Example: membrane pressure sensor
18. Piezoresistive Pressure Sensors : Case Study
Introduction
Piezoresistance
Motorola MAP sensor
Piezoresistive pressure sensor design
Capacitive pressure sensor design
„
„
„
„
The piezoresistive straing gauge are usually made of doped poly-Si and are
designed in pairs with a readout cicuitry (usually a Wheatstone bridge).
While strain-pressure reponses of the membrane have been modelled in
previous chapter, practical devices are usually rather calibrated, and their
response stored on-chip in a look-up table.
The response of the device to applied pressure is related through the
mechanical response of membrane, piezoresistive response of transducer:
Vout ∝ ΔR ∝ Π ( P − P0 )
In this design, a capacitive bridge can be formed wit two reference
capacitors, and the output voltage is related to the deflection of the
membrane Δx and hence the differential pressure (P - P0) through:
Vout ∝ ΔC ∝ Δx ∝ ( P − P0 )
„
By controlling the background pressure P0 it is possible to fabricate the
following types of pressure sensors:
„
„
„
An absolute pressure sensor that is referenced to vacuum (P0 = 0)
A gauge-type pressure sensor that is referenced to atmoshperic
pressure (P0 = 1 atm)
A differential sensor where P0 is maintained at a known value
1
Piezoresistive vs capacitive approaches
Piezoresistive pressure sensors
„
„
This chapter introduces piezoresistive devices through the
specific case study of membrane pressure sensors
While other approaches such as capacitive effects can be
used for such applications, silicons also possess the
property of piezoresistance whose implementations as
transduction mechanism in membrane is somewhat more
straightforward
Piezoresistive designs are the most employed because of its low cost,
robustness, and ease of circuit integration
Piezoresistivity
9. Piezoresistive Pressure Sensors : Case Study
Introduction
Piezoresistance
Motorola MAP sensor
„
„
Piezoresistivity is the dependence of electrical resistivity on strain
Such an effect is related to the rearrangement of energy bands of a
solid under applied strain (above)
2
Piezoresistivity
Analytic formulation in cubic materials
„
Assuming effect is linear, the relationships between electric field
and current density is given by:
ε = [ρe + Π ⋅ σ ] ⋅ J
where ρe is the resistivity tensor, Π is the piezoresistive tensor, σ is
the stress tensor, and J is the current density
„
„
„
Note: while ε and J are vectors, ρe and σ are second rank tensors,
while Π is a fourth rank tensor
However, in a cubic crystal the resistivity tensor is diagonal and
characterized by a unique diagonal value ρe
In addition, as previously described, the stress tensor can be
reduced to six independent elements and re-annoted as such:
Such strain will modify both the bandgap as well as the effective
masses (and thus mobilities) ascribed to the bands
The effect is isotropic in as much as a given strain may increase
resistivity along one direction while decreasing it along others
„
„
Analytic formulation in cubic materials (ctnd.)
„
Thus, the above equation can be written along the three principal
directions of the cubic lattice
ε3
= [1 + π11σ3 + π12 (σ1 + σ 2 )]J 3 + π 44 (τ13J1 + τ 23J 2 )
ρe
where the three independent piezoresistive coefficients are:
ρ eπ 11 = Π 1111
ρ eπ 12 = Π 1122
ρ eπ 44 = 2Π 2323
σ 23 = σ32 = τ 23
σ 22 = σ 2
σ31 = σ13 = τ31
σ33 = σ3
σ12 = σ 21 = τ12
Longitudinal and transverse piezoresistance
σt
σl
J
ε1
= [1 + π11σ1 + π12 (σ 2 + σ3 )]J1 + π 44 (τ12 J 2 + τ13J 3 )
ρe
ε2
= [1 + π11σ 2 + π12 (σ1 + σ3 )]J 2 + π 44 (τ12 J1 + τ 23J 3 )
ρe
σ11 = σ1
σl
σt
„
„
„
If a relatively long and narrow resistor is defined in a planar
structure, then the primary current density and electric field are
both along the long axis of the resistor.
Structures are usually designed so that one of the axes of the
principle in-plane stress is also along the resistor axis.
This simplifies the set of equations to following simplified
formulation:
ΔR Δρ
=
= πl σ l + π t σ t
R
ρ
where R is the resistance of the resistor and the subscripts l & t
refer to transverse and longitudinal stresses along the resistor
axis.
3
Longitudinal and transverse piezo... (ctnd.)
„
„
The orientations of resistor is not necessarily aligned with
crystalline orientations of device
Longitudinal and transverse piezo... (ctnd.)
„
Piezoresistors are often oriented in the [110] directions.
„
The directional cosines are:
The general expressions πl and πt are related to the original tensor
through:
πl = π11 − 2(π11 − π12 − π 44 )(l12 m12 + l12 n12 + m12 n12 )
and
π t = π12 + (π11 − π12 − π 44 )(l12l 22 + m12 m 22 + n12 n 22 )
⎛ 1 1 ⎞
(l1 , m1, n1 ) = ⎜
,
,0 ⎟
⎝ 2 2 ⎠
Thus:
1
(π11 + π12 + π 44 )
2
1
= (π11 + π12 − π 44 )
2
πl,110 =
where (l1, m1, n1) are the directional cosines between the
longitudinal resistor direction and the crystal axis and (l2, m2, n2)
is the direction cosines between the transverse direction and the
crystal axes
„
⎛ 1 1 ⎞
(l 2 , m 2 , n 2 ) = ⎜ −
,
,0 ⎟
2 2 ⎠
⎝
π t ,110
Note: by this definition l1l2 + m1m2 + n1n2 = 0 given that these
direction cosines are orthogonal to each other.
Piezoresistive coefficients in Si
Design example
„
The resistors are fabrictated along the [110] directions
4
Design example (ctnd.)
Design example (ctnd.)
„
„
For n-type resistors:
For p-type resistors:
πl = 71.8 ×1011 Pa -1
1
1
2
2
1
1
π t = (π 11 + π 12 − π 44 ) = (−102.2 + 53.4 − 13.6) = −17.6 × 1011 Pa -1
2
2
π l = (π 11 + π 12 + π 44 ) = (−102.2 + 53.4 − 13.6) = −31.2 × 1011 Pa -1
π t = −66.3 ×1011 Pa -1
thus, p-type is better suited to perform piezoresistive
readout in this direction
Design example (ctnd.)
Numerical example
„
„
„
„
„
Which is better, a longitudinal or a transverse resistor?
The transverse resistor is fully plunged in region of maximum
strain, but will also be greatly affected by placement error
The longitudinal resistor spans over a wider region of stresses. It
will be less sensitive, but will be less prone to alignment errors
from device to device
The n-type cantilever is 200 μm long 20 μm wide and 5 μm thick. It is bent
by a point load on its end. The p=type piezoresistors are 20 μm long and
2μm wide. A force of 10 μN is applied at extremity of device. Calculate
change of resistance
From Senturia, section 9.3:
⎛ 4L3 ⎞
⎛
⎞
4(200 ×10−6 )3
⎟F = ⎜
⎟10 ×10 −6 = 0.8μm
w max = ⎜
⎜ EWH3 ⎟
⎜ (160 ×109 )(20 ×10 −6 )(5 ×10 −6 )3 ⎟
⎝
⎝
⎠
⎠
„
The cantilever deflection at any point x is given by:
w (x) =
FL 2 ⎛
x ⎞
x ⎜1 −
⎟
2EI ⎝ 3L ⎠
5
Numerical example (ctnd.)
Alternate design
The radius of curvature is then given by:
„
1
∂ 2w
F
=
= (L − x )
ρ( x ) ∂x 2
EI
„
Since σ = − zE ρ , the stress at surface (z =-H/2) is given by:
„
Given that I = WH 12 :
σ( x )
3
σ( x ) =
6F
H 2w
EH FH
=
(L − x )
2ρ
2I
( L − x ) = 1.2 × 1011 ( L − x )
It therefore spans from σ = 24 MPa at x = 0 to 21.6 MPa at x = 20 μm
„
An average stress of σ = 22.8 MPa is therefore used to calculate change of
resistance. We finally get:
„
ΔR
= (67.6 × 10 −11 )( 22.8 × 106 ) = 1.54 %
R0
„
Alternate design (cntd)
„
Resistors R1 and R3 experience stresses σ in their longitudinal
direction and a stress υσ in their transverse direction:
Δ R 1 ΔR 3
=
= πl σ + π t υσ = (67.6 × 10 −11 )σ
R1
R3
(note: the Poisson ratio in the [110] direction is ν = 0.064)
All four resistors are aligned along one of the [110] directions, and
aligned with the principle axes of stresses
Alternate design (ctnd.)
„
Resistors R2 and R4 experience stresses σ in their transverse
direction and a stress υσ in their longitudinal direction:
Δ R 2 ΔR 4
=
= υσπl + σπ t = −61.7 × 10 −11 σ l
R2
R4
6
Alternate design (ctnd.)
Averaging over doping variations
ρ0(z)
surface diffusion
z
z
„
Connecting the four resistors in a Wheatstone bridge configuration,
we get:
V0
R1 R3 − R2 R4
α1 + α 2
=
≈
Vs ( R1 + R2 )( R3 + R4 ) 2(1 + α 1 − α 2 )
where:
Real life piezoresistors will present some degree of non-uniformity
with respect to the doping levels and the stress distribution they are
subjected to.
For instance, creation of piezoresistor through surface diffusion
doping will create a depth-dependent profile of the unstrained
resistivity ρ0, as seen above.
„
„
α1 = (π l + υπ t )σ
α 2 = (υπ l + π t )σ
Averaging over doping variations (cntd)
Averaging over stress variations
substrate
doped
resistor
L
z=0
Ec
σ
z=H
Ef
R0
ρ
Ev
z=0
„
z=zj
To calculate the nominally unstrained resistance R0 , we evaluate the
zj
following integral:
„
σ( z ) =
„
1
W
=
∂z
R 0 ∫0 Lρe,0 (z)
where zj is the junction depth equal to the edge of the space-charge
junction.
The neutral axis is at z = H/2 thus:
„
E ( H 2 − z)
ρ
Transverse stresses are neglected, thus, the stress-induced
resistivity change is:
Thus:
ρe (z) = ρe,0 (z)[1 + πl σl (z)]
zj
1
W
=
[1 − πlσ l (z)]∂z
R ∫0 Lρe,0 (z)
where we assumed that: [1 + πl σ]
−1
≈ [1 + πl σ]
7
Averaging over stress variations (cntd)
z=0
σ
z=H
9. Piezoresistive Pressure Sensors : Case Study
ρ
„
9.1 Introduction
9.2 Piezoresistance
9.3 Motorola MAP sensor
Rearranging:
zj
⎡
⎤
W
R = R 0 ⎢1 + R 0 ∫
πlσl (z)∂z ⎥
Lρe,0 (z)
⎢
⎥
0
⎣
⎦
Motorola MAP sensor
„
The Motorola MAP sensor has been developped to measure the
absolute pressure in the intake of automobile engines
Motorola MAP sensor
„
The Motorola MAP sensor has been developped to measure the
absolute pressure in the intake of automobile engines
8
Motorola MAP sensor: transistor fabrication
Motorola MAP sensor: transistor fabrication
Motorola MAP sensor: diaphragm fabrication
Motorola MAP sensor: diaphragm fabrication (cntd.)
9
Design of piezoresistor
Design of piezoresistor (ctnd.)
I
1
WR
+
V2
LR
-
1
„
Vo
-
„
The resistor is probed in a four-point configuration as follows:
piezoresistor in the [100] direction
Design of piezoresistor (ctnd.)
„
+
2
3
„
2
3
The current density J in resistor is J = J1 along resistor axis but is J = 0 along the
other direction.
Design of piezoresistor (ctnd.)
„
However the field in the transverse direction is not zero due to the piezoresistive
properties of the materials
V1 =
Using the piezoresistive equations:
ε2
= [1 + π11σ 2 + π12 (σ1 + σ3 )]J 2 + π 44 (τ12 J1 + τ 23J 3 )
ρe
„
LR
0
0
∫ ε1∂x1 = ∫ ρe (1 + π11σ1 + π12σ2 )J1
LR
⎤
0
⎦
∫ (π11σ11 + π12σ12 )∂x1 ⎥⎥
The transverse voltage is given by:
V2 =
since J2 = J3 = 0 and σ3 = τ13 = τ23 = 0, we obtain:
WR
WR
0
0
∫ ε 2∂x 2 = ∫ ρe π44τ12J1∂x 2 = ρe π44τ12 J1WR
⎛W ⎞
V2 = π 44 τ12 ⎜⎜ R ⎟⎟V1
⎝ LR ⎠
ε1 = ρe (1 + π11σ1 + π12σ 2 )J1
ε 2 = ρe π 44 τ12 J1
ε3 = 0
LR
⎡
1
V1 = ρe L R J1 ⎢1 +
⎢⎣ L R
ε1
= [1 + π11σ1 + π12 (σ 2 + σ3 )]J1 + π 44 (τ12 J 2 + τ13J 3 )
ρe
ε3
= [1 + π11σ3 + π12 (σ1 + σ 2 )]J 3 + π 44 (τ13J1 + τ 23J 2 )
ρe
The voltage across the piezoresistor is given by:
„
Thus the transverse voltage V2 only depends on the shear stress present in
regions between the two taps.
10
Stress analysis
„
Stress analysis
A uniform plate under uniform pressure possess a displacement function
of:
C ⎡
⎛ 2πy ⎞⎤
⎛ 2πx ⎞⎤ ⎡
w ( x, y) = 1 ⎢1 + cos⎜
⎟⎥
⎟⎥ ⎢1 + cos⎜
4 ⎣
⎝ L ⎠⎦
⎝ L ⎠⎦ ⎣
with C1 is the displacement in center of plate, and is given by:
P=
π 4 EH 3
6(1 − υ2 )L4
„
„
C1
2
The radius of curvature in the middle of the edge is given by
1 ∂ 2w
=
ρ x ∂x 2
„
2
x = L 2, y = 0
⎛ 2π ⎞ C
=⎜ ⎟ 1
⎝ L⎠ 2
1
π2
A numerical analysis of that structure would actually yield:
2
which is a factor of two off from our analytical solution. This numerical
result is used for the rest of the analysis.
EH
σx =
2ρ x
σx =
„
⎛L⎞
σ x = 0.294⎜ ⎟ P
⎝H⎠
The related surface stress at that location is:
Thus:
Using υ = 0.06 for the [110] direction:
⎛L⎞
σ x = 0.606⎜ ⎟ P
⎝H⎠
where P is the applied pressure.
„
Since we are dealing with a plate, the y direction stress at center of
the edge is:
σ y = υσ x
2
⎛L⎞
(1 − υ2 )⎜ ⎟ P
⎝H⎠
Stress analysis (ctnd.)
„
„
„
„
Now, lets calculate the shear stress on the resistors arranged as above.
The axial stresses in the x and y directions add up to a shear stress
2
given by:
σx − σy
⎛L⎞
τ12 =
= 0.141⎜ ⎟ P
2
H
⎝ ⎠
2
Thus:
⎛W ⎞
V2
⎛L⎞ ⎛W ⎞
= π 44 τ12 ⎜⎜ R ⎟⎟ = 0.141π 44 ⎜ ⎟ ⎜⎜ R ⎟⎟ P
V1
⎝ H ⎠ ⎝ LR ⎠
⎝ LR ⎠
With π 44 = 138 × 10 −11 Pa −1 , L H = 50, and WR L R = 1 / 5
we get:
V2
mV
= 0.096
V1
V ⋅ kPa
11