piezoresistive sensors 11-4-2011

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Piezoresistive Sensors Origin and expression of piezoresistivity The resistance of a resistor:  A  L  R   The resistance changes with the changes in geometry and resistivity. The change in resistance is much greater due to the change in resistivity.  A  A l l  R  R     

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8/3/2019 Piezoresistive Sensors 11-4-2011

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Piezoresistive Sensors

Origin and expression of piezoresistivity

The resistance of a resistor:

 A

 L R  

The resistance changes with the changesin geometry and resistivity.

The change in resistance is much greaterdue to the change in resistivity.

 A

 A

l

l

 R

 R

  

  

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Expression of piezoresistivity

Resistivity of a semiconductor:

*

11

m

t q

qN 

qn

 

     

Both t and m* are a function of atomic spacing.

 

 

  

  

 N 

 N 

V  )(

l

l

 A

 A

    

 G

 N 

 N 

 R

 R

))(

2(

Gauge factor

 )100...20(

 R

 RFor semiconductors

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Single Crystal Silicon

Recall Ohm’s law: 

 

 

 

 

 

 

 

 

 

 

 

 

 z

 y

 x

 z

 y

 x

i

i

i

 E 

 E 

 E 

345

426

561

      

      

      

 J  E 

 E  J 

  

 

Since piezoresistive coefficients depend onthe crystallographic directions:

Changes of 1 through 6 are related to the six stress components:

 

 

 

 

 

 

 

 

 

 

 

 

6

5

4

3

2

1

44

44

44

111212

121112

121211

06

05

04

03

02

01

0000000000

00000

000

000

000

 /  / 

 / 

 / 

 / 

 / 

 

 

 

 

 

 

 

 

 

   

   

   

    

    

    

    

    

    

For silicon, when the axesare aligned to <100>

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Piezoresistance Tensor, 

There are three independent piezoresistive coefficients:

441211 ,,    

These are influenced by the doping type,doping concentration N, and temperature.

when N and T.

Material  Resistivity, Ω-cm  π11, 10-11/Pa  π12, 10

-11/Pa  π44, , 10-11/Pa 

p silicon  7.8  6.6  -1.1  138.1 n silicon

 11.7

 -102.2

 53.4

 -13.6

 p germanium  1.1  -3.7  3.2  96.7 n germanium  1.5  -2.3  -3.2  -138.1 

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Gauge factor, G

Longitudinal gauge factor, Gl,transverse gauge factor, Gt.

G is determined by multiplying the effective with E

in the direction of applied strain.

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Effective piezoresistive coefficient

eff , varies with crystallographic direction.

eff  in the (100) plane for p-si eff  in the (100) plane for n-si

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Example I

A longitudinal piezoresistor is embedded on the top surfaceof a silicon cantilever near the anchored base. The cantileverPoints in the <110> direction. The piezoresistor is p-typedoped with resistivity of 7.8 cm. Young’s modulus is 168GPa.Find the longitudinal gauge factor of the piezoresistor.

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Polycrystalline Silicon

Advantages:• Ability to deposit on different substrates• G not dependent on the orientation of resistor• Easy doping

Disadvantages:• G is much smaller.• G is influenced by the growth and annealing conditions.

Gauge factor

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Stress Analysis - Cantilevers

The magnitude of stress at any point on the cross sectionis linearly proportional to the distance to the neutral axis.

The magnitude of stress reaches maximum at the top and

bottom surfaces.

 

The maximum strain is found to be at the surfacenear the fixed end.

 EI 

FLt 

 EI 

t  x M 

22

)(max  

F

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Example II

Consider a fixed-free cantilever beam made of single-crystalsilicon with its length pointing in the <100> direction. Ten points(A through J) are identified on the cantilever. The length, l,width, w, and thickness, t, are 100m, 10 m, and 6 m,respectively. If a 1mN force acts at the end of the cantilever,

what is the max? At which point does max occur?

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Example III

A fixed-free cantilever is made of single crystal silicon. Thelongitudinal axis points in the [100] direction. The resistorhas a longitudinal gauge factor of 50. The length, l,width, w, and thickness, t, are 200m, 20 m, and 5 m,respectively. If a force of 100 N is applied at the end of the

cantilever, what is the percentage change of resistance?

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Example IV

A single crystal cantilever beam has a diffused piezoresistor.The longitudinal gauge factor is 20 and the transverse gaugefactor is 10. Find the change in the resistance of the piezoresistorwhen the maximum longitudinal strain near the sensor is 1%.

F

Top view

Side view

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Stress in the Membrane

Stress analysis is more complicated.The maximum stress is located at the center of the outsideedges and in the center of the plate.

2

2

|max

308.0

 pa

edge

 2

2

|max

139.0

 pacenter 

 

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Read-out circuits

A Wheatstone bridge is used to read the small changesin resistance.

ino V  R R

 R

 R R

 RV  )(

43

4

21

2

When the bridge is balanced, R1R3=R2R4,

Vo = 0

The change in Vo induced by the change in R

can be expressed as

)(4

1 4312

 R

 R

 R

 R

 R

 R

 R

 R

in

o

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Placing resistors

The resistors need to be carefully placed. Two should betransverse resistors and the other two be longitudinal ones.

Material  Resistivity, Ω-cm  π11, 10-11/Pa  π12, 10

-11/Pa  π44, , 10-11/Pa 

p silicon  7.8  6.6  -1.1  138.1 n silicon  11.7  -102.2  53.4  -13.6 

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Applications  – Inertia Sensors

)

2

(LlF  M 

Assume the force isconcentrated at the centerof the proof mass.

The maximum of longitudinalstrain occurring at the base of the cantilever is

33max

)

2

(6

122

)

2

(

2 Ewt 

t  LlF 

wt  E 

t  LlF 

 EI 

 Mt 

 

The relative change in resistance:

a Ewt 

 LlGm

G R

 R2max

)

2

(6

 

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Inertia Sensors

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Pressure Sensors (Case Study)

The Omega PX409 pressure transducer can measure pressureranges from 0 – 6.9kPa to 0 – 34.5MPa. The DC excitationvoltage is between 5 – 10V. The output signals can eitherbe voltage (1 – 100mV, 0 – 5V) or current (4 – 20mA).

The (100) n-Si diaphragm is a square, 1.2 mm on each side and80m thick. The sensor has a 0-1MPa full scale input and0-100mV full scale output, a 10 VDC excitation.

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Pressure Sensors (Case Study)

We will begin find the stress and strain in the resistors.

Resistors 1 and 3 experience the longitudinal stress.Resistors 2 and 4 experience the transverse stress.

The longitudinal gauge factor is GL.

2

2308.0

 pac  

2

2308.0

 Et 

 pa

 E 

cc

  

The transvers gauge factor is GT.

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Pressure Sensors (Case Study)

The output voltage change:

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Pressure Sensors (Case Study)

With L = -T  44 /2, Vin = 10V, and p=100kPa

mV 

V o

7.95

10)1080(

)102.1(101308.0101.1382

2

1

26

23511