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EE 452 and EE 830 Assignment 1, 1999 Solutions 4. Choose a thermocouple: Type K is most linear Type E has the greatest sensitivity Types B, C, J, R and S do not cover the necessary range I will choose Type K because of its better linearity. [Note: when making an arbitrary choice, it is a good idea to explain why that choice has been made. It will make your past work a much better reference for future work.] From the thermocouple reference tables for Type K, the output voltage at -100 ºC is -3.554 mV, and at +500 ºC, it is +20.644mV. Our desired output is -1000 mV and +5000 mV, respectively. For a single gain to apply over the entire range, we have to solve the following simultaneous equations: -1000 mV = G * (-3.554 mV) + Vo and +5000 mV = G * (20.644 mV) + Vo, where G is the gain and Vo is the offset. This results in a gain of 248 and an offset of -121 mV. Note some of you chose to have zero offset, but then your output wasn't -1.00 V and +5.00 V at -100 and + 500 ºC, respectively. That is, without other techniques for correcting the nonlinearity, a choice must be made as to how the error will be incorporated into the overall response. If you are adding an offset voltage, then you should show how it is to be implemented. That is, you can't just put in a power supply as you might in the lab. Temper ature degrees celsius -100 -50 0 50 100 150 200 250 300 350 400 450 500 Thermo couple Desired Expected Desired output Error in temperature Output output output referred to the input due to nonlinearity mV V V mV degrees celsius -3.544 -1.889 0 2.023 4.069 6.138 8.138 10.153 12.209 14.293 16.397 18.516 20.644 -1.00 -0.50 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50 4.00 4.50 5.00 -1.00 -0.59 -0.12 0.38 0.89 1.40 1.90 2.40 2.91 3.42 3.95 4.47 5.00 -3.544 -1.528 0.488 2.504 4.520 6.536 8.552 10.569 12.585 14.601 16.617 18.633 20.649 0 10 12 11 10 10 10 10 9 7 5 2 0 You were asked to use an instrumentation amplifier. Usually the second (differential stage) is set for unity gain to improve the common-mode rejection performance. In this application, that is not a major consideration. A gain of 248 is too high to accomplish in a single stage. Therefore, the differential stage can be designed for a gain that is easy to achieve with fixed value resistors, such as a gain of ten. Any combination of resistors of R/10R will then be possible over a range of about 1 k to 10 M. From a noise performance perspective, it is better to keep the resistance values small, say 1 k and 10 k. Note: one of the resistors in the differential stage (usually the one shown in the figure) must be adjustable. This is done so that the resistor ratios of the two halves of the differential stage can be set to be equal. The common-mode rejection performance degrades substantially for small differences in the resistor ratios. The first stage of the instrumentation amplifier can be used to obtain a gain of 24.8 by having fixed feedback resistors and a fixed/variable gain-controlling resistor. Therefore, I arbitrarily chose a value of 22 k for the feedback resistors, which means the gain control resistor must be 1.849 k. This can be achieved with a 1.6-k, fixed, 5% resistor in series with a 500- potentiometer, or a 1.78-k, fixed, 1-% resistor in series with a 100- potentiometer. Note: an online table of standard resistor values can be found at: http://www.action-electronics.com/resist1p.htm Note also: a comparable table of standard capacitor values can be found at: http://www.action-electronics.com/capacito.htm In this case, note that the capacitor values are not all available over the full range of decades as is the case for resistors. Figure 1 in the accompanying pdf link shows the instrumentation amplifier circuit. 5. Using the formula provided in class, the shunt resistor for the range 0 to 40 ºC for this thermistor is 9.136 k. I used an actual value of 9.09 k as that is the closest 1-% standard value. The table below shows the resulting linearized resistance versus temperature and the calculated input deviation with and without the shunt resistor. Temperature Resistance Rsh//Rt "linearized" input "linearized" input degrees ohms ohms temperature deviation temperature deviation celsius without without with shunt with shunt shunt shunt resistor resistor -10 42,000 7473 -28.3 -18.3 -8.3 1.7 -5 34,000 7172 -13.2 -8.2 -4.6 0.4 0 27,000 6800 0.0 0.0 0.0 0.0 5 22,000 6432 9.4 4.4 4.5 -0.5 10 18,000 6040 17.0 7.0 9.3 -0.7 15 15,000 5660 22.6 7.6 14.0 -1.0 20 12,000 5172 28.3 8.3 20.0 0.0 25 10,000 4762 32.1 7.1 25.0 0.0 30 8,300 4339 35.3 5.3 30.2 0.2 35 6,900 3923 37.9 2.9 35.3 0.3 40 5,800 3541 40.0 0.0 40.0 0.0 45 4,900 3184 41.7 -3.3 44.4 -0.6 50 4,200 2873 43.0 -7.0 48.2 -1.8 From this table, it can be seen that the maximum input deviation without the shunt resistor is 8.3 ºC, whereas with the shunt resistor it is 0.7 ºC. The nonlinearity is 21% and 1.75% respectively. Thus, the linearization technique allows us to achieve 1-degree accuracy. This level of accuracy holds to at least -5 and +45 ºC. 9. Many of you developed a circuit in which the current from the AD590 was sensed by a series resistor, and the voltage across the resistor was used directly as the input to a comparator. While these circuits would work in principle, the change in voltage is so small that it makes the setting of the hysteresis on the comparator very sensitive to actual component values. The other point of interest is that many comparators can sink much more current (when their output is at 0 V) than they can source (when their output is +Vcc). Therefore, the preferred design would be to have the drive current to the transistor (or directly to the relay) occur when the comparator output is low. The circuit of Figure 2 attached shows one possible configuration based on a comparator with hysteresis. The first stage is a current to voltage converter for the AD590. However, because of the large offset current (290 A), the summing node (at the inverting terminal) is used to subtract 300 A. This means the output of the first operational amplifier is positive for temperatures below 27 ºC. (If the comparator cannot withstand negative input voltages and if higher temperatures are expected, then the subtracted current would have to be increased.) The comparator with hysteresis is set up with a simple reference voltage, basically a voltage divider. One of the three resistor values must be chosen arbitrarily (R1 = 100 k). Then two simultaneous equations can be set up at the inverting node of the comparator. One equation is for the output voltage of +15 V and a desired node voltage of 3.0 V; the other equation for the output voltage of 0 V and a desired node voltage of 1.5 V. Solution of the two simultaneous equations yields R2 = 80.6 k and R3 = 12.7 k. All these resistor values are the closest standard 1% values. 12. The main point of this question was to get you thinking about using self-heating for a functional purpose rather than as a side-effect to be minimized. Once an appropriate voltage has been generated, a comparator can be used to generate a control signal as in the previous question. The first decision to be made is the degree of self-heating to be used. To maximize the voltage difference between air versus liquid, a relatively large self-heating temperature should be used when the sensor is in air, say T = 50 ºC. This means that 400 mW must be dissipated. Since there is an 8:1 ratio in dissipation constant, T should be about 6 ºC, when the sensor is in the liquid. The sensor has a resistance of 19 ohms at 50 ºC. For 400-mW of heat to be generated with this resistance, a current of 145 mA is required. If we assume a 15-V supply is available, then we need a total resistance of 103 . We can use a 100- (5%) resistor for this. When the sensor is in the liquid, the resistance will be 83 . Thus the voltage across the sensor will go from 0.24 V when it is in air to 6.8 V when it is in the liquid.