Temperature Sensing How does the temperature sensor work and how can it be used to control the temperature of a refrigerator? Temperature Sensing page: 1 of 22 Contents Initial Problem Statement 2 Narrative 3-11 Notes 12 Solutions 13-21 Appendix 22 MEI 211
Temperature Sensing Initial Problem Statement A household refrigerator is a device to keep food products within a fixed temperature range. The temperature should be low enough to preserve the food as long as possible, but not too low that the liquids in the refrigerator freeze. To do this the refrigerator uses a temperature sensor to indicate when to turn on or off the cooling system. How does the temperature sensor work and how can it be used to control the temperature of a refrigerator? Temperature Sensing page: 2 of 22 MEI 211
Narrative Introduction Temperature, T Hint Discussion Look at the following graph of a temperature measurement made inside a refrigerator. Describe what is happening. What temperature range would be important for a refrigerator? How could you control the device to achieve an appropriate temperature? How could you design the refrigerator to be as efficient as possible? Figure 1. Think about what happens when the cooling system of a refrigerator is turned on. Hint Time, t Think about what happens when the cooling system of a refrigerator is turned off. To prevent the cooling system turning on and off repeatedly, the control logic will usually turn off the cooling when the lowest allowable temperature has been reached and will not turn it back on until the highest allowable temperature has been reached. This means the temperature is controlled within a range rather than at a single value. This method of using different control points is called introducing hysteresis into the system and is very common in many control applications. One possible temperature sensor is the thermistor. This is a device that has a resistance to the flow of electrical current that is temperature dependent. There are two kinds of thermistor, the positive temperature coefficient (PTC) thermistor and the negative temperature coefficient (NTC) thermistor. The response coefficient of a device tells you whether an increase in the input leads to an increase or decrease in the output. Temperature Sensing page: 3 of 22 MEI 211
Activity 1 Look at the following two graphs. Which do you think has a positive temperature coefficient and which has a negative temperature coefficient? Discussion What do you think positive temperature coefficient and negative temperature coefficient means? Resistance graph (a) Temperature Figure 2. Resistance graph (b) Temperature The most appropriate type of thermistor used for temperature sensing is the NTC thermistor. In all the following text the term thermistor will specifically mean NTC thermistor. Temperature Sensing page: 4 of 22 MEI 211
2. The characteristics of a thermistor Thermistors are usually constructed using a semiconductor material. A typical thermistor is shown below. Multimedia Figure 3. The movie Temperature Sensing Video is available to demonstrate the behaviour of a thermistor. The resistance of a thermistor varies with temperature as You may sometimes see this T T R = Re written as where R = R exp T T R is the resistance of the thermistor at the temperature being measured R is the resistance of the thermistor at a calibration temperature T B is a characterising parameter published by the manufacturer See "Calibration" T is the temperature being measured on page 6. T is the calibration temperature used when measuring R T T Note, be very careful when using the expression R = Re as the temperature in this expression is in Kelvin (K), not Celsius. See "The Kelvin temperature scale" on page 7. Temperature Sensing page: 5 of 22 MEI 211
Calibration The behaviour of a device such as a thermister depends upon the materials and construction details used in its manufacture. For a given design to be useful it must give repeatable results which are valid over a range of temperatures. The behaviour is characterised by calibrating the design. For the thermistor two characterising values are used at a given calibration temperature. The first characterising value is the resistance, R, at a specified calibration temperature, T. Usually the design is such that R and T are standard. A typical value of R would coincide with a standard resistor value; for example 15 kω, while a typical value of T would coincide with an expected temperature; for example 25 C. The second characterising value is the parameter B. This is determined by how the resistance changes from the resistance at the calibration temperature as the temperature changes from the calibration temperature. Hint Discussion Look at the expression that shows how the resistance of a thermistor varies with temperature, R = Re T T What would the resistance, R, be if the measuring temperature, T, is the freezing point of water? (Use R = 15 (Ω), T = 25 ( C) and B = 35.) Is there a problem with the device or the equation? The freezing point of water is C. Hint You could start with T = 2 C and evaluate R. Then reduce T in steps to, for example, 1 C, 1 C,.1 C and note what happens in the brackets. Temperature Sensing page: 6 of 22 MEI 211
Hint Activity 2 After production a thermistor is found to have a characteristic resistance, R, of 15 Ω at the characterising temperature, T, of 25 C. A second calibrating measurement is made. It is found that at a temperature, T, of 21 C the resistance, R, is 17 793 Ω. Calculate the characterising parameter, B. Remember to convert C to Kelvin by adding 273. The Kelvin temperature scale Discussion What are the units of B? The Kelvin temperature measures the so-called absolute temperature. The temperature of absolute zero is defined as the point at which the motion of atoms due to thermal energy ceases. Its value is determined to be -273.16 C (although in this resource a value of -273 is used). The average temperature of outer space is 2.73 K. To convert from C to Kelvin, you just add 273 so that C is 273 K and 1 C is 373 K Activity 3 Using the value of B just calculated fill in the following table and plot the results to show how resistance varies with temperature in C over the given range. Use B rounded to the nearest integer, as this is the value that would published by the manufacturer. Convert R in to kilohm to 1 d.p. Verify that the graph fits the definition of an NTC thermistor, i.e. has a negative gradient. T ( C) T (K) R (Ω) Convert R to kω for plotting -5 5 1 15 2 25 3 35 You could try to share the work out in a group to complete this table, or use a spreadsheet. Temperature Sensing page: 7 of 22 MEI 211
Resistance, R (kω) 7. 6. 5. 4. 3. 2. 1.. -1-5 5 1 15 2 25 3 35 4 Temperature, T ( C) Activity 4 Figure 4. The thermistor is used in a device to measure the temperature of a room. A resistance of 27 kω is recorded. Use your graph to determine the temperature to the nearest ½ C. Substitute your answer back into the equation R = Re T T and find the corresponding value of R to check for consistency. Temperature Sensing page: 8 of 22 MEI 211
3. Using a thermistor in a device When using a thermistor in a practical device it is obviously not convenient for it to report the resistance as you either have to refer that to a conversion table or a graph. Instead the resistance is converted to a temperature which can be displayed directly or used to trigger a control mechanism. Activity 5 Figure 4. Look at the expression relating measured resistance to the temperature: R = Re T T Rearrange this expression to make T the subject of the equation so that a measured resistance will translate directly to a temperature. Temperature Sensing page: 9 of 22 MEI 211
Activity 6 You are testing a thermistor to be used in a device to measure the temperature for control purposes. A resistance of 15.29 kω is recorded. T T Figure 5. Use the expression R = Re to determine what the temperature should be. Recall that for the device you have determined that Hint (a) (b) (c) R T = 15 ( in kω) = 298 ( in K) B = 374 Discussion A calibrating thermometer determines that the actual temperature is 25 C. If the temperature given by your expression does not agree with the calibration temperature what are the possible explanations? the analysis? the temperature sensor? both the analysis and the sensor? the resistance reading? the calibration thermometer? Hint (d) (e) Multimedia The specification for the thermistor is that the characteristic resistance, R, is 15 kω at a calibrating temperature, T, of 25 C. Looking at the above experiment, which was set up at 25 C, what would you expect the resistance to read? The movie resource Temperature Sensing Video is available to demonstrate the reading of the resistance for this experiment. Temperature Sensing page: 1 of 22 MEI 211
4. Calculating the accuracy of the device Most devices are manufactured to within specified tolerances. For calibration instruments these tolerances are usually small so that accurate readings are given. For more general devices however, more relaxed tolerances can be used as long as the impact of these on a design are assessed as being appropriate. Hint Activity 7 The tolerance on the thermistor being used is stated as being ±5%. This is allowed variation of the calibration resistance, R, at the calibration temperature, T. Assuming B is fixed, what is the range of temperatures that could be displayed when the R = 15.29 kω? Consider what the tolerance affects in the expression for temperature: B T = R B + R ln T Discussion Hint What is the accuracy of the thermistor you are using? Activity 8 What is the smallest possible value of R? What is the largest possible value of R? Assume the calibration thermometer reading of 25. C is correct. Show this value and the expected range of variation due to the tolerance on a number line. Is the calculated reading of 24.5 C consistent with the accuracy of the thermistor? Discussion Would this device be suitable for your refrigerator control mechanism? How could you compensate for any bias in the readings due to the manufacturing tolerance? Temperature Sensing page: 11 of 22 MEI 211
Notes Semiconductor resistance Semiconductors are materials in which electrons are not usually free to conduct electricity. As the temperature increases it give some electrons enough energy to become available to conduct so a small current flows. Increasing the temperature further makes more electrons available and a larger current flows. If there is a fixed potential across the semiconductor then Ohm s law states that the potential difference, V, the current, I, and the resistance, R, are related as V = IR As the amount of current that can flow increases with increasing temperature, it can be seen that the resistance must fall with increasing temperature. Temperature Sensing page: 12 of 22 MEI 211
Solutions Introduction Discussion solution The graph initially shows the temperature inside the refrigerator is falling. After a period of time this stops abruptly and the temperature starts to rise, although the increase in temperature is slower than the decrease. Eventually, the temperature reaches an upper limit (the same temperature as that recorded when t = ) and the temperature starts to fall again. It continues to fall until it again reaches a lower limit, at which point the graph repeats. Household refrigerators usually operate between about 1 C 5 C. If they reach C or below, the liquid products, such as your milk, will start to freeze. At higher temperatures biological activity is higher so food will rot more quickly. Refrigerators are typically cooled using a liquid/gas expansion system. When they are running they remove heat from inside the refrigerator and transfer it to the outside. This lowers the temperature inside the refrigerator and increases it outside (this is why you can t cool your kitchen by leaving the refrigerator door open). If the system were left running the temperature inside would fall until it reached the limit of the cooling system, which is below freezing. To prevent this, the system is turned off when a desired temperature has been reached. Insulation prevents heat from entering the refrigerator and warming it up. However, no insulation is perfect and a small amount of heat enters the refrigerator and slowly heats up the interior. An efficient design would be one where the time taken for the inside to heat up to the maximum allowable temperature was very long. The time taken to cool the inside must be shorter than the time taken for it to heat up, otherwise the cooling system would never be able to cool the inside! Activity 1 solution A positive temperature coefficient (PTC) thermistor has a resistance that increases as the temperature increases. The gradient of the curve of resistance against temperature is therefore positive. This is the case for graph (a). A negative temperature coefficient (NTC) thermistor has a resistance that decrease as the temperature increases. The gradient of the curve of resistance vs. temperature is therefore negative. This is the case for graph (b). Discussion The sign of the temperature coefficient tells you the sign of the gradient. Temperature Sensing page: 13 of 22 MEI 211
2. The characteristics of a thermistor Discussion solution The freezing point of water is C. Substituting T = into the expression for the resistance gives T T R = R e = 15 e See "The Kelvin temperature scale" on page 7. ln(x) is the inverse function of e x ; ln(e x ) = x 35 35 25 The term 35 cannot be evaluated (try it on your calculator) so the expression cannot be evaluated! The above is in error though. It assumes that the temperature is measured in C, whereas in fact it should be measured in Kelvin for this application. Activity 2 solution A possible error here is not to convert from C to Kelvin. The following variables have been identified R = 15 T = 298 R = 17793 T = 294 Substituting values into the expression for resistance: R= R e T T 294 298 17 793 = 15e 298B 294B 17793 294 298 = e 15 17793 4B B ln = = 15 87 612 2193 17793 B = 2193ln 15 374. 1 d.p. = ( ) Discussion solution B must be measured in K and this ensures the expression in the exponential is dimensionless. Temperature Sensing page: 14 of 22 MEI 211
Plotting these results Activity 3 solution Resistance, R (kω) 7. 6. 5. 4. 3. 2. 1. T ( C) T (K) R (Ω) Convert R to kω for plotting -5 268 61126 61.1 273 4734 47.3 5 278 371 37. 1 283 29174 29.2 15 288 23192 23.2 2 293 18582 18.6 25 298 15 15. 3 33 12194 12.2 35 38 998 1.. -1-5 5 1 15 2 25 3 35 4 Temperature, T ( C) Figure 6. Temperature Sensing page: 15 of 22 MEI 211
Activity 4 solution An accurate plot will show that 27 kω corresponds to 11.5 C. Substituting this back into the expression for resistance gives R = R e = 15e T T 374 284. 5 374 298 = 27. 21 ( kω) The calculated value of 27.21 kω is close to the stated reading of 27 kω so the answer is consistent with the calculation. Note, in this solution both values of resistance are based in kω, so that 15 should be used not 15,! Temperature Sensing page: 16 of 22 MEI 211
3. Using a thermistor in a device Activity 5 solution R ln R Activity 6 solution R= R e R = e R T T B B T T B B = T T B R B = ln + T R T B T = R B + R ln T The temperature is given by the expression: B T = R B + R ln T Substituting in the known values of gives: R = 15 ( in kω) R = 15. 29 ( in kω) T = 298 ( in K) B = 374 B T = R B + R ln T 374 = 15. 29 + 374 ln 15 298 374 =. 19 + 12. 55 297. 5 1 d.p. ln(x) is the inverse function of e x ; ln(e x ) = x = ( ) Temperature Sensing page: 17 of 22 MEI 211
Remember, this value is in Kelvin. Subtracting 273 to obtain a value in C gives: T = 297. 5 273 = 24. 5 ( C) ( 1 d.p. ) Discussion solution This value does not agree with the calibration thermometer. The mathematical analysis is correct so the difference must be due to other factors: The thermistor will not be manufactured perfectly. There will be a tolerance for the design which leads to a limited practical accuracy. The resistance reading may be in error. As the meter is designed for accurate measurement of resistance it is expected that any error in the reading will be small. The calibration thermometer may have an error. The specification for the device should be checked to determine the possible error on the calibration device. Discussion solution The experiment was performed at the same temperature as the device characterisation temperature so you would expect to read a resistance of 15 kω. Temperature Sensing page: 18 of 22 MEI 211
4. Calculating the accuracy of the device Activity 7 solution The reference resistance is R = 15 kω. The tolerance states that this can vary by ±5%. 5 5% of R = 15 = 75. 1 The smallest possible value of R is: R = 15 75. = 1425. Substituting this into the temperature expression: B T = R B + R ln T with the known values of gives: R = 14. 25 R = 15. 29 ( in kω) ( in kω) T = 298 ( in K) B = 374 B T = R B + R ln T 374 = 15. 29 ln + 14. 25 374 298 374 =. 7 + 12. 55 296. 3 1 d.p. = ( ) Remember, this value is in Kelvin. Subtracting 273 to obtain a value in C gives: T = 296. 3 273 = 23. 3 ( C) ( 1 d.p. ) Temperature Sensing page: 19 of 22 MEI 211
The largest possible value of R is: R = 15 + 75. = 1575. Substituting this into the temperature expression: with the known values of gives: T = B R B + R ln T R 15. 75 in kω R = 15. 29 T = 298 ( in kω) ( in K) B = 374 B T = R B + R ln T 374 = 15. 29 ln 15. 75 374 298 374 =. 29 + 12. 55 298. 7 1 d.p. = ( ) = ( ) Remember, this value is in Kelvin. Subtracting 273 to obtain a value in C gives: T = 298. 7 273 = 25. 7 ( C) ( 1 d.p. ) The range of temperature readings for this device is from 23.3 C to 25.7 C. If the calibration resistance, R is exactly 15 kω the temperature for the recorded resistance is 24.5 C. The thermistor is therefore accurate to ±1.2 C. Temperature Sensing page: 2 of 22 MEI 211
Activity 8 solution A comparison of the predicted temperature of 24.5 C and the calibration thermometer reading of 25. C shows that the result is within the range of accuracy of the thermistor from the stated tolerance. This can be seen by considering the values on a number line: Tolerance lower limit (-1.2 C) Calculated value Discussion solution Actual value Range of accuracy Figure 6. Tolerance upper limit (+1.2 C) The total range of possible values is 2.4 C. This is about half the total allowable range for the temperature inside a refrigerator. The device could be used but care would have to be taken to avoid over or under cooling. To allow for a systematic bias due to the tolerance a control bias could be added. This would add or subtract an amount set by the user in order to control the average temperature of the refrigerator. This is usually installed as a temperature control knob inside the refrigerator. Temperature Sensing page: 21 of 22 MEI 211
Appendix mathematical coverage PL objectives Use algebra to solve engineering problems Be able to evaluate expressions. Understand and be able to work with percentages. Change the subject of a formula. Know how to check answers by substitution. Be able to plot data. Be able to draw graphs by constucting a table of values. Be able to construct and use conversion graphs. Be able to extract information from a graph. Solve problems using laws of logarithms. Solve problems involving exponential growth and decay. Temperature Sensing page: 22 of 22 MEI 211