Lecture 23 Piezoresistive Pressure Sensorocw.snu.ac.kr/sites/default/files/NOTE/5374.pdf · 2018. 1. 30. · Sensing Pressure-Sensing pressure directlypiezoelectric material : transduce

Lecture 23Lecture 23

Piezoresistive Pressure SensorPiezoresistive 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

마이크로시스템기술개론 MEMS_Lect23_1

Sensing PressureSensing Pressure

- Sensing pressure directly

gg

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.


PiezoresistancePiezoresistance

- 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 llapplication 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)

[ ]JE e σρ ⋅Π+=

: the electric field intensity: the resistivity tensor, : the full fourth-rank piezoresistive tensor: the full second-rank stress tensor, : the current density.

( )

eρ Πσ J

E

Summation convention( ) jkijkijei JE llσρ Π+= ,

jkjkjjejkjkjjejkjkjje JJE,JJE,JJE llllll σρσρσρ 33,322,211,1 Π+=Π+=Π+=

Cubic material resistivity : only diagonal terms exist.

⎞⎛ 00

jkjkjjejkjkjjejkjkjje llllll 33,322,211,1

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛=

e

e

e

eρρ

ρρ

000000


⎠⎝ eρ

Analytic Formulation in Cubic Materials (II)Analytic Formulation in Cubic Materials (II)

jkjke JJE llσρ 111 Π+=

①)(3132121111

→Π+Π+Π

Π+Π+Π=Π

J

JJJJ kkkkkkjkjk llllllll

σσσ

σσσσ

The Second term

③)(②)(①)(

3313321321131

2312321221121

1311321121111

→Π+Π+Π+

→Π+Π+Π+

→Π+Π+Π=

JJJ

llllll

llllll

llllll

σσσσσσσσσ

331133321132311131231123221122211121131113121112111111① σσσσσσσσσ Π+Π+Π+Π+Π+Π+Π+Π+Π⇒

Applying 612531423333222111 →→→→→→

363464565464262666565666161

313414515414212616515616111

② σσσσσσσσσσσσσσσσσσ

Π+Π+Π+Π+Π+Π+Π+Π+Π⇒

Π+Π+Π+Π+Π+Π+Π+Π+Π=

Applying 612,531,423,333,222,111 →→→→→→

353454555454252656555656151

363464565464262666565666161

③

②

σσσσσσσσσσσσσσσσσσ




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

⎟⎞

⎜⎛ ΠΠΠ 121212 000

⎟⎟⎟⎟⎟

⎜⎜⎜⎜⎜

ΠΠΠΠΠΠΠ

=Π44

111212

121112

00000000000

⎟⎟⎟⎟

⎠⎜⎜⎜⎜

⎝ ΠΠ

Π

44

44

44

000000000000000

312212111

312454212666111

0000000

000000σσσ

σσσσσσσσσ

+++++Π++Π+

Π+Π+Π=Π+⋅+⋅+⋅+Π+⋅+⋅+⋅+Π⇒

②

①

3454442654461

12441244644

3454264466441

0000000222

0000000

σσσσσσσσστσσ

σσσσσσσσσ

⋅+⋅+Π+⋅+⋅+⋅+Π+⋅+⋅⇒

Π=Π=Π=⋅+⋅+⋅+⋅+⋅+Π+⋅+Π+⋅⇒

③

②


13441344544 222 τσσ Π=Π=Π=

Analytic Formulation in Cubic Materials (IV)Analytic Formulation in Cubic Materials (IV)

Applying (18 6)

22,, 44232344112212111111 eee πρπρπρ Π=Π=Π=Π=Applying (18.6)

)()(22)(

3132124413122121111

3134421244131221211111

JJJJJJJJE

eee

e

ττπρσπσπσπρρττσσσρ

+++++=Π+Π+Π+Π+Π+=

[ ][ ]

)()(1 31321244132121111 JJJE e ττπσσπσπρ +++++=∴[ ][ ] )()(1

)()(1

22311344321123113

11232344213122112

JJJEJJJE

e

e

ττπσσπσπρττπσσπσπρ

+++++=+++++=


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 ofthe 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

ttRσπσπ += ll

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 21

21

21

21

21

2144121111 nmnm ++−−−= lll πππππ

- General expressions of andlπ tπ

n

crystal resistor

where (ℓ1, m1, n1) is the set of direction cosines between

))((

))((22

21

22

21

22

2144121112

11111144121111

nnmm

nmnm

t ++−−+= ll

lll

πππππ

πππππm

l

yaxis

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))( zn

transverse resistor direction (subscript 2) and the crystal axis.

- In many silicon micromachined devices, resistors are oriented along [110] directions in (100) wafers.

)( ym

)( xl

transverse

longitudinal

[110]

)0,21,21(),,(),0,21,21(),,( 222111 −== nmnm ll

Therefore

oriented along [110] directions in (100) wafers. longitudinal

Therefore

1

)(21)0041)((2 44121144121111110, ππππππππ ++=++−−−=l


)(21)04141)(( 44121144121112110, ππππππππ −+=++−−+=t

Piezoresistive Coefficients of SiliconPiezoresistive 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

11π 44π12π

yp

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 i1019cm-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.

231)6134532102(1)(1+++n-type piezoresistor

6.17)6.134.532.102(21)(

21

2.31)6.134.532.102(2

)(2

441211

441211

−=++−=−+=

−=−+−=++=

ππππ

ππππ

t

l

- p-type piezoresistors have a larger sensitivity and and coefficients havelπ tπ

p-type piezoresistor

22

3.66)1.1381.16.6(21 ,8.71)1.1381.16.6(

21

−=−−==+−= tππ l


- p-type piezoresistors have a larger sensitivity and and coefficients have opposite signs and almost equal magnitudes.

lπ tπ

Structural 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 R1 experiences a longitudinal stress es sto 1 e pe e ces a o g tud a st essσ , it must simultaneously experience a transverse stress ν σ (ν is the Poisson ratio).- The total change in resistance for R1 would beThe total change in resistance for R1 would be

)(1

1 +=+=Δ σνππσπσπ llll tttRR

p-type

064.01

=νlσ

in the [110] direction of (100) plane.

ll σππ

112

11

1

11111

1070461)8710640366(

10556.67103.66 ,108.71

−

−−−

××+Δ

×=Δ

→×−=×=

RRR

tlσ


ll σσ 11

2

2 10704.61)8.71064.03.66( ×−=×+−=R

Structural Examples (Structural Examples (IV)IV)

RRRRRR )1(,)1( 242131 αα −==+==

- Wheatstone-bridge circuit

- α1 and α2 represent the product of the effective

piezoresistive coefficient and the stress.

oo RRRRRR )1(,)1( 242131 αα+

4231423141 ,

))((==

++−

=+

−+

= SSSo RRRRVRRRR

RRRRVRR

RVRR

RV Q

21

21

21

21

21

21

43214321

2)1()1()1()1(

))((

αααα

αααα

−++

=−++−−+

=+−

=

++++

SVRRRR

RRRRRRRR

Therefore,

21

21

2 αααα−+

+=

S

o

VV

- 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 gdirection normal to the surface.

- The resistor can be thought of as a stack of slices, each slice having a slightly different dopingdifferent doping.

- The slices are connected electrically in parallel.

dzWG jz

∫==1 dz

zLG

R oeo

o∫==

0, )(ρ

where ρ e,o(z) is the unstrained doping-dependent electrical resistivity, L is the length of the resistor, and zj is depth of the resistor.g , j 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 zerostructure, 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

ρρθρθθρε zHzHz −

=−−+

=2)2()( ρ

HZ =

Neutral f

The longitudinal stress due to bending

Strain at z

ρρθ

ρσ zHEz −

⋅=2)(l

θ⎟⎠⎞

⎜⎝⎛

2H

surface

Stress at z

)( dzW

- Since this is a cantilever, we can ignore transverse stresses.

The conductance g(z) of a single differential slice of thickness dz

)()(

zLzg

eρ=

where ρ e(z) is the strained resistivity which varies with depth due both to

doping and to stress

[ ])(1)()( 0, zzz ee llσπρρ +=

dzWdzW

doping and to stress.

- The stress-dependence


- The conductance[ ] [ ])(1

)()(1)()(

,,z

zLdzW

zzLdzWzg

oeoell

ll

σπρσπρ

−≅+

=

Averaging over Stress and Doping Variations (Averaging over Stress and Doping Variations (III)III)- The total conductance

[ ]dzzLW

Rjz )(1

)(1

0 llσπ−= ∫

- Let the first term be Ro

zLR oe )(0,ρ∫

111 W

-Since A is small,

)1(1)()(

110

,

ARR

dzzzL

WRR o

o

z

oeo

j −=−= ∫ llσπρ

))(1()1(10 dzzWRRARRRR jz

llσπ∫+=+≅=

- If we assume that zj is much less than H/2 , we can replace σ by EH/2ρ .

Since A is small, ))()(

1()1(1 0

,00 dzz

zLRRARR

ARRR

oeoooo llσπ

ρ∫++≅−

But there still may be some z-dependence in the π coefficient because of doping dependence.

- Equivalent importance is the variation of stress along the length of a q p g gresistor 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 ExampleA Numerical Exampleil 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 ㎛).

⎟⎟⎞

⎜⎜⎛

⎟⎟⎞

⎜⎜⎛

=xxww 13

2

- Interconnect regions : wide, more heavily doped than the piezoresistor portions.- Assuming that the cantilever tip is displaced downward by an amount wmax

⎟⎟⎠

⎜⎜⎝

−⎟⎟⎠

⎜⎜⎝

=cc LL

ww3

12 max

where x is measured from the support, Lc is the length of the cantilever.- The magnitude of the radius curvature and σ

− σ = 30 MPa at x = 0 and 27 MPa at x = 20 μm when w = 1 μm and E= 160 GPa.

3maxmax

2

2

2)(3

2,)(31

c

c

c

c

LxLwEHEH

LxLw

dxwd −

==−

==ρ

σρ l

σ = 30 MPa at x = 0 and 27 MPa at x = 20 μm when wmax = 1 μm and E= 160 GPa.

020500)10366(1052810871 11611 =××−+×××=+=Δ −−RR σπσπ

- Since the piezoresistance is a linear function of stress, the average stress (28.5 MPa) is used.


0205.00)103.66(105.28108.71 =××−+×××=+=Δ ttRR σπσπ ll

- 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 ittthe 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 axistwo : the current along the resistor axis.the other two : transverse voltage taps, connected

to a high-impedance op-amp input so as todraw no currentdraw 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

)3118()1( JE

- 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

112442

12121111

==

++=

EJE

JE

e

e

τπρσπσπρ


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

)3418()(11 ⎥⎤

⎢⎡

∫∫ RLRL dJLdEV

The change in voltage due to a length averaging of the piezoresistance effect

)34.18()(110 121211110 111 ⎥

⎦

⎤⎢⎣

⎡++== ∫∫ R

RRe

R dxL

JLdxEV σπσπρ

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

RWEV 22 =)35.18()/()/( 1124411244112442 VLWWLVWJV RRRReeRe τπρτπρτπρ =⋅==

- The transverse voltage depends only on the shear stress τ12 in the region betweenthe 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.

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎥⎦

⎤⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+=

Ly

Lxcw ππ 2cos12cos1

4ˆ 1

: the displacement function, L : the edge-length of the diaphragm,c1 : the displacement at the center of the diaphragm.

⎦⎣ ⎠⎝⎦⎣ ⎠⎝ LL4w

- A load-deflection equation

314142

3

20

)1()(

)1(c

LEHfCc

LEHC

LHCP ssbr ⎥

⎦

⎤⎢⎣

⎡

−+

⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎥⎦

⎤⎢⎣

⎡

−+⎥⎦

⎤⎢⎣⎡=

νν

νσ

Cr : the coefficient of the residual-stress termCb : the coefficient of the plate-bending termCs : 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=0For a small-amplitude loading Cs term ignored.

4

- The energy-method solution

64π=bC

34EHπ


142 )1(6c

LEHPν

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 shearx

A

L⎟⎞

⎜⎛ 0L

나타내기 위하여 x 방향 과 y 방향 stress를 45 방향의 shear stress로 나타낸다.

1 2cos12cos14

ˆ)1( ⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎥⎦

⎤⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+=

Ly

Lxcw ππ

⎟⎠

⎜⎝

0,2

22

1 2cos12sin24

ˆ4

⎤⎡ ⎞⎛⎞⎛⎞⎛

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛ −=

∂∂

⎦⎣ ⎠⎝⎦⎣ ⎠⎝

Ly

Lx

Lc

xw

LLπππ

( )2

12

1

0,22

21

0,22

2

24

2cos12cos44

ˆ

⎟⎞

⎜⎛⎟

⎞⎜⎛

⎥⎦⎤

⎢⎣⎡

⎟⎠⎞

⎜⎝⎛+⎟

⎠⎞

⎜⎝⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛−=

∂∂

====

cc

Ly

Lx

Lc

xw

yLxyLx

ππ

πππ

x 방향으로의 곡률

( ) 12

1 22

2144

⎟⎠⎞

⎜⎝⎛=⋅−⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

Lc

Lc ππ

22 2ˆ1 ⎞⎛∂ cw π


1

0,22

22

1⎟⎠⎞

⎜⎝⎛=

∂∂

===

Lc

xw

yLxx

πρ

Stress Analysis (Stress Analysis (III)III)(2) A점에서의 x 방향 stress σx

LcEHEH

xx

21 2

222⎟⎠⎞

⎜⎝⎛==

πρ

σ

LLEH

PEH

Lc

x

2422

34

42

1

6)1(6

)1(6

⎞⎛⎞⎛

−=

⎠⎝

πν

ρ

이므로

(3) y 방향 stress σy

PHLP

EHL

LEH

x2

234

42

2

2)1(6)1(6

⎟⎠⎞

⎜⎝⎛−=⎟

⎟⎠

⎞⎜⎜⎝

⎛ −⋅= ν

ππνπσ

실리콘(100)면에서 [110]방향으로의 Poisson ratio ν =0.06따라서, LL 226 ⎞⎛⎞⎛

xy σνσ =

- Clark과 Wise의 full numerical simulation 결과

,P

HLP

HL

x2

2 606.0)06.01(6⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛−=

πσ

앞으로 이 식을 사용.

PHL

x

2294.0 ⎟

⎠⎞

⎜⎝⎛=σ


이 식 사용

Stress Analysis (Stress Analysis (IV)IV)(4) Shear stress 구하기 )( yx σσ −xσ(4) Shear stress 구하기

LL

xxxyx

22

12

06012

122

⎞⎛⎞⎛

−=

−=

−= σννσσσσ

τyσ−

2y

yσ

- 출력 : (18.35) 식에서 V2를 구할 수 있다.

PHLP

HL 138.0294.0

206.01

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛⋅

−=

22

xσ−

* 책에서는 0.138 대신에

0.141을 사용.

10138

10190.0138.010138

11144

29

211

12441

2

−−

−−

×=

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛×=⎟⎟

⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛××=⎟⎟

⎠

⎞⎜⎜⎝

⎛=

Pa

PLW

HLP

LW

HL

LW

VV

R

R

R

R

R

R

여기서, π

τπ

입력 를 k 출력전압을 f 전압을 정의하면

)1052.9(1052.9515010190.0

51,50:

18829

1

2

44

−−−− ×==×=×××=

==−

PaKPKPPVV

LWHL

pp

RR 실제예

입력 P 를 kPa로, 출력전압을 mV로, reference 전압을 V 로 정의하면,1

kPakPa mV

PVPamV 1a

)(10)10(

,,)(1013

13

2

13

12

′′×′′=

×=

−

−

pp

pp

KPVPKV

KVPVPKV

된다. 이 은 이고,은때 이 이며,

이다. 는 은 는 때 이 되고, 가

mVkPa

mV/V/kPa

mV/V/kPa mVmV

481005

1052.9

)()()10(2

116

×=″

″′″=′′=

−p

ppp

pp

VV

K

KVPKVPK

되고출력이의이면압력이입력만약

따라서,

이다.은여기서,


.가능하다 활용이 이어서mV/V/kPa

mVkPa

1.0

481005 11

=″=

pK

VV

되고,출력이 의이면,압력이입력 만약,

SignalSignal--Conditioning and Calibration (Conditioning and Calibration (I)I)

- Supply voltage : 5 V, KP = 0.096 mV/V-kPa

- 입력 압력이 영이라도 오프셋(offset)이 존재.

- 오프셋은 온도가 올라갈수록 증가한다. 왜냐하면, 소자의

저항이 온도에 따라 증가하기 때문이다.

- 온도가 증가하면 압저항 감도가 감소하기 때문에 기울기가

감소한다.

- 오프셋과 감도의 온도 의존성이 서로 반대이므로 매우

좁은 영역에서 특성곡선이 교차하는 선회중심점(pivot

i t)이 나타난다point)이 나타난다.

- 신호처리와 눈금 맞추기(calibration)의 목적은 온도와 무관하게 오프셋 출력 전압과( )

압력에 대한 출력(span)을 일정하게 하는 것이다.


SignalSignal--Conditioning and Calibration (Conditioning and Calibration (II)II)

- Motorola pressure sensor

- 그림에서 압저항을 제외한

모든 저항은 온도 의존성이

없다고 가정.

- 눈금 맞추기 과정

(1) 스팬의 온도 보상

( ) i i 과정 셋 전압 정(2, 3) 두 trimming 과정으로 오프셋 전압 조정

(4) 오프셋의 온도 보상(temperature compensation of the offset, TCO)

(5) 전 영역 출력에 대한 overall gain(5) 전 영역 출력에 대한 overall gain

- 7개의 trimmable 저항이 있다.

- V1~V4, Vex 와 supply voltage VD, output Vout, 접지의 8개 단자가 필요.


1 4, ex 와 supp y o tage D, output out, 접지의 8개 단자가 필

Documents

Lecture 23 Piezoresistive Pressure Sensorocw.snu.ac.kr/sites/default/files/NOTE/5374.pdf · 2018. 1. 30. · Sensing Pressure-Sensing pressure directlypiezoelectric material : transduce