Physics 610/Chemistry 678 Semiconductor Processing and Characterization Quiz I July 18, 2014

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1 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, k 0, 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, k 0 > 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 k 0 value. If k 0 < 1, then the zone gets increasingly dopant-rich; the opposite is true for k 0 > 1. The float-zone method can clean wafers if k 0 < 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.

2 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, k e, 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 As atoms/cm 3 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) k 0 = C s C l = for As The concentration of dopant in the solid phase of silicon is given by C s = k 0 C 0 (1 M M 0 ) k 0 1. M 0 = 2M ingot We want a concentration of C s = atoms of As/cm 3 halfway through the final grown ingot, i.e. when M M 0 = 1/4. The thee initial concentration of As atoms in the melt should be, C 0 = C 0 = C s k 0 (1 M k 0 1 ) M cm 3 0.3(1 0.25) 0.3 1

3 C 0 = As atoms/cm 3 b. Calculate the mass of the silicon charge (M 0). (5 points) Final mass of ingot = M ingot = Volume density = A h ρ Si = (πr 2 h)ρ, and M 0 = 2M ingot. With r = 15 cm, h = 100 cm, and ρ = 2.33 g/cm 3, M 0 = kg c. Calculate how many grams of As should be added to the silicon charge? (5 points) mass As = C 0 2A h MW Ar /N A Where MW Ar = g/mol is the molecular weight of arsenic, and N A = atoms/mole is Avogadro s number. We have, mass Ar = 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 C 1/2 = atoms/cm 3 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 k e instead of k 0 in the formula for C s (M).

4 2. You have a n-type wafer with 5 x As atoms/cm 3 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) = x2 + Ax B The rate growth coefficients defined by A and B, and τ are given in the reference data: τ 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. t wet (300nm) = hr = 8.75 min t dry (300nm) = 2.24 hr = 134 min. b. How long does it take to grow the second 300 nm? (5 points) t wet (600nm) t wet (300nm) = hr = 23.8 min t dry (600nm) t dry (300nm) = 6.27 hr = 376 min c. How long does it take to grow the third 300 nm? (5 points)

5 t wet (900nm) t wet (600nm) = hr = 38.8 min t dry (900nm) t dry (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 C m that is then cooled below the liquidus line. Find the solid:liquid weight ratio in terms of C m, and the concentration of the dopant (i.e. excess As or Ga) in the solid, C s, and liquid, C l. 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 M l be the weight of the liquid and M s 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, or So, (M l + M s )C m = M l C l + M s C s M l (C m C l ) = M s (C s C m ) M s M l = C m C l C s C m

6 The following graph only gives information about atomic percent, but the above equation used compositions, C m, C s, and C l, 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. C l and C s are given the intersections horizontal lines at a fixed temperature with the liquidus and solidus curves, respectively, and C m by the vertical line (at 25 atomic %) From the graph, we see that M s M l (400 C) = 1 and M s M l (800 C) = 2 3

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