23a HW p17 WB strain

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2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges HW problem 17: Wheatstone bridge circuits with strain gauges (100 points) From the very beginning of the course, we are already familiar with voltage dividers. Wheatstone bridge circuit is simply a pair of voltage dividers, with the output voltage measured between the dividers’ middle nodes. Despite its simplicity, Wheatstone bridge offers extremely valuable advantages: 1. Quite often, the sensor signals are weak. For example, electric resistance of a strain gauge varies due to the load only by ~0.1%. Wheatstone bridge allows to amplify these small signals for more accurate calibration and measurements. 2. The electric resistance of a sensor depends on several factors, including temperature. The variations of resistance due to the variations of temperature may be large, ~20% or more. Wheatstone bridge allows to compensate these huge variations due to temperature in order to focus on the small variations due to the load. In this problem, we will see both advantages of Wheatstone bridge circuits in action. For the theoretical background, see file “Old problem solved- 18a HW p6” posted on Canvas. I will frequently refer to this file as “Old problem.” In all parts of this problem, use the following numerical values:  The source voltage V S = 12 V  In all circuits, the fixed resistances in all circuits R = 120 Ω .  The resistance of a strain gauge at room temperature without load also equals R = 120 Ω.  The magnitude of change of resistance of a strain gauge due to the load is ∆ R = 1 Ω . This change is positive if the gauge is stretched; it is negative if the gauge is compressed. Note that the resistance of a “dummy” gauge does not change due to loading.  The change of gauge resistance due to temperature is ∆ R T =+ 16 Ω regardless of whether the gauge is unloaded, stretched, or compressed. Part One (10 points) Electric resistances of strain gauges Calculate the electric resistances in Ω: Fill the table below. Show your work to get credit. Table One. Electric resistances of strain gauges Dummy gauge at room temperature, or constant resistance 120 Ω Dummy gauge at high temperature 136 Ω Stretched gauge at room temperature 121 Ω Stretched gauge at high temperature 137 Ω Compressed gauge at room temperature 119 Ω Compressed gauge at high temperature 135 Ω 1. R = 120 Ω 2. R = 120 + 16 = 136 Ω © 2023 Alexander Ganago Page 1 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges 3. R = 120 + 1 = 121 Ω 4. R = 120 + 16 + 1 = 137 Ω 5. R = 120 – 1 = 119 Ω 6. R = 120 + 16 – 1 = 135 Ω Part Two: Compare a voltage divider with a Wheatstone quarter-bridge circuit (20 points) Consider a cantilever with one strain gauge, mounted on top: see “Old problem” page 2. In this Part, all calculations should be done for room temperature conditions. Circuit diagram #1 Circuit diagram #2 V SG is the voltage across the strain gauge R SG V F is the voltage across the fixed resistor R Usually, a Wheatstone bridge circuit includes 4 resistors. Here, it includes only one gauge; three other resistors are fixed. This configuration is called a quarter bridge . In the following parts, we will work with half-bridges , each of which includes two gauges. Full-bridge circuits with 4 gauges are also used in practice, but we do not consider them in this HW Problem. Calculate the output voltages of circuits #1 and #2. Fill the table below. Show your work. Table Two. Output voltages of a voltage divider and a Wheatstone bridge circuit Without load With load The difference, mV Circuit diagram #1 Voltage divider 6 V 6.0249 V 24.9 mV Circuit diagram #2 Wheatstone bridge 0 V 0.0249 V 24.9 mV Circuit Diagram #1 V OD = V S * R SG /(R + R SG ) Without load V OD = 12 * 120/(120 + 120) = 6 V © 2023 Alexander Ganago Page 2 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges With load V OD = 12 * 121/(120 + 121) = 6.0249 V Difference = 6.0249 – 6 = 0.0249 V Circuit Diagram #2 V OB = V S * (R SG /(R + R SG ) – R/(R + R)) Without load V OB = 12 * (120/(120 + 120) – 120/(120 + 120)) = 0 V With load V OB = 12 * (121/(120 + 121) – 120/(120 + 120)) = 0.0249 V Difference = 0.0249 – 0 = 0.0249 V I assume that you have obtained very similar results for circuits #1 and #2. In other words, at this step, the advantage of a Wheatstone bridge compared with a voltage divider is not evident yet. Since the changes of voltages due to the load are quite small, we need to amplify the circuit’s output signals. The output voltage V_OD or V_OB serves as the amplifier’ input voltage V ¿ . The amplifier’s gain should be chosen so that the amplifier’s output should not exceed 10 V V OUT = ( Gain ) ×V ¿ ≤ 10 V Calculate the maximal possible amplifier’s gain for both circuits. Write your results in the table below. Show your work. Next, calculate the difference in mV between the voltages at the amplifier’s output due to the load. Write your results in the table below. Show your work. Circuit Diagram #1 Gain = 10/6.0249 = 1.6598 ∆V = 1.6598 * 0.0249 = 0.0413 V Circuit Diagram #2 Gain = 10/0.0249 = 401.6064 ∆V = 401.6064 * 0.0249 = 10 V Maximal output voltage of the circuit with strain gauge Maximal voltage at the output of the amplifier Maximal gain of the amplifier At the output of the amplifier, ∆V due to the load Circuit 1: Voltage divider 6.0249 V 10 V 1.6598 41.3 mV Circuit 2: Quarter bridge 0.0249 V 10 V 401.6064 10 V I expect that now you observe the striking difference between the voltage divider and the Wheatstone quarter-bridge: the bridge provides much larger signal, which is easier to work with. © 2023 Alexander Ganago Page 3 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

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2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges For the following parts of the problem: We work at the interface between the “electric world” and the “non-electric world”: o In the “electric world”, we learn what circuits to build. Here, we have already learned that a Wheatstone quarter-bridge has advantages vs. a single voltage divider. o In the “non-electric world”, o We may compare the positions of strain gauges on the cantilever (compare position #1 on page 2 and position #2 on page 4 of the “Old problem” file). o Also, for the half-bridges (circuits with two strain gauges each), we can compare the orientation of the rosettes in relation to the load. o Eventually, we will look for the optimal combinations of circuits and mounts, which compensate the changes of the gauges’ electric resistance due to temperature, and emphasize the changes of the electric resistance dues to load. In the “electric world”, we keep using the same circuit diagrams as in the “Old problem” fil:. Strain gauges are made from foil (aluminum or stainless steel), shaped as rosettes: for details, see the manufacturers’ web sites such as Omega.com . The electric resistance changes due to strain applied along the long axis of the rosette, as shown on the sketch below. We consider two gauges bonded to two surfaces of the same cantilever beam. When a downward force is applied to the beam, the gauge on the top surface is stretched; the gauge on the bottom surface is compressed. © 2023 Alexander Ganago Page 4 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges In each of the following parts, the calculations remain almost as simple as in Part Two above. Our goal is to find the optimal circuit and the optimal mounting of strain gauges, which ensure that strongest output signals free from distortions. Each part of the problem has its special focus. At the bottom line, the question is very practical: o When the part of machinery is both strained and heated, how to obtain the output signal, which is due only to the strain? We formulate our criteria for a good combination of Mounting variant and half-bridge circuit: (1) Zero output at zero load, at both temperatures. (2) The same output voltage due to load, at both temperatures. We will use these criteria to judge whether the given combination is good or bad. © 2023 Alexander Ganago Page 5 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges Part Three: Mounting Variant One; Electric circuit #3 (20 points) Electric resistances (feel free to use your results from Part One) Gauge #1 Gauge #2 No load; Room Temperature 120 Ω 120 Ω No load; High Temperature 136 Ω 136 Ω Loaded; Room Temperature 121 Ω 120 Ω Loaded; High Temperature 137 Ω 136 Ω Output voltages: Write the equation in the terms of V S , R, R SG 1 ,R SG 2 V 3 = V S * (R SG1 /(R + R SG1 ) - R SG2 /(R + R SG2 )) No load; Room Temperature 0 V No load; High Temperature 0 V Loaded; Room Temperature 24.90 mV Loaded; High Temperature 21.89 mV 1. V 3 = 12 * (120/(120 + 120) - 120/(120 + 120)) = 0 V 2. V 3 = 12 * (136/(120 + 136) - 136/(120 + 136)) = 0 V 3. V 3 = 12 * (121/(120 + 121) - 120/(120 + 120)) = 0.02490 4. V 3 = 12 * (137/(120 + 137) - 136/(120 + 136)) = 0.02189 © 2023 Alexander Ganago Page 6 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

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2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges Is this combination good? Justify your answer. Yes, the combination is good because there is no output voltage when there is no load for both temperatures. The output voltage when there is a load is also similar for both temperatures with a percent difference of around 12%. Part Four: Mounting Variant Two; electric circuit #3 (20 points) Electric resistances (feel free to use your results from Part One) Gauge #1 Gauge #2 No load; Room Temperature 120 Ω 120 Ω No load; High Temperature 136 Ω 136 Ω Loaded; Room Temperature 121 Ω 119 Ω Loaded; High Temperature 137 Ω 135 Ω Output voltages: Write the equation in the terms of V S , R, R SG 1 ,R SG 2 V 3 = V S * (R SG1 /(R + R SG1 ) - R SG2 /(R + R SG2 )) No load; Room Temperature 0 V No load; High Temperature 0 V Loaded; Room Temperature 50.00 mV Loaded; High Temperature 43.95 mV © 2023 Alexander Ganago Page 7 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges 1. V 3 = 12 * (120/(120 + 120) - 120/(120 + 120)) = 0 V 2. V 3 = 12 * (136/(120 + 136) - 136/(120 + 136)) = 0 V 3. V 3 = 12 * (121/(120 + 121) - 119/(120 + 119)) = 0.05000 V 4. V 3 = 12 * (137/(120 + 137) - 135/(120 + 135)) = 0.04395 V Is this combination good? Justify your answer. Yes, the combination is good because there is no output voltage when there is no load for both temperatures. The output voltage when there is a load is also similar for both temperatures with a percent difference of around 12%. Part Five: Mounting Variant Two; Electric circuit #4 (20 points) Electric resistances (feel free to use your results from Part One) Gauge #1 Gauge #2 No load; Room Temperature 120 Ω 120 Ω No load; High Temperature 136 Ω 136 Ω Loaded; Room Temperature 121 Ω 119 Ω Loaded; High Temperature 137 Ω 135 Ω Output voltages: Write the equation in the terms of V S , R, R SG 1 ,R SG 2 V 4 = V S * (R SG1 /(R + R SG1 ) - R/(R + R SG2 )) No load; Room Temperature 0 V No load; High Temperature 0.75 V © 2023 Alexander Ganago Page 8 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

2023 Winter EECS 314 HW p17 Wheatstone bridges with strain gauges Loaded; Room Temperature -0.21 mV Loaded; High Temperature 0.75 V 1. V 4 = 12 * (120/(120 + 120) - 120/(120 + 120)) = 0 V 2. V 4 = 12 * (136/(120 + 136) - 120/(120 + 136)) = 0.75 V 3. V 4 = 12 * (121/(120 + 121) - 120/(120 + 119)) = -2.08 * 10 -4 V 4. V 4 = 12 * (137/(120 + 137) - 120/(120 + 135)) = 0.75 V Is this combination good? Justify your answer. No, this combination is not good because there is a non-zero output voltage when there is no load at a high temperature. The output voltages when there is a load are also significantly different. Part Six: Conclusions (10 points) Out of the combinations, which you studied in Parts Three, Four, and Five, o Which one is the worst and why? The worst combination was the one in part five because the load alone does not affect the output voltage as shown by the non-zero output voltage at high temperature and no load. o Which one is the best and why? The best combination was the one in part four because the output voltages were zero when there was no load at both temperatures. In addition, the output voltage rises consistently with load as there is no dummy gauge. o Is this best combination ideal? If not, what improvements can you suggest? No, it is not ideal because with the same load, the output voltage of a circuit at a higher temperature is smaller than the output voltage of one at a room temperature. This means both the load and the temperature are affecting the output voltage rather than just the load alone. I would recommend experimenting with different temperatures to minimize the effects of temperature on the circuit. © 2023 Alexander Ganago Page 9 of 9 Last printed 3/27/23 5:19:00 PM File: 86951ef77c2bfba69dfc3d14c6f9f484e860451a.docx

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