Lecture 2 MEMS

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Lecture 2Lecture 2

MEMS

Electrostatic Comb Drive Fabrication: Step 1

SiO2LPCVD Nitride

The fabrication process begins with a standard blanket n+ diffusion of

Silicon substrate

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e ab cat o p ocess beg s w t a sta da d b a et d us o o

the substrate to define the substrate ground plane. Next, the wafer is

passivated with a layer of 15 nm thick LPCVD (Low Pressure Chemical

Vapor Deposition) nitride deposited on top of a layer of 50 nm thick

thermally grown Si02

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The nitride and silicon dioxide layers are then patterned and etched to

open contact windows to the substrate ground plane


Polysilicon ground plane

A l f 30 thi k i it h h d d l ili i

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A layer of 30 nm thick in situ phosphorus-doped polysilicon is

deposited by LPCVD process at 650 “C. The polysilicon layer is then

patterned and etched dry. This layer serves as a second electrode plane

and the interconnection to the n+ diffusion.

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PSG

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A 2 m-thick phosphosilicate glass (PSG) layer is then deposited by

LPCVD process. This layer will act as the sacrificial layer to define

the gap between the structure and the ground plane


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The PSG layer is then patterned and etched selectively to define the

anchors of the microstructures.

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Structural polysilicon

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The 2 m-thick polysilicon structural layer is then deposited by

LPCVD (undoped) process.


2nd PSG layer

The structural layer is doped by depositing another layer of 30 nm

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thick PSG using LPCVD process and then annealing at 950 “C for

one hour.

Doping is done to reduce resistivity of the comb structure and

annealing is necessary to reduce the residual stress

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The top PSG layer is then stripped by wet etching in 10:1diluated

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hydrofluoric acid. The polysilicon layer is then patterned using a

photoresist to define the plates, beams and electrostatic comb drive

and sense structures.

Electrostatic Comb Drive Fabrication: Step 9Fixed electrodeMovable electrode

Anchor

After patterning, the polysilicon layer is anisotropically etched using

CCl4 plasma (reactive-ion etching) to achieve nearly vertical

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sidewalls. Wafer is then immersed in 10:1 diluted HF to etch away

the sacrificial PSG layer. The wafer is rinsed repeatedly with DI

water for at least 30 minutes after the micromachining step is

completed and then dried in a standard spin dryer.

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Material PropertiesCrystal Structures

• In a crystalline material, the atoms are situatedin a repeating or periodic three dimensionalarray over large atomic distances.

• Properties of the crystalline solid depend onthe crystalline structure of the material.

• Lattice means a three dimensional array ofpoints coinciding with atom positions.

• The basic atomic pattern that repeats itself in a Face centeredcubic lattice

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crystalline material is called a unit cell.

• Face centered cubic lattice (fcc) refers to aunit cell where atoms are located at each ofthe corners and the centers of all the cubefaces.

cubic lattice

Miller Indices

• Miller Indices are a symbolic vectorrepresentation for the orientation of anatomic plane in a cubic crystal lattice andare defined as the reciprocals of thef i l i hi h h l kfractional intercepts which the plane makeswith the crystallographic axes.

• To determine Miller indices of a plane takethe following steps:

1. Determine the intercepts of the plane alongeach of the three crystallographic directions

a→ lattice constant

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each of the three crystallographic directions

2. Take the reciprocals of the intercepts

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

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Crystal Directions• Many planes in an crystalline solid are

equivalent.

• Six equivalent faces of a cubic latticeare expressed as {100}.

• A direction in a lattice is expressed as aset of three integers with the samerelationship as the components of avector in that direction and expressedas [100].

• They are determined in the same way

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y yas the Miller indices are determined.

• Equivalent directions are grouped andexpressed as <100> directions.

• In a cubic lattice, a direction [hkl] isperpendicular to a plane (hkl).

Miller Indices Determination: (100) Plane

h k l

Intercept length 1 ∞ ∞

Unit cell length=a

Intercept length 1 ∞ ∞

Take reciprocal 1/1 1/∞ 1/∞

Cleared fraction 1 0 0

Miller indice: (100)

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Miller Indices Determination: (110) Plane

h k l

Intercept length 1 1 ∞

Unit cell length=a

Intercept length 1 1 ∞

Take reciprocal 1/1 1/1 1/∞

Cleared fraction 1 1 0


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Miller Indices Determination:More Examples

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Silicon Crystal Structure

• The Si diamond lattice is composed of two interpenetrating fcclattices, one displaced 1/4 of a lattice constant in each direction from the other

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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

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Thin Film vs. Bulk Solid

• The properties of a material are determined by thecooperative effect of a huge number of similarparticles (atoms, lattice, etc.) in a 3D arrangement.

• Many physical properties of materials require a

Thin film or surface layer

• Many physical properties of materials require alarger ensemble of atoms for a meaningfuldefinition, independent of the amount of material,for example, density, the thermal expansioncoefficient, hardness, color, electrical and thermalconductivity.

With lid t i l th ti f f

The thin film layer may be a few atoms or a

Bulk

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

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

few atoms or a few lattice constant thick

Effects of Bonding Forces on Material Properties• The properties of a material are controlled by the bond strengths

between the particles.

• In classical technology and usually also in microtechnology, a separation between the bonding forces in the bulk material and the surface forces has some significancethe surface forces has some significance.

• Both internal and external bonds are based on interatomicinteractions, and the chemical bonds.

• Besides the spatial separation of a material, the orientation of the internal and the surface bonds also determine the properties of materials or of material compounds.

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p

• Conventional technology uses materials with isotropic properties. Isotropic means that these properties are approximated as being similar in all spatial orientations of the solid.

• Materials may exhibit 30% variation of Young’s modulus depending on crystal orientation.

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Various Stages of Solidification of a Crystalline Material

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Density Fluctuation in crystalline Thin Films• The macroscopic model of ideal isotropy is also not valid

for monocrystalline materials such as silicon, gallium arsenide or other typical microelectronic materials.

• A monocrystalline solid excludes the statistical distribution f i t t i di t d f b d i t tiof inter-atomic distances and of bond orientations.

• It includes elementary cells consisting of a few atoms, and a randomly oriented plane results in a density fluctuating with the angle of this plane.

• In addition, the bond strength between atoms is localized and is determined from its orientation

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and is determined from its orientation.

• Such elementary cells create the solid in a periodic arrangement in an identical orientation.

• So the anisotropy of the particle density and bond strength on the atomic scale is transformed into macroscopic dimensions.

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Origin of Anisotropy• 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 ith th t l l tti di i th hi h i twith the crystal lattice dimensions, they possess a high anisotropy

even for a material with macroscopic isotropy.

• The anisotropy of a monocrystalline material is determined by the anisotropic electron configuration and the electronic interactions between the atoms of the crystal.

• It is based on the arrangement of the locations of the highest

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It is based on the arrangement of the locations of the highest occupation probability of the electrons, especially of the outer electrons responsible for chemical bonds.

• 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 on it.

• Non-crystalline materials created by surface deposition processes canalso show anisotropy.

• Almost all thin layers prepared by evaporation or sputtering exhibitanisotrop d e to the preferred positioning b an initial n cleation

Origin of Anisotropy

anisotropy due to the preferred positioning by an initial nucleationand a limited surface mobility of the particles, which results in grainboundaries and the overall morphology of the layer.

• Even spin-coated polymer layers have such anisotropic properties,because the shear forces induced by the flow of the thin film lead to apreferred orientation of the chain-like molecules parallel to the

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substrate plane.

• The transition from an almost isotropic to an anisotropic situation ispartly based on the downscaling of the dimensions. For example, amaterial consists of many small crystals, so these statisticallydistributed crystals appear in total as an isotropic material.

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Stress

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

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

• Each surface has two types of forces, normal (σx, σy, σz) andshear (τxy, τyz, τzx).

• The convention is that the shear force is given a subscript inwhich the first element refer to the surface and the second refer

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which the first element refer to the surface, and the second referto the direction.

Stresses on a Differential Volume

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Two-dimensional (View of the Differential Volume)

• Stresses on opposite surfacesare equal in magnitude andacting in opposite directionwhen the differential volumeis in static equilibrium.

• The condition of zero nettorque requires the followingrelationship between theshear forces:

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Tensors

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

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

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

• A third rank tensor is a relationship between a first rank tensor and a

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

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

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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

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• 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).

Different Axial Stress States

• Uniaxial stress: σ 0 0

0 0 0

0 0 0

• Biaxial stress:

0 0 0

σ1 0 0

0 σ2 0

0 0 0

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• Hydrostatic stress:

σave 0 0

0 σave 0

0 0 0ave

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Pure Shear Stress

• All stress states can berepresented in terms of tensileforces (which may be

Principal di

45 degrees t ti f th

negative, i.e. compressive)only, provided that thecoordinate system is correctlychosen.

• Some solids are weaker undertensile loads than shear loads.

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coordinate System (Only normal stress)

rotation of the coordinate system (Only shear stress)

tensile loads than shear loads.

• These materials therefore tendto break along the principalaxes of the stress.

Stress in Different Coordinate Systems

• In a coordinate system that is rotated through an angle θ, theuniaxial normal stress is transformed into a combination of

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normal and shear stresses.

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Strain

• A solid body under mechanical stress will deform.A solid body under mechanical stress will deform.

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

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

• Strain is a dimensionless variable

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• Strain is a dimensionless variable.

Axial Strain & Shear Strain

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Poisson Ratio

• 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.

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p p y

• The Poisson ratio is dimensionless, and has a value between 0.2 and 0.3 for most materials, although highly structured materials can have Poisson ratios outside this range (even negative values are possible).

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

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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)

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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

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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 of an element changes as a consequence of the strain. g q

The volume change is:

• The volume expansion is proportional to (1-2ν ), which means that materials with ν =0 5 does not change their volume under uniaxial

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materials with ν 0.5 does not change their volume under uniaxialstress.

• 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.

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Shear and Bulk modulus of Elasticity

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

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

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

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

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ratio by:

Isotropic Elasticity in Three Dimensions

• The complete stress-strain relations for an isotropic elastic solidcan be derived by combining the results for normal and shearstresses in three dimensions

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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

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each stress and can be expressed as a linear function of each stress

• A material which exhibits symmetry with respect to three mutually orthogonal planes is called an orthotropic material.

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

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 linear thermal expansion.p

• 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.

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

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Elastic Constants for Anisotropic Materials

• A generalized expression of elastic constants relating stress and strainfor isotropic, orthotropic, or anisotropic materials can be developed.

• Since both stress and strain are second rank tensors, the most generall ti hi b t t d t i i f th k t ith 34relationship 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 21parameters to contend with, and these 21 components can be writtenas the elements of a square 6 x6 symmetric matrix.

• The six independent components of stress and strain (having axes

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• The six independent components of stress and strain (having axesalong the symmetry axes of the material) are organized into a columnvector array and the elastic constants are written in a symmetricmatrix

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:The generalized Hooke s law:

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• The elements of this matrix are called the stiffness matrix that relates stress to strain at a point in a material (isotropic or orthotropic).

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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’smodulus E, Poisson ratio (ν ), and the shear modulus (G). It can beshown that in isotropic materials, these three constants are related as

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Stress in Thin Films

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

• The coefficient of linear thermal expansion of a material is definedas:

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as:

• The strain caused by thermal expansion is then simply:

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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 expansioncoefficients of the film and the substrate creates stress and strain.

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

• Where Td is the deposition temperature and Tr is the operatingtemperature (which is often room temperature). The film then getsthi t i i th l d t th f t th t it i tt h d t th

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this same strain in the plane due to the fact that it is attached to thesubstrate:

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

• The difference between the strains the film has with and withouttt h t t th b t t i ll d th th l i t h t iattachment to the substrate is called the thermal mismatch strain.

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

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complicated in the edge regions.

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

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Thin Film In-Plane Stress and Strain

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

• In homogeneous films on cubic or isotropic substrates the twoinplane stresses are equal:

Th i E/(1 ) i ll d h bi i l d l Th i l l

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

Determination of Stress in Thin films

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

• One of the most definitive means to measure the stress in a thin filmi b h f h h d d l d i his by the use of the M-test method developed in the MIT.

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

• Since the pull-in voltage is a direct consequence of the thin filmstress, using the expressions outlined in

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P. M. Osterberg and S. D. Senturia, “M-TEST: A Test Chip for MEMSMaterial Property Measurement Using Electrostatically Actuated TestStructures”, Journal of Microelectromechanical Systems, Vol. 6, No. 2,pp. 107-118, Jun. 1997, the stress can be determined.

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Other Sources of Thin Film Stress• Intrinsic stress

• Chemical reactions

• Doping (by diffusion or ion implantation)

• Lattice mismatch• Lattice mismatch

• Rapid deposition ( evaporation or sputtering)

• Residual stress :

• Thermal mismatch

• Intrinsic stress

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• Stress gradient

• Intrinsic stress can sometimes be annealed out almost, completely,where some amount of thermal mismatch stress is unavoidable whenworking with materials with different coefficient of thermalexpansion.

Thin Film Stress Gradients

• The stress gradients are often as important as average stress in thinfilms.

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

• Annealing is very effective in removing a variety of gradients in thin

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Annealing is very effective in removing a variety of gradients in thinfilms.

• If stress gradients are not removed, they can lead to curvature offreestanding thin films.

• A freestanding film with a stress gradient will curve towards the sidethat is in tensile stress.

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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 stressStrain caused by in plane stress

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

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• The total perpendicular strain is then:

Stress in Thin Films – Edge Regions

• In the edge regions, the stress cannot be constant, because at the endof 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

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plane stress is transformed into shear stress that is terminated at thesubstrate.

• The termination of the stresses at the film/substrate interface leads toa peeling force that can detach the film from the substrate if the filmis in tensile stress. Compressive stress don’t lead to this type ofdetachment failure.

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Lecture 2 MEMS