CHAPTER 2. Structural Issues of Semiconductors

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1 CHAPTER 2 Structural Issues of Semiconductors OUTLINE 1.0 Energy & Packing 2.0 Materials & Packing 3.0 Crystal Structures 4.0 Theoretical Density, r 5.0.Polymorphism and Allotropy 6.0 Close - Packed Crystal Structures 7.0 Structure of Compounds: NaCl 8.0 X-Ray Diffraction From Crystals 9.0 Crystallographic Points, Directions and Planes: Miller Indices 2 1

2 1.0 Energy & Packing i. Non-Dense Random Packing Energy typical neighbor bond length typical neighbor bond energy r ii. Dense Regular Packing Energy typical neighbor bond length typical neighbor bond energy r Dense, regular-packed structures tend to have lower energy Materials & Packing 1. Crystalline materials... Si the atoms pack in periodic, 3D arrays in a long range dense, regular packing typical of: -metals -many ceramics -some polymers 2. Noncrystalline materials... "Amorphous" atoms have no periodic packing in long range occurs for: -complex structures -rapid cooling Oxygen crystalline SiO2 noncrystalline SiO2 Adapted from Fig. 3.18(b),Callister 6e. 4 2

3 Examples Crystalline materials: Quartz(SiO2) 4- Si0 4 tetrahedron Si 4+ O 2- Basic Unit 5 Examples Noncrystalline materials(amorphous): Glass Amorphous structure occurs by adding impurities (Na +,Mg 2+,Ca 2+, Al 3+ ) Impurities: interfere with formation of crystalline structure. Na + Si 4+ O 2- (soda glass) 6 3

4 Type of Crystals: 1. Single Crystals: When the regular arrangement of atoms is perfect and extends throughout the specimen without interruption, the result is single crystal. They are difficult to grow. GRAPHITE 7 Some engineering applications require single crystals: 1)diamond single crystals for abrasives 3)silicon single crystals are grown for quantum electronics 2)turbine blades Fig. 8.30(c), Callister 6e. 8 4

5 2. Polycrystalline Minerals and Materials A collection of many small crystals or grains Random orientations 9 Polycrystalline Minerals and Materials 10 5

6 Most engineering materials are polycrystalline. 1 mm Callister 6e. Nb-Hf-W plate with an electron beam weld. Each "grain" is a single crystal. If crystals are randomly oriented, overall component properties are not directional. Crystal sizes range from 1 nm to 2 cm (i.e., from a few atoms to millions of atomic layers). 11 Single vs Polycrystals Single Crystals 1. Properties vary with direction: anisotropic Example: the modulus of elasticity (E) in BCC iron: E (diagonal) = 273 GPa Polycrystals 1. Properties may/may not vary with direction. 1. If grains are randomly oriented: isotropic (E poly iron = 210 GPa) 2. If grains are textured:anisotropic 200 mm E (edge) = 125 GPa Data from Table 3.3, Callister 6e. Adapted from Fig. 4.12(b), Callister 6e. 12 6

7 3.0 Crystal Structures Diffraction pattern of electron beams produced for a single crystal of Ga As (Gallium Arsenide) using Transmission Electron Microscope 13 Definitions about Crystal Structures Lattice: 3D array of regularly spaced points (a) Hard sphere representation: atoms denoted by hard, touching spheres (b) Reduced sphere representation (c)unit cell: basic building block unit (such as a flooring tile) that repeats in space to create the crystal structure 14 7

8 Definition of Crystal Systems Lattice Parameters: The unit cell geometry is defined in terms of six parameters. the three edge lengths a, b, c, and the three interaxial angels α, β, γ Seven possible combinations of a, b, c & α, β, γ, resulting in seven crystal systems 15 Crystal Systems 16 8

9 Atomic Packing Factor APF = Volume of atoms in unit cell* Volume of unit cell *assume hard spheres Coordination Number The number of closest atoms to a selected atom in a unit cell 17 Cubic Crystal Structures 1. Simple Cubic (SC) 2. Face Centered Cubic ( FCC ) 3. Body Centered Cubic ( BCC ) 18 9

10 Simple Cubic Structure (SC) Coordination # = 6(# nearest neighbors) Rare due to poor packing (only Po has this structure) Close-packed directions are cube edges. Click on image to animate (Courtesy P.M. Anderson) 19 Volume of Atoms In Unit Cell(SC) Step1: R=0.5a Step2: Volume of unit cell= a3 a R=0.5a Step3: Volume of atom = 4 3 π (0.5a) 3 close-packed directions Step4: atoms unit cell = 8 x 1/8 = 1 atom/unit cell Fig. 3.19,Callister 6e

11 Atomic Packing Factor: SC a R=0.5a Volume of atoms in unit cell* APF = Volume of unit cell *assume hard spheres atoms unit cell APF = 1 volume 4 atom 3 π (0.5a)3 a 3 volume unit cell APF for a simple cubic structure = Face Centered Cubic Structure (FCC) Close packed directions are face diagonals. Coordination # = 12 Click on image to animate Note: All atoms are identical; the facecentered atoms are shaded differently only for ease of viewing. This structure is exhibited by metals Ag, Al, Au, Cu, Ir, Ni, Pb, Pd, Pt, and Rh, as well as the noble gases Ar, Kr, Ne, and Xe. This arrangement (but without actual contact by atoms of the same element) is also assumed by the anions (or, less often, the cations) in numerous ionic compounds

12 Atomic Packing Factor: FCC Close-packed directions length = 4R = 2 a Unit cell c ontains: 6 x 1/2 + 8 x 1/8 = 4 atoms/unit cell Adapted from Fig. 3.1(a), Callister 6e. atoms unit cell APF = π ( 2a/4)3 a 3 volume unit cell volume atom APF for a body-centered cubic structure = Body Centered Cubic Structure (BCC) Close packed directions are cube diagonals. Coordination # = 8 Click on image to animate The metals Ba, Cr, Cs, α-fe, K, Li, Mo, Na, Rb, Ta, V, and W adopt this structure. If the atom in the center of the cube is different than those at the corners, the cesium chloride structure results

13 Atomic Packing Factor: BCC Close-packed directions length = 4R = 3 a Unit cell contains: x 1/8 = 2 atoms/unit cell atoms unit cell APF = π ( 3a/4)3 a 3 volume unit cell volume atom APF for a body-centered cubic structure = Theoretical Density, ρ Density = mass/volume mass = (number of atoms per unit cell) x (mass of each atom) mass of each atom = Molar Mass /Avogadro s number # atoms/unit cell Atomic Molar Mass weight (g/mol) Volume/unit cell (cm 3 /unit cell) ρ = n A V c N A Avogadro's number (6.023 x atoms/mol) 26 13

14 Theoretical Density, ρ # atoms/unit cell Atomic Molar Mass weight (g/mol) Volume/unit cell (cm 3 /unit cell) ρ = n A V c N A Avogadro's number (6.023 x atoms/mol) Example: Copper crystal structure = FCC: 4 atoms/unit cell molar mass = g/mol (1 amu = 1 g/mol) -7 atomic radius R = nm (1 nm = 10 cm) Vc = a 3 ; For FCC, a = 4R/ 2 ; Vc = 4.75 x cm 3 Compare to actual:ρcu = 8.94 g/cm 3 Result: theoretical ρcu = 8.89 g/cm 3 27 Characteristics of Selected Elements at 20 0 C Element Aluminum Argon Barium Beryllium Boron Bromine Cadmium Calcium Carbon Cesium Chlorine Chromium Cobalt Copper Flourine Gallium Germanium Gold Helium Hydrogen Symbol Al Ar Ba Be B Br Cd Ca C Cs Cl Cr Co Cu F Ga Ge Au He H Molarmass (g/mol) Density (g/cm 3 ) Crystal Structure FCC BCC HCP Rhomb HCP FCC Hex BCC BCC HCP FCC Ortho. Dia. cubic FCC Atomic radius (nm) Callister 6e

15 Densities of Material Classes ρ metals > ρ ceramics > ρ polymers Why? Metals have... close-packing (metallic bonding) large atomic mass Ceramics have... less dense packing (covalent bonding) often lighter elements Polymers have... poor packing (often amorphous) lighter elements (C,H,O) Composites have... ρ (g/cm 3 ) Metals/ Alloys Platinum Gold, W Tantalum Silver, Mo Cu,Ni Steels Tin, Zinc Graphite/ Ceramics/ Semicond Polymers Composites/ fibers Based on data in Table B1, Callister *GFRE, CFRE, & AFRE are Glass, Carbon, & Aramid Fiber-Reinforced Epoxy composites (values based on 60% volume fraction of aligned fibers in an epoxy matrix). Zirconia Titanium Al oxide Diamond Si nitride Aluminum Glass-soda Concrete Silicon PTFE Magnesium Graphite intermediate values Data from Table B1, Callister 6e. Silicone PVC PET PC HDPE, PS PP, LDPE Glass fibers GFRE* Carbon fibers CFRE* Aramid fibers AFRE* Wood Polymorphism and Allotropy Polymorphism Same compound occurring in more than one crystal structure. Crystal structure depends on Temperature and Pressure. Allotropy Polymorphism in elemental solids (e.g., sulphur,carbon, oxygen) HEATING AND COOLING OF AN IRON WIRE Demonstrates "polymorphism" Temperature, C Liquid BCC Stable The same atoms can have more than one crystal structure. 914 Tc 768 heat up FCC Stable BCC Stable cool down longer shorter! longer! magnet falls off shorter 30 15

16 Shape memory alloys materials that "remember" their original shape and return to it when heated, even if apparent residual deformation was introduced below a certain temperature. The original shape can be set easily by heat treatment. These alloys are used in orthopedic implants and other medical applications, as well as in air conditioners and other home appliances, automobiles, etc Close - Packed Crystal Structures Close packed planes of atoms:planes having a maximum atom density

17 Close - Packed Crystal Structures Close packed planes of atoms:planes having a maximum atom density. Compare with open pack Structure of Compounds: NaCl Compounds: Often have similar close-packed structures. Structure of NaCl Click on image to animate Click on image to animate 34 17

18 8.0 X-ray Diffraction From Crystals There are many experimental techniques that can be used to study the properties and structure of solids. These include: Diffraction based techniques. X-ray :X-rays interact with electrons in matter. X-rays are scattered by the electron clouds of atoms. Neutron Electron Microscopy. Thermal Analysis. Spectroscopy. 35 X-ray Diffraction From Crystals X Rays λ : nm ν:10 18 Hz (visible light Hz) W. H. Bragg (father) and William Lawrence Bragg (son) developed a simple relation for scattering angles, nowadays known as Bragg s law

19 X-ray Diffraction From Crystals nλ = SQ + QT nλ = d hkl sinθ + d hkl sinθ=2 d hkl sinθ nλ = 2 d hkl sinθ 37 X-ray Diffraction From Crystals 38 19

20 A Modern Diffractometer 39 Powder Diffraction Powder diffraction experiment requires only as small quantity of a mineral: mg Sample preparation is very simple and fast Reliable accurate results are obtained in a relatively short time:10 minutes to 2 hours. The ideal powder size is 5-10 microns

21 Applications of XRD Identification of unknowns. Phase purity. Determination of lattice parameters. Determination of crystal size. Variable temperature studies. Structure determination and refinement Crystallographic Points, Directions And Planes: Miller Indices 42 21

22 Crystallographic Directions It is a vector between two points. 1. The length of the vector projection on the three axis is determined as a, b, c. 2. These three numbers are multiplied or by divided by a common number to reduce to the smallest integers. 3. Three indices are written as [u, v, w] 4. Negative indices are represented by a bar over the index. 43 Crystallographic Directions And Planes: Miller Indices 44 22

23 Crystallographic Directions And Planes: Miller Indices 1. If the plane passes through the selected origin, either another parallel plane or a new origin must be established. 2. Find intersects of the plane with the three axis 3. Take the reciprocals. A plane that parallels and axis has an infinite intercept 4. If necessary find the smallest integers 5. Write the indices as (h k l) 45 Miller Indices of Planes:Examples 46 23

24 Miller Indices of Planes:Examples 47 FCC Unit Cell With (110) Plane 48 24

25 49 Problem 2.10 Below is a unit cell for a hypothetical metal. a) To which crystal system does this unit cell belong? b) What would this crystal structure be called? c) Calculate the density of the material, given that its molar mass is 141 g/mol. (a) The tetragonal crystal system since a = b= 0.35 nm, c = 0.45 nm, and α = β = γ = 90. (b) The crystal structure can be called body-centered tetragonal. (c) BCC, n = 2 atoms/unit cell. A=141g/mol V C = (3.5 x 10 8 cm)2(4.5 x 10 8 cm) =5.51 x cm 3 /unit cell ρ = na / V C N A = (2 atoms/unit cell)(141 g/mol) /(5.51 x cm 3 /unit cell )(6.023 x atoms/mol) ρ = 8.49 g/cm

26 Problem 2.11 What are the indices for the direction indicated by the two vectors in the sketch? For direction 1, the projection on the x-axis is zero (since it lies in the y-z plane), while projections on the y- and z-axes are b/2 and c, For direction 1 For direction 2 51 Problem about direction(callister 3.52,p.88) Determine the indices for the directions shown. Direction A : We position the origin of the coordinate system at the tail of the direction vector; then in terms of this new coordinate system [ -110] direction 52 26

27 Problem 2.13 What are the indices for the two planes drawn in the sketch below? Plane 1 :(020) Plane 2 : 53 Problem2.12 Sketch within a cubic unit cell the following planes: 54 27

28 Problem 2.15 Determine the Miller indices for the planes shown in the following unit cell. Plane A Plane B 55 Problem 2.16 Determine the Miller indices for the planes shown in the following unit cell

29 Problem 2.17 Determine the Miller indices for the planes shown in the following unit cell

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