NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p1
Lecture 9-1 Thermal Actuators
!! Thermal Actuator fundamental
t
W
Δl
l
CTE: Coefficient of Thermal Expansion (linear range)
CTE : α ≡
Δl 1
l ΔT
(9-1)
where: l is the length, ΔT is the temperature difference
Notice:
a.! CTE is approximately constant for a considerable rage of
temperature (in general, the coefficient increases with an
increase of temperature)
b.! For homogeneous and isotropic material, the coefficient
applied to all dimensions (directions).
c.! CTE of some materials:
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p2
From Eq. (9-1), for the thermal induced strain:
ε therm
Δl
≡
= αΔT
l
(9-2)
For a beam clamped between two walls, the induced stress
and total force are:
Ftherm = Aσ = AEε therm = WtEαΔT
(9-3)
Heat
For a silicon beam, α=2.6×10-6 m/m°C, l=100 µm, W=t=10 µm,
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p3
and ΔT=100 °C, E~1011 Pa, Cp,si=0.714 KJ/Kg°C
F=WtEαΔT=2.6 mN
Δl=lαΔT=26 nm
"! large force, small displacement!!
Input energy:
F
Win= Cp,simΔT=714 pJ
x
Output energy:
ΔL
ΔL
0
0
Wout = ∫ Fdx = ∫
AEx
AE
dx =
(ΔL) 2 = 26.45
L
2L
Efficiency: 3.7 %
=>
Low efficiency!!
pJ
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p4
How to magnify the displacement?
1.! Extending arm
x
X
Displacement increases by X/x, but force decreases by
x/X (for rigid vertical beam, and heating only the
right side horizontal bar)
2.! Different beam width
Hot Arm
A
Si
Flexure beam
Cold Arm
Power Dissipation=I2R => P ∝ R,
here
R=
ρA
L
and I=constant
=> Hot arm has much larger resistance
A’
Metal
Poly1
Poly2
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p5
Hot
θ/2
Deflection
L
R
Cold
θ
Arc length = Rθ = Hot Arm length
Chord length L= (R sin(θ/2))×2= cold arm length
ΔL = Rθ − 2 R sin
ΔL
=
L
θ − 2 sin
2 sin
θ
θ
2
θ
2 = αΔT
2
Use Taylor expansion =>
D = L × sin
θ
θ = 24αΔT
θ
θ
= (2 R ⋅ sin ) sin
2
2
2
θ
θ
(2 R ⋅ sin ) sin
D
2
2 ≈ 12 ≈ 160
=
ΔL
θ
θ θ3
R(θ − 2( − ))
2 48
=> Magnify displacement to 160 times !!
for ΔT = 100°C
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p6
3.! Bimorph beam
(ii) Composite plate
Al
Al
Si
T
T
Al: EAl=73Gpa;
Si: Esi=10e11Pa;
Al=22.5e-6
m/m/
Si=2.3e-6
m/m/
Analysis:
Assume
1.Al much thinner than Si
2.Biaxial Stress
3.Plane Remain plane (No warp)
3D isotropic system:
y
x
x
y
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p7
1
σ x − ν (σ y + σ z )
E
1
ε y = σ y − ν (σ z + σ x )
E
1
ε z = σ z − ν (σ x + σ y )
E
εx =
where
v
[
]
[
]
[
]
:Poisson’s ratio
Biaxial stress =>
σz =0 σx =σ y
1 −ν
2νσ
ε
=
ε
=
σ
ε
=
−
y
x( ORG )
z
=> x
E
E
E
σ =
ε x # Replace
=> x
1 −ν
E
with
E
1 −ν
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p8
Al film
Δl
l
tf
Ff
ts
σf
Al Film
tf
Mf
σs
z
ts/2
Ms
ρ
σs
Si
θ
what’s is
ρ
? and
Mf
?
(1)!Geometry
Si deformation at Boundary= Al film deformation
! + !β s ΔT + Δ! = ! + !β f ΔT − !ε f
Δ! !
! ts
=
Δ
!
=
ts
Here:
=>
ρ
ρ2
2
) 1 −ν &
εf ='
$ σf
( E %f
t s ( 1 −ν %
+&
# σ f = (β f − β s )ΔT
2ρ ' E $ f
(1)
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p9
(2)!Moment balance
M f = σ f ⋅W ⋅ t f ⋅
ts
M s = w∫ t s2
−
2
ts
2
(2)
2
E
z
)
&
zσ (z )dz = w∫ t s2 '
$ dz
−
2 ( 1 −ν % ρ
ts
(3)
3
) E & wts
='
$
( 1 − ν % s 12 ρ
M f = Ms
2
) E & ts
σf ='
$
( 1 − ν % s 6 ρt f
(4)
(4)=>(1)
ts ( 1 − ν % ( E % ts2
+&
= (β f − β s )ΔT
# &
#
2 ρ ' E $ f ' 1 − ν $ s 6 ρt f
t s ( 1 − ν % ( E % t s2
+&
# &
#
2 ' E $ f ' 1 − ν $ s 6t f
ρ=
(β f − β s )ΔT
(3)!Force
(5)
3
( E % wt s
Mf =&
#
' 1 − ν $ s 12 ρ
3
M ( E % wts
F ≈ s =&
#
! ' 1 − ν $ s 12 ρ!
Force generated by thermal deflection
(6)
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p10
(4)!Displacement
d 2z 1
=
2
dx
ρ
1 2
z=
x + c1 x + c2
2ρ
dz
=0
at x=0, z = 0
dx
x2
z=
2ρ
for
x=!
!2
z=
2ρ
(7)
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p11
Example
E(Pa)
β (m/m/
T(µm)
Si
10e11
2.3e-6
Al
73e9
22.5e-6
2
0.1
T=100
Neglecting poisson’s ratio
=0
w=2µm
l=10
m
From equation (5)
(
)
2 × 10 −6 ) 106 & 2 × 10 −6
$
+ ''
9 $
2
73 × 10 % 6 ⋅ 0.1 × 10 − 6
(
ρ=
= 0.005m
(22.5 − 2.3)10− 6 × 100
From equation (6)
−6 3
(
2 ×10 )(2 ×10 )
F ≈ (10 )
12 ⋅ 0.005 ⋅10 ⋅10
−6
11
−6
= 2.67 ×10 −6 N
From equation (7)
2
(
−6 2
)
!
10 × 10
z=
=
= 10 nm
2ρ
2 × 0.005
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p12
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p13
NTHU ESS5850
F. G. Tseng
Micro System Design
Fall/2016, 9-1, p14