Aplication Note Thermal Relay, Rumus

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Application Note Temperature Considerations for DC Relays Relays and temperature are intertwined. When a relay is exposed to various temperatures, its operating characteristics change dependent upon the temperature. The most notable changes occur in the pick-up voltage (VPI) and coil resistance (RC). The coil winding of a relay is produced with copper wire and thus the coil resistance varies with the temperature coefficient of copper. For the temperature range that a relay will normally be expo
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  1 Application Note It is not a critical matter which reference temperature is used. Theinternational was selected as 23 ° C ( & 3 ° C) to encompass both of theprevious standards and thus appease everyone. It is recommended thatthis value be used whenever specifying new products since it is notmore than 1.2% from either the 20 or 25 ° C reference and will providefuture consistency if adopting ISO countries begin to utilize it. Theequivalent values can be calculated from this reference for the others.Not that almost all European specifications still use the 20 ° C referencewhile many U.S. firms are beginning to utilize the 23 ° C reference.While temperature changes affect relay parameters, the power dissipatedwithin the relay also affects the temperature in most applications. Thepower dissipated within the relay may be broken down into two majorcomponents. The first is heat generated in the relay coil when voltage isapplied to it. This heat creates a temperature rise (or increase) in therelay coil and package. The amount of temperature rise created isdependent upon several factors such as the volume of copper wire used,insulation thickness, insulation type, bobbin material, bobbin thickness,terminal size, conductor size, and several other factors that are designrelated. Each of these factors will either enhance or resist the flow ofgenerated heat out of the coil assembly and into the ambient air. For agiven relay design, these factors can be summed together into a valuecalled the “coil to ambient thermal resistance” of the relay. Thedimensions of such a value are ° C/Watt. The thermal resistance isanalogous to the electrical resistance and the temperature rise createdby coil power dissipation follows the equation: Eqn. 3 T RC = θ   CA x P D where:T RC =Temperature rise caused by coil dissipation θ CA =Thermal resistance from coil to ambientP D =Final steady-state power dissipated in coilFor normal relay temperature ranges, this relationship is nearly linearand consistent under the following conditions:1.The relay is in still air and not subjected to significant air flow orthe value of θ CA was determined with an air flow identicalto the end application (difficult to simulate). For pc board relays,the still air assumption is often valid because of the end productenclosure.2.All power calculations deal with the coil resistance at the finalcoil temperature (T C ) attained. If only room temperature coilresistance were used, the resulting non-linearity would resultin significant errors at higher temperatures.3.The value for thermal resistance is determined from test datawhere the relay carried no load current.We now have the information necessary to calculate the final coiltemperature from data book parameters under no load conditions for arelay. Let us try an example.Given the following:T 0 =20 ° CV 0 =V PI = 6.8 voltsR 0 =90 OhmsV A =13.5 volts (V A = applied coil voltage) θ CA =40 ° C/WT A =85 ° C (T A = ambient temperature)I L =0 Amperes (I L = load current) Temperature Considerations for DC Relays Relays and temperature are intertwined. When a relay is exposed tovarious temperatures, its operating characteristics change dependentupon the temperature. The most notable changes occur in the pick-upvoltage (V PI ) and coil resistance (R C ). The coil winding of a relay is producedwith copper wire and thus the coil resistance varies with the temperaturecoefficient of copper. For the temperature range that a relay will normallybe exposed to, the change in copper follows the form of: Eqn. 1 R 1 =R 0 x (1 + A x (T 1 - T 0 ))where:R 1 =Resistance at temperature T 1 R 0 =Resistance at temperature T 0 A=Slope of a line from a point ( - 234.5,0) throughthe point (T 0 , 1)(A = 0.003929 at T 0 = 20 ° C)T 1 =New temperature of interestT 0 =Reference temperature (20 ° C is typically used forthis value)Now that we can calculate the coil resistance at a new temperaturegiven a value at some known reference temperature, let us look at thepick-up voltage. For a DC relay, the magnetic force developed isproportional to the Ampere-turns developed in the coil. Since themechanical forces are fairly constant over the normal temperature range(and the number of turns is fixed), we can also deduce that the pick-upcurrent (I PI ) will be constant. If pick-up current is constant and coilresistance varies, it follows that pick-up voltage (V PI = I PI x R C ) variesdirectly as the coil resistance. This leads to a simple mathematical methodto determine coil resistance and pick-up voltage at any temperature if areference point is known.For example:Assume that a relay has the following parameters at 20 ° C (T 0 ).R C =90 OhmsV PI =6.5 voltsCalculate the new coil resistance at 105 ° C (T 1 )From Eqn. 1 we find:R 1 =90 Ω x (1 + 0.003929/  ° C x (105 ° C - 20 ° C))=90 Ω x (1.334)=120.1 Ω Eqn. 2 To find the new pick-up voltage, we replace R 1 and R 0 with V 1 and V 0 respectfully to find:V 1 =6.5 volts x (1.334)=8.67 voltsTo find the factor, A, the following equation is provided:A=1 / (T 0 + 234.5)For three common reference temperatures, A is as follows:European & Asian at 20 ° C:A=0.003929International (IEEE) at 23 ° C:A=0.003883United States at 25 ° C:A=0.003854  2 Application Note Specifications and availability subject to change without notice.13C3226Printed U.S.A.IH/12-00Tyco Electronics Corporation – P&B, Winston-Salem, NC 27102Technical Support Center: 1-800-522-6752, www.pandbrelays.comstudies imply that the contact power dissipation may be treated as aseparate heat source that adds heat into the relay package. Its effect oncoil temperature is dependent upon many factors including package size,contact to coil distance, contact terminal size, connecting wire size, sharedthermal paths, etc. Again, these factors can be lumped into a contact tocoil thermal resistance. This leads to an equation similar to Eqn. 3. Eqn. 5 T RL = θ CC x P K = θ CC x R K I L   2 where:T RL =Temperature rise in coil caused by the load current θ CC =Thermal resistance from contact to coilP K =Power dissipated in contactsR K =Contact circuit resistanceI L =Load current flowing through the contact circuitAs an alternative possibility, and in an effort to provide a best fit curve toearlier test data, the following equation has yielded good approximations. Eqn. 6 T RL =K RL x I L1.85 This formula has been empirically derived from several test results andhas successfully predicted final coil temperature rise caused by contactloads. The value K RL can be derived from a two step temperature test.First determine temperature rise with no contact load and then measureunder the same conditions with a contact load. The coil temperaturerise minus the part caused by the coil power dissipation yields a T RL andI L combination that may be used to solve Eqn. 6 for K RL .The final coil temperature is then found by adding the respectivecomponents to obtain: Eqn. 7 T C =T A + T RC + T RL =T A + θ CA x (V A2 / R C ) + K RL x I L1.85 This formula also requires solution by iteration. Since the only differencehere is the added T RL term, the following example is left to the reader.All conditions the same as in the previous example except with:I L =20 AmperesK RL =0.029The answers should be 113.0 Ohms, 8.54 volts, T C = 146.5 ° C, R 1 = 134.73Ohms, and V 1 = 10.18 volts.The reader should now be able to determine the steady-statecharacteristics for any temperature and voltage combination given theappropriate relay data. It must be stressed that the values obtainedhere apply to DC relays operated continuously at these values.Intermittent duty (with short, i.e. less than 1 minute, “on” times andlonger “off” times) may result in substantially lower temperatures.Therefore if a specific known duty cycle is given for the relay operation,testing at these conditions could yield acceptable results for final coiltemperature when the continuous duty temperatures calculated herewould not. The methods discussed here are applicable to standard DCrelays and while the coil resistance formula will work for Polarized DCrelays (one that utilizes a permanent magnet) and AC relays as well, thepick-up voltage equations will not work in such cases. With a polarizedDC relay the temperature induced change in magnetic force of the magnetmust be considered. This is normally such that it reverses part of thechange in pick-up voltage caused by the copper wire resistance. In thecase of AC relays, the inductance contributes a significant portion of thecoil impedance and is related to the turns in the coil. since the inductancevaries only slightly with temperature, the pick-up voltage exhibits lessvariation over temperature than for DC relays.Determine the following:1.“Cold start” pick-up voltage (with the coil previouslyunenergized) and coil resistance at T A 2.Final steady-state coil temperature (T C ) and resistance for V A 3.“Hot start” pick-up voltage (after coil energized at V A ) at T A andV A First we solve Eqn. 1 for R 1 at 85 ° CR 1 =90 x (1 + 0.003929 x (85 - 20))=90 x (1.2554)=113.0 OhmsAgain we find V 1 at 85 ° C by using the same factorV 1 =6.8 x (1.2554)=8.54 voltsNow the difficult part, finding T C with 13.5 volts applied to the coil.From Eqn. 3, and realizing that T C = T A + T RC and P D = V A   2  /R C : Eqn. 4 T C = θ   CA x V A   2  /R C = T A Now we have a problem. As we have already seen, R C changes withtemperature. Since we are calculating temperature, we have twovariables. The easiest approach to use here is simple iteration. Let usstart by using the initial coil resistance at the ambient temperature ofinterest:T C1 =(40 x ((13.5) 2 / 113)) + 85=64.5 + 85=149.4 o CWe must now calculate a new value of R C using T C1 and Eqn. 1.R C1 =90 x (1 + 0.003929 x (149.5 - 20))=90 x (1.5088)=135.8 OhmsNow using Eqn. 4 again;T C2 =(40 x ((13.5) 2 / 135.8)) + 85=53.7 + 85=138.7 ° CAgain we would calculate a new value of R C at T C2 and repeat the processuntil a suffient accuracy is obtained. With several iterations, the answerto this example becomes:T C =140 ° CNow that we have the final coil temperature, we can find the coilresistance with Eqn. 1.R C =90 x (1 + 0.003929 x (140 - 20))=90 x (1.4715)=132.4 OhmsThe “hot start” pick-up voltage is found using the same factor:V 1 =6.8 x (1.4715)=10.0 voltsThe only remaining piece to the puzzle is how a contact load currentaffects the temperature of the relay coil and thus its parameters. Past
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