lecture+3 material+properties

RESEARCH CENTRE FOR INTEGRATED MICROSYSTEMS - UNIVERSITY OF WINDSOR

06-88-552 ADVANCED TOPICS IN MICROELECTROMECHANICAL SYSTEMS (MEMS) WINTER, 2011

Research Centre for Integrated MicrosystemsUniversity of Windsor

Material Properties

Sazzadur Chowdhury, Ph.D.

Lecture 3

RESEARCH CENTRE FOR INTEGRATED MICROSYSTEMS - UNIVERSITY OF WINDSORRESEARCH CENTRE FOR INTEGRATED MICROSYSTEMS - UNIVERSITY OF WINDSOR


Types of Solids

Material classification according to atomic arrangements



Various Stages of Solidification of a Crystalline Material

Courtesy: Callister, Jr.

Monocrystal Polycrystal



Undoped Polysilicon

• The physical properties of undoped polysilicon depends on:– Type of deposition method, like LPCVD, PECVD, etc.– Nucleation– Growth of silicon grains– TDep<600 Amorphous– Tdep >600 Crystalline

Undoped polysilicon

Individual crystalsthat have nucleatedwith a randomorientation are called grains



Unit Cells

• In a crystalline material, the atoms are situated in a repeating or periodic three dimensional array over large atomic distances

• Lattice means a three dimensional array of points coinciding with atom positions

• The basic atomic pattern that repeats itself in a crystalline material is called a unit cell

• Properties of the crystalline solid depend on the crystalline structure of the material



Crystal Planes

• A set of three integers (h, k, l) are used to represent the orientation of an atomic plane in a cubic unit cell

• These are called Miller indices, (100), (010), (001)• Equivalent plane: {100}



Crystal Directions

• A direction in a lattice is expressed as a set of three integers with the same relationship as the components of a vector in that direction and expressed as [100].

• Equivalent directions are grouped and expressed as <100> directions.

• In a cubic lattice, a direction [hkl] is perpendicular to a plane (hkl)



Miller Indices Determination1. Determine the intercepts (or projection) of the plane along each of the

three crystallographic directions

2. Take the reciprocals of the intercepts

3. If fractions result, multiply each by the denominator of the smallest fraction

4. If a plane doesn’t have an intercept with an axis or parallel to a plane passing through the axis the intercept is taken as infinity.

5. If an intercept is on the negative branch of an axis, a minus sign is placed above the Miller index for that axis

a lattice constant



Miller Indices Determination: (100) Plane

Unit cell side length=a

001Cleared fraction

1/ 1/ 1/1Take reciprocal

1Intercept length

lkh

Miller Indice: (100)




011Cleared fraction

1/ 1/11/1Take reciprocal

11Intercept length

lkh

Miller Indices: (110)





111Cleared fraction

1/11/11/1Take reciprocal

111Intercept length

lkh



Miller Indices Determination: More Examples




Why Silicon

• Silicon is abundant, inexpensive, and can now be produced and processed controllably to unparalleled standards of purity and perfection

• Excellent mechanical characteristics• Definition and reproduction of the device shapes are performed using

photolithography process, a technology that has been matured over the years

• Batch fabrication capability• Intrinsic mechanical stability and feasibility of integrating electronics on

silicon makes it an excellent choice for a substrate for mechanical sensors and actuators



Silicon Crystal Structure: Diamond Lattice

• The Si crystal structure is known as diamond lattice• The diamond lattice consists of two interpenetrating fcc lattices,

one displaced 1/4 of a lattice constant in each direction from the other

• Each site is tetrahedrally coordinated with four other sites in the other sublattice

• Silicon unit cell has a side length, a of 5.4309 Å



Si Unit Cell

http://jas.eng.buffalo.edu/

Tetrahedral bond



Production of Polycrystalline Silicon

• Most common method of producing polycrystalline silicon is by the reduction of trichlorosilane using hydrogen and then thermal decomposition of silane

• Chemically pure SiHCl3 mixed with H2 are introduced into the large quartz glass bell jar in which silicon seed filaments are electrically heated to about 1200°C. The silicon compounds grow on the surface of silicon filaments.

SiHCl3 + H2 -> Si + 3HCl

• The rods grow radially over periods ranging from 10 to 30 days.



Single Crystal Silicon: Czochralski Method

• High purity polycrystalline silicon is melted down in a crucible using a graphite heating system.

• A small piece of solid silicon (the seed) is placed on the molten liquid in an inert argon gas atmosphere.

• As the seed is slowly rotated and pulled from the melt, the liquid cools.

• The surface tension between the seed and the molten silicon causes a small amount of the liquid to rise with the seed and cool into a single crystalline ingot with a specified diameter.

Courtesy: MEMC



Single Crystal Silicon Ingot

Courtesy: M. Madou



Single Crystal Silicon: Float Zone Method

• The FZ method for producing single crystal Si takes place in an inert gaseous atmosphere, keeping a polycrystalline rod and a seed crystal vertically face to face.

• Both are partially melted by high frequency inductive heating at the (molten zone) liquid phase.

• At the next step, this molten zone is gradually moved upwards rotating with the seed crystal until the entire polycrystalline rod has been converted to single crystal.

• Advantage there is no physical contact with a crucible

Courtesy: M. Madou



Wafer Preparation from Ingot

1. Grounding

2. Lapping

3. Etching

4. Edge beveling

5. Polishing

6. Cleaning

Steps:



Czochralski Method



Mechanical Properties of Single Crystal Silicon

• Specific strength (sy/r) better than most common engineering materials

• Knoop hardness • Silicon 850 kg/mm2

• steel 820 kg/mm2

• Quartz 800 kg/mm2

• Coefficient of thermal expansion• Si 2.5 10-6/(°C)-1

• Pyrex 3.3 10-6/(°C)-1

• Thermal conductivity• Si 141 W/m-K• Al 222 W/m-K• Epoxy 0.19 W/m-K



Thin Film vs. Bulk Solid

• The properties of a material are determined by the cooperative effect of a huge number of similar particles (atoms, lattice, etc.) in a 3D arrangement.

• Many physical properties of materials require a larger ensemble of atoms for a meaningful definition, independent of the amount of material, for example, density, the thermal expansion coefficient, hardness, color, electrical and thermal conductivity.

• With solid materials, the properties of surfaces may differ from the bulk conditions.

• In the classical case, the number of surface atoms and molecules is small compared with the number of bulk particles.

• This ratio is inverted in the case thin flims with dimensions in the range of a few microns or nanometers.

The thin film layer may be a few atoms or a few lattice constant thick

Bulk

Thin film or surface layer



Density Fluctuation in crystalline Thin Films

• Density is orientation dependent• In addition, the bond strength between the atoms is

localized and is determined from its orientation. • A classification of isotropic is justified as long as the

individual crystals are much smaller than the smallest dimension of a technical structure created by the material.

• If the dimensions of the deposited thin films are of comparable scale with the crystal lattice dimensions, they possess a high anisotropy even for a material with macroscopic isotropy.

• The length, strength and direction of the bonds as well as the number of bonds per atom in a material therefore determine the integral properties of the material and the spatial dependence of these properties.



Etch Rate at Different Planes

• In {100} and {110} planes, two bonds are directed back into crystal while two bonds “dangle”.

• In {111} planes, three bonds directed to the crystal and one dangle.• As a results, the {111} plane needs more chemical energy to be etched

compared to {100} or {110} plane.• However, not adequate explanations are available so far



Plane Intersections

tan()=L/a=1/√2

=arctan (1/√2)=35.26

Complementary angle=54.74

• {111} planes and {110} planes can intersect each other at 35.26, 90, and 144.74.

• {100} and {110} planes can intersect each other are at 45 or 90

• (100} and {111} planes can intersect each other at 54.73



Plane Intersections

Every {111} plane intersects a (100) surface along one of the <110> directions



Anisotropic Silicon Etch: (100) Plane

Proper alignment leads to {111} sidewalls, and (100) bottom surfaces



Anisotropic Silicon Etch : Vertical Sidewall

There are {100} planes perpendicular to the wafer surface (at an angle of 45° with the wafer flat i.e. the {110} direction



Stress

• Stress is the force per unit area that is acting on a surface of a differential volume element of a solid.

• It is assumed that the differential volume is in static equilibrium, i.e. there are no significant forces or torques created by gravity, electric fields, magnetic fields, or inertial forces.



Stresses on a Differential Volume

• Each surface has two types of forces, normal (x, y, z) and shear (xy, yz, zx).

• The convention is that the shear force is given a subscript in which the first element refer to the surface, and the second refer to the direction.



Two-dimensional View of the Differential Volume

• Stresses on opposite surfaces are equal in magnitude and acting in opposite direction when the differential volume is in static equilibrium.

• The condition of zero net torque requires the following relationship between the shear forces:

yzzy

zxxz

yxxy



Stress in Different Coordinate Systems

• In a coordinate system that is rotated through an angle θ, the uniaxial normal stress is transformed into a combination of normal and shear stresses.

sinFF

cosFF

V

N

2cosAF

sincosAF

Normal force:

Shear force:

Normal stress:

Shear stress:



Pure Shear Stress

0 0 00 00 0 σ-

0 0 00 0 σ0 σ 0

Principal coordinate system

(Only normal stress)

45 degrees rotation of the coordinate

system

(Only shear stress)

• All stress states can be represented in terms of tensile forces (which may be negative, i.e. compressive) only, provided that the coordinate system is correctly chosen.

• This transformation can be verified by applying the laws of rotation of a second-rank tensor



Tensors

• A tensor expresses the relation between material response or force with respect to the axes of its underlying symmetry to the axes of response or force in a laboratory frame.

• A first rank tensor is a spatial vector: its three components refer to the axes of some reference frame.

• A second rank tensor has 9 components, like a matrix. Each component is associated with two axes: one from the set of the reference frame axes and one from the material frame axes.

• A third rank tensor is a relationship between a first rank tensor and a second rank tensor, and so on.

• An N-rank tensor will have 3N components, but there may be symmetry relations that reduce the number of independent components considerably.



3-D Stress Tensor

•The six elements of stress can be represented as a second rank tensor

• The stress tensor is symmetric so it is uniquely identified by six numbers. Of these six, only three are linearly independent.

• This follows from the fact that symmetric, second-rank tensors have only non-zero elements along its diagonal when the coordinate system is chosen to coincide with the principal axes of the stress tensor (rotation eliminates shear).

zyzxz

yzyxy

xzxyx

zzyzx

yzyyx

xzxyx

σ ττ τ στ τ τσ

σ ττ τ στ τ τσ

σT3



Different Axial Stress States

0 0 00 0 00 0 σ

0 0 00 00 0 σ

2

1

ave

ave

ave

0 0 00 00 0 σ

•Uniaxial stress:

•Biaxial stress:

•Hydrostatic stress:



Strain

• A solid body under mechanical stress will deform.

• The deformation can be quantified in terms of the displacement vector, u(x).

• The strain tensor is defined in terms of the partial derivatives of the displacement.

• Strain is a dimensionless variable.



Axial Strain & Shear Strain

yuy

y

zuz

z

yu

zu zy

yz

xu

zu zx

zx

Axial strain

Shear strain

Displacement U is a vector function of original position x

xu

x)x(u)xx(u

lim xxx

xx

0

xu

yu

xu

yu yxyx

yxxy

0,

lim



Poisson Ratio

• The Poisson ratio is dimensionless, and has a value between 0.2 and 0.3 for most materials.

• Under uniaxial stress, the volume element expands in the direction of the stress and contracts in the directions orthogonal to the stress.

• The contraction is proportional to the elongation, and the proportionality constant is the Poisson ratio v.

xy v



Elasticity Curve

• In MEMS we want to operate in the linear region of the stress vs. strain characteristics.

• This ensures that no work is done on the compliant structure, which then can undergo large numbers of deformations without changing its performance.

• Brittle materials (silicon, silicon dioxide, silicon nitride) are therefore often preferable to ductile and viscoelastic materials (metals, plastics)



Mechanical Properties of Single Crystal Silicon

• Single crystal silicon is very brittle. It will yield and fracture when stress is beyond the proportional limit (yield point)

• Silicon does not exhibit plastic deformation or creep below 800° C

• With 108 cyclic load Silicon does not fail• Young’s modulus E (111) of silicon is 190 GPa,

comparing to 206-235 for stainless steel• Yield strength of Aluminum is 35 MPa, 1400

MPa for some steels and 2800-6800 MPA for silicon

• Silicon has a lower density (2.32 g/cm3) than Aluminum (2.71 g/cm3) but surpasses the yield strength of steel

• Above 800°C silicon shows considerable plasticity



Volume Change Under Stress

• The volume expansion is proportional to (1-2v ), which means that materials with ν =0.5 does not change their volume under uniaxial stress.

• Materials with Poisson’s ratios close to 0.5 are called incompressible.

• Most materials have Poisson’s ratios less than 0.5, and experience some volume increase under uniaxial stress.

• The volume of an element changes as a consequence of the strain.The volume change is:

)v(zyxV)v(zyx)v(zyxzyxV

zyx)v(z)v(y)(xV

x

xxx

xxx

21

111



Shear and Bulk modulus of Elasticity

• Shear modulus G expresses the ratio of shear stress and shear strain

xyxy G

• Shear modulus is related to the Young’s modulus and the Poisson ratio by:

)v(EG

12

• Bulk modulus, K is related to the Young’s modulus and the Poisson ratio by:

)v(EK

213

• Bulk modulus is analogous of Young’s modulus for an object subject to hydrostatic pressure (identical normal stress in all directions).



Isotropic Elasticity in Three Dimensions

)](v[E zyxx 1

)](v[E xzyy 1

)](v[E yxzz 1

xyxy G 1

yzyz G 1

zxzx G 1

• The complete stress-strain relations for an isotropic elastic solid can be derived by combining the results for normal and shear stresses in three dimensions



Isotropic, Orthotropic, and Cubic Materials

• The stress-strain models presented so far apply only to homogeneous and isotropic materials.

• Homogeneous refers to the fact that the elastic properties do not change from point to point in the body.

• Isotropic means that the properties do not vary with respect to directions.

• Many engineering materials are not isotropic. Their elastic properties vary depending on directions. They may also be inhomogeneous.

• For inhomogeneous materials, in general, each strain is dependent on each stress and can be expressed as a linear function of each stress

• an orthotropic material has at least 2 orthogonal planes of symmetry, where material properties are independent of direction within each plane. (e.g. Rochelle salt and fiber-reinforced composites.)

• If the properties of an orthotropic material are identical in all three directions, the material is said to have a cubic structure.

• In a polycrystalline material, the individual grains may be anisotropic, but if the material as a whole is comprised of many randomly oriented grains, then its measured mechanical properties will be an average of the properties over all possible orientations of the individual grains.



Plane Stress

• Plane stress is a special case that occurs very frequently in thin film materials used in MEMS devices.

• A thin film deposited on a thick substrate develop some stress due to deposition condition or method applied or due to the different coefficient of linear thermal expansion.

• All of the stresses, except the edge regions lie in the plane since the top surface is stress free.

• There is no in-plane shear stress.

]v[E yxx 1 ]v[

E xyy 1

• Biaxial plane stress occurs when the two in-plane stress components are equal. yxyx

v

E1

vE

1Biaxial modulus



Elastic Constants for Anisotropic Materials

• A generalized expression of elastic constants relating stress ands strain for isotropic, orthotropic, or anisotropic materials can be developed.

• Since both stress and strain are second rank tensors, the most general relationship between stress and strain is a fourth rank tensor, with 34 or 81 components.

• Due to symmetry, in every real material there is a maximum of 21 parameters to contend with, and these 21 components can be written as the elements of a square 6 x6 symmetric matrix.

• The six independent components of stress and strain (having axesalong the symmetry axes of the material) are organized into a column-vector array and the elastic constants are written in a symmetric matrix.



Generalized Hooke’s Law

• Stress and strain both have at most six components (only three of them are independent)

• It is therefore convenient to write them as vectors with six elements.• The generalized Hooke’s law:

• This matrix is called the stiffness matrix that relates stress to strain at a point in a material (isotropic or orthotropic).

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

γ

γ

γ

ε

ε

ε

ε

ε

ε

ε

ε

ε

C C C C CC

C C C C CC

C C C C CC

C C C C CC

C C C C CC

C C C C CC

σ

σ

σ

σ

σ

σ

τ

τ

τ

σ

σ

σ

C

6

5

4

3

2

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

6

5

4

3

2

1



Compliance Matrix

• The inverse of the stiffness matrix is the compliance matrix, which relates strain to stress. These two are expressed in terms of Young’s modulus E, Poisson ratio (ν ), and the shear modulus (G). It can be shown that in isotropic materials, these three constants are related as

zx

yz

xy

z

y

x

zx

yz

xy

z

y

x

τ

τ

τ

σ

σ

σ

σ

σ

σ

σ

σ

σ

S S S S SS

S S S S SS

S S S S SS

S S S S SS

S S S S SS

S S S S SS

ε

ε

ε

ε

ε

ε

γ

γ

γ

ε

ε

ε

S

6

5

4

3

2

1

666564636261

565554535251

464544434241

363534333231

262524232221

161514131211

6

5

4

3

2

1

)1(2 vEG



Stress in Thin Films

• In general the thin-film and substrate have different linear thermal expansion coefficients.

• The coefficient of linear thermal expansion of a material is defined as:

• The strain caused by thermal expansion is then simply:

dTd x

T

TTTTTT TxTxx )()()()( 000



Thin Film on Thick (Rigid) Substrate

• When a thin film is deposited on a thick substrate at elevated temperatures, and subsequently cooled and operated at much lowertemperatures the difference between the thermal expansion coefficients of the film and the substrate creates stress and strain.

• The strain of the substrate in one direction along the plane of its surface can be expressed as:

• Where Td is the deposition temperature and Tr is the operating temperature (which is often room temperature). The film then gets this same strain in the plane due to the fact that it is attached to the substrate:

TTT TSrdTSS )(

TTSSattachedf ,



Thermal Strain Mismatch

• If the film were unattached, however, its strain would be

• The difference between the strains the film has with and withoutattachment to the substrate is called the thermal mismatch strain

• The thermal mismatch leads to stress in the film. The stress is complicated in the edge regions.

• In the center of the film, far away from the edges, the film is strained symmetrically in the plane, and there is nothing to support stress in the direction perpendicular to the film.

TTT TfrdTffreef )(,

TT

TSTfmismatchf

TSTffreefSfreefattachedfmismatchf

)(

))((,

,,,,



Thin Film In-Plane Stress and Strain

• In the center, the strain can be relate to the stress in the principal coordinate system in the following way:

• In homogeneous films on cubic or isotropic substrates the two in-plane stresses are equal:

• The ratio, E/(1-v ) is called the biaxial modulus. The in-plane plane thermal-mismatch stress is:

xyzxyyyxx vvE

vE

vE

1 ;1 ;1

planeinplanein

planeinplaneinplaneinyx

vE

vE

1

1

Tv

Ev

ETSTfmismatchfmismatchf

)(

11 ,,



Determination of Stress in Thin films

• It is not possible to measure the thin film stress experimentally when a film is being deposited.

• One of the most definitive means to measure the stress in a thinfilm is by the use of the M-test method developed in the MIT.

• In the M-Test method, the pull-in voltage is determined using empirical methods.

• Since the pull-in voltage is a direct consequence of the thin film stress, using the expressions outlined inP. M. Osterberg and S. D. Senturia, “M-TEST: A Test Chip for MEMS Material Property Measurement Using Electrostatically Actuated Test Structures”, Journal of Microelectromechanical Systems, Vol. 6, No. 2, pp. 107-118, Jun. 1997, the stress can be determined.



Other Sources of Thin Film Stress

• Intrinsic stress• Chemical reactions• Doping (by diffusion or ion implantation)• Lattice mismatch• Rapid deposition ( evaporation or sputtering)

• Residual stress :• Thermal mismatch• Intrinsic stress• Stress gradient

• Intrinsic stress can sometimes be annealed out almost, completely, where some amount of thermal mismatch stress is unavoidable when working with materials with different coefficient of thermal expansion



Thin Film Stress Gradients

• The stress gradients are often as important as average stress in thin films.

• Stress gradients are caused by variations of film composition caused by all the different effects that can cause intrinsic stress.

• Annealing is very effective in removing a variety of gradients in thin films.• If stress gradients are not removed, they can lead to curvature of

freestanding thin films.• A freestanding film with a stress gradient will curve towards the side that

is in tensile stress.



Perpendicular Strain

• In the case of thermally induced stress and strain in thin films, the perpendicular strain has two components:

• Strain caused by thermal expansion/contraction

• Strain caused by in-plane stress

• The part caused by the in-plane stress is:

• The total perpendicular strain is then:

T)(v

vmismatch,fv

EEv

Ev

Ev

TSTf

mismatch,fplaneinz

12

12

22

Tv

vTSTfTSz

)(

12



Stress in Thin Films – Edge Regions

• In the edge regions, the stress cannot be constant, because at the end of the film there is nothing to support the in-plane stress.

• Instead we get a transition region or edge region in which the in-plane stress is transformed into shear stress that is terminated at the substrate.

• The termination of the stresses at the film/substrate interface leads to a peeling force that can detach the film from the substrate if the film is in tensile stress. Compressive stress doe not lead to this type of detachment failure.



Reading Assignments

• Chapter 8 Senturia

• Chapter 5 Richard Budynas, “ Advanced Strength and Stress Analysis”, McGraw-Hill, New York, 1997

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