑦 (300𝑛𝑚) = 2.24 ℎ𝑟 = 134 𝑚𝑖𝑛
.
b. How long does it take to grow the second 300 nm? (5 points)
𝑡𝑤𝑒𝑡 (600𝑛𝑚) − 𝑡𝑤𝑒𝑡 (300𝑛𝑚) = 0.396 ℎ𝑟 = 23.8 𝑚𝑖𝑛
𝑡𝑑𝑟𝑦 (600𝑛𝑚) − 𝑡𝑑𝑟𝑦 (300𝑛𝑚) = 6.27 ℎ𝑟 = 376 𝑚𝑖𝑛
c. How long does it take to grow the third 300 nm? (5 points)
𝑡𝑤𝑒𝑡 (900𝑛𝑚) − 𝑡𝑤𝑒𝑡 (600𝑛𝑚) = 0.646 ℎ𝑟 = 38.8 𝑚𝑖𝑛
𝑡𝑑𝑟𝑦 (900𝑛𝑚) − 𝑡𝑑𝑟𝑦 (600𝑛𝑚) = 10.3 ℎ𝑟 = 616 𝑚𝑖𝑛
d. Sketch a graph showing the thickness-dependence of the oxide growth time. (5 points)
e. Briefly explain the trend you observe in growth times for a-c in terms of a microscopic
picture of the growth process. (5 points)
The time required to grow the same thickness increases because the oxidant must diffuse through
the previously grown oxide.
3. Suppose a melt of GaAs has an initial composition (weight percent scale) of 𝐶𝑚 that is then
cooled below the liquidus line. Find the solid:liquid weight ratio in terms of 𝐶𝑚 , and the
concentration of the dopant (i.e. excess As or Ga) in the solid, 𝐶𝑠 , and liquid, 𝐶𝑙 . Find this ratio at
400°C and 800°C if the atomic percent of As in a melt of GaAs is 25%. (20 points)
Let 𝑀𝑙 be the weight of the liquid and 𝑀𝑠 be the weight of the GaAs solid. The amount of dopant in the initial
melt must be equal to the total amount of dopant found in final liquid and solid phases—conservation of mass.
Thus,
(𝑀𝑙 + 𝑀𝑠 )𝐶𝑚 = 𝑀𝑙 𝐶𝑙 + 𝑀𝑠 𝐶𝑠
or
𝑀𝑙 (𝐶𝑚 − 𝐶𝑙 ) = 𝑀𝑠 (𝐶𝑠 − 𝐶𝑚 )
So,
𝑀𝑠 𝐶𝑚 − 𝐶𝑙
=
𝑀𝑙 𝐶𝑠 − 𝐶𝑚
The following graph only gives information about atomic percent, but the above equation used compositions,
𝐶𝑚 , 𝐶𝑠 , and 𝐶𝑙 , measured in weight percent. However, the molecular weights of As and Ga only differ by 7%, so
we can approximate the atomic percent as weight percent. 𝐶𝑙 and 𝐶𝑠 are given the intersections horizontal lines at
a fixed temperature with the liquidus and solidus curves, respectively, and 𝐶𝑚 by the vertical line (at 25
atomic %) From the graph, we see that
𝑀𝑠
𝑀𝑙
(400°C) = 1
and
𝑀𝑠
𝑀𝑙
(800°C) =
2
3