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# RESISTIVE SENSORS

RESISTIVE SENSORS. Dept. of Biomedical Engineering 2003200449 YOUNHO HONG. Stress (axial), Strain. y. A=xy. On the surface, the average force per unit area is denoted as σ. x. M. : “stress” [N/m ²]. F=mg. L. F. : “strain” [unitless ]. F. δ. Stress-Strain Curve.

## RESISTIVE SENSORS

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1. RESISTIVE SENSORS Dept. of Biomedical Engineering 2003200449 YOUNHO HONG

2. Stress (axial), Strain y A=xy On the surface, the average force per unit area is denoted as σ x M : “stress” [N/m²] F=mg L F : “strain” [unitless] F δ

3. 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

4. Stress-Strain Curve Ductile material (ex. Al, steel) For elastic region ( linear region ) [σ ≤ σ PL] σ = E ε (E : Young’s modulus) A : Al B : steel σ PL cf) Less ductile materials such as aluminum and medium to high carbon steels do not have a well-defined yield point. 1 : ultimate stress 2 : yield stress 3 : rupture 4 : elastic region 5 : plastic region

5. Cantilever L σ = E ε If a material of a cantilever is a aluminum, A and L are almost constant. Al F Al F δ F L+ ε F = αε

6. Strain gage (electrical wire) - + V A ρ L resistivity is low

7. Strain gage L+ ∆L If A, L, ρ change at the same time, L D D- ∆D Poisson’s ratio

8. 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

9. 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&2 : L => L + ∆L Gage 3&4 : L => L - ∆L Gage 1 Top view Gage 2 f = ε AE ε = (1/AE)f Gage 3 Bottom view Gage 4

10. Problems and Solutions Vo = Av(Va-Vb)

11. 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 Gage 3 Gage 2 Bottom view Gage 4

12. 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

13. 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. • 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. p-type Si strain-gage S1&S2 : G=100 n-type Si strain-gage S3&S4 : G=-100

14. Problems and Solutions (c) Define the Voltage Gain of the op-amp. Input voltage is DC 1V. Output voltage changes 0-1V. Av = 100000