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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.

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## RESISTIVE SENSORS

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**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/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) 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**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**Strain gage**(electrical wire) - + V A ρ L resistivity is low**Strain gage**L+ ∆L If A, L, ρ change at the same time, 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**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**Problems and Solutions**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 Gage 3 Gage 2 Bottom view Gage 4**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. • 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**Problems and Solutions**(c) Define the Voltage Gain of the op-amp. Input voltage is DC 1V. Output voltage changes 0-1V. Av = 100000

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