Piezoelectric Sensor Gurkan Erdogan March 28, 2008
Content • Piezoelectric Sensor Design – Strain to Voltage Equation
• Interface Circuit for – Decreasing HP Filter Cut-off Frequency – Adding Offset Voltage • Wireless Transmission
(-6V
+6V)
(-3V (0V
+3V) +6V)
– Voltage to Frequency Conversion (expects 0V 8V) – Antenna Transmission (what is the loss here?) – Frequency to Voltage Conversion
• S626 Data Acquisition
Piezoelectric Sensor Geometry Strain to Voltage Equations
Sensor Equation
D(3×1) = d
d (3×6 )
σ (6×1)
D(3×1) d (d3×6 )
Electric Displacement Coulomb meter 2 Direct Piezoelectric Coefficient Matrix Coulomb Newton or meter Volt
σ (6×1)
Stress Vector
Newton meter 2
Why do we need to use double layer? •
When force is applied to a long piezoelectric cantilever beam, one side is in tension while the other side will be in compression. No electrical output can be obtained from this homogenous body by bending.
Bimorphs •
Bimorphs made with two halves of separate beams with electrodes in between, on the top and bottom surfaces a) series connection: If the beams are poled in the opposite direction then on the application of a force 'F' the voltage generated on the outer electrodes will be additive b) parallel connection:If the beams are poled in the same direction, the additive output can be obtained by connecting the outer electrodes and the center electrode
Approximation of Normal Stress • There are two ways to approximate the stress distribution. – Calculate an average stress – Maximum stress at the root of the cantilever beam is assumed to be the same throughout the beam surface.
1st Way: From Strain to Voltage v& p =
D& 3 = d 31σ&11 i = q& = ∫∫ D& 3 dA3 v& p C p = ∫∫ d 31σ&11 dl c dbc
=
q = vin C p
ε = scσ Ycε = σ
d 31Yc Cp
∫∫ ε&
11
dl c dbc
d 31Ycbc ε&11dl c C p l∫c
d 31Ycbc l c 1 ε&11dl c = ∫ Cp l c lc =
= ∫∫ d 31Ycε&11dl c dbc ε&11 =
Sq Cp Cp Sq
ε&11
v& p
S q = d 31Yc bc l c 1 ε&11 = ∫ ε&11dl c lc l c
Real Values for Piezo-Crystal ε 0 = 106 − 113 [ pF m] dielectric tc = 40 [µm]
C p = 1.38 [nF ]
C N 2 S q = 2.48 × 10 −6 m = C 2 N m
Yc = 2 − 4 [GPa ] bc = 12 [mm]
l c = 30 [mm]
1.38 ×10 −9 C V 1 −3 F = = . 56 × 10 = = C S q 2.48 × 10 −6 C V
2st Way: Maximum Stress M ( x ) = F (L − x ) Fx 2 (3L − x ) w(x ) = 6 EI y σ (x ) = M (x ) Ix ymax I 1T = FL I 2 3EI 1 T = wmax 3 L L I 2 3ET = wmax 2 2L
Q = ∫∫ D3 dA = ∫∫ d 31σ 11 dA 6 L2 3L2 = d 31 F WL′ = d 31 2 F 2 4WT 2T kε (WL′) Q C= o C T 3L2 T 3L = d 31 2 F F = g 31 ( ) 2 T k ε WL 2 WT o
V=
Interface Circuit
Why do we use Interface Circuit? Two Main Parts for Two Main Reasons; • Extra Capacitor for Decreasing the Cut-off Freq. – Piezo Sensor coupled with a load resistor acts like a high pass filter. In order to read low frequency signal we need to decrease the cutoff frequency.
• Voltage Divider for Adding Offset Voltage – We can only transmit positive voltage signal so we need to add an offset voltage to the circuit. Maybe we could have done this with an OpAmp.
Elements of the Circuits • • • • • •
Vp Cp RL Ce Vos R1,R2
: : : : : :
Voltage Generated in Piezo Sensor Piezoelectric Strip Capacitor Load Resistor Extra Capacitor Offset Voltage Resistors of Voltage Divider
Simple Connection with a Load Resistor Time Domain
iC (t ) = iR (t ) C p [v&P (t ) − v&L (t )] =
vL (t ) RL
C p RL v&L (t ) + vL (t ) = C p RL v&P (t )
τ HP = C p RL
τ HP v&L (t ) + vL (t ) = τ HP v&P (t )
Frequency Domain C p = 2.6 [nF ] RL = 22 [M Ω] f cHP =
1 ≅ 2.8 [Hz ] 2π × τ HP
τ HP sVL (s ) + VL (s ) = τ HP sVP (s ) VL (s )(τ HP s + 1) = τ HP sVP (s ) τ s V (s ) H (s ) = L = HP VP (s ) τ HP s + 1
Connecting an External Capacitor - 1 Time Domain
iC p (t ) = iCe (t ) + iRL (t ) C p [v&P (t ) − v&L (t )] = Ce v&L (t ) +
(C
p
vL (t ) RL
+ Ce )RL v&L (t ) + vL (t ) = RLC p v&P (t )
τˆHP v&L (t ) + vL (t ) = τ HP v&P (t ) C p = 2.6 [nF ] Ce = 22 [nF ]
RL = 22 [M Ω]
τˆHP = (C p + Ce )RL fˆcHP =
1 ≅ 0.3 [Hz ] 2π × τˆHP
Frequency Domain
τˆHP sVL (s ) + VL (s ) = τ HP sVP (s ) VL (s )(τˆHP s + 1) = τ HP sVP (s ) V (s ) τ s H (s ) = L = HP VP (s ) τˆHP s + 1
Connecting an External Capacitor - 2
VL H = 20 log10 VP VL = 10 ≅
−19.5 20
VP 10
VP
= −19.5
Adding an Off-Set Voltage - 1
Time Domain
[
]
C p v& p (t ) − v& L (t ) = iC p (t ) C e v& L (t ) = iCe (t )
v L (t ) = i R2 (t ) × R2
Frequency Domain
[
]
C p s V p − VL = I C p Ce sVL = I Ce VL = I R2 × R2
vos (t ) − v L (t ) = i R1 (t ) × R1
Vos − VL = I R1 × R1
iC p (t ) + i R1 (t ) = iCe (t ) + i R2 (t )
I C p + I R1 = I Ce + I R2
Adding an Off-Set Voltage - 2 I C p + I R1 = I Ce + I R2
Cont. in Frequency Domain
[
]
C p s V p − VL + C p sV p − C p sVL +
Vos − VL V = Ce sVL + L R1 R2
1 1 1 Vos − VL = Ce sVL + VL R1 R1 R2
1 1 1 C p s + + Ce s + VL = C p sV p + Vos R1 R2 R1 1 1 (C p + Ce )s + VL = C p sV p + Vos RL R1 (RL (C p + Ce )s + 1)VL = RLC p sVp + RL Vos R1
(τˆHP s + 1) VL = τ HP sV p + RL Vos R1
VL =
τ HP s
(τˆHP s + 1)
Vp +
RL 1 Vos R1 (τˆHP s + 1)
Adding an Off-Set Voltage – 3 AC Equivalent
1 1 1 = + RL R1 R2
R1 R2 RL = R1 + R2
τˆHP = (C p + Ce )RL VL =
τ HP s
(τˆHP s + 1)
Vp
Exact same thing as back in slide 11
Wireless Transmitter 390 Ω
1000 pF
G + 12 V
G
G
G G
3 .6 V
G
G
250 kΩ
G
100 MΩ 1000 pF
100 MΩ
1000 Ω
G V piezo
+ parallel180k
Voltage to Frequency Conversion AD 654
Voltage to Frequency Conversion •
Output Frequency Range
– pin 1:
f = 0−5
[kHz]
•
Digital Ground
– pin 2:
VDGND = 0
[volt ]
•
Timer Resistor
– pin 3:
RT = 250
Vin = Vmin : Vmax
•
Input Voltage Range
– pin 4:
•
Positive Voltage Supply
– pin 5:
•
Timer Capacitor
– pin 6&7: (see below)
•
Positive Voltage Supply
– pin 8:
(
= 0.1 : Vs+ − 4 = ~ 0− ~ 8
[kΩ]
)
[volt ] Vs− = 0
[volt ]
CT = 0.64
[nF ]
Vs+ = 12
[volt ]
Conversion Equation Q CT = VCT
Vin IT = RT
Q = VCT CT f out Q = f out VCT CT I T = f out VCT CT Vin = = f out VCT CT RT
VCT = 10 V RT = 250 kΩ
f out
Vin 1 = VCT RT CT
f out
Vin+ 1 = 10 RT CT
CT = 1000 pF
τ = RT CT = 250 µ sec
AD654 Block Diagram VS = +12 Volt
CT = 1 picoFarad VLOGIC = + 4.9 Volt
VIN = + 5 Volt
RPU = 980 Ohm
R1 + R2 = RT = 250 kiloOhm
VS = 0 Volt
f out
Vin 1 = = 400 × Vin 3 −9 10 250 ×10 × 1×10
(
) (
)
FOUT vs. VIN(piezo)
Wireless Receiver + Frequency to Voltage Conversion AD 650
F/V Converter
Complete Circuit
F/V Converter Equations t os − 3 × 10 −7 Cos = 6.8 × 10 3
RINT
max Vout = max f in × α × tos
C INT =
Mechanical Response Time # of Time Constants × R INT
F/V Conversion – Ripples (1) 1 Vout (s ) C INT = 1 I IN (s ) s + RINT CINT
v&out (t ) +
vout (t ) i (t ) = RINT C INT C INT v&out (t ) = − v(t ) = e
−
vout (t ) i (t ) + RINT CINT C INT t
τ INT
t
v(0) + ∫ e
−
(t −τ ) τ INT
0
α i (τ ) = 0 x(t ) = e
1 i (τ ) dτ C INT
0 ≤ τ ≤ tos tos < τ ≤ T
A(t − t o )
t
x(to ) + ∫ e A(t −τ ) B u (τ ) dτ to
F/V Conversion – Ripples (2) vmax = e
−
t os
τ INT
t os
vmin + ∫ e
−
(t os −τ ) τ INT
0
v(t ) = e
−
t
τ INT
t
v(0) + ∫ e
−
(t −τ ) τ INT
0
α 0 ≤ τ ≤ tos i (τ ) = 0 tos < τ ≤ T
1 i (τ ) dτ C INT
=e =e
=e
vmin = vmax e
−
−
−
−
t os
τ INT
vmin + e
t os
τ INT
vmin + e
t os
τ INT
vmin + e
−
t os
τ INT
=e
−
t os
τ INT
vmin + e
α C INT
−
t os
τ INT
α C INT
−
t os
τ INT
(T −tos ) τ INT
1 α dτ C INT
−
t os
τ INT
t os
∫e
τ τ INT
dτ
0 t os
∫e
τ τ INT
dτ
0
τ α τ INT τ INT e C INT
τ INT α
t os
τ INT
e C INT t t − os − os = e τ INT vmin + RINT α 1 − e τ INT
t os
0
− 1
F/V Conversion – Ripples (3) vmin = vmax e vmax = e
−
t os
τ INT
−
(T −tos ) τ INT
t − os τ vmin + RINT α 1 − e INT
(T −tos )
t − os τ vmax = e vmax e + RINT α 1 − e INT (T −t os ) t t − os − − os vmax 1 − e τ INT e τ INT = RINT α 1 − e τ INT −
t os
τ INT
vmax = RINT α
−
1− e 1− e
vmin = RINT α
1− e 1− e
vmax − vmin
τ INT
−
−
−
−
vmax − vmin
vmax − vmin
t os
τ INT
vmax − vmin
T
τ INT tos
τ INT T
τ INT
−
e
(T −tos ) τ INT
t (T −tos ) t (T −t os ) − os − − os − τ τ τ INT INT INT 1− e e −e e τ INT = RINT α − T T − − τ INT τ INT 1− e 1− e
(T −t os ) t T − os − − 1 − e τ INT − e τ INT + e τ INT = RINT α T − 1 − e τ INT t (T −tos ) T T T T − os − − τT − e INT + e τ INT e τ INT + e τ INT e τ INT − e τ INT e τ INT = RINT α T T T − τ INT τ INT τ INT −e +e e (T −tos ) tos τT τ τ − e INT + e INT + e INT − 1 = RINT α T 1 − eτ INT
