ee c245 - me c218 introduction to mems design …...5 ee c245 – me c218 fall 2003 lecture 9...

EE C245 – ME C218 Fall 2003 Lecture 9

EE C245 - ME C218Introduction to MEMS Design

Fall 2003

Roger Howe and Thara SrinivasanLecture 9

Energy Methods II

2EE C245 – ME C218 Fall 2003 Lecture 9

Today’s Lecture• Mechanical structures under driven harmonic motion

� develop analytical techniques for estimating the resonant frequency

• Lumped mass-spring case: ADXL-50 example• Rayleigh-Ritz method and examples:

lateral resonator, double-ended tuning fork• Viscous damping• Damped 2nd order mechanical systems:

transfer functions

• Reading:Senturia, S. D., Microsystem Design, Kluwer Academic

Publishers, 2001, Chapter 9, pp. 260-264.


Estimating Resonant FrequenciesSimple harmonic motion

k Mx(t) = xocos(ωt)

Potential energy:

Kinetic energy:

)(cos21)(

21)( 222 tkxtkxtW o ω==

)(sin21)(

21)( 2222 tMxtxMtK o ωω== &


Energy ConservationIgnoring dissipation � Wmax = Kmax

22max

2max 2

121

oo xMKkxW ω===

Resonant frequency:

Mk=ω

Applications: lumped masses that dominate thesuspension’s mass


Example: ADXL-50

Suspension beam: L = 260 μm, h = 2.3 μm, W = 2 μmAverage residual stress through thickness: σr = 50 MPa


Lumped Spring-Mass Approximation

• Mass = M = 162 nano-grams, 60% is from the capacitive sense fingers

• Suspension: four tensioned beams … include both bending and stretching terms

F/4

F/4Bending compliance kb

-1

Stretching compliance kst-1

A.P. Pisano, BSAC Inertial SensorShort Courses, 1995-1998


ADXL-50 Suspension Model• Bending contribution:

• Stretching contribution:

NmEWh

LWhELkkk ccb μμ /2.4

)12/(3)2/(2)/1/1( 3

3

3

31 ==⎥

⎦

⎤⎢⎣

⎡=+=−

NmWhLSLk

rst μμ

σ/14.1/1 ===−

S

Sθ

Fy = S sinθ ≈ S(x/L)• Total spring constant: add bending to stretching

mNkkk stb μμ /5.4)88.024.0(4)(4 =+=+=



ADXL-50 Resonant Frequency

kHzkgxmN

Mkf 5.26

10162/48.4

21

21

12 === −ππ

Lumped mass-spring approximation:

Data sheet: f = 24 kHz

Note: we have not included the frequency-pullingeffect of the DC bias voltage on the sense capacitor plates



Distributed Mechanical Structures• Vibrating structure’s displacement function can be

separated into two parts:

)cos()(ˆ),( txytxy ω=

• We can use the static displacement of the structure as a trial function and find the strain energy Wmax atthe point of maximum displacement (t = 0, π/ω, …)

y(x)

v(x,t) = -ωy(x)sin(ωt) = 0


Maximum Kinetic Energy• At times t¸= π/(2ω), 3π/(2ω), …, then the displacement

of the structure is y(x ) = 0.• The velocity of the structure is maximum and all its

energy is kinetic (since W = 0)

v(x,t)̧

y(x,t)̧ = 0

v(x,t)̧ = -ωy(x)sin(ωW̧) = -ωy(x)


Finding the Maximum Kinetic Energy• A differential length dx along the beam has a kinetic

energy dK due to its transverse motion:

dx

Wh

v(x,t)̧ = - ωy(x)dK = (1/2)(dm)v2(x,t´)

dm = ρ(Whdx)

• Total maximum kinetic energy:

∫∫ =′=LL

dxxyWhtxWhdxvK0

22

0

2max )(ˆ

21),(

21 ωρρ


The Raleigh-Ritz Method• The maximum potential and maximum kinetic

energies must be equal �

max0

22max )(ˆ

21 WdxxyWhK

L

== ∫ ωρ

• The resonant frequency of the beam is therefore:

∫= L

dxxyWh

W

0

22

max

)(ˆ21 ωρ

ω


Example: Folded-Flexure Resonator

Lumped masses (shuttle, truss)need to be included in thekinetic energy expression

h

W

hT


Kinetic Energy• Shuttle mass = Ms, truss mass = Mt

∫++=L

ttss dxxyWhvMvMK0

2222max )(ˆ

21

21

21 ωρ

• Use static deflection as estimate of mode shape

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=≅

32

23)()(ˆLx

LxYxyxy o

• Magnitude of shuttle velocity is vs =ωYo

• Magnitude of truss velocity is vt = ωYo/2


Resonant Frequency• Find maximum potential energy from spring constant

and shuttle deflection:

2max 2

1oyYkW =

• Rayleigh-Ritz equation:

( ) )35/12(4/ bts

y

MMMk

++=ω

both trusses all 8 beams

3

3

3

224L

EWhLEIk z

y ==


Double-Ended Tuning Forks

Trey Roessig, Ph.D., ME Dept., UC Berkeley, 1998


Mode Shapes for Clamped-Clamped Beams

( ) ))]cos()(cosh()sin([sinh( βεβεαβεβεκεφ −+−=

Mode α β κ [ε = x/L]First -1.018 4.730 -0.618Second -0.999 -7.853 -0.663Third -1.000 -10.996 0.711

φ1(ε = 0.5) = 1

φ3(ε = 0.5) = 1

φ2,max= 1


Applying Rayleigh-Ritz to DETFs

Case 1: σr = 0 Case 2: σr = 30 MPamodel comb drive as a point masswith m = 1 pgm

Dimensions: L = 200 μm, h = W = 2 μm, E = 150 GPa, ρ = 2.3 gmcm-3

Use first mode shape for both cases:analyze a single tine Trey Roessig, Ph.D., ME Dept.,

UC Berkeley, 1998


DETF Results

∫ =⎟⎟⎠

⎞⎜⎜⎝

⎛1

0

2

21

2

6.198εεφ d

dd

Bending:

Axial load: ∫ =⎟⎠⎞⎜

⎝⎛1

0

21 85.4ε

εφ d

dd

Kinetic energy (distributed along beam): ∫ =1

0

21 397.0)( εεφ d

eff

eff

rz

Mk

mdWhL

ddd

LWhd

dd

LEI

=+

⎟⎠⎞⎜

⎝⎛+⎟⎟⎠

⎞⎜⎜⎝

⎛

=

∫

∫ ∫1

0

21

21

1

0

1

0

21

2

21

2

32

1

)5.0(φεφρ

εεφσε

εφ

ω

Trey Roessig, Ph.D., ME Dept.,UC Berkeley, 1998

Case 1: 412 kHz

Case 2: 339 kHz


Resonant SensorsThe double-ended tuning fork is an excellent sensor for axial stress, since it converts it into a shift in frequency (or in oscillation period) �these are easily and accurately measured.

Therefore, we can use the DETF as a building block for a variety ofresonant sensors, such as accelerometers (T. A. Roessig, Ph.D. ME, 1998)and gyroscopes (A. A. Seshia, Ph.D. EECS, 2002)

Rayleigh-Ritz Method yields analytical expressions for frequencyshift due to axial stress (residual or applied):

∫ ⎟⎠⎞⎜

⎝⎛=⎟

⎟⎠

⎞⎜⎜⎝

⎛==

− 0

1

21

1

2/1

1 121

21 ε

εφ

ωσσσω d

dd

LWh

ddk

Mk

Mk

dd

dd

r

eff

eff

eff

eff

eff

rr

4.85

design insight!


Finite Element Analysis

Mark Lemkin, Ph.D., ME Dept., UC Berkeley, 1997


Damping• Many sources: ignore internal damping and acoustic radiation from

the anchors � focus on drag from the surrounding gas

Gap dimension can beon the order of mean-freepath � continuum modeldoesn’t apply

David Horsley, Ph.D., ME Dept., UC Berkeley, 1998


Couette Damping

William Clark, Ph.D., EECS Dept., UC Berkeley, 1997

Damping coefficient: b

b = μA / yo

Drag force = Fd = -bvy

Effective viscosity at reduced pressure (p << 50 Torr):

μ= (μp) pyo = (3.7 x 10-2) pyo

units of μp are kg-s-1-m-2 Torr-1

for p in Torrunits fixed


Squeeze-Film Damping

• Plates slide in y direction

Note that it’s just μ, not μp

b ≈ 7μAzo2 / yo

3


Mass-Spring-Damper System

k Mx(t) = Xcos(ωt)

b

)(2

2

tFkxdtdxb

dtxdM =++

Sinusoidal, steady-state response: X(t) = X ejωt

F(t) = Fcos(ωt)

tjtjtjtj FekXebXejXeM ωωωω ωω =++− 2


Second-Order Resonant Transfer Function

( )

1

2

1

1

1ωω

ωω

ω

Qj

kFXjH

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

==−

Solve for the ratio of the phasor displacement X to the phasor driving force F

Above analysis is for the light damping case (Q >> 1)Mk=1ω b

MQ 1ω=


Limiting Cases

Low frequency: 1ωω <<

High frequency: 1ωω >>

Resonant frequency: 1ωω =

Peak width (-3 dB): Q/1ωω =Δ


Magnitude Bode Plot

ω1 ω

| H |dB

0

20

40

60

ω1 + Δωω1 - Δω

-20

k-1 = 1 m/NQ = 1000


Phase Bode Plot

Phase(H)

-90

-45

ω1 ω0

ω1 + Δωω1 - Δω

-135

-180

ee c245 - me c218 introduction to mems design …...5 ee c245 – me c218 fall 2003 lecture 9...

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