Transcript RESISTIVE SENSORS

RESISTIVE SENSORS
Dept. of Biomedical Engineering
2003200449
YOUNHO HONG
Stress (axial), Strain
y
x
A=xy
M
On the surface,
the average force per unit area is
denoted as σ
: “stress” [N/m²]
F=mg
L
F
: “strain” [unitless]
F
δ
Stress-Strain Curve
Brittle Materials ( ex. glass, concrete)
# Do not have a yield point
# ultimate strength and breaking strength
are the same
nonlinear
over a wide range
1 : ultimate stress
2 : rupture
Stress-Strain Curve
Ductile material (ex. Al, steel)
A : Al
B : steel
σ PL
1 : ultimate stress
2 : yield stress
3 : rupture
4 : elastic region
5 : plastic region
For elastic region ( linear region )
[σ ≤ σ PL]
σ = E ε (E :Young’s modulus)
cf) Less ductile materials such as aluminum
and medium to high carbon steels
do not have a well-defined yield point.
Cantilever
L
Al
σ=Eε
F
Al
δ
F
F
L+ ε
F = αε
If a material of a cantilever
is a aluminum,
A and L are almost constant.
Strain gage
(electrical wire)
+
A
V
ρ
-
L
resistivity is low
Strain gage
If A, L, ρ change at the same time,
L
L+ ∆L
D
D- ∆D
Poisson’s ratio
Gage factor
for metal strain gage, G : ~1.6
for semi-conductor strain gage, G : 100~170
for a bit of changes of resistance,
use Bridge Circuit method
100 Ω -> 100.01 Ω
100 Ω -> 101 Ω
+
IA
-
OUT
Problems and Solutions
(3) Four metal strain gages which gage factor is 10 are attached
on a plain. By forcing F to the plain, Gage1 and 2 are expanded as long as ∆L,
whereas Gage3 and 4 are shorten in the same length. It has a relation that ∆L/L = kf ,
k is constant. Design a bridge circuit getting output voltage in proportion to
F, describe output voltage as F. Voltage source of the bridge circuit is dc 5[V].
Gage 1
Top view
Bottom view
Gage 2
Gage 3
Gage 4
Gage 1&2 : L => L + ∆L
Gage 3&4 : L => L - ∆L
f = ε AE
ε = (1/AE)f
Problems and Solutions
R3
R1
R2
R4
5V
+
IA
-
OUT
Vo = Av(Va-Vb)
Problems and Solutions
(4) Consider to design a system measuring force by using both two P-type Si strain
gages which gage factor is 100 and two N-type Si strain gages which one is -100.
p-type Si strain-gage S1&S2 : G=100
n-type Si strain-gage S3&S4 : G=-100
Gage 1
Top view
Bottom view
Gage 3
Gage 2
R3
R1
R2
R4
E(v )
+
IA
Gage 4
-
OUT
Problems and Solutions
(b) Assuming that both top and bottom of cantilever is changed in the same length
in case that forced. By forced F, maximum change of the length of strain gage is
+0.05%, resistor is 200 without any load. Specify gain in order output to vary
in the range between -5V to +5V.
R=200Ω
-5 ≤ Vo ≤ 5
Vo.max = Av*5*100*0.0005
= Av*0.25 = 5V
Av = 20
(c) Derive to calibrate this kind of instrument.
Vo
# Use least square method to find
the calibration equation.
Change f by using different metal,
and measure Vo
f
Problems and Solutions
(6) Four metal strain gages are attached on the diaphragm below. Two of them which
are p-type Si strain gages have 100 gage factor and the others which are n-type Si strain
Gages have -100 gage factor. When the diaphragm is pressed, each of strain gages has
the same strain and sensitivity is (1/100000)%/mmHG. When It isn’t pressed, resistance
is 50. Assume the relation between pressure and strain is linear.
(a) How much does each resistance of p-type and n-type Si strain gages change,
when the pressure is changed ?
The sensitivity is (1/100000)%/mmHG and the resistance is 50 when pressure is zero.
So, when the pressure is 500mmHG, the resistance of p-type is 50 + 0.00005 and
the resistance of n-type is 50-0.00005.
(b) Design Bridge circuit with four strain gages. Make the positions of strain gages.
R3
R1
E(v )
R2
R4
+
IA
-
OUT
p-type Si strain-gage S1&S2 : G=100
n-type Si strain-gage S3&S4 : G=-100
Problems and Solutions
(c) Define the Voltage Gain of the op-amp. Input voltage is DC 1V. Output voltage changes 0-1V.
R3
R1
R2
R4
1V
+
IA
OUT
-
Av = 100000
Thank for your attention