Lecture 23 Lecture 23
Piezoresistive Pressure Sensor Piezoresistive Pressure Sensor
• Sensing Pressure
• Piezoresistance
• Piezoresistance
• Analytic Formulation in Cubic Materials
d l d
• Longitudinal and Transverse Piezoresistance
• Piezoresistive Coefficients of Silicon
• Structural Examples
• Averaging over Stress and Doping Variations
• A Numerical Example
Sensing Pressure Sensing Pressure
- Sensing pressure directly
g g
piezoelectric material : transduce normal stress into voltage.
- Sensing deformation of diaphragm
capacitance change, some optical signature, change in current in a tunneling tip.
- Sensing change in resistance with stress (strain) piezoresistive
- Sensing change in resonant frequency.
diaphragm displacement creates a changing stress in the resonant structure, thereby shifting its resonant frequency.
Piezoresistance Piezoresistance
- Piezoresistivity : the dependence of electrical resistivity on strain.
- The resistivity of a material depends on the internal atom positions and theirThe resistivity of a material depends on the internal atom positions and their motions.
- Strains change these arrangements and, hence, the resistivity.g g , , y - Metal.
Changing the internal atomic positions by applying stresses to the metal distorts the energy bands slightly, resulting in small changes in the amount of conduction that results from an applied field.
- Semiconductor
When the internal atomic positions in a semiconductor are changed by the
li i f h b d d i b ll
application of a stress, these band-edge energies move by small amounts.
But even small shifts can have enormous effects on the conductivity properties.
properties.
The effect can be considerably larger than in metals.
Analytic Formulation in Cubic Materials (I) Analytic Formulation in Cubic Materials (I)
- Piezoresistive effect : a fourth-rank tensor
(resistivity : a second-rank tensor, stress : a second-rank tensor)
[ ] J
E = ρ
e+ Π ⋅ σ
: the electric field intensity
: the resistivity tensor, : the full fourth-rank piezoresistive tensor : the full second-rank stress tensor, : the current density.
( )
ρ
eΠ
σ
JE
Summation convention
(
e ij ijk k)
ji
J
E = ρ
,+ Π
lσ
lj k jk j
j e j
k jk j
j e j
k jk j
j
e
J J , E J J , E J J
E
1= ρ
,1+ Π
1 lσ
l 2= ρ
,2+ Π
2 lσ
l 3= ρ
,3+ Π
3 lσ
lCubic material resistivity : only diagonal terms exist.
⎞
⎛ 0 0
j k jk j
j e j
k jk j
j e j
k jk j
j
e,1 1 l l 2 ,2 2 l l 3 ,3 3 l l
1
⎟ ⎟
⎟
⎠
⎞
⎜ ⎜
⎜
⎝
⎛
=
e e
e
ρ
eρ ρ
ρ
0 0
0 0
0 0
⎠
⎝ ρ
eAnalytic Formulation in Cubic Materials (II) Analytic Formulation in Cubic Materials (II)
j k jk
e
J J
E
1= ρ
1+ Π
1 lσ
l) ① (
3 13
2 12
1 11
1
→ Π
+ Π
+ Π
Π + Π
+ Π
= Π
J J J
J
J j k k k k k k
k
jkl l l l l l l l
σ σ
σ
σ σ
σ σ
The Second term
) ③ (
) ② (
) ① (
3 3 133 2
132 1
131
2 3 123 2
122 1
121
1 3 113 2
112 1
111
→ Π
+ Π
+ Π
+
→ Π
+ Π
+ Π
+
→ Π
+ Π
+ Π
=
J J J
l l l
l l
l
l l l
l l
l
l l l
l l
l
σ σ
σ
σ σ
σ
σ σ
σ
33 1133 32
1132 31
1131 23
1123 22
1122 21
1121 13
1113 12
1112 11
①⇒ Π1111σ +Π σ +Π σ +Π σ +Π σ +Π σ +Π σ +Π σ +Π σ Applying 11→1 22→2 33→3 23→4 31→5 12→6
3 63 4
64 5
65 4
64 2
62 6
66 5
65 6
66 1
61
3 13 4
14 5
15 4
14 2
12 6
16 5
15 6
16 1
11
② σ σ σ σ σ σ σ σ σ
σ σ
σ σ
σ σ
σ σ
σ
Π + Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
⇒
Π + Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
=
Applying 11→1, 22→2, 33→3, 23→4, 31→5, 12→6
3 53 4
54 5
55 4
54 2
52 6
56 5
55 6
56 1
51
3 63 4
64 5
65 4
64 2
62 6
66 5
65 6
66 1
61
③
②
σ σ
σ σ
σ σ
σ σ
σ
σ σ
σ σ
σ σ
σ σ
σ
Π + Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
⇒
Π + Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
+ Π
⇒
Analytic Formulation in Cubic Materials (III) Analytic Formulation in Cubic Materials (III)
- Using cubic symmetry for piezoresistive coefficients and using the intrinsic symmetry of the stress tensor (p.194).
For cubic material
⎟⎞
⎜⎛Π12 Π12 Π12 0 0 0
⎟⎟
⎟⎟
⎟
⎜⎜
⎜⎜
⎜
Π Π
Π Π
Π Π
Π
= Π
44 11
12 12
12 11
12
0 0
0 0
0
0 0
0
0 0
0
⎟⎟
⎟⎟
⎜⎜ ⎠
⎜⎜
⎝ Π
Π Π
44 44
44
0 0
0 0
0
0 0
0 0
0
0 0
0 0
0
3 12 2
12 1
11
3 12 4
5 4
2 12 6
6 6
1 11
0 0
0 0
0 0
0
0 0
0 0
0 0
σ σ
σ
σ σ
σ σ
σ σ
σ σ
σ
+ +
+ +
+ Π
+ +
Π +
Π + Π
+ Π
=
Π +
⋅ +
⋅ +
⋅ + Π
+
⋅ +
⋅ +
⋅ + Π
⇒
②
①
3 4
5 44 4
2 6
5 44 6
1
12 44 12
44 6
44
3 4
5 4
2 6
44 6
6 44 1
0 0
0 0
0 0
0
2 2
2
0 0
0 0
0 0
0
σ σ
σ σ
σ σ
σ σ
σ
τ σ
σ
σ σ
σ σ
σ σ
σ σ
σ
⋅ +
⋅ + Π
+
⋅ +
⋅ +
⋅ + Π
+
⋅ +
⋅
⇒
Π
= Π
= Π
=
⋅ +
⋅ +
⋅ +
⋅ +
⋅ + Π
+
⋅ + Π
+
⋅
⇒
③
②
13 44 13
44 5
44 2 2
2Π
σ
= Πσ
= Πτ
=
Analytic Formulation in Cubic Materials (IV) Analytic Formulation in Cubic Materials (IV)
Applying (18 6)
2 2
,
,
12 1122 44 2323 441111
11 e e
e
π ρ π ρ π
ρ = Π = Π = Π = Π
Applying (18.6)
) (
) (
2 2
) (
3 13 2
12 44
1 3 12 2
12 1
11 1
3 13 44 2
12 44 1
3 12 2
12 1
11 1
1
J J
J J
J J
J J
E
e e
e e
τ τ
π ρ σ
π σ
π σ
π ρ ρ
τ τ
σ σ
σ ρ
+ +
+ +
+
=
Π +
Π +
Π + Π
+ Π
+
=
[ ]
[ 1
11 1 12(
2 3) ]
1 44(
12 2 13 3)
1
J J J
E ρ
e= + π σ + π σ + σ + π τ + τ
∴
[ ]
[ 1 1
11 3 12( (
1 2) ) ]
3 44( (
13 1 23 2) )
3
1 12 3
23 44
2 1
3 12
2 11 2
J J
J E
J J
J E
e e
τ τ
π σ
σ π
σ π ρ
τ τ
π σ
σ π
σ π ρ
+ +
+ +
+
=
+ +
+ +
+
=
Longitudinal and Transverse Piezoresistance (I) Longitudinal and Transverse Piezoresistance (I)
- If a relatively long, relatively narrow resistor is defined in a planar structure, the primary current density and electric field are both along the long axis of the primary current density and electric field are both along the long axis of the resistor.
- This axis need not coincide with the cubic crystal axes.
Th f it i t k h t t f th i i ti
- Therefore, it is necessary to know how to transform the piezoresistive equations to an arbitrary coordinate system.
- The structures are typically designed so that one of the axes of principal in- plane stress is also along the resistor axis.
R +
Δ
where R is the resistance of the resistor, and the subscripts ℓ and t refer to
t
R = π l σ l + π t σ
where R is the resistance of the resistor, and the subscripts ℓ and t refer to longitudinal and transverse stresses with respect to the resistor axis.
Longitudinal and Transverse Piezoresistance (II) Longitudinal and Transverse Piezoresistance (II)
) )(
(
2
11 12 44 12 12 12 12 12 1211
− − −
m+
n+
m n= l l
l
π π π π
π
- General expressions of and
π
lπ
tn
crystal resistor
where (ℓ1, m1, n1) is the set of direction cosines between
) )(
(
) )(
(
2 2 2 1 2
2 2 1 2
2 2 1 44
12 11
12
1 1 1
1 1
1 44
12 11
11
n n m
m
n m n
m
t
= + − − l l + +
l
l
l
π π
π π
π
π π
π π
π
ml
axisy
transverse(2) longitudinal(1)
the longitudinal resistor direction (subscript 1) and the crystal axis.
(ℓ2, m2, n2) is the set of direction cosines between the
transverse resistor direction (subscript 2) and the crystal
longitudinal(1)
) n( z
transverse resistor direction (subscript 2) and the crystal axis.
- In many silicon micromachined devices, resistors are oriented along [110] directions in (100) wafers.
) m( y
) l( x
transverse longitudinal
[110]
) 0 , 2 1
, 2 1
( ) , ,
( ), 0 , 2 1
, 2 1
( ) , ,
( l
1m
1n
1= l
2m
2n
2= −
Therefore
oriented along [110] directions in (100) wafers. longitudinal
Therefore
1
) 2 (
) 1 0 0 4 1 )(
(
2
11 12 44 11 12 4411 110
,
π π π π π π π
π
l= − − − + + = + +
) 2 (
) 1 0 4 1 4 1 )(
(
11 12 44 11 12 4412 110
,
π π π π π π π
π
t= + − − + + = + −
Piezoresistive Coefficients of Silicon Piezoresistive Coefficients of Silicon
- The coefficients of silicon depend strongly on the doping type, a reflection of the fact that the detailed valance-band and conduction-band structures in
π
silicon are different.
Typical room-temperature piezoresistance coefficients for n- and p-type silicon
Type Resistivity
Units ohm-cm 10-11 Pa-1 10-11 Pa-1 10-11 Pa-1
n-type 11.7 -102.2 53.4 -13.6
π
11π
12π
44yp
p-type 7.8 6.6 -1.1 138.1
- These coefficients are weak functions of doping level for doping below about 1019 3 b t th d k dl t hi h d i
1019cm-3 but then decrease markedly at high doping.
- The coefficients decrease with increasing temperature, dropping about 0.7 of their room-temperature value at 150℃.
At hi h d i th t t d d f th i i ti ffi i t
- At higher doping, the temperature dependence of the piezoresistive coefficients becomes small.
- For a wide temperature range, there may be a design advantage in sacrificing piezoresistive sensitivity in exchange for small temperature dependences by piezoresistive sensitivity in exchange for small temperature dependences by using heavily doped piezoresistors.
Structural Examples ( Structural Examples (I) I)
- Micromachined piezoresistive accelerometer structure.
- Assuming that the resistor orientations are along [110] direction.
2 31 )
6 13 4 53 2
102 1 (
)
1 ( + + +
n-type piezoresistor
6 . 17 )
6 . 13 4 . 53 2
. 102 2 (
) 1 2 (
1
2 . 31 )
6 . 13 4 . 53 2
. 102 2 (
) 2 (
44 12
11
44 12
11
−
= +
+
−
=
− +
=
−
=
− +
−
= +
+
=
π π
π π
π π
π π
t l
- p-type piezoresistors have a larger sensitivity and
π
l andπ
t coefficients have p-type piezoresistor2 2
3 . 66 )
1 . 138 1
. 1 6 . 6 2( 1
, 8 . 71 ) 1 . 138 1
. 1 6 . 6 2(
1 − + = = − − =−
= πt
πl
- p-type piezoresistors have a larger sensitivity and and coefficients have opposite signs and almost equal magnitudes. l
π π
tStructural Examples (
Structural Examples (II) II)
- The transverse resistor orientation has the potential for the largest response because, if it can be placed at exactly the right point, the entire resistor will experience the maximum bending stress.
- This orientation is very susceptible to manufacturing variations that can arise from small photographic alignment errors.o s a p otog ap c a g e t e o s
- The longitudinal resistor must extend over some finite length along the cantilever and, for alignment reasons, may also extend onto the support.
- Not every part of the resistor experiences the maximum stress and some loss of - Not every part of the resistor experiences the maximum stress, and some loss of sensitivity will result.
- Which orientation is best? That is a system issue.
Th t i t di t t ifi ti f iti it d
- The system requirements dictate specifications for sensitivity, accuracy and precision.
- If the ultimate in sensitivity is required, it may be necessary to use the transverse orientation.
- A longitudinal may allow a less robust compensation circuit and a less costly manufacturing and calibration procedure.
Structural Examples (
Structural Examples (III) III)
- All resistor axes are along one of the <110>
directions.
- The longitudinal stress on R1 and R3 is the transverse stress at R2 and R4, and vice versa.
- If resistor es sto R11 experiences a longitudinal stress e pe e ces a o g tud a st ess σz , it must simultaneously experience a
transverse stress ν σz (ν is the Poisson ratio).
- The total change in resistance forThe total change in resistance for RR11 would bewould be
) (
1
1
= + = +
Δ π
lσ
lπ
tσ
tπ
lνπ
tσ
l RR
p-type
064 . 0
1
ν =
σ
lin the [110] direction of (100) plane.
l
l π σ
π
2 11
11 1
11 1 11
10 704
61 )
8 71 064
0 3 66 (
10 556
. 67 10
3 . 66 ,
10 8 . 71
−
−
−
−
×
× Δ +
× Δ =
→
×
−
=
×
= R
R R
t
σ
ll
l σ
σ 11
2
2 = (−66 .3 + 0.064 × 71.8) = −61 .704 ×10 R
Structural Examples (
Structural Examples (IV) IV)
R R
R R
R
R
1=
3= ( 1 + α
1) ,
2=
4= ( 1 − α
2)
- Wheatstone-bridge circuit
- α1 and α2 represent the product of the effective piezoresistive coefficient and the stress.
o
o
R R R
R R
R
1 3( 1 + α
1) ,
2 4( 1 α
2)
4 2
3 1
4 2 3
1 4
1
,
) )(
( = =
+ +
= −
− +
= +
S S So V R R R R
R R
R R
R R R
V R R R
V R R R
V R
Q
2 1
2 1
2 1
2 1
2 1
2 1
4 3
2 1
4 3
2 1
2 )
1 ( ) 1
(
) 1
( ) 1
(
) )(
(
α α
α α
α α
α α
− +
= +
− + +
−
−
= + +
= −
+ +
+ +
VS
R R
R R
R R
R R
R R
R R
Therefore,
2 1
2 1
2 α α α α
− +
= +
S o
V V
- Since α1 and α2 are typically small (on the order of 0.02 or less), and differ from each other by only 10 %, this bridge gives an optimally large output
i h l li i
without a large nonlinearity.
Averaging over Stress and Doping Variations ( Averaging over Stress and Doping Variations (I) I)
- Piezoresistors are typically formed by diffusion, hence, have nonuniform doping.
They also span a finite area on the device, hence, have nonuniform stress.
- A good approximation is to begin with the doping variation which occurs in a g pp g p g direction normal to the surface.
- The resistor can be thought of as a stack of slices, each slice having a slightly different doping
different doping.
- The slices are connected electrically in parallel.
W dz G = ∫
z j1 =
z dz G L
R
o=
o= ∫
0 e o,
( )
ρ
where ρ e,o(z) is the unstrained doping-dependent electrical resistivity, L is the length of the resistor, and g , zjj is depth of the resistor.p
- Recall that in pure bending, the stress goes from tensile at one surface to compressive at the other. If the piezoresistor goes all the way through the structure the stress goes response will average to zero
structure, the stress goes response will average to zero.
- At the other extreme, if zj is very small compared to the structural thickness, one can use the surface stress to the estimate the response.
- In practical structures, zj may not negligibly be small, so stress-averaging may be necessary.
Averaging over Stress and Doping Variations (
Averaging over Stress and Doping Variations (II) II)
=0
Z Thickness H
- We consider first a case of a cantilever in pure bending, with no variation of bending stress along the axial direction.
-The longitudinal stress due to bending
ρ ρθ
ρθ θ
ε z = ( ρ + H 2 − z ) − = H 2 − z )
( ρ
H Z =
Neutral f
The longitudinal stress due to bending Strain at z
ρ ρθ
σ z = E ⋅ H 2 ρ − z )
l
(
θ
⎟⎠
⎜ ⎞
⎝
⎛ 2 H
surface
Stress at z
)
( W dz
- Since this is a cantilever, we can ignore transverse stresses.
The conductance g(z) of a single differential slice of thickness dz
) ) (
( z L z
g
ρ
e=
where ρ e(z) is the strained resistivity which varies with depth due both to doping and to stress
[ 1 ( ) ]
) ( )
( z
e,0z z
e
ρ π
lσ
lρ = +
dz W dz
W
doping and to stress.
- The stress-dependence - The conductance
[ ] [ 1 ( ) ]
) ( )
( 1
) ) (
(
, ,
z z L
dz W z
z L
dz z W
g
o e o
e l l
l l
σ ρ π
σ π
ρ + ≅ −
=
Averaging over Stress and Doping Variations (
Averaging over Stress and Doping Variations (III) III)
- The total conductance
[
z]
dzL W R
zj
) ( ) 1
( 1
0 −πlσl
=
∫
- Let the first term be Ro z
L
R
∫
0 ρe,o( )1 1
1 W
-Since A is small,
) 1
1 ( )
) ( ( 1
1
0 ,
A R R
dz z z
L W R
R o o
z
o e o
j = −
−
=
∫ ρ π
lσ
l) ) ( 1
( )
1 1 (
0
W z dz
R R
A R R
R
R = ≅ + = + ∫
zjπ
lσ
l- If we assume that zj is much less than H/2 , we can replace σz by EH/2ρ .
Since A is small,
( ) )
) 1 (
( )
1
1
0(
0 ,0
z dz
z R L
R A
R A R
R R R
o e o
o o
o
π
lσ
l∫ ρ +
+
− ≅
But there still may be some z-dependence in the πz coefficient because of doping dependence.
- Equivalent importance is the variation of stress along the length of a q p g g resistor that is in the longitudinal orientation
- We can think of cutting the resistor along its length into segments, and evaluate the resistance of each segment and sum the result, since the evaluate the resistance of each segment and sum the result, since the segments are electrically in series.
A Numerical Example A Numerical Example
il l h 200 id h 20 hi k
- n- type cantilever : length 200 ㎛, width 20 ㎛, thickness 5 ㎛.
- Point load at the free end.
- A longitudinal p-type piezoresistor.
-The piezoresistor is formed into two parallel
narrow longitudinal strips (20 ㎛ in length, 2 ㎛ in width, and a depth into the substrate of zj =0.2 ㎛).
⎟⎟⎞
⎜⎜⎛
⎟⎟⎞
⎜⎜⎛
= x x
w
w 3 2 1
- Interconnect regions : wide, more heavily doped than the piezoresistor portions.
- Assuming that the cantilever tip is displaced downward by an amount wmax
⎟⎟⎠
⎜⎜⎝ −
⎟⎟⎠
⎜⎜⎝
=
c
c L
w L
w 1 3
2 max
where x is measured from the support, Lc is the length of the cantilever.
- The magnitude of the radius curvature and σz
− σz = 30 MPa at x = 0 and 27 MPa at x = 20 μm when w = 1 μm and E= 160 GPa.
3 max max
2 2
2
) (
3 , 2
) (
3 1
c c c
c
L
x L w
EH EH
L
x L w
dx w
d = − = = −
= σ ρ
ρ l
σz = 30 MPa at x = 0 and 27 MPa at x = 20 μm when wmax = 1 μm and E= 160 GPa.
0205 0
0 ) 10 3 66 ( 10
5 28 10
8
71 ×
11× ×
6+ − ×
11× =
= +
=
Δ
R Rπ σ π σ
− −- Since the piezoresistance is a linear function of stress, the average stress (28.5 MPa) is used.
0205 .
0 0 ) 10 3 . 66 ( 10
5 . 28 10
8 .
71 × × × + − × × =
= +
=
Δ
R Rπ
lσ
lπ
tσ
t- 2 % change in resistance due to bending.
The Motorola MAP Sensor / Process Flow ( The Motorola MAP Sensor / Process Flow (II))
- Motorola manifold-absolute-pressure (MAP) sensor
- It uses piezoresistance to measure diaphragm bending.
- It integrates the signal-conditioning and calibration circuitry onto the same chip as diaphragm.
- Unusual respects of Motorola MAP Sensor 1. High-volume fully-integrated silicon
pressure sensor
2. Bipolar transistors instead of MOS.
(111) wafer → (100) wafer used.
3. Only one piezoresistor, oriented in a [100]
direction and located near the edge of the diaphragm, to sense the bending stress.
- Typical bipolar process
- n-epi layer on (100) p-type waferp y ( ) p yp
- deep p–type diffusion through the n-epi layer.
Process Flow (
Process Flow (II) II)
- The bipolar transistors require a p+ diffusion to form the base followed by an n+ diffusion to form th itt
the emitter.
- The lightly p-doped piezoresistors require their own process step.
Th b diff i hi hl d d
- The p+ base diffusion : highly-doped interconnect for the piezoresistors.
- The n+ emitter diffusion : collector contact.
Th i l ll t
- The n-epi layer : collector
- The transistors : the three op-amps needed for the signal-conditioning and calibration circuitry.
C Si ll thi fil i t t i i d i - Cr, Si alloy thin-film resistors : trimming during the calibration step.
- Passivation overcoat (oxide) over the electronics portion of the chip.
The n epita ial la e diaph agm (thickness nifo mit p n j nction etch stop) - The n-epitaxial layer : diaphragm (thickness uniformity, p-n junction etch stop) - Glass-frit bonding.
Details of the Diaphragm and Piezoresistor ( Details of the Diaphragm and Piezoresistor (I) I)
- Typical diaphragm : 1000 ㎛ × 1000 ㎛, thickness 20 ㎛.
- Four contacts of the p-type piezoresistor two : the current along the resistor axis two : the current along the resistor axis.
the other two : transverse voltage taps, connected to a high-impedance op-amp input so as to draw no current
draw no current.
- The dark-shaded contact regions : low resistance
→ Wide and Pt
Di h d [110] di ti
- Diaphragm edge : a [110] direction
- “1” axis : the resistor, “3” axis : normal to the diaphragm.
1
3
) 31 18 ( )
1
( J
E
- J1≠0, J2=0, J3=0, σ3=0, τ13=0, τ12=,τ23=0 in (18.3), (18.4) and (18.5)
Electric field 2
) 33 . 18 ( 0
) 32 . 18 (
) 31 . 18 ( )
1 (
3
1 12 44 2
1 2 12 1
11 1
=
=
+ +
=
E
J E
J E
e e
τ π ρ
σ π σ
π ρ
Assumption : J1 does not vary with depth.
Details of the Diaphragm and Piezoresistor (
Details of the Diaphragm and Piezoresistor (II) II)
- V1 : voltage along the length of the piezoresistor
) 34 18 ( )
1 (
1 ⎥⎤
⎢⎡
∫
∫
LR E d L J LR dV
The change in voltage due to a length averaging of the piezoresistance effect
) 34 . 18 ( )
1 (
1 0 11 1 12 2 1
0 1 1 1
1 ⎥
⎦
⎢ ⎤
⎣
⎡ + +
=
=
∫ ∫
RR R
e
R dx
J L L dx
E
V ρ π σ π σ
The unstrained resistivity of the piezoresistor
averaging of the piezoresistance effect.
Negligible because of a few percent resistance changes.
- The transverse voltage (WR : the width of the piezoresistor)
Th t lt d d l th h t i th i b t
WR
E V2 = 2
) 35 . 18 ( )
/ (
) /
( 1 44 12 1
12 44 1
12 44
2 J W V L W W L V
V = ρeπ τ R = ρeπ τ ρe R ⋅ R = π τ R R
- The transverse voltage depends only on the shear stress τ12 in the region between the voltage taps.
- The placement of the voltage taps determines where the sensitive region is.
Stress Analysis ( Stress Analysis (I) I)
- An approximate model of the bending of a plate under the effects of a pressure load.
⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
= L
y L
x
w c π 2π
cos 2 1
cos 4 1
ˆ 1
: the displacement function, L : the edge-length of the diaphragm, c1 : the displacement at the center of the diaphragm.
⎦
⎣ ⎝ ⎠
⎦
⎣ ⎝ L ⎠ L
4 wˆ
- A load-deflection equation
3 4 1 4 1
2 3 2
0
) 1
( ) ( )
1
( c
L f EH
C L c
C EH L
C H
P r b s s ⎥
⎦
⎢ ⎤
⎣
⎡ + −
⎪⎭
⎪⎬
⎫
⎪⎩
⎪⎨
⎧ ⎥
⎦
⎢ ⎤
⎣
⎡ + −
⎥⎦⎤
⎢⎣⎡
= ν ν
ν σ
Cr : the coefficient of the residual-stress term Cb : the coefficient of the plate-bending term
Cs : the coefficient of the large-amplitude in-plane stretching term.
) (
)
( ⎭ ⎣ ⎦
⎩ ⎣ ⎦
Cs : the coefficient of the large amplitude in plane stretching term.
For bulk silicon, Cr=0
For a small-amplitude loading Cs term ignored.
4
- The energy-method solution
4 6
π
b = C
3 4EH
π
4 1 2) 1
(
6 c
L P EH
ν π
= −
Stress Analysis (
Stress Analysis (II) II)
Calculation of the shear stress at the diaphragm edge - Calculation of the shear stress at the diaphragm edge (1) A 점(L/2,0)에서 x 방향으로의 곡률을 구한다.
(2) Diaphragm 두께에 비교하여 압저항 막이 매우 얇다고 가정하고, 곡률로부터 표면의 stress를 구한다
y L 곡률로부터 표면의 stress를 구한다.
(3) y 방향 stress는 x 방향 stress에 Poisson ratio 를 곱한다.
(4) 마지막으로 압저항 축과 그것에 수직인 transverse 축 방향을
나타내기 위하여 x 방향 과 y 방향 stress를 45˚ 방향의 shear x
A
⎟ L
⎜ ⎞
⎛ L 0
나타내기 위하여 x 방향 과 y 방향 stress를 45 방향의 shear stress로 나타낸다.
1 2
cos 2 1
cos 4 1
ˆ ) 1
( ⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
= L
y L
x
w c π π
⎟⎠
⎜⎝ ,0 2
2 2
1 2
cos 2 1
2 sin 4
ˆ 4
⎤
⎡ ⎛ ⎞
⎞
⎞ ⎛
⎛
⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
⎟⎠
⎜ ⎞
⎝
⎟ ⎛
⎠
⎜ ⎞
⎝⎛ −
∂ =
∂
⎦
⎣ ⎝ ⎠
⎦
⎣ ⎝ ⎠
L y L
x L
c x w
L L
π π
π
( )
1 22 1
0 , 2 2
2 1
0 , 2 2
2
2 4
cos 2 2 1
4 cos 4
ˆ
⎟⎞
⎜⎛
⎟⎞
⎜⎛
⎥⎦⎤
⎢⎣⎡
⎟⎠
⎜ ⎞
⎝ + ⎛
⎟⎠
⎜ ⎞
⎝
⎟ ⎛
⎟
⎠
⎞
⎜⎜
⎝
⎛−
∂ =
∂
=
= =
=
c c
L y L
x L
c x
w
y L y x
L x
π π
π π
π
x 방향으로의 곡률
( )
12
1 2
2 2 4 1
4 ⎟
⎠
⎜ ⎞
⎝
= ⎛
⋅
⎟ −
⎟
⎠
⎞
⎜⎜
⎝
⎛−
= L
c L
c π π
2 2
2 ˆ
1 ∂ w c1 ⎛ π ⎞
0 , 2 2
2 2
1 ⎟
⎠
⎜ ⎞
⎝
= ⎛
∂
= ∂
=
= L
c x
w
y L x x
π ρ
Stress Analysis (
Stress Analysis (III) III)
(2) A점에서의 x 방향 stress σx
L c
EH EH
x x
2 1 2 2 2
2 ⎟
⎠
⎜ ⎞
⎝
= ⎛
= π
σ ρ
L L
EH EH P c L
x
4 2 2 2
3 4
4 2 1
6 )
1 ( 6 ) 1
( 6
⎞
⎞ ⎛
⎛
= −
⎠
⎝ π
ν ρ
이므로
(3) y 방향 stress σy
H P P L
EH L L
EH
x 2
2 3
4
4 2 2
2
) 1
6 ( )
1 (
6 ⎟
⎠
⎜ ⎞
⎝
− ⎛
⎟ =
⎟
⎠
⎞
⎜⎜
⎝
⎛ −
⋅
= ν
π π
ν σ π
실리콘(100)면에서 [110]방향으로의 Poisson ratio ν =0.06
따라서, L 2 L 2
6 ⎛ ⎞ ⎛ ⎞
x
y ν σ
σ =
- Clark과 Wise의 full numerical simulation 결과 ,
H P P L
H L
x 2
2 (1 0.06 ) 0.606
6 ⎟
⎠
⎜ ⎞
⎝
= ⎛
⎟⎠
⎜ ⎞
⎝
− ⎛
= π σ
앞으로 이 식을 사용.
H P L
x
2
294 .
0 ⎟
⎠
⎜ ⎞
⎝
= ⎛ σ
이 식 사용
