KOREA UNIVERSITY. Ch. 4 Basics of device fabrication Reference: S. M. Sze, Semiconductor Devices, ISBN Photonics Laboratory

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1 Ch. 4 Basics of device fabrication Reference: S.. Sze, Semiconductor Devices, ISBN : We will understand the common techniques for growing single crystals. (a) The starting materials (e.g., silicon dioxide for a silicon wafer) are chemically processed to form a high-purity polycrystalline semiconductor from which single crystals are grown. (b) The single-crystal ingots are shaped to define the diameter of the material and sawed into wafers. (c) These wafers are etched and polished to provide smooth, specular surfaces on which devices will be made. A technology closely related to crystal growth involves the growth of single-crystal semiconductor layers upon a single-crystal semiconductor substrate. This is called epitaxy. (1) Crystal growth from the melt There are basically two techniques for crystal growth from the melt: the Czochralski technique and the Bridgman technique. A substantial percentage (~9%) of the silicon crystals for the semiconductor industry are prepare by the Czochralski technique; virtually all the silicon used for fabricating integrated circuits is prepared by this technique. ost gallium arsenide, on the other hand, is grown by the Bridgman technique. However, the Czochralski technique is becoming more popular for the growth of large-diameter gallium arsenide. 1

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3 () Starting materials The starting material for silicon is a relatively pure form of sand (SiO ) called quartzite. This is placed in a furnace with various forms of carbon (coal, coke, and wood chips). This process produces metallurgical-grade silicon with a purity of about 98%. SiC ( solid ) SiO ( solid ) Si( solid ) SiO( gas) CO( gas). Next, the silicon is pulverized and treated with hydrogen chloride (HCl) to form trichlorosilane (SiHCl 3 ): o 3 C Si( solid ) 3HCl( solid ) SiHCl3( gas) H( gas). The trichlorosilane is a liquid at room temperature (boiling point 3 o C). Fractional distillation of the liquid removes the unwanted impurities. The purified SiHCl 3 is then used in a hydrogen reduction reaction to prepare the electronic-grade silicon (EGS): SiHCl 3( gas) H( gas) Si( solid ) 3HCl( gas). This reaction takes place in a reactor containing a resistance-heated silicon rod, which serves as the nucleation point for the deposition of silicon. The EGS, a polycrystalline material of high purity, is the raw material used to prepare device quality, single-crystal silicon. Pure EGS generally has impurity concentrations in the parts-per-billion range. 3

4 (3) The Czochralski technique The Czochralski technique for silicon crystal growth uses an apparatus called a puller. The puller has three main components: (1) a furnace, which includes a fused-dilica (SiO ) crucible, a graphite susceptor, a rotation mechanism, a heating element, and a power supply; () a crystalpulling mechanism, which includes a seed holder and a rotation mechanism; and (3) an ambient control, which includes a gas source, a flow control, and a exhaust system. In the crystal-growing process, polycrystalline silicon is placed in the crucible and the furnace is heated above the melting temperature of silicon. A suitably oriented seed crystal (e.e., <111>) is suspended over the crucible in a seed holder. The seed is inserted into the melt. Part of it melts, but the tip of the remaining seed crystal still touches the liquid surface. It is then slowly withdrawn. Progressive freezing at the solid-liquid interface yields a large single crystal. A typical pull rate is a few millimeters per minute. 4

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6 (4) Unitary diagrams These diagrams show the phase change in a single element as a function of temperature and pressure. The common point, referred to as a triple point, is invariant for the system and defines the temperature and pressure at which solid, liquid, and gaseous phases are all in equilibrium with one another. 6

7 (5) Binary diagrams These phase diagrams show the relationship between two components as a funciton of temperature. The second variable, pressure, is usually set at 1 atm. In this way a relatively complex three-dimensional representation is avoided. 7

8 (6) The lever rule At any temperature, the equilibrium composition of the two single phases that make up a twophase region may be determined as follows. Consider a melt of initial composition C (the percentage weight of B in the melt). Let this melt be cooled from some temperature T 1 to a tempeerature T, corresponding to a point in the two-phase region: W L =weight of liquid at this temperature W s =weight of solid (in the β phase, for this example) C L, C s =composition of the liquid and solid, respectively (percentage amount of B by weight) Then W L C L and W s C s are the weights of B in the liquid and solid, respectively. But total weight of B is (W L +W s )C. Hence by the conservation of matter W W s L C C s C C L l s where l and s are the length of the twolines measured from the C the two- phase region.this is known as the lever rule. ordinate to the boundaries of 8

9 (7) The phase rule The correct interpretation of phase diagrams is greatly helped by knowledge of the phase rule. This rule, which is based on thermodynamic consideration, states that for any system in thermal equilibrium, the sum of the number of phases P and the number of degrees of freedom F is related to the number of components C by P + F = C +. Here the degrees of freedom are the number of variables that can be independently changed while still preserving a specific phase. Ex) For a single-component diagram of the type shown in Fig..1 P + F = 3 (a) For water in its liquid phase (P=1), F=; we have degrees of freedom, i.e., both pressure and temperature can be independently changed, and still maintain water in this liquid phase. (b) Along B, however, P= (liquid and vapor) so that we have 1 degree of freedom (F=1). Now either pressure or temperature (but not both) can be independently varied while this two-phase coexistence is still preserved. (c) At, P=3, and F=. i.e., there is a unique temperature-pressure combination where water coexists in all three phases. For a binary phase diagram, F + P = C + = 4. 9

10 (8) Distribution of dopant In crystal growth, a known amount of dopant is added to the melt to obtain the desired doping concentration in the grown crystal. For silicon, boron and phosphorus are the most common dopants for p- and n-type materials, respectively. As a crystal is pulled from the melt, the doping concentration incorporated into the crystal (solid) is usually different from the doping concentration of the melt (liquid) at the interface. The ratio of these two concentrations is defined as the equilibrium segregation coefficient k : k c c s l where C s and C l are respectively the equilibrium concentrations of the dopant in the solid and liquid near the interface. Consider a crystal being grown from a melt having an initial weight with an initial doping concentration C in the melt (i.e., the weight of the dopant per 1 gram melt). A given point of growth when a crystal of weight has been grown, the amount of dopant remaining in the melt (by weight) is S. 1

11 1 ) (1 ) (1 / 1 / 1 ) ( ) (1 ) ln(1 ln ln ) ( k s k s s l k S C l s l s C k C C k C C k C C C C S C S k k C S d k S ds d k C C d S ds S C C d ds where C s and C l are respectively the equilibrium concentrations of the dopant in the solid and liquid near the interface. Consider a crystal being grown from a melt having an initial weight with an initial doping concentration C in the melt (i.e., the weight of the dopant per 1 gram melt). A given point of growth when a crystal of weight has been grown, the amount of dopant remaining in the melt (by weight) is S. 11

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13 Problem 1) A silicon ingot, which should contain 1 16 boron atoms/cm 3, is to be grown by the Czochralski technique. What concentration of boron atoms should be in the melt to give the required concentration in the ingot? If the initial load of silicon in the crucible is 6 kg, how many grams boron (atomic weight 1.8) should be added? (9) Wafer shaping and material characterization After a crystal is grown, (a) the first shaping operation is to remove the seed and the other end of the ingot, which is last to solidify. (b) The next operation is to grind the surface so that the diameter of the material is defined. (c) After that, one or more flat regions are ground along the length of the ingot. These regions, or flats, mark the specific crystal orientation of the ingot and the conductivity type of the material. The largest flat, the primary flat, allows a mechanical locator in automatic processing equipment to position the wafer and to orient devices relative to the crystal in a specific manner. Other smaller flats, called secondary flats, are ground to identify the orientation and conductivity type of the crystal. 13

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16 1) Once these operations have been done, the ingot is ready to be sliced by diamond saw into wafers. Slicing determines four wafer parameters: surface orientation, thickness, taper (i.e., wafer thickness variations from one end to another), and bow (i.e., surface curvature of the wafer, measured from the center of the wafer to its edge). ) After slicing, both sides of the wafer are lapped using a mixture of Al O 3 and glycerine to produce a typical flatness uniformity within μm. The lapping operation usually leaves the surface and edges of the wafer damaged and contaminated. The damaged and contaminated regions can be removed by chemical etching. The final step of wafer shaping is polishing. Its purpose is to provide a smooth, specular surface where device features can be defined by lithographic processes. 3) The oxygen and carbon concentrations are substantially higher in Czochralski crystals than in float zone crystals due to the dissolution of the silica crucible (for oxygen) and transport to the melt from the graphite susceptor (carbon) during crystal growth. Especially, the precipitates of oxygen due to the solubility effect, can be used for gettering. Gettering is a general term meaning a process that removes harmful impurities or defects from the region in a wafer where devices are fabricated. 16

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19 (1) Vapor-phase epitaxy Epitaxial processes are differentiated from the melt growth processes in that the epitaxial layer can be grown at a temperature substantially below the melting point (typically 3 to 5% lower). Four silicon sources have been used for vapor phase epitaxial growth. They are silicon tetrachloride (SiCl4), dichlorosiliane (SiHCl), trichlorosilane (SiHCl3), and silane (SiH4). Silicon tetrachloride has been the most studied and has the widest industrial use. The typical reaction temperature is 1 o C. SiCl SiCl 4 4 ( gas) H ( gas) ( gas) Si( solid ) Si( solid ) 4HCl( gas) SiCl ( gas) 19

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21 The dopant is introduced at the same time as the silicon tetrachloride during epitaxial growth. Gaseous diborane (B H 6 ) is used as the p-type dopant, while phosphine (PH 3 ) and arsine (AsH 3 ) are used as n-type dopants. Gas mixtures are ordinarily used with hydrogen as the diluent to allow reasonable control of flow rates for the desired doping concentration. The dopant chemistry shows dopant being adsorbed on the surface, decomposing, and being incorporated into the growing layer. 1

22 (11) olecular-beam epitaxy (BE) BE is an epitaxial process involving the reaction of one or more thermal beams of atoms or molecules with a crystalline surface under ultrahigh vacuum conditions(~1-1 Torr). BE can achieve precise control in both chemical compositions and doping profiles. Single crystal multilayer structures with dimensions of the order of atomic layers can be made using BE. The ideal gas law states that PV RT N kt where P is the pressure,v is the volume of one mole of gas, R is the gas constant (1.98 cal/mol - K, or 8 atm - cm T is the absolute temperature in K, N Boltzmann constant (1.381 can use theideal law to calculate the molecular concentration n (the number of N A P 18 P 3 n molecules/cm V kt T where P is in Torr.The density of a gas is given by theproduct of its molecular weight and its concentration : d molecular weight The diffusivities of gases at room temperature are on the order of 1cm /s and increase with absolute temperature approximately as T A. P kt -3. the Avogadro constant (6.1 molecules/mole), and k is the J/K).Since real gases behave more and more like the ideal gas as the pressureis lowered, we d A 3 molecules per unit volume) : 3 /mole - K),

23 The gas molecules are in constant motion and their velocities are temperature dependent.the distribution of is described by the axwell - Boltzmann distributin law, 1 dn 4 m mv f exp v v n dv kt kt where m is the mass of a molecule. If there are n molecules in the volume, there will be dn molecules having a speed between v and v dv. The average speed is v av An important kt. m parameter for vacuum technology is the molecular impingemen t rate, that is, on a unit area per unit time. 1/ 1 dn x m mv exp. x fv v x x n dv x kt kt The molecular impingemen t rate is given by v x vfvdv f dv v dn x where P is the pressurein Torr and is the molecular weight. Therefore, at 3K and 1 theimpingemen t rate is / m kt vxn 1/ v x 14 mv exp x dv kt molecules/cm x n which states that for a given speed v, kt m P(mkT) -s for oxygen ( 3). 1/ how many molecules impinge P T Torr pressure, velocities 3

24 -6 Ex) Find the time required to form a monolayer of oxygen at pressuresof 1,1, and 1 The effictive molecular diameter of oxygen is 3.64 A. Assuming close packing, the number N of mlecules per unit area for oxygen is 8.61 The time required to form a monolayer (assuming 1% sticking) is obtained from theimpingemen t rate: t t Ns N s s at 1 6 hr at mkt P 6 Torr Torr s at 1Torr 14 cm o Torr. s 4

25 Another important parameter is the mean free path. During their motion, the molecules suffer collisions among themselves. The average distance traversed by all the molecules between successive collisions with each other is defined as the mean free path. A molecule having a diameter d and a velocity v will move a distance vδt in the time δt. The molecule suffers a collision with another molecule if its center is anywhere within the distance d of the center of another molecule. Therefore, it sweeps out (without collision) a cylinder of diameter d. The volume of the cylinder is V (d) vt Since there are n molecules/cm, the volume associated with one molecule is on the average 1/n cm. When the volume V is equal to1/ n, it must contain on the average a collision has occurred. If t is the average 1 d v n and the mean free path is then 1 kt P v n nd d dt A more rigorous derivation gives kt Pd time between collisions, one other molecule; thus, 3 51 o cm for air molecules (equivalent molecular diameter of 3.7 A) at room temerature. P( in Torr) 5

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27 Problem. Assume an effusion oven geometry of area A=5 cm and a distance L between the top of the oven and the gallium arsenide substrate of 1 cm. Calculate the BE growth rate for the effusion oven filled with gallium arsenide at 9 o C. 6

28 1) Thermal oxidation Semiconductors can be oxidized by various methods. These include thermal oxidation, electrochemical anodization, and plasma reaction. Among these methods thermal oxidation is by far the most important for silicon devices. It is the key process in modern silicon integrated-circuit technology. The silicon-silicon dioxide interface moves into the silicon during the oxidation process. The following chemical reactions describe the thermal oxidation of Si(solid) O ( gas) SiO Si( solid ) H O( gas) SiO ( solid ) ( solid ) H ( gas). silicon in oxygen or water vapor: Problem 3) If a silicon oxide layer of thickness x is grown from the thermal oxidation, what is the thickness of silicon being consumed?

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31 13) etallization etallization refers to the formation of metal films used for interconnections, ohmic contacts, and rectifying metal-semiconductor contacts. etal films can be formed by various methods, the most important being physical vapor deposition and chemical vapor deposition. We shall now consider the thickness of the film deposited from an evaporation source.

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33 Consider a small sphereds1 evaporation material in all directions at a rate of m (g/s). Such an evaporating source is called a point source. The amount of m dm d. 4 A related case is the evaporation from a plane source of area ds1 from which the material is evaporated from one side at a rate ofm (g/s). The amount of material passing through a solid angle d in a direction forming an angle with thenormal to the surface per unit time is given by thecosine law : m dm cos d. If the material arrives at a small area ds on a surface whose normal is inclined at an angle to the direction of the vapor stream, d ds cos. r material passing through a solid angle d in any direction per unit time is then

34 m cos dm for the point source and ds 4 r m cos cos dm for the plane source. ds r Assuming the material has a density (g/cm the volume of material deposited on ds dm lds The thickness of the film at locations corresponding to the area ds m l 4 cos r mh l 4 ( H L ) mh l ( H L m cos l d Especially, d d d d. for the point source and cos for the plane source. r for discrete devices, a practical arrangemen t of ) 3/ m 4 H m H d d d 1 [1 (L/H)] 1 [1 (L/H) must be lds ] 3 3/ ) and the thickness of the film deposited per unit time is l (cm/s),. for the point source and for the plane source. is then given by thesource and a parallel - plane receiving surface,

35 Problem 4) A silicon wafer is placed at a perpendicular distance of 3cm from a plane source. If the total deposited mass is 1 g and the density is.7 g/cm 3, what is the film thickness at L=? If the variation in film thickness must be less than 1%, how large can the wafer be? Aluminum etallization Aluminum and its alloy are used extensively for metallization in integrated circuits. Because aluminum and its alloys have low resistivities (i.e.,.7μω-cm for Al and up to 3.5 μω-cm its alloys), these metals satisfy the requirements of low resistance. However, the use of aluminum in integrated circuits at shallow junctions often creates problems such as spiking and eletromigration. Hence, during annealing silicon will dissolve into the aluminum. The amount of silicon dissolved will depend not only on the solubility at the annealing temperature but also on the volume of aluminum to be saturated with silicon.

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37 Assuming Vol where S is the solubility of silicon in aluminum at theannealing uniformly over the contact area A, the depth towhich silicon would be consumed is b Al and Si Al Dt( HZ) S Si Si HZ Al Dt S. A are the densities of aluminum and silicon, the volume of silicon consumed is, temperature. If the consumption takes place Problem 5) For T=5 o C, t=3min, ZL=16μm, Z=5μm, and H=1μm. Find the depth b, assuming uniform dissolution.