Mems

Front 1

bulk micromachining

Back 1

Bulk micromachining processes involve selectively removing the bulk (silicon substrate) material in order to form certain 3D features or mechanical elements, such as beams and membranes. Unlike surface micromachining, which uses a succession of thin film deposition and selective etching, bulk micromachining defines structures by selectively etching inside a substrate. Whereas surface micromachining creates structures on top of a substrate, bulk micromachining produces structures inside a substrate.

Front 2

surface micromachining

Back 2

surface micromachining builds microstructures by deposition and etching of different structural layers on top of the substrate. Generally polysilicon is commonly used as one of the layers and silicon dioxide is used as a sacrificial layer which is removed or etched out to create the necessary void in the thickness direction. Added layers are generally very thin with their size varying from 2-5 um.

Front 3

sacrificial layer

Back 3

an underlying place-holding thin film layer, instead of the substrate underneath.

Front 4

integration

Back 4

Integration refers to the act of combining mechanical and electrical functionalities.

Front 5

packaging

Back 5

Packaging refers to the act of placing loose (diced) chips into human- or machine handlable modules that can be directly assembled on circuit boards and into systems.
Thus, packaging refers to dicing, die assembly, encapsulation, and testing.

Front 6

classes of integration

Back 6

1. @ wafer level
2. @ package(or chip) level
3. @ the board level

Front 7

advantage of wafer level integration

Back 7

minimum electromagnetic or other noises: because the distance between mechanical components and circuits is close

Front 8

disadvantage of wafer level integration

Back 8

1. chip foot print mismatch
2. material complexity and possible reduction of yield
3. possible cost increase due to complexity for fabrication and manufacturing

Front 9

encapsulation

Back 9

The act of sealing the MEMS devices into a controlled environment is referred to as encapsulation (or sealing, vacuum packaging)

Front 10

consideration of deposition processes

Back 10

1. ultimate thickness
2. deposition rate and control factors
3. temperature of the process
4. deposition profile

Front 11

consideration of etching processes

Back 11

1. etch rate
2. end point detection
3. etch rate selectivity
4. processing temperature
5. etch uniformity across a wafer
6. sensitivity to overtime etch
7. safety and cost of etchants
8. surface finish and defects

Front 12

Ideal Process Rule(IPR)

Back 12

IPR #1: Any layer deposited on a wafer should not physically, thermally, or chemically damage or compromise layers already on the wafer.
IPR #2: Any etching agent for removing one layer of material should ideally not attack other materials at all.

Front 13

Types of process-related uncertainties

Back 13

1. Deposition rates and etch rates are variable.
2. Uncertainty of determining the endpoint of process
3. Material characteristics may be uncertain or variable.

Front 14

levers for changing conductivity of semiconductor

Back 14

1) intentionally induced impurities
2) externally applied electric field
3) charge injection
4) ambient light
5) temperature variation

Front 15

the bandgap of the semiconductor material

Back 15

The statistically minimal energy needed to excite a covalently bonded electron to become a free charge carrier.

Front 16

the bandgap of silicon at room temperature

Back 16

1.11 eV, or 1.776*10^-19 J

Front 17

semiconductor materials

Back 17

1. Silicon
2. Germanium(Ge)
3. Silicon Germanium(Si_1-xGe_x)
4. Gallium Arsenide(GaAs)
5. Gallium Nitride(GaN)
6. Silicon Carbide(SiC)

Front 18

intrinsic semiconductor material

Back 18

A perfect semiconductor crystal having no impurities and lattice defects in its crystal structure. Thus, electrons and holes are created through thermal or optical excitation in this material.

Front 19

electron-hole pair generation

Back 19

When a valence electron receives enough energy, either thermally or optically, it is freed from the silicon atom and leaves a hole behind. This event is called electron–hole pair generation.
(occurs in intrinsic semiconductor materials)

Front 20

extrinsic semiconductor material

Back 20

Most semiconductor pieces are not perfect intrinsic material, however. They often have impurities atoms in them—by accident or by design. The impurity atoms contribute additional electrons or holes, generally in an unbalanced manner. The intentional introduction of impurities, called doping, would turn an intrinsic material into an extrinsic semiconductor material.

Front 21

ionization

Back 21

A dopant atom generally displaces a host atom when introduced into the crystal lattice. A dopant atom with more electrons in its outmost shell than a host atom is able to introduce, or “donate” its extra electrons to the bulk. This step, called ionization, occurs readily under room temperature.

Front 22

n-type material

Back 22

When the electron concentration is greater than the hole concentration, the electron is called the majority carrier and the semiconductor material is called an n type material.

Front 23

p-type material

Back 23

When the hole concentration is greater than that of electrons, the hole is the majority carrier and the semiconductor material is called p type.

Front 24

the concentration of carriers in doped semiconductor

Back 24

n0*p0 = ni^2

Front 25

p0 + Nd+ = n0 + Na-

Back 25

Since the doping process involves injecting neutral atoms into a neutral bulk, charge neutrality of the bulk is always maintained. The concentration of negative charges in a bulk is made up of that of electrons and ionized acceptor atoms . The concentration of positive charges in a bulk consists of that of holes and ionized donor atoms.
Think about doping p(or n) type material into a bulk. Then, the overall electric charge should be maintained stably because the bulk and dopant are neutral in the first time.

Front 26

n0^2 - n0*(Nd - Na) - ni^2 = 0
p0^2 - p0*(Na - Nd) - ni^2 = 0

Back 26

The magnitude of n0 and p0 can be found by solving these second order quadratic equations, once the concentration of dopants are known.

Front 27

overall conductivity of semiconductor

Back 27

calculating electrical conductivity when carrier concentrations of both types (electrons and holes) are known. The overall conductivity of a semiconductor is the sum of conductivities contributed by these two types individually.

Front 28

carrier mobility, u

Back 28

u = Vavg/E,
[(m/s)/(V/m) = m^2/(V*s)]
Vavg : A charge carrier reaches a statistical average velocity between collision events.
E : the magnitude of the local electric field (E)

The values of the mobility are influenced by the doping concentration, temperature, and crystal-orientation, in a complex manner

Front 29

conductivity of a bulk semiconductor, sigma

Back 29

1. sigma = 1/lo, lo : the bulk resistivity
2. E = lo*J, J : the resultant current density(current divided by the cross-sectional area)
3. sigma = J/E

Front 30

sigma_n = n*u_n*q
sigma_p = p*u_p*q

Back 30

sigma_n
= J/E = (I/A)/E
= [Q/(A*t)]/E
= [(n*q*d*A)/(A*t)]/E
= (n*q*d/t)/E
= (n*q*V)/E
= n*q*u_n
n : # of electron ions from dopant
q : a charge of unit carrier(1.6*E^-19C)
t : mean free time
d : mean free path
A : cross section area

sigma_p : likewise

Front 31

the total conductivity, sigma_total

Back 31

lo_total
= 1/sigma_total
= 1/(sigma_n+sigma_p)
= 1/[q*(u_n*n+u_p*p)]

Front 32

units of
1) resistivity
2) mobility
3) conductivity
of semiconductors

Back 32

1) = E/J = [V/m]/[A/m^2]
= [Ohm*m]
2) = Vavg/E = [m/s]/[V/m]
= [m^2/(V*s)]
3) = 1/resistivity
= [1/(Ohm*m)]

Front 33

R = V/I = ?

Back 33

= (E*L)/(J*A)
= (E*L)/[(E/lo)*(w*t)]
= lo*L/(w*t)
= lo_s*L/w
w : width, t : thickness, L : length of resistor
lo_s : sheet resistivity[Ohm/unit area]

Front 34

cubic lattice family

Back 34

In a cubic lattice, materials properties exhibit rotational symmetry. Hence (010) and (001) planes in the lattice are equivalent to (100) plane in terms of material properties.

Front 35

How to represent a family of equivalent planes?

Back 35

a set of integers are enclosed in braces {} instead of parentheses ().

Front 36

How to represent a family of equivalent directions?

Back 36

a set of integers are enclosed in brackets <> instead of [].

Front 37

<100>-oriented wafers

Back 37

1. used for metal-oxide-semiconductor(MOS) electronics devices due to its low density of interface states.
2. prevalently used for bulk silicon micromachining of MEMS devices.

Front 38

<111>-oriented wafers

Back 38

1. commonly used for bipolar junction transistors because of high mobility of charge carriers in the <111> direction.
2. occasionally used for MEMS application.

Front 39

Poisson's ratio, v

Back 39

the ratio between transverse and longitudinal elongations
ex) longitudinal stress along the x-axis produces
i) a longitudinal elongation in the direction of stress,
ii) a reduction of cross-sectional area
v = |s_y/s_x| = |s_z/s_x|
s_x : x-directional strain
s_y : y-directional strain
s_z : z-directional strain

Front 40

modulus of elasticity, E

Back 40

a.k.a. Young's modulus
1) the general relation between stress and strain over a wider range of deformation
2) Atoms are held together with atomic forces. If one imagines inter-atomic force acting as springs to provide restoring force when atoms are pulled apart or pushed together, the modulus of elasticity is the measure of the stiffness of the inter-atomic spring near the equilibrium point.
3) (stress) = E*(strain)

Front 41

shear stress, (tau)

Back 41

1) has no tendency to elongate or shorten the element in the x,y,z directions.
2) produces a change in the shape of the element

Front 42

shear strain, (gamma)

Back 42

1) defined as the extent of rotational displacement
2) (gamma) = (dX)/L
dX : angular displacement
L : perpendicular length to dX

Front 43

shear modulus of elasticity, G

Back 43

1) G = (shear stress)/(shear strain)
= (tau)/(gamma)
2) depends on the material, not the shape and dimensions of an object

Front 44

E, G, v relation

Back 44

G = E/[2*(1+v)]

Front 45

tensile test I : strain ranges
(Generic stress-strain relationship)

Back 45

1) elastic regime :
(stress) = E*(strain) applicable
2) perfect plasticity or yielding
3) strain hardening
4) necking

Front 46

tensile test II : stress points
(Generic stress-strain relationship)

Back 46

1) proportional limit
2) yield stress
3) fracture point

Front 47

elastic deformation regime

Back 47

At low levels of applied stress and strain, the stress value increases proportionally with respect to the developed strain, with the proportional constant being the Young’s modulus. This segment of the stress-strain curve is called the elastic deformation regime.

Front 48

plastic deformation regime

Back 48

As the stress exceeds a certain level, the material enters the plastic deformation regime. In this regime, the amount of stress and strain does not follow a linear relationship anymore. Furthermore, deformation cannot be fully recovered after the external loading is removed.

Front 49

yield strength

Back 49

1) the stress at a yield point
2) Before the yield point is reached, the material remains elastic.

Front 50

ultimate strength(fracture strength)

Back 50

At the fracture point, the specimen suffers from irreversible failure.

Front 51

stress-strain relations of brittle materials(i.e. silicon), soft rubbers, general elastic materials

Back 51

Figure 3.13
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Front 52

common qualitative phrases to describe materials

Back 52

1) strong
2) ductile
3) resilient
4) tough

Front 53

strength

Back 53

A material is strong if it has high yield strength or ultimate strength. By this account, silicon is even stronger than stainless steel.

Front 54

ductility

Back 54

It is a measure of the degree of plastic deformation that has been sustained at the point of fracture. A material that experiences very little or no plastic deformation upon fracture is termed brittle. Silicon is a brittle material, which fails in tension with only little elongation after the proportional limit is exceeded. Ductility may be expressed quantitatively as either percent elongation or percent reduction in area.

Front 55

toughness

Back 55

Toughness is a mechanical measure of the ability of a material to absorb energy up to fracture. For a static situation, toughness may be ascertained from the result of the tensile stress-strain test. It is the area under the stress-strain curve up to the point of fracture. For a material to be tough, it must display both strength and ductility.

Front 56

resilience

Back 56

Resilience is the capacity of a material to absorb energy when it is deformed elastically and then, upon unloading, to have this energy recovered.

Front 57

Why are the experimental values of fracture stress, Young's modulus, and fracture strain of SCS, poly-silicon, Si relatively scarce and scattered?

Back 57

It is scarce because the accurate measurement of miniature sample specimen is more challenging than for macroscopic samples. It is scattered because the materials properties are affected by a variety of subtle factors that are often unreported or not easily traceable, such as exact material growth conditions, surface finish, and thermal treatment history.

Front 58

size of specimen dependency

Back 58

Certain measured material properties, such as fracture strength, quality factor, and fatigue lifetime, depend on the size of specimen.
For example, one study found that the fracture strength is size dependant, being 23–38 times larger than that of a millimeter-scale sample [16]. The fracture behavior of silicon, for example, is governed by the presence of flaws and preexisting cracks. For small single crystal silicon structures, the devices may exhibit large elastic strength and strain than what is predicted for bulk materials due to the lack of flaws in the small volume of silicon involved.

Front 59

Single Crystalline Silicon(SCS)
mechanical properties
1) Young's modulus
2) Shear modulus
3) Possion's ratio

Back 59

1) In the {100} plane:
168 GPa @ [110] direction,
130 GPa @ [100] direction
In the {110} plane:
187 GPa @ [111] direction(greatest)
2) 79.92 GPa(ref. website)
3) from 0.055 to 0.36

Front 60

Polysilicon thin film mechanical properties
1) Young's modulus
2) Poisson's ratio
3) fracture strength
4) fracture strain

Back 60

1) the Young’s modulus depends on the exact process conditions of the materials, which differs from laboratory to laboratory due to subtle changes of growth conditions.
* LPCVD polysilicon : from 120 GPa to 175 GPa(avg. 160 GPa)
2) from 0.15 to 0.36(widely 0.22)
3) from 1.0 to 3.0 GPa
4) depends on temperature(0.7%-1.6%)

Front 61

Silicon Nitride thin film mechanical properties
1) Young's modulus
2) fracture strength
3) fracture strain

Back 61

1) 254 GPa
2) 6.4 GPa
3) 2.5 %

Front 62

shock loading to MEMS applications

Back 62

1) occurs during fabrication, operation, deployment
2) increasing shock tolerance : use ductile materials such as polymers

Front 63

fatigue in MEMS devices

Back 63

Microstructures may fail under repeated loading even if the magnitude of the stress is below the fracture strength
Ex) Repeated small angle bending of a paper clip may cause it to develop surface cracks and eventually break over certain number of cycles.

Front 64

MEMS structure of SCS: size and fatigue relation

Back 64

Single crystal silicon material and micro mechanical structures in general exhibit good fatigue life because, as the size of device becomes smaller, the number of crack-initiating sites on a given structure decreases.

Front 65

life cycles of SCS & poly-Si MEMS devices

Back 65

SCS : from E6 to E11 cycles
Poly-Si : up to E11 cycles

Front 66

general stress-strain relations

Back 66

1) stress-strain are tensors.(expressed in matrix form)
2) a unit cube from inside a material : 6 faces, 12 shear stress, 6 normal stress

Front 67

independent shear stress components I

Back 67

12 -> 6
why? each pair of shear stress components acting on parallel faces but along the same axis have equal magnitude and opposite directions for force balance

Front 68

independent shear stress components II

Back 68

6 -> 3
Based on torque balance, two shear stress components acting on two facets but pointing towards a common edge have the same magnitude. Specifically, τxy= τyx, τxz= τzx, and τzy= τyz. In other words, equal shear stresses always exist on mutually perpendicular planes.

Front 69

independent normal stress components

Back 69

3
Under equilibrium conditions, the normal stress components acting on opposite facets must have the same magnitude and point to opposite directions.

Front 70

General matrix equation between stress and strain

Back 70

T1, T2, T3 : three independent normal stress components
T4, T5, T6 : three independent shear stress components
s1, s2, s3 : three independent normal strain components
s4, s5, s6 : three independent shear strain components
(eqn. 3.27, 3.28, 3.29, 3.30)

Front 71

1) eqn. 3.27
2) eqn. 3.29

Back 71

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Front 72

The coefficient matrix, C, of SCS with coordinate axes along <100> directions

Back 72

a.k.a. the stiffness matrix
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Front 73

3 possible B.C.'s for beams lying in single planes

Back 73

1) The fixed boundary condition
: restricts both linear & rotational DOFs.(i.e. the anchored end of a diving board)
2) The guided boundary condition
: allows two linear DOFs but restricts the rotational DOF
3) The free boundary condition
: provides both linear & rotational DOFs.(i.e. the free end of a diving board)

Front 74

a fixed-free beam

Back 74

a.k.a. cantilever
: a beam fixed at one end and free at another

Front 75

most frequently encountered MEMS beams

Back 75

1) a fixed-free beam(cantilevers)
2) a fixed-fixed beam(bridges)
3) a fixed-guide

Front 76

pure beam bending situation

Back 76

The lower part of the beam is in tension and the upper part is in compression.
Thus, somewhere between the top and bottom of the beam is a surface in which longitudinal lines do not change in length.

Front 77

1) neutral surface
2) neutral axis

Back 77

1) Somewhere between the top and bottom of the beam is a surface in which longitudinal lines do not change in length.
2) The intersection between the neutral surface with any cross-sectional plane.

Front 78

distribution of stress-strain for a beam with symmetry and material homogeneity

Back 78

1) The magnitude of stress and strain at any interior point is linearly proportional to the distance between this point and the neutral axis;
2) On a given cross section, the maximum tensile stress and compressive stress occur at the top and bottom surfaces of the cantilever;
3) The maximum tensile stress and the maximum compressive stress have the same magnitude;
4) Under pure bending, the magnitude of the maximum stress is constant through the length of the beam.

Front 79

ideal pure bending beam case
1) max. stress
2) max. strain

Back 79

Ref. eqn (3.33, 3.34, 3.35)
1) max stress = (M*I*t)/2
2) max strain = (M*t)/(2*E*I)
I : the moments of inertia
t : the thickness of beam
E : Young's modulus

Front 80

The general method for calculating the curvature of beam under small displacement(2nd order differential equation)

Back 80

Ref. eqn 3.36
E*I*(d^2y/dx^2) = M(x)
M(x) : represents the bending moment at the cross section at location x
y : the displacement at location x
Thus, x & y relation can be gotten!

Front 81

Solving eqn 3.36
E*I*(d^2y/dx^2) = M(x)

Back 81

1. Find the moment of inertia with respect to the neutral axis;
2. Find the state of force and torque along the length of a beam;
3. Identify boundary conditions. Two boundary conditions are necessary to deterministically find a solution.

Front 82

Rectangle cross section cantilever beam's moment of inertia, I

Back 82

I = (w*t^3)/12
w : width of cross section
t : thickness of cantilever

Front 83

Circular cross section cantilever beam's moment of inertia, I

Back 83

I = (pi*R^4)/4
R : radius of cantilever

Front 84

Circular beam with end-point force exertion case:
the moment of inertia and the max stress & strain

Back 84

M = [Int{Int(dF*y)}]
dF = Stress(y)*dA
dA = r*d(Theta)*dr
y = r*sin(Theta)
Stress(y) = Stress_max*(y/R)
Thus,
I = (pi*R^4)/4
Stress_max = 4*M/(pi*R^3)
Strain_max = 4*M/(pi*R^3*E)

R : radius of beam
y : distance from the neutral axis, the moment arm of dF
M : the moment at the tip of beam

Front 85

Applying point force, F, to a rectangular cross-sectional, fixed-free beam
1) Theta
2) D_vertical
3) k_vertical

Back 85

I = w*t^3/12
1) Theta = F*l^2/(2*E*I)
2) D_vertical = F*l^3/(3*E*I)
3) k_vertical = F/D_vertical
= 3*E*I/(l^3)
= E*w*t^3/(4*l^3)

I : the moment of inertia
w,t,l : beam's length dimensions