MSE 355 Sensors, Measurements and Data Acquisition System PDF

MSE 355 Sensors, Measurements and Data Acquisition System Assoc. Prof. Dr. Mohamed Atef Ismail Kamel [email protected] MSA University - Faculty of Engineering Mechatronics Systems Engineering (MSE) Program...

MSE 355 Sensors, Measurements and Data Acquisition System Assoc. Prof. Dr. Mohamed Atef Ismail Kamel [email protected] MSA University - Faculty of Engineering Mechatronics Systems Engineering (MSE) Program Fall Semester, 2023 Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 1 / 40  Force, Tactile, and Pressure Sensors Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 2 / 40 Outlines 1 Force Measurements 2 Tactile Sensors 3 Pressure Sensors 4 Recommended Readings and Practice Problems Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 3 / 40  Force Measurements Outlines 1 Force Measurements 2 Tactile Sensors 3 Pressure Sensors 4 Recommended Readings and Practice Problems Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 3 / 40  Force Measurements Force Sensors Force can me measured in many ways: Measuring the slight deformation caused by the force (using strain gauge). Measuring the acceleration (F = ma). Measuring the displacement of a spring under action of force (F = kx). Measuring the pressure produced by a force (F = P A). Use a piezoelectric transducer (induced voltage based on the applied force). All the previous methods do not measure the force directly. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 4 / 40  Force Measurements Strain Gauges It is the main tool in sensing force. Although, as their name implies, measure strain, the strain can be related to stress, force, torque, displacement, acceleration, or position. With proper application of transduction methods, it can even be used to measure temperature, level, and many other related quantities. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 5 / 40  Force Measurements Strain Gauges: Principle of Operation If the object is put under tension, the gauge will stretch and elongate the wires. The wires not only get slightly longer but also thinner. Both actions cause the total wire resistance to rise. The resistance of a wire can be calculated as: ρL R= A For small deformation ε: ∆R =gε R g is the gauge sensitivity, also named gauge factor. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 6 / 40  Force Measurements Strain Gauges: Principle of Operation Also, the relationship between resistance and strain can also be obtained as: R(ε) = Ro (1 + g ε) Ro is the no strain resistance. Given a wire of length l and apply a force F F σ= =Eε A Then, the force F can be calculated as: F =EεA Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 7 / 40  Force Measurements Strain Gauges: Types There are many types of strain gauges. The most common types are: Wire–bonded (resistive) strain gauge. Semiconductor strain gauge. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 8 / 40  Force Measurements Wire–Bonded Strain Gauge It is built out of a thin layer of conducting mate- rial deposited on an insulating substrate (plastic, ceramic, etc.) and etched to form a long, mean- dering wire. Constantan (an alloy made of 60% copper and 40% nickel) is the most common material because of its low temperature coefficient of resistance. Strain gauges may also be used to measure mul- tiple axis strains by simply using more than one gauge, or by producing them in standard configu- rations. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 9 / 40  Force Measurements Wire–Bonded Strain Gauge Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 10 / 40  Force Measurements Semiconductor Strain Gauge It operates in the same way as the resistive strain gauge. One major difference is that the gauge factor for semiconductors is much higher than for metals. Also, the change in conductivity due to strain is much larger than in metal. However, it is more sensitive to temperature vari- ations (require temperature compensation). The most common material is silicon because of its properties and ease of production. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 11 / 40  Force Measurements Semiconductor Strain Gauge dR Generally, the typical behavior is: R = g1 ε + g2 ε2. Although it is nonlinear, but its higher sensitivity is an advantage. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 12 / 40  Force Measurements Strain Gauges: Sources of Error Strain gauges are subject to a variety of errors. The first is due to temperature, as the resistance is affected by temperature changes: R0 (1 + α [T − T0 ]). Solution: Select strain gauges made from materials with low temperature coefficients of resistance. Use temperature commentators. The second is due to is due to the strain itself, which, over time, tends to permanently deform the gauge. Solution: Periodic re-calibration. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 13 / 40  Force Measurements Example A wire–bonded strain gauge is made with the dimensions shown. The gauge is 5 µm thick. The sensor is made of constantan to reduce temperature effects. 1 Calculate the resistance of the sensor at 25◦ C without strain. 2 Calculate the resistance of the sensor if force is applied longitudinally causing a strain of 0.001. 3 Estimate the gauge factor from the calculations in (1) and (2). Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 14 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain l R(T ) = (1 + α [T − T0 ]) σA Conductivity and temperature coefficients of resistance for selected materials (at 20◦ C unless otherwise indicated) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 15 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain l R(T ) = (1 + α [T − T0 ]) σA Conductivity and temperature coefficients of resistance for selected materials (at 20◦ C unless otherwise indicated) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 15 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain L R(T ) = (1 + α [T − T0 ]) σA Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 16 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain L R(T ) = (1 + α [T − T0 ]) σA The length l: Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 16 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain L R(T ) = (1 + α [T − T0 ]) σA The length l: l = 10 × 25 × 10−3 + 9 × 0.9 × 10− 3 = 0.2581 m. The cross-sectional area A: Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 16 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain L R(T ) = (1 + α [T − T0 ]) σA The length l: l = 10 × 25 × 10−3 + 9 × 0.9 × 10− 3 = 0.2581 m. The cross-sectional area A: A = 0.2 × 10−3 × 5 × 10−6 = 1 × 10−9 m2. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 16 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain: Given: l σ = 2×106. R25 = (1 + α [T − T0 ]) σA α = 0.00001. 0.2581 T0 = 20◦ C. l = 0.2581 m. = (1 + 0.0001 [25 − 20]) = 129.05 Ω 2 × 106 × 1 × 10−9 A = 1 × 10-9 m2. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 17 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain: Given: l σ = 2×106. R25 = (1 + α [T − T0 ]) σA α = 0.00001. 0.2581 T0 = 20◦ C. l = 0.2581 m. = (1 + 0.0001 [25 − 20]) = 129.05 Ω 2 × 106 × 1 × 10−9 A = 1 × 10-9 m2. 2) Get the resistance for 0.001 strain: ∆L ε= → ∆L = 0.001 × 0.2581 = 0.0002581 m L Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 17 / 40  Force Measurements Example 1) Get the resistance 25◦ C without strain: Given: l σ = 2×106. R25 = (1 + α [T − T0 ]) σA α = 0.00001. 0.2581 T0 = 20◦ C. l = 0.2581 m. = (1 + 0.0001 [25 − 20]) = 129.05 Ω 2 × 106 × 1 × 10−9 A = 1 × 10-9 m2. 2) Get the resistance for 0.001 strain: ∆L ε= → ∆L = 0.001 × 0.2581 = 0.0002581 m L As the gauge factor is unknown, the resistance can be calculated as: l8 R= σA8 where l8 and A8 are the new dimensions of the gauge due to force applying. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 17 / 40  Force Measurements Example 2) Get the resistance for 0.001 strain: Given: σ = 2×106. l8 = l + ∆l = 0.2581 + 0.0002581 = 0.2583581 m α = 0.00001. V lA 0.2581 × 1 × 10−9 T0 = 20◦ C. l = 0.2581 m. A8 = = = = 9.99 × 10−10 m2 l8 l8 0.2583581 A = 1 × 10-9 m2. l8 0.2583581 ∆l = 0.0002581 m. R= = = 129.308 Ω σA8 2 × 106 × 9.99 × 10−10 Rno strain = 129.05 Ω. 3) Get the gauge factor: ∆R ∆R R − Rno strain 129.308 − 129.05 =gε→g= = = =2 R εR εR 0.001 × 129.05 Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 18 / 40  Force Measurements Example A wire–bonded strain gauge is made with platinum. It has a temperature coefficient of resistance α = 0.00385 Ω◦ C and a gauge factor 8.9. Also, it has a nominal resistance 350 Ω at 20◦ C. The sensor is exposed to temperature variation between −50 and 200◦ C. 1 Calculate the maximum resistance expected for a maximum strain of 2%. 2 Calculate the change in resistance due to temperature and the maximum error due to temperature changes. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 19 / 40  Force Measurements Example 1) For a maximum strain of 2% at nominal temper- Given: ature: R0 = 350 Ω. T0 = 20◦ C. ∆R = g ε → ∆R = R g ε = 350 × 8.9 × 0.02 = 62.3 Ω α = 0.00385. R g = 8.9 ε = 2% Temp. variation: −50 to 200◦ C. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 20 / 40  Force Measurements Example 1) For a maximum strain of 2% at nominal temper- Given: ature: R0 = 350 Ω. T0 = 20◦ C. ∆R = g ε → ∆R = R g ε = 350 × 8.9 × 0.02 = 62.3 Ω α = 0.00385. R g = 8.9 The maximum resistance of 2% strain at nominal ε = 2% temperature: Temp. variation: −50 to 200◦ C. R = R0 + ∆R = 350 + 62.3 = 412.3 Ω Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 20 / 40  Force Measurements Example 1) For a maximum strain of 2% at nominal temper- Given: ature: R0 = 350 Ω. T0 = 20◦ C. ∆R = g ε → ∆R = R g ε = 350 × 8.9 × 0.02 = 62.3 Ω α = 0.00385. R g = 8.9 The maximum resistance of 2% strain at nominal ε = 2% temperature: Temp. variation: −50 to 200◦ C. R = R0 + ∆R = 350 + 62.3 = 412.3 Ω At 20◦ C and no strain: Rno strain = 350 Ω. At 20◦ C and 2% strain: R = 412.3 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 20 / 40  Force Measurements Example At −50◦ C with no strain: Given: R0 = 350 Ω. R = R0 (1 + α [T − T0 ]) T0 = 20◦ C. R−50 = 350 (1 + 0.00385 [−50 − 20]) = 255.675 Ω α = 0.00385. g = 8.9 At −50◦ C with 2% strain: ε = 2% Temp. variation: −50 to R−50 = 412.3 (1 + 0.00385 [−50 − 20]) = 301.185 Ω 200◦ C The error at −50◦ C with 2% strain: 301.185 − 412.3 error = = −0.2695 = 26.95% 412.3 Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 21 / 40  Force Measurements Example At 200◦ C with no strain: Given: R0 = 350 Ω. R = R0 (1 + α [T − T0 ]) T0 = 20◦ C. R200 = 350 (1 + 0.00385 [200 − 20]) = 592.55 Ω α = 0.00385. g = 8.9 At 200◦ C with 2% strain: ε = 2% Temp. variation: −50 to R200 = 412.3 (1 + 0.00385 [200 − 20]) = 698.02 Ω 200◦ C. The error at 200◦ C with 2% strain: 698.02 − 412.3 error = = −0.693 = 69.3% 412.3 Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 22 / 40  Force Measurements Example At 200◦ C with no strain: Given: R0 = 350 Ω. R = R0 (1 + α [T − T0 ]) T0 = 20◦ C. R200 = 350 (1 + 0.00385 [200 − 20]) = 592.55 Ω α = 0.00385. g = 8.9 At 200◦ C with 2% strain: ε = 2% Temp. variation: −50 to R200 = 412.3 (1 + 0.00385 [200 − 20]) = 698.02 Ω 200◦ C. The error at 200◦ C with 2% strain: 698.02 − 412.3 error = = −0.693 = 69.3% 412.3 Conclusion: Due to temperature change, error in strain measurement occur. This error can be compensated using bridge circuits. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 22 / 40  Force Measurements Bridge Circuit Also known as the Wheatstone bridge. The output voltage is of the bridge is:   R1 R2 Vo = Vin − R1 + R3 R2 + Rx The bridge is balanced (Vo = 0) if R1 R2 = R3 Rx Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 23 / 40  Force Measurements Temperature Compensation of Strain Gauges The output voltage is of the bridge is:   R1 R2 Vo = Vin − R1 + R3 R2 + Rx The bridge sensitivity (the change in the output voltage to the change of the resistances) can be calculated by taking the first derivatives dVo R3 dVo R2 = Vin = Vin dR1 (R1 + R3 )2 dRx (R2 + Rx )2 dVo Rx dVo R1 = −Vin = −Vin dR2 (R2 + Rx )2 dR3 (R1 + R3 )2 Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 24 / 40  Force Measurements Temperature Compensation of Strain Gauges Summing and cross-multiplying gives the following dVo R3 dR1 − R1 dR3 Rx dR2 − R2 dRx = − Vin (R1 + Z3 )2 (R2 + Rx )2 If dR1 = dR3 and dR2 = dRx , the change in output is zero. In strain measurement: Suppose two sensors S1 and S3 are used, S3 is not exposed to force (it is exposed only to temperature as S1 ). R2 and Rx are two identical resistors. Under these conditions and no force is applied: the output is zero even if the tempera- ture has been changed. Once a force is applied, the output changes since the resistance of S1 changes. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 25 / 40  Force Measurements Temperature Compensation of Strain Gauges Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 26 / 40  Force Measurements Temperature Compensation of Strain Gauges As the output voltage is of the bridge is:     R1 R2 R1 Rx − R2 R3 Vo = Vin − = Vin R1 + R3 R2 + Rx (R1 + R3 ) (R2 + Rx ) If all the gauges and resistors have the same nominal resistance R0 , then Vo = 0 if there is no change in resistance. If R1 is the resistance of the active strain gauge, and a force applied to the system, R1 becomes R + ∆R, then:   ∆R ∆R Vo = Vin u Vin 4R + 2∆R 4R Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 27 / 40  Force Measurements Example A wire–bonded strain gauge is made with platinum. It has a temperature coefficient of resistance α = 0.00385 Ω◦ C and a gauge factor 8.9. Also, it has a nominal resistance 350 Ω at 20◦ C. The sensor is exposed to temperature variation between −50 and 200◦ C. A bridge circuit is designed to compensate the temperature effect while the maximum strain is 2% 1 Design a bridge circuit to compensate the temperature effect. 2 Find the output of the bridge with a reference voltage of 10 V for the range (0 to 2% strain). Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 28 / 40  Force Measurements Example Step 1: Design the bridge circuit: Given: As the sensor has nominal resistance 350 Ω, then the R0 (no strain) = 350 Ω. designed bridge consists of: R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. R0 (2% strain)= 413.2 Ω. R−50 (2% strain) = 301.185 Ω. R200 (2% strain) = 698.02 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 29 / 40  Force Measurements Example Step 1: Design the bridge circuit: Given: As the sensor has nominal resistance 350 Ω, then the R0 (no strain) = 350 Ω. designed bridge consists of: R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Two identical strain gauges Z1 and Z2 with R0 = R0 (2% strain)= 413.2 Ω. 350 Ω at T0 = 20◦ C. R−50 (2% strain) = 301.185 Ω. Z1 is an active strain gauge that placed in the R200 (2% strain) = 698.02 Ω. direction of applied force. Z2 is a dummy strain gauge placed perpendicu- lar to the direction of applied force (not exposed to normal stress). Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 29 / 40  Force Measurements Example Step 1: Design the bridge circuit: Given: As the sensor has nominal resistance 350 Ω, then the R0 (no strain) = 350 Ω. designed bridge consists of: R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Two identical strain gauges Z1 and Z2 with R0 = R0 (2% strain)= 413.2 Ω. 350 Ω at T0 = 20◦ C. R−50 (2% strain) = 301.185 Ω. Z1 is an active strain gauge that placed in the R200 (2% strain) = 698.02 Ω. direction of applied force. Z2 is a dummy strain gauge placed perpendicu- lar to the direction of applied force (not exposed to normal stress). Two identical resistors Z3 and Z4 with nominal re- sistance 350 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 29 / 40  Force Measurements Example At no strain and T = 20◦ C: Given: Z1 = Z2 = Z3 = Z4 = 350 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. Vo = 0. R200 (no strain) = 592.55 Ω.   Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 30 / 40  Force Measurements Example At no strain and T = 20◦ C: Given: Z1 = Z2 = Z3 = Z4 = 350 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. Vo = 0. R200 (no strain) = 592.55 Ω. At no strain and T = −50◦ C:   Z1 Z4 − Z3 Z2 Z1 = Z2 = 255.675 Ω. Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z3 = Z4 = 350 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 30 / 40  Force Measurements Example At no strain and T = 20◦ C: Given: Z1 = Z2 = Z3 = Z4 = 350 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. Vo = 0. R200 (no strain) = 592.55 Ω. At no strain and T = −50◦ C:   Z1 Z4 − Z3 Z2 Z1 = Z2 = 255.675 Ω. Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z3 = Z4 = 350 Ω. Vo = 0. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 30 / 40  Force Measurements Example At no strain and T = 20◦ C: Given: Z1 = Z2 = Z3 = Z4 = 350 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. Vo = 0. R200 (no strain) = 592.55 Ω. At no strain and T = −50◦ C:   Z1 Z4 − Z3 Z2 Z1 = Z2 = 255.675 Ω. Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z3 = Z4 = 350 Ω. Vo = 0. At no strain and T = 200◦ C: Z1 = Z2 = 592.55 Ω. Z3 = Z4 = 350 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 30 / 40  Force Measurements Example At no strain and T = 20◦ C: Given: Z1 = Z2 = Z3 = Z4 = 350 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. Vo = 0. R200 (no strain) = 592.55 Ω. At no strain and T = −50◦ C:   Z1 Z4 − Z3 Z2 Z1 = Z2 = 255.675 Ω. Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z3 = Z4 = 350 Ω. Vo = 0. At no strain and T = 200◦ C: Z1 = Z2 = 592.55 Ω. Z3 = Z4 = 350 Ω. Vo = 0. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 30 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω.   Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. Vo = 0.041 V. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. R0 (no strain) = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω.   Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. Vo = 0.041 V. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. At 2% strain and T = −50◦ C: R0 (no strain) = 350 Ω. Z1 = 301.185 Ω Z3 = Z4 = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Z2 = 255.675 Ω.   Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. Vo = 0.041 V. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. At 2% strain and T = −50◦ C: R0 (no strain) = 350 Ω. Z1 = 301.185 Ω Z3 = Z4 = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Z2 = 255.675 Ω. Vo = 0.04086 V.   Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. Vo = 0.041 V. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. At 2% strain and T = −50◦ C: R0 (no strain) = 350 Ω. Z1 = 301.185 Ω Z3 = Z4 = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Z2 = 255.675 Ω. Vo = 0.04086 V.   At 2% strain and T = 200◦ C: Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z1 = 698.02 Ω Z3 = Z4 = 350 Ω. Z2 = 592.55 Ω. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Force Measurements Example At 2% strain and T = 20◦ C: Given: R0 (2% strain)= 413.2 Ω. Z1 = 413.2 Ω. Vo = 0.041 V. R−50 (2% strain) = 301.185 Ω. Z2 = Z3 = Z4 = 350 Ω. R200 (2% strain) = 698.02 Ω. At 2% strain and T = −50◦ C: R0 (no strain) = 350 Ω. Z1 = 301.185 Ω Z3 = Z4 = 350 Ω. R−50 (no strain) = 255.675 Ω. R200 (no strain) = 592.55 Ω. Z2 = 255.675 Ω. Vo = 0.04086 V.   At 2% strain and T = 200◦ C: Z1 Z4 − Z3 Z2 Vo = Vin (Z1 + Z2 ) (Z3 + Z4 ) Z1 = 698.02 Ω Z3 = Z4 = 350 Ω. Z2 = 592.55 Ω. Vo = 0.04086 V. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Tactile Sensors Outlines 1 Force Measurements 2 Tactile Sensors 3 Pressure Sensors 4 Recommended Readings and Practice Problems Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 31 / 40  Tactile Sensors Tactile Sensors They are low-force sensors. They sense the presence of force, and react cor- responding to its magnitude. The most common example is the keyboards, In many tactile sensing applications it is often im- portant to sense a force distribution over a spec- ified area (such as the hand of a robot or the gripper of a robotic arm. Usually, they are made from piezoelectric films which respond with an electrical signal in response to deformation. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 32 / 40  Tactile Sensors Conductive–Foam Tactile Sensor The sensor used in membrane keypads. It is a soft foam rubber saturated with very small carbon particles. When the foam is squeezed, the carbon particles are pushed together, and the resistance of the material decreases. Due to its simplicity, it is used widely in many applications such as robot tactile sensors. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 33 / 40  Tactile Sensors Force Sensitive Resistive (FSR) Tactile Sensor The resistance of the material is pressure dependent. It is made from conducting base, an elastomer, and an electrode (made from the ma- terial of the based). Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 34 / 40  Pressure Sensors Outlines 1 Force Measurements 2 Tactile Sensors 3 Pressure Sensors 4 Recommended Readings and Practice Problems Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 34 / 40  Pressure Sensors Pressure Sensors Pressure sensors usually consist of two parts: The first converts pressure to a force or displacement. The second converts the force or displacement to an electrical signal. Pressure measurement can be done in three ways: Gauge Pressure: The difference between the measured pressure and ambient pressure. Differential Pressure: The difference in pressure between two places where neither pressure is necessarily atmospheric. Absolute Pressure: The differential pressure sensor where one side is referenced at 0 psi Pressure sensors can be: mechanical. magnetic, and piezoelectric. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 35 / 40  Pressure Sensors Bourdon Tube It is a short bent tube, closed at one end. When the tube is pressurized, it tends to straighten out. This motion (displacement of the tube) is propor- tional to the applied pressure. If a position sensor or an LVDT is connected to the tube, the displacement can be converted to electrical signal. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 36 / 40  Pressure Sensors Bourdon Tube It is a short bent tube, closed at one end. When the tube is pressurized, it tends to straighten out. This motion (displacement of the tube) is propor- tional to the applied pressure. If a position sensor or an LVDT is connected to the tube, the displacement can be converted to electrical signal. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 36 / 40  Pressure Sensors Bellows It uses a small metal bellows to convert pressure into linear motion. As the pressure inside increases, the bellows ex- pands. This motion is detected with a position sensor such as a potentiometer. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 37 / 40  Pressure Sensors Piezoelectric Pressure Sensor It uses the piezoresistive property of silicon to mea- sure the pressure. It has the advantage of “no moving parts”. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 38 / 40  Pressure Sensors Q&A and Discussion Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 39 / 40  Recommended Readings and Practice Problems Outlines 1 Force Measurements 2 Tactile Sensors 3 Pressure Sensors 4 Recommended Readings and Practice Problems Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 39 / 40  Recommended Readings and Practice Problems Recommended Readings and Practice Problems Recommended Readings: Ida, N., Sensors, Actuators and Their Interfaces: A multidisciplinary introduction. Chapter 6 – Sections 6.3 and 6.5. Chapter 11 – Section 11.6. Practice Problems: Ida, N., Sensors, Actuators and Their Interfaces: A multidisciplinary introduction. Chapter 6. Assoc. Prof. Dr. M. A. Kamel MSE 355 Year 3 – Fall Semester 40 / 40