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? semiconductor components industries, llc, 2001 november, 2001 rev. 1 1 publication order number: an569/d an569/d transient thermal resistance - general data and its use prepared by: bill roehr and bryce shiner on semiconductor applications engineering introduction for a certain amount of dc power dissipated in a semiconductor, the junction temperature reaches a value which is determined by the thermal conductivity from the junction (where the power is dissipated) to the air or heat sink. when the amount of heat generated in the junction equals the heat conducted away, a steadystate condition is reached and the junction temperature can be calculated by the simple equation: t j p d r jr t r (1a) where: t j = junction temperature; t r = temperature at reference point; p d = power dissipated in the junction; r q jr = steadystate thermal resistance from junction to the temperature reference point. power ratings of semiconductors are based upon steadystate conditions, and are determined from equation (1a) under worst case conditions, i.e.: p d(max) t j(max) t r r jr(max) (1b) t j(max) is normally based upon results of an operating life test or serious degradation with temperature of an important device characteristic. t r is usually taken as 25 c, and r q jr can be measured using various techniques. the reference point may be the semiconductor case, a lead, or the ambient air, whichever is most appropriate. should the reference temperature in a given application exceed the reference temperature of the specification, p d must be correspondingly reduced. equation (1b) does not exclusively define the maximum power that a transistor may handle. at high power levels, particularly at high voltages, second breakdown may occur at power levels less than that given by equation (1b). (second breakdown is a result of current concentrating in a small area which causes the transistor to lose its ability to sustain a collectoremitter voltage. the voltage drops to a low value generally causing the circuit to deliver a very high current to the transistor resulting in a collector to emitter short.) second breakdown can be prevented by operating within the safe area given on a manufacturer's data sheet. needless to say, abiding by the safe area is most important in avoiding a catastrophic failure. dc safe area and the thermal resistance allow the designer to determine power dissipation under steadystate conditions. steadystate conditions between junction and case are generally achieved in one to ten seconds while minutes may be required for junction to ambient temperature to become stable. however, for pulses in the microsecond and millisecond region, the use of steadystate values will not yield true power capability because the thermal response of the system has not been taken into account. note, however, that semiconductors also have pulse power limitations which may be considerably lower or even greater than the allowable power as deduced from thermal response information. for transistors, the second breakdown portion of the pulsed safe operating area defines power limits while surge current or power ratings are given for diodes and thyristors. these additional ratings must be used in conjunction with the thermal response to determine power handling capability. to account for thermal capacity, a time dependent factor r(t) is applied to the steadystate thermal resistance. thermal resistance, at a given time, is called transient thermal resistance and is given by: r jr(t) r(t) r jr (2) the mathematical expression for the transient thermal resistance has been determined to be extremely complex. the response is, therefore, plotted from empirical data. curves, typical of the results obtained, are shown in figure 1. these curves show the relative thermal response of the junction, referenced to the case, resulting from a step function change in power. observe that during the fast part of the response, the slope is 1/2 for most of the devices; (i.e., t j t ), a characteristic generally found true of metal package devices. the curves shown are for a variety of transistor types ranging from rather small devices in to5 packages to a large 10 ampere transistor in a t o3 package. observe that the total percentage difference is about 10:1 in the short pulse ( t ) region. however, the values of thermal resistance vary over 20:1. as an aid to estimating http://onsemi.com application note
an569/d http://onsemi.com 2 response, appendix c provides data for a number of packages having different die areas. many on semiconductor data sheets have a graph similar to that of figure 2. it shows not only the thermal response to a step change in power (the d = 0, or single pulse curve) but also has other curves which may be used to obtain an effective r(t) value for a train of repetitive pulses with different duty cycles. the mechanics of using the curves to find t j at the end of the first pulse in the train, or to find t j(pk) once steadystate conditions have been achieved, are quire simple and require no background in the subject. however, problems where the applied power pulses are either not identical in amplitude or width, or the duty cycle is not constant, require a more thorough understanding of the principles illustrated in the body of this report. use of transient thermal resistance data part of the problem in applying thermal response data stems from the fact that power pulses are seldom rectangular, therefore to use the r(t) curves, an equivalent rectangular model of the actual power pulse must be determined. methods of doing this are described near the end of this note. before considering the subject matter in detail, an example will be given to show the use of the thermal response data sheet curves. figure 2 is a representative graph which applies to a 2n5632 transistor. pulse power, p d = 50 watts duration, t = 5 milliseconds period, t p = 20 milliseconds case temperature, t c = 75 c junction to case thermal resistance, r q jc = 1.17 c/w the temperature is desired, a) at the end of the first pulse b) at the end of a pulse under steadystate conditions. for part (a) use: t j r(5 ms)r jc p d t c the term r(5 ms) is read directly from the graph of figure 2 using the d = 0 curve, t j 0.49 1.17 50 75 28.5 75 103.5 the peak junction temperature rise under steady conditions is found by: t j r(t, d)r jc p d t c 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 4 6 5 7 8 to3 & to66 packages are all copper or have a copper slug under die. 2 1 3 4 6 5 7 8 to61 to3, to66 case 77 to3, to66 case 77 to5 (kovar) to5 (steel) to5 (steel) 10,000 3,600 3,600 16,800 8,000 3,600 3,600 14,400 case die size (sq. mil) 0.01 0.02 0.05 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) d = 0.5 0.2 0.1 0.05 0.02 0.01 single pulse figure 1. thermal response, junction to case, of various semiconductor types for a step of input power figure 2. thermal response showing the duty cycle family of curves
an569/d http://onsemi.com 3 d = t/ t p = 5/20 = 0.25. a curve for d = 0.25 is not on the graph; however, values for this duty cycle can be interpolated between the d = 0.2 and d = 0.5 curves. at 5 ms, read r(t) 0.59. t j 0.59 1.17 50 75 34.5 75 109.5 c the average junction temperature increase above ambient is: t j(average) t c r jc p d d (1.17)(50)(0.25) 14.62 c (3) note that t j at the end of any power pulse does not equal the sum of the average temperature rise (14.62 c in the example) and that due to one pulse (28.5 c in example), because cooling occurs between the power pulses. while junction temperature can be easily calculated for a steady pulse train where all pulses are of the same amplitude and pulse duration as shown in the previous example, a simple equation for arbitrary pulse trains with random variations is impossible to derive. however, since the heating and cooling response of a semiconductor is essentially the same, the superposition principle may be used to solve problems which otherwise defy solution. time time time time p 1 p 2 p 3 p 4 p in p in t j p 1 p 2 p 3 p 4 p 1 p3 p 4 p 2 t 0 t 1 t 2 t 3 t 4 t 5 t 6 t 7 (a) input power (b) power pulses separated into components (c) t j change caused by components (d) composite t j figure 3. application of superposition principle p pk 1 peak power (watts) 50 40 30 20 10 0 t 0 t 1 t 2 t 3 t 4 t 5 0 1.0 2.0 3.0 4.0 t, time (ms) p 1 p 2 p 3 p o t t 0 t 1 t 2 t 3 t 5 t 4 t 6 t 7 t 9 t 8 t 5 t 2 t (conditions for numerical examples p o = 5 watts t = 5 ms t = 20 ms figure 4. nonrepetitive pulse train (values shown apply to example in appendix i) figure 5. a train of equal repetitive pulses using the principle of superposition each power interval is considered positive in value, and each cooling interval negative, lasting from time of application to infinity. by multiplying the thermal resistance at a particular time by the magnitude of the power pulse applied, the magnitude of the junction temperature change at a particular time can be obtained. the net junction temperature is the algebraic sum of the terms. the application of the superposition principle is most easily seen by studying figure 3. figure 3a illustrates the applied power pulses. figure 3b shows these pulses transformed into pulses lasting from time of application and extending to infinity; at t o , p 1 starts and extends to infinity; at t 1 , a pulse (p 1 ) is considered to be present and thereby cancels p 1 from time t 1 , and so forth with the other pulses. the junction temperature changes due to these imagined positive and negative pulses are shown in figure 3c. the actual junction temperature is the algebraic sum as shown in figure 3d. problems may be solved by applying the superposition principle exactly as described; the technique is referred to as method 1, the pulsebypulse method. it yields satisfactory results when the total time of interest is much less than the time required to achieve steadystate conditions, and must be used when an uncertainty exists in a random pulse train as to which pulse will cause the highest temperature. examples using this method are given in appendix a under method 1.
an569/d http://onsemi.com 4 figure 6. model for a repetitive equal pulse train t t p o p avg t n th pulse n + 1 pulse for uniform trains of repetitive pulses, better answers result and less work is required by averaging the power pulses to achieve an average power pulse; the temperature is calculated at the end of one or two pulses following the average power pulse. the essence of this method is shown in figure 6. the duty cycle family of curves shown in figure 2 and used to solve the example problem is based on this method; however, the curves may only be used for a uniform train after steadystate conditions are achieved. method 2 in appendix a shows equations for calculating the temperature at the end of the n th or n + 1 pulse in a uniform train. where a duty cycle family of curves is available, or course, there is no need to use this method. temperature rise at the end of a pulse in a uniform train before steadystate conditions are achieved is handled by method 3 (a or b) in the appendix. the method is basically the same as for method 2, except the average power is modified by the transient thermal resistance factor at the time when the average power pulse ends. a random pulse train is handled by averaging the pulses applied prior to situations suspected of causing high peak temperatures and then calculating junction temperature at the end of the n th or n + 1 pulse. part c of method 3 shows an example of solving for temperature at the end of the third pulse in a three pulse burst. handling nonrectangular pulses the thermal response curves, figure 1, are based on a step change of power; the response will not be the same for other waveforms. thus far in this treatment we have assumed a rectangular shaped pulse. it would be desirable to be able to obtain the response for any arbitrary waveform, but the mathematical solution is extremely unwieldy. the simplest approach is to make a suitable equivalent rectangular model of the actual power pulse and use the given thermal response curves; the primary rule to observe is that the energy of the actual power pulse and the model are equal. experience with various modeling techniques has led to the following guidelines: ? for a pulse that is nearly rectangular, a pulse model having an amplitude equal to the peak of the actual pulse, with the width adjusted so the energies are equal, is a conservative model. (see figure 7a). ? sine wave and triangular power pulses model well with the amplitude set at 70% of the peak and the width adjusted to 91% and 71%, respectively, of the baseline width (as shown in figure 7b). figure 7. modeling of power pulses ??????????? ??????????? ??????????? ??????????? ??????????? ?????? ?????? ?????? ?????? ?????? t 0 t 2 t 1 p 1 p 2 a a (c) (b) (a) t 1 p 1 p 1 t 1 = a t t 0.7 p p 0.7 p p p p p p 0.91 t 0.71 t p 1 (t 1 t 0 ) + p 2 (t 2 t 1 ) = a ? a power pulse having a sin 2 shape models as a triangular waveform. power pulses having more complex waveforms could be modeled by using two or more pulses as shown in figure 7c. a point to remember is that a high amplitude pulse of a given amount of energy will produce a higher rise in junction temperature than will a lower amplitude pulse of longer duration having the same energy. as an example, the case of a transistor used in a dc to ac power converter will be analyzed. the idealized waveforms of collector current, i c , collector to emitter voltage, v ce , and power dissipation p d , are shown in figure 8. a model of the power dissipation is shown in figure 8d. this switching transient of the model is made, as was suggested, for a triangular pulse. for example, t j at the end of the rise, on, and fall times, t 1 , t 2 and t 3 respectively, will be found. conditions: to3 package, r q jc = 0.5 c/w, i c = 60 a, v ce(off) = 60 v t a = 50 c t f = 80 m s, t r = 20 m s v ce(sat) = 0.3 v @ 60 a frequency = 2 khz \ t = 500 m s p on = (60)(0.3) = 18 w p f = 30 30 = 900 w = p r
an569/d http://onsemi.com 5 figure 8. idealized waveforms of i c , v ce and p d in a dc to ac inverter time time t (time) t (time) (b) (a) (d) (c) power dissipation collector current collectoremitter voltage v ce i c p d p d t t off t on t f t r p f pr p on t on 0.7 t r 0.7 t f t 0 t 1 t 2 t 3 p on 0.7 p f 0.7 p r 0.7 p f t 1 t 2 t 3 assume that the response curve in figure c3 for a die area of 58,000 square mils applies. also, that the device is mounted on an ms15 heat sink using dow corning dc340 silicone compound with an air flow of 1.0 lb/min flowing across the heatsink. (from ms15 data sheet, r q cs = 0.1 c/w and r q sa = 0.55 c/w.) procedure: average each pulse over the period using equation 13 (appendix a, method 2), i.e., p avg 0.7p r 0.71 t r p on t on 0.7p f 0.71 t f (0.7)(900)(0.71) (20) 500 (18) (150) 500 (0.7)(900)(0.71) 80 500 17.9 5.4 71.5 94.8 w from equation 14, method 2a: t 1 [p avg (0.7p r p avg ) r(t 1 t o )]r jc at this point it is observed that the thermal response curves of figure c3 do not extend below 100 m s. heat transfer theory for one dimensional heat flow indicates that the response curve should follow the t law at small times. using this as a basis for extending the curve, the response at 14.2 m s is found to be 0.023. we then have: t 1 [94.8 (630 94.8)0.023](0.5) (107.11)(0.5) 53.55 c for t 2 we have, by using superposition: t 2 [p avg p avg r(t 2 t o ) 0.7p r r(t 2 t o ) 0.7p r r(t 2 t 1 ) p on r(t 2 t 1 )]r jc [p avg (0.7p r p avg ) r(t 2 t o) (p on 0.7p r ) r(t 2 t 1 )]r jc [94.8 (630 94.8) r(164 s) (18 630) r(150 s)](0.5) [94.8 (535.2)(0.079) (612)(0.075)](0.5) [94.8 42.3 45.9](0.5) (91.2)(0.5) 45.6 c for the final point t 3 we have: t 3 [p avg p avg r(t 3 t o ) 0.7p r r(t 3 t o ) 0.7p r r(t 3 t 1 ) p on r(t 3 t 1 ) [p avg (0.7p r p avg ) r(t 3 t o) (p on 0.7p r ) r(t 3 t 1 ) (0.7p r p on) [94.8 (535.2) r(221 s) ( 612) r(206.8 s) (612) r(56.8 s)](0.5) [94.8 (535.2)(0.09) (612)(0.086) [94.8 481.7 52.63 27.54](0.5) (117.88)(0.5) 58.94 c p on r(t 3 t 2 ) 0.7p f r(t 3 t 2 )]r jc r(t 3 t 2 )]r jc (612)(0.045)](0.5) the junction temperature at the end of the rise, on, and fall times, t j1 , t j2 , and t j3 , is as follows: t j1 t 1 t a r ca p avg r ca r cs r sa 0.1 0.55 t j1 53.55 50 (0.65)(94.8) 165.17 c
an569/d http://onsemi.com 6 t j2 t 2 t a r ca p avg 45.6 50 (0.65)(94.8) 157.22 c t j3 t 3 t a r ca p avg 58.94 50 (0.65)(94.8) 170.56 c t j(avg) p avg (r jc r cs r sa ) t a (94.8)(0.5 0.1 0.55) 50 (94.8)(1.15) 50 159.02 c inspection of the results of the calculations t 1 , t 2 , and t 3 reveal that the term of significance in the equations is the average power. even with the poor switching times there was a peak junction temperature of 11.5 c above the average value. this is a 7% increase which for most applications could be ignored, especially when switching times are considerably less. thus the product of average power and steadystate thermal resistance is the determining factor for junction temperature rise in this application. summary this report has explained the concept of transient thermal resistance and its use. methods using various degrees of approximations have been presented to determine the junction temperature rise of a device. since the thermal response data shown is a step function response, modeling of different wave shapes to an equivalent rectangular pulse of pulses has been discussed. the concept of a duty cycle family of curves has also been covered; a concept that can be used to simplify calculation of the junction temperature rise under a repetitive pulse train. safe area and surge ratings must also be observed. it is possible to have t j well below t j(max) as calculated from the thermal response curves, yet have a hotspot in the semiconductor which will cause a failure. appendix a methods of solution in the examples, a type 2n3647 transistor will be used; its steadystate thermal resistance, r q jc , is 35 c/w and its value for r(t) is shown in figure a1. definitions: p 1 , p 2 , p 3 p n = power pulses (watts) t 1 , t 2 , t 3 t n = junction to case temperature at end of p 1 , p 2 , p 3 p n t 0 , t 1 , t 2 , t n = times at which a power pulse begins or ends r(t n t k ) = transient thermal resistance factor at end of time interval (t n t k ) table 1. several possible methods of solutions 1. junction temperature rise using pulsebypulse method a. temperature rise at the end of the n th pulse for pulses with unequal amplitude, spacing, and duration. b. temperature rise at the end of the n th pulse for pulses with equal amplitude, spacing, and duration. 2. temperature rise using average power concept under steadystate conditions for pulses of equal amplitude, spacing, and duration a. at the end of the n th pulse. b. at the end of the (n + 1) pulse. 3. temperature rise using average power concept under transient conditions. a. at the end of the n th pulse for pulses of equal ampli- tude, spacing and duration. b. at the end of the n + 1 pulse for pulses of equal amplitude, spacing and duration. c. at the end of the n th pulse for pulses of unequal amplitude, spacing and duration. d. at the end of the n + 1 pulse for pulses of unequal amplitude, spacing and duration. method 1a finding t j at the end of the nth pulse in a train of unequal amplitude, spacing, and duration general equation: t n n i 1 p i [r(t 2n 1 t 2i 2 ) (11) r(t 2n 1 t 2i 1 )]r jc where n is the number of pulses and p i is the peak value of the i th pulse. to find temperature at the end of the first three pulses, equation 11 becomes: t 1 p 1 r(t 1 )r jc (11a) t 2 [p 1 r(t 3 ) p 1 r(t 3 t 1 ) (11b) p 2 r(t 3 t 2 )]r jc t 3 [p 1 r(t 5 ) p 1 r(t 5 t 1 ) p 2 r(t 5 t 2 ) (11c) p 2 r(t 5 t 3 ) p 3 r(t 5 t 4 )]r jc example: conditions are shown in figure 4 as: p 1 = 40 w t 0 = 0 t 3 = 1.3 ms p 2 = 20 w t 1 = 0.1 ms t 4 = 3.3 ms p 3 = 30 w t 2 = 0.3 ms t 5 = 3.5 ms therefore, t 1 t 0 = 0.1 ms t 3 t 1 = 1.2 ms t 2 t 1 = 0.2 ms t 5 t 1 = 3.4 ms t 3 t 2 = 1 ms t 5 t 2 = 3.2 ms t 4 t 3 = 2 ms t 5 t 3 = 2.2 ms t 5 t 4 = 0.2 ms
an569/d http://onsemi.com 7 0.01 0.02 0.05 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) figure 1. 2n3467 transient thermal response procedure: find r(t n t k ) for preceding time intervals from figure 2, then substitute into equations 11a, b, and c. t 1 p 1 r(t 1 )r jc 40 0.05 35 70 c t 2 [p 1 r(t 3 ) p 1 r(t 3 t 1 ) p 2 r(t 3 t 2 )]r jc [40(0.175) 40(0.170) 20(0.155)]35 t 3 [p 1 r(t 5 ) p 1 r(t 5 t 1 ) p 2 r(t 5 t 2 ) p 2 r(t 5 t 3 ) p 3 r(t 5 t 4 )]r jc [40(0.175 0.170) 20(0.155)]35 [0.2 3.1]35 115.5 c [40(0.28) 40(0.277) 20(0.275) 20(0.227) 30(0.07)]35 [40(0.28 0.277) 20(0.275 0.227) 30(0.07)]35 [0.12 0.96 2.1] 35 3.18 35 111.3 c 2relative amounts of temperature residual from p 1 , p 2 , and p 3 respectively are indicated by the terms in brackets. note, by inspecting the last bracketed term in the equations above that very little residual temperature is left from the first pulse at the end of the second and third pulse. also note that the second pulse gave the highest value of junction temperature, a fact not so obvious from inspection of the figure. however, considerable residual temperature from the second pulse was present at the end of the third pulse. method 1b finding t j at the end of the nth pulse in a train of equal amplitude, spacing, and duration the general equation for a train of equal repetitive pulses can be derived from equation 11. p i = p d , t i = t, and the spacing between leading edges or trailing edges of adjacent pulses is t . general equation: t n p d r jc n i 1 r[(n i) t] r[(n i) ] (12) expanding: t n p d r jc r[(n 1) t] r[(n 1) ] (12a) r[(n 2) t] r[(n 2) ] r[(n 3) t] r[(n 3) ] r[(n i) t] r[(n i) ] r[(t)] for 5 pulses, equation 12a is written: t 5 p d r jc [r(4 t) r(4 ) r(3 t) r(3 ) r(2 t) r(2 ) r( t) r( ) r(t)] example: conditions are shown in figure 5 substituting values into the preceding expression: t 5 (5)(35)[r(4.20 5) r(4.20) r(3.20 5) r(3.20) r(2.20 5) r(2.20) r(20 5) r(20) r(5)] t 5 (5)(35)[0.6 0.76 0.73 0.72 0.68 0.66 0.59 0.55 0.33] (5)(35)(0.40) t 5 70.0 c note that the solution involves the difference between terms nearly identical in value. greater accuracy will be obtained with long or repetitive pulse trains using the technique of an average power pulse as used in methods 2 and 3. method 2 average power method, steadystate condition the essence of this method is shown in figure 6. pulses previous to the n th pulse are averaged. temperature due to
an569/d http://onsemi.com 8 the n th or n + 1 pulse is then calculated and combined properly with the average temperature. assuming the pulse train has been applied for a period of time (long enough for steadystate conditions to be established), we can average the power applied as: p avg p d t (13) method 2a finding temperature at the end of the n th pulse applicable equation: t n [p avg (p d p avg )r(t)]r jc (14) or, by substituting equation 13 into 14, t n t 1 t r(t)
p d r jc (15) the result of this equation will be conservative as it adds a temperature increase due to the pulse (p d p avg ) to the average temperature. the cooling between pulses has not been accurately accounted for; i.e., t j must actually be less than t j(avg) when the n th pulse is applied. example: find t n for conditions of figure 5. procedure: find p avg from equation (13) and substitute values in equation (14) or (15). t n [(1.25) (5.0 1.25)(0.33)](35) 43.7 43.2 86.9 c method 2b finding temperature at the end of the n + 1 pulse applicable equation: t n 1 [p avg (p d p avg )r(t ) (16) p d r(t) p d r( )]r jc or, by substituting equation 13 into 16, t n 1 t 1 t r(t ) r(t) r( )
p d r jc (17) example: find t n for conditions of figure 5. procedure: find p avg from equation (13) and substitute into equation (16) or (17). t n 1 [(1.25) (5 1.25)(0.59) (5)(0.33) (5)(0.56)](35) 80.9 c equation (16) gives a lower and more accurate value for temperature than equation (14). however, it too gives a higher value than the true t j at the end of the n + 1 pulse. the error occurs because the implied value for t j at the end of the n th pulse, as was pointed out, is somewhat high. adding additional pulses will improve the accuracy of the calculation up to the point where terms of nearly equal value are being subtracted, as shown in the examples using the pulsebypulse method. in practice, however, use of this method has been found to yield reasonable design values and is the method used to determine the duty cycle of family of curves e.g., figure 2. note that the calculated temperature of 80.9 c is 10.9 c higher than the result of example 1b, where the temperature was found at the end of the 5th pulse. since the thermal response curve indicated thermal equilibrium in 1 second, 50 pulses occurring 20 milliseconds apart will be required to achieve stable average and peak temperatures; therefore, steady stateconditions were not achieved at the end of the 5th pulse. method 3 average power method, transient conditions the idea of using average power can also be used in the transient condition for a train of repetitive pulses. the previously developed equations are used but p avg must be modified by the thermal response factor at time t (2n1) . method 3a finding temperature at the end of the nth pulse for pulses of equal amplitude, spacing and duration application equation: t n t rt (2n 1) 1 t r(t)
p d r jc (18) conditions: (see figure 5) procedure: at the end of the 5th pulse (see figure 7). t 5 [5 20 r(85) (1 5 20)r(5)](5)(35) [(0.25)(0.765) (0.75)(0.33)](175) 77 c this value is a little higher than the one calculated by summing the results of all pulses; indeed it should be, because not cooling time was allowed between p avg and the n th pulse. the method whereby temperature was calculated at the n + 1 pulse could be used for greater accuracy. method 3b finding temperature at the end of the n + 1 pulse for pulses of equal amplitude, spacing and duration application equation: t n 1 t r(t 2n 1) 1 t r(t ) r(t)
(19) t r( )
p d r jc example: conditions as shown in figure 5. find temperature at the end of the 5th pulse. for n + 1 = 5, n = 4, t 2n1 = t 7 = 65 ms, t 5 5 20 r(65 ms) 1 5 20 r(25 ms) r(5ms)
t r(20 ms)
(5)(35)
an569/d http://onsemi.com 9 t 5 [(0.25)(0.73) (0.75)(0.59) 0.33 0.55](5)(35) t 5 70.8 c the answer agrees quite well with the answer of method 1b where the pulsebypulse method was used for a repetitive train. method 3c finding t j at the end of the nth pulse in a random train the technique of using average power does not limit itself to a train of repetitive pulses. it can be used also where the pulses are of unequal magnitude and duration. since the method yields a conservative value of junction temperature rise it is a relatively simple way to achieve a first approximation. for random pulses, equations 14 through 17 can be modified. it is necessary to multiply p avg by the thermal response factor at time t (2n 1) . p avg is determined by averaging the power pulses from time of application to the time when the last pulse starts. application equations: general: p avg n i 1 p i t (2i 1) t (2i 2) t (2n) t (2i 2) (110) for 3 pulses: p avg p 1 t 1 t 0 t 4 t 0 p 2 t 3 t 2 t 4 t 2 (111) example: conditions are shown in figure 4 (refer to method 1a). procedure: find p avg from equation 13 and the junction temperature rise from equation 14. conditions: figure 4 p avg 40 0.1 3.3 20 1 3 1.21 6.67 7.88 watts t 3 [p avg r(t 5 ) (p 3 p avg )r(t 5 t 4 )]r jc [7.88(0.28) (30 7.88) 0.07]35 [2.21 1.56]35 132 c this result is high because in the actual case considerable cooling time occurred between p 2 and p 3 which allowed t j to become very close to t c . better accuracy is obtained when several pulses are present by using equation 110 in order to calculate t j t c at the end of the n th + 1 pulse. this technique provides a conservative quick answer if it is easy to determine which pulse in the train will cause maximum junction temperature. method 3d finding temperature at the end of the n + 1 pulse in a random train the method is similar to 3c and the procedure is identical. p avg is calculated from equation 110 modified by r(t 2n 1 ) and substituted into equation 16, i.e., t n 1 [p avg r(t 2n 1 ) (p d p avg )r(t 2n 1 t 2n 2 ) p d r(t 2n 1 t 2n ) p d r(t 2n 1 t 2n 1 )]r jc the previous example can not be worked out for the n + 1 pulse because only 3 pulses are present. table 2. summary of numerical solution for the repetitive pulse train of figure 5 temperature obtained, c temperature desired pulseby pulse average power nth pulse average power n + 1 pulse at end of 5th pulse 70.0 (1b) 77 (3a) 70.8 (3b) steady state peak 86.9 (2a) 80.9 (2b) note: number in parenthesis is method used. appendix b thermal response measurements to measure the thermal response of a semiconductor, a temperature sensitive parameter of the device is used as an indicator of device temperature. other methods are impractical on a completely assembled device. if the parameter varies linearly with temperature, finding thermal response is greatly simplified since the measured parameter value will be directly proportional to temperature. the forward voltage drop of a diode or scr or the collectorbase junction of a transistor, when operated a low currents, has a linear voltage change with temperature and is a good choice for the temperature sensitive parameter. since the collectorbase junction of a transistor is reversed biased when power is dissipated, a measurement using the forward v cb must be made during the cooling interval. a suitable measurement procedure is to dissipate power long enough to achieve thermal equilibrium while periodically switching the circuit to shutoff the power, forwardbias the collectorbase junction, and monitor its change in voltage as the device cools off. the switching must be performed rapidly or the initial part of the response will not be seen. a circuit that can be used to measure thermal resistance or thermal response of a power transistor is shown in figure b1. with the transistor under test (t.u.t.) in the circuit, the powersense switch (sw 2 ) is set in the sense position which removes the emitter current supply and the base return from the transistor. after the collector supply is set at a suitable value, the sense supply is adjusted to allow a lowlevel sense current through the basecollector junction. note that the sense current flows in the opposite direction from that of the collector current; thus, a reversing switch on the collector current meter is needed to set the sense current level.
an569/d http://onsemi.com 10 10 k 0.5 w input from timer pulse generator input response resistance 300 4.0 w q1 mps6534 50 4.0 w 1:1 0 v +5.0 v 300 1.0 w 30 1.0 w q2 2n5478 1n4935 100 m f 3.5 7.0 w q3 2n5302 1n4935 30 50 w 20 2.0 w 0.1 m f emitter supply 20 v sense power sw2 sense power sw2 tut r e ** * q3 2n5478 1n4935 100 1.0 w 20 20 w 1.0 k 0.5 w to scope v sense supply v cc + 10 v * i v cc collector supply * 100 20 w 2see text * aux. supply 33 v * a bypass capacitor may be needed with long supply leads. ** r e 20 v/i c (amps) figure b1. thermal resistance/response test fixture for npn power transistors up to 10 amperes collector current switch sw 2 is then set at the power position and the emitter supply is connected. with sw 1 in the response position, a positive voltage pulse, slightly greater than the auxiliary supply, from a timer circuit is needed to cycle the t.u.t. the timer must provide a pulse of sufficient length to allow the t.u.t. to achieve thermal equilibrium. at this time, transistors q1 and q2 are off and q3 and q4 are on. when the output of the timer changes to a low state, q1 and q2 turn on and q3 and q4 turn off. thus, the only current flowing through the transistor is the collectorbase sense current which allows the collectorbase voltage to be monitored by an oscilloscope. the response curve is photographed at different sweep rates to facilitate plotting. the period of the timer output pulse must be variable in order to allow thermal equilibrium to be achieved for different types of transistors. also, the cooling time must be at least equal to the heating time to allow the junction temperature to cool to its original value. case mounted semiconductors should be mounted on a heat sink that will hold the case at a nearly constant temperature. in other words, the thermal resistance from the transistor case to the heat sink should be negligible compared to the junctiontocase thermal resistance. if this precaution is not taken, the results will be influenced by the response (the change in temperature) of the heat sink. with sw 1 in the thermal resistance position, steadystate thermal resistance can be measured. for this measurement, the t.u.t. is on most of the time; a narrow pulse is used to turn the t.u.t. off to allow sensing of the junction temperature. the forward voltage drop vs. temperature must be known in order to calculate thermal resistance. figure b2 is a thermal resistance/response test fixture for scrs and rectifiers. operation of this text fixture is similar to the npn test fixture except that switch sw 3 is added to provide gate drive to turn the scr on. when testing scrs, the current sense level must be above the holding current.
an569/d http://onsemi.com 11 27 4.0 w input from pulse generator 51 4.0 w mps6534 10 k input from timer 1n4935 33 v 300 response sw1 resistance 300 2.0 w 30 1n4935 2n5477 bias supply 16 v 75 10 w 3.0 5.0 w 100 m f 1n4935 drive supply 10 v 10 10 m f 1n4935 5.0 50 w (4) 2n5683 0.002 (4) 1n2839 1n4001 r f * to scope i sense power sw2 sense i f 1.0 k sense supply 10 v 1n4001 gate supply 16 v sw3 200 dut *r f 10 v/i f figure b2. thermal resistance/response test fixture for scr's and rectifiers up to 50 amperes forward current appendix c transient thermal response curves the data presented in this appendix is for the purpose of enabling a reasonable estimate of transient thermal resistance to be made in situations where data is not readily available. as can be seen by studying figure 1, the normalized response values cover a range of about 10:1 in the time region between 100 m s and 10 ms for the whole gamut of power transistor packages and die areas offered by on semiconductor. in this section, data is supplied for a number of packages, each having several different die sizes, so that a closer estimate of response can be made. each figure indicates some representative part numbers for the given curves; obviously, space limitations prevent listing all parts for which the curves apply. as an additional aid, figure c11 may be used to estimate die area from thermal resistance information. figure c11 is not applicable to darlington transistors and other dual devices or integrated circuits, as the ratio of active transistors and other dual devices or integrated circuits, as the ratio of active transistor area to total die area is different from that of the transistors used to prepare the figure. measurement of thermal response can hardly be regarded as an exact science. as a result of switching, transients occur which make it very difficult to ascertain response below 100 m s. one dimensional heat flow theory predicts that temperature will be proportional to the square root of time; even though this relationship is not always observed, it is generally used to extrapolate data below 100 m s. 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 1 2 2 1 10,000 2n5974 2n5977 die size (sq. mil) 27,200 2n5980, 2n5986 2n5983, 2n5989 device type figure c1. case 90 and 199 (thermopad) thermal response
an569/d http://onsemi.com 12 figure c2. case 90 and 199 (thermopad) thermal response 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 4 6 5 1,200 2,000 3,600 6,300 8,100 16,900 mje170, mje180, mje210, mje200 2n4918, 2n4921 2n5190, 2n5193 mcr106 2n6342 die size (sq. mil) device type 1 2 3 4 5 6 thermowatt case 221 all other curves are for case 77 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 2 1 3 3,600 16,800 58,000 2n3021, 2n3740, 2n4898 2n3789, 2n5875, 2n3713, 2n5879, 2n5881 2n6274, 2n5683, 2n5685, 2n6309, 2n5983 2n4398, 2n5885, 2n5301, 2n6029 die size (sq. mil) device type figure c3. to3 and to66 (diamond) thermal response. data applies to packages which are all copper or contain a copper slug under the die 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 2 1 6,200 22,500 2n5344 2n6233 die size (sq. mil) device type figure c4. to66 thermal response, all steel package
an569/d http://onsemi.com 13 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 2 1 10,000 58,000 2n5304 2n3487, 2n1724 die size (sq. mil) device type figure c5. to61 (11/16 hex, stud mount) thermal response figure c6. to5 (solid kovar header) thermal response figure c7. to5 (solid steel header) thermal response 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 2 1 3 1,200 3,600 8,100 2n3498, 2n3053, 2n3252, 2n3444 2n3506 mcr320 die size (sq. mil) device type 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 4 2 1 3 4 850 1,225 3,600 14,400 mm4005, 2n4406 mm3006 2n3719, 2n5334 no standard devices die size (sq. mil) device type
an569/d http://onsemi.com 14 figure c8. shell header to5, to18 and to46 thermal response, applies to all commonly used die 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) to46 to18 to5 package figure c9. case 152 (uniwatt) thermal response figure c10. to92 (unibloc) thermal response, applies to all commonly used die 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized) 2 1 3 2 1 3 625 1,225 2,025 mpsu05, mpsu07 mpsu03 mpsu10 die size (sq. mil) device type 0.1 0.2 0.5 1.0 2.0 5.0 10 20 50 100 200 500 1000 2000 5000 10,000 1.0 0.7 0.5 0.3 0.2 0.1 0.07 0.05 0.03 0.02 0.01 t, time (ms) r(t), transient thermal resistance (normalized)
an569/d http://onsemi.com 15 figure c11. typical thermal resistance as a function of case material and die area. data applies to solid header parts only. use copper curve for aluminum and steel packages with a copper slug under the die, which is the standard on semiconductor design. 2.0 1.0 0.5 0.2 0.1 r q jc , thermal resistance junction to case ( c/w) 0.1 0.2 0.5 1.0 2.0 5.0 10 20 die area 1000 (mil 2 ) 50 100 20 10 5.0 100 50 2 1 3 kovar header steel header copper header 2 1 3 references 1. gutzwiller, f. w., and sylvan, t. p. asemiconductor ratings under transient loads,o aiee transactions, volume 79, part 1, pp. 699 to 706. (communications and electronics) january, 1961. 2. lockett, r. a., bell, h. a., priston, r. athermal resistance of low power semiconductor devices under pulse conditions,o mullard technical communications, no. 76, july, 1965.
an569/d http://onsemi.com 16 on semiconductor and are trademarks of semiconductor components industries, llc (scillc). scillc reserves the right to make changes without further notice to any products herein. scillc makes no warranty, representation or guarantee regarding the suitability of its products for any particular purpose, nor does scillc assume any liability arising out of the application or use of any product or circuit, and specifically disclaims any and all liability, including without limitation special, consequential or incidental damages. atypicalo parameters which may be provided in scill c data sheets and/or specifications can and do vary in different applications and actual performance may vary over time. all operating parameters, including atypicalso must be validated for each customer application by customer's technical experts. scillc does not convey any license under its patent rights nor the rights of others. scillc products are not designed, intended, or authorized for use as components in systems intended for surgical implant into t he body, or other applications intended to support or sustain life, or for any other application in which the failure of the scillc product could create a sit uation where personal injury or death may occur. should buyer purchase or use scillc products for any such unintended or unauthorized application, buyer shall indem nify and hold scillc and its of ficers, employees, subsidiaries, affiliates, and distributors harmless against all claims, costs, damages, and expenses, and re asonable attorney fees arising out of, directly or indirectly, any claim of personal injury or death associated with such unintended or unauthorized u se, even if such claim alleges that scillc was negligent regarding the design or manufacture of the part. scillc is an equal opportunity/affirmative action employ er. publication ordering information japan : on semiconductor, japan customer focus center 4321 nishigotanda, shinagawaku, tokyo, japan 1410031 phone : 81357402700 email : r14525@onsemi.com on semiconductor website : http://onsemi.com for additional information, please contact your local sales representative. an569/d literature fulfillment : literature distribution center for on semiconductor p.o. box 5163, denver, colorado 80217 usa phone : 3036752175 or 8003443860 toll free usa/canada fax : 3036752176 or 8003443867 toll free usa/canada email : onlit@hibbertco.com n. american technical support : 8002829855 toll free usa/canada

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