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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 1
EE C245 – ME C218Introduction to MEMS Design
Fall 2007
Prof. Clark T.-C. Nguyen
Dept. of Electrical Engineering & Computer SciencesUniversity of California at Berkeley
Berkeley, CA 94720
Lecture 14: Beam Bending
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 2
Announcements
•Move midterm date to Thursday, Nov. 1 (1 week later than original tentative date)
•Website issues
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 3
Lecture Outline
• Reading: Senturia Chpts. 8, 9• Lecture Topics:
Material PropertiesYoung’s modulusYield strengthQuality factor
On-chip Measurement of Material PropertiesAnisotropic Material PropertiesBeam Bending
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 4
Quality Factor (or Q)
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 5
Lr hWr
VPvi
Clamped-Clamped Beam μResonator
Smaller mass higher freq. range and lower series Rx
Smaller mass higher freq. range and lower series Rx(e.g., mr = 10-13 kg)(e.g., mr = 10-13 kg)
Young’s Modulus
DensityMass
Stiffness
ωωο
ivoi
203.121
rr
ro L
hEmkf
ρπ==
Frequency:
Q ~10,000
vi
Resonator Beam
Electrode
VP
C(t)
dtdCVi Po =
Note: If VP = 0V device off
Note: If VP = 0V device off
io
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 6
Quality Factor (or Q)•Measure of the frequency selectivity of a tuned circuit
• Definition:
• Example: series LCR circuit
• Example: parallel LCR circuit
dB3CyclePer Lost EnergyCyclePer Energy Total
BWfQ o==
( )( ) CRR
LZZQ
o
o
ωω 1
ReIm
===
( )( ) LGG
CYYQ
o
o
ωω 1
ReIm
===
BW-3dB
fo f
4
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 7
Selective Low-Loss Filters: Need Q
• In resonator-based filters: high tank Q ⇔ low insertion loss
• At right: a 0.1% bandwidth, 3-res filter @ 1 GHz (simulated)
heavy insertion loss for resonator Q < 10,000 -40
-35
-30
-25
-20
-15
-10
-5
0
998 999 1000 1001 1002
Frequency [MHz]
Tran
smis
sion
[dB
]
Tank Q30,00020,00010,0005,0004,000
Increasing Insertion
Loss
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 8
•Main Function: provide a stable output frequency• Difficulty: superposed noise degrades frequency stability
ωωο
ivoi
A
Frequency-SelectiveTank
SustainingAmplifier
ov
Ideal Sinusoid: ( ) ⎟⎠⎞⎜
⎝⎛= tofVotov π2sin
Real Sinusoid: ( ) ( ) ( )⎟⎠⎞⎜
⎝⎛⎟
⎠⎞⎜
⎝⎛ ++= ttoftVotov θπε 2sin
ωωο
ωωο=2π/TO
TO
Zero-Crossing Point
Tighter SpectrumTighter Spectrum
Oscillator: Need for High Q
Higher QHigher Q
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 9
• Problem: IC’s cannot achieve Q’s in the thousandstransistors consume too much power to get Qon-chip spiral inductors Q’s no higher than ~10off-chip inductors Q’s in the range of 100’s
•Observation: vibrating mechanical resonances Q > 1,000• Example: quartz crystal resonators (e.g., in wristwatches)
extremely high Q’s ~ 10,000 or higher (Q ~ 106 possible)mechanically vibrates at a distinct frequency in a thickness-shear mode
Attaining High Q
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 10
Energy Dissipation and Resonator Q
supportviscousTEDdefects QQQQQ11111
+++=
Anchor LossesAnchor Losses
Material Defect Losses
Material Defect Losses Gas DampingGas Damping
Thermoelastic Damping (TED)Thermoelastic Damping (TED)
Bending CC-Beam
CompressionHot Spot
Tension Cold Spot Heat Flux(TED Loss)
Elastic Wave Radiation(Anchor Loss)
At high frequency, this is our big problem!
At high frequency, this is our big problem!
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 11
Thermoelastic Damping (TED)
•Occurs when heat moves from compressed parts to tensioned parts → heat flux = energy loss
QfT
21)()( =ΩΓ=ς
pCTETρ
α4
)(2
=Γ
⎥⎦
⎤⎢⎣
⎡
+=Ω 222)(
fffff
TED
TEDo
22 hCKf
pTED ρ
π=
Bending CC-Beam
CompressionHot Spot
Tension Cold Spot Heat Flux(TED Loss)
ζ = thermoelastic damping factorα = thermal expansion coefficientT = beam temperatureE = elastic modulusρ = material densityCp = heat capacity at const. pressureK = thermal conductivityf = beam frequencyh = beam thicknessfTED = characteristic TED frequency
h
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 12
TED Characteristic Frequency
• Governed byResonator dimensionsMaterial properties
22 hCKf
pTED ρ
π=
ρ = material densityCp = heat capacity at const. pressureK = thermal conductivityh = beam thicknessfTED = characteristic TED frequency
Critical D
amping
Fac
tor,
ζ
Relative Frequency, f/fTED
Q
[from Roszhart, Hilton Head 1990]
Peak where Q is minimizedPeak where Q is minimized
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 13
Q vs. Temperature
Quartz Crystal Aluminum Vibrating Resonator
Q ~5,000,000 at 30K
Q ~5,000,000 at 30K
Q ~300,000,000 at 4KQ ~300,000,000 at 4K
Q ~500,000 at 30K
Q ~500,000 at 30K
Q ~1,250,000 at 4KQ ~1,250,000 at 4K
Even aluminum achieves exceptional Q’s at
cryogenic temperatures
Even aluminum achieves exceptional Q’s at
cryogenic temperatures
Mechanism for Q increase with decreasing temperature thought to be linked to less hysteretic motion of material defects
less energy loss per cycle
Mechanism for Q increase with decreasing temperature thought to be linked to less hysteretic motion of material defects
less energy loss per cycle
[from Braginsky, Systems With
Small Dissipation]
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 14
Output
Input
Input
Output
Support Beams
Wine Glass Disk Resonator
R = 32 μm
Anchor
Anchor
Resonator DataR = 32 μm, h = 3 μmd = 80 nm, Vp = 3 V
Resonator DataR = 32 μm, h = 3 μmd = 80 nm, Vp = 3 V
-100
-80
-60
-40
61.325 61.375 61.425
fo = 61.37 MHzQ = 145,780
Frequency [MHz]
Unm
atch
ed T
rans
mis
sion
[dB
]
Polysilicon Wine-Glass Disk Resonator
[Y.-W. Lin, Nguyen, JSSC Dec. 04]
Compound Mode (2,1)
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 15
-100-98-96-94-92-90-88-86-84
1507.4 1507.6 1507.8 1508 1508.2
1.51-GHz, Q=11,555 Nanocrystalline Diamond Disk μMechanical Resonator
• Impedance-mismatched stem for reduced anchor dissipation
•Operated in the 2nd radial-contour mode•Q ~11,555 (vacuum); Q ~10,100 (air)• Below: 20 μm diameter disk
PolysiliconElectrode R
Polysilicon Stem(Impedance Mismatched
to Diamond Disk)
GroundPlane
CVD DiamondμMechanical Disk
Resonator Frequency [MHz]
Mix
ed A
mpl
itude
[dB
]
Design/Performance:R=10μm, t=2.2μm, d=800Å, VP=7Vfo=1.51 GHz (2nd mode), Q=11,555
fo = 1.51 GHzQ = 11,555 (vac)Q = 10,100 (air)
[Wang, Butler, Nguyen MEMS’04]
Q = 10,100 (air)
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 16
Disk Resonator Loss Mechanisms
Disk Stem
Nodal Axis
Strain Energy Flow No motion along
the nodal axis,but motion along the finite width
of the stem
No motion along the nodal axis,
but motion along the finite width
of the stem
Stem Heightλ/4 helps
reduce loss, but not perfect
λ/4 helps reduce loss,
but not perfect
Substrate Loss Thru Anchors
Substrate Loss Thru Anchors
Gas Damping
Gas Damping
Substrate
Hysteretic Motion of
Defect
Hysteretic Motion of
Defect
e-
Electronic Carrier Drift Loss
Electronic Carrier Drift Loss
(Not Dominant in Vacuum)
(Dwarfed By Substrate Loss)
(Dwarfed By Substrate Loss)
(Dominates)
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 17
MEMS Material Property Test Structures
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 18
Stress Measurement Via Wafer Curvature
• Compressively stressed film →bends a wafer into a convex shape
• Tensile stressed film → bends a wafer into a concave shape
• Can optically measure the deflection of the wafer before and after the film is deposited
• Determine the radius of curvature R, then apply:
RthE
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2′=σ
σ = film stress [Pa]E′ = E/(1-ν) = biaxial elastic modulus [Pa]h = substrate thickness [m]t = film thicknessR = substrate radius of curvature [m]
Si-substrate
Laser
Detector
MirrorxScan the
mirror θ
θ
xSlope = 1/R
h
t
R
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 19
MEMS Stress Test Structure
• Simple Approach: use a clamped-clamped beam
Compressive stress causes bucklingArrays with increasing length are used to determine the critical buckling load, where
Limitation: Only compressive stress is measurable
2
22
3 LEh
criticalπσ −=
E = Young’s modulus [Pa]I = (1/12)Wh3 = moment of inertiaL, W, h indicated in the figure
L
W
h = thickness
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 20
More Effective Stress Diagnostic
• Single structure measures both compressive and tensile stress
• Expansion or contraction of test beam → deflection of pointer
• Vernier movement indicates type and magnitude of stress
Expansion → Compression
Contraction → Tensile
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 21
Output
Input
Input
Output
Support Beams
Wine Glass Disk Resonator
R = 32 μm
Anchor
Anchor
Resonator DataR = 32 μm, h = 3 μmd = 80 nm, Vp = 3 V
Resonator DataR = 32 μm, h = 3 μmd = 80 nm, Vp = 3 V
-100
-80
-60
-40
61.325 61.375 61.425
fo = 61.37 MHzQ = 145,780
Frequency [MHz]
Unm
atch
ed T
rans
mis
sion
[dB
]
Q Measurement Using Resonators
[Y.-W. Lin, Nguyen, JSSC Dec. 04]
Compound Mode (2,1)
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 22
Folded-Beam Comb-Drive Resonator
• Issue w/ Wine-Glass Resonator: non-standard fab process• Solution: use a folded-beam comb-drive resonator
8.3 Hz
3.8500,342
=Q
fo=342.5kHzQ=41,000
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 23
Comb-Drive Resonator in Action
• Below: fully integrated micromechanical resonator oscillator using a MEMS-last integration approach
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 24
Folded-Beam Comb-Drive Resonator
• Issue w/ Wine-Glass Resonator: non-standard fab process• Solution: use a folded-beam comb-drive resonator
8.3 Hz
3.8500,342
=Q
fo=342.5kHzQ=41,000
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 25
Measurement of Young’s Modulus
• Use micromechanical resonatorsResonance frequency depends on EFor a folded-beam resonator: 213)(4
⎥⎥⎦
⎤
⎢⎢⎣
⎡=
eqo M
LWEhfResonance Frequency =
LW
h = thickness
Young’s modulus
Equivalent mass
• Extract E from measured frequency fo
•Measure fo for several resonators with varying dimensions
• Use multiple data points to remove uncertainty in some parameters
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 26
Anisotropic Materials
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 27
Elastic Constants in Crystalline Materials
• Get different elastic constants in different crystallographic directions → 81 of them in all
Cubic symmetries make 60 of these terms zero, leaving 21 of them remaining that need be accounted for
• Thus, describe stress-strain relations using a 6x6 matrix
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
xy
zx
yz
z
y
x
xy
zx
yz
z
y
x
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
γγγεεε
τττσσσ
665646362616
565545352515
464544342414
363534332313
262524232212
161514131211
Stiffness CoefficientsStresses Strains
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 28
Stiffness Coefficients of Silicon
• Due to symmetry, only a few of the 21 coefficients are non-zero
•With cubic symmetry, silicon has only 3 independent components, and its stiffness matrix can be written as:
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
=
⎥⎥⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢⎢⎢
⎣
⎡
xy
zx
yz
z
y
x
xy
zx
yz
z
y
x
CC
CCCCCCCCCC
γγγεεε
τττσσσ
66
55
44
332313
232212
131211
000000000000000000000000
whereC11 = 165.7 GPaC12 = 63.9 GPaC44 = 79.6 GPa
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 29
Young’s Modulus in the (001) Plane
[units = 100 GPa]
[units
= 1
00 G
Pa]
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 30
Poisson Ratio in (001) Plane
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 31
Anisotropic Design Implications
• Young’s modulus and Poisson ratio variations in anisotropic materials can pose problems in the design of certain structures
• E.g., disk or ring resonators, which rely on isotropic properties in the radial directions
Okay to ignore variation in RF resonators, although some Q hit is probably being taken
• E.g., ring vibratory rate gyroscopes
Mode matching is required, where frequencies along different axes of a ring must be the sameNot okay to ignore anisotropic variations, here Ring Gyroscope
Drive AxisDrive Axis
Sense
Axis
Sense
Axis
Wine-Glass Mode Disk
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 32
Bending of Beams
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 33
Beams: The Springs of Most MEMS
• Springs and suspensions very common in MEMSCoils are popular in the macro-world; but not easy to make in the micro-worldBeams: simpler to fabricate and analyze; become “stronger” on the micro-scale → use beams for MEMS
Comb-Driven Folded Beam Actuator
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 34
Bending a Cantilever Beam
•Objective: Find relation between tip deflection y(x=L) and applied load F
• Assumptions:1. Tip deflection is small compared with beam length2. Plane sections (normal to beam’s axis) remain plane and
normal during bending, i.e., “pure bending”3. Shear stresses are negligible
F
x′ L
x
Clamped end condition:At x=0:
z=0dz/dx = 0
Free end condition
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 35
Reaction Forces and Moments
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 36
Sign Conventions for Moments & Shear Forces
(+) moment leads to deformation with a (+) radius of curvature
(i.e., upwards)
(-) moment leads to deformation with a (-) radius of curvature (i.e., downwards)
(+) shear forces produce clockwise
rotation
(-) shear forces produce counter-clockwise rotation
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EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 37
Beam Segment in Pure Bending
Small section of a beam bent in response to a tranverse load R
EE C245: Introduction to MEMS Design Lecture 14 C. Nguyen 10/11/07 38
Beam Segment in Pure Bending (cont.)