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1 Thermistor cracking van der Veen, M. Published: 01/01/1989 Document Version Publisher s PDF, also known as Version of Record (includes final page, issue and volume numbers) Please check the document version of this publication: A submitted manuscript is the author's version of the article upon submission and before peer-review. There can be important differences between the submitted version and the official published version of record. People interested in the research are advised to contact the author for the final version of the publication, or visit the DOI to the publisher's website. The final author version and the galley proof are versions of the publication after peer review. The final published version features the final layout of the paper including the volume, issue and page numbers. Link to publication Citation for published version (APA): Veen, van der, M. (1989). Thermistor cracking. (Opleiding wiskunde voor de industrie Eindhoven : student report; Vol. 8902). Eindhoven: Technische Universiteit Eindhoven. General rights Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of accessing publications that users recognise and abide by the legal requirements associated with these rights. Users may download and print one copy of any publication from the public portal for the purpose of private study or research. You may not further distribute the material or use it for any profit-making activity or commercial gain You may freely distribute the URL identifying the publication in the public portal? Take down policy If you believe that this document breaches copyright please contact us providing details, and we will remove access to the work immediately and investigate your claim. Download date: 23. Oct. 2018

2 Opleiding Wiskunde voor de Industrie Eindhoven REPORT THERMISTOR CRACKING Marco van der Veen ECMI March 1989 ;~ ~ 2.~113 I8IJ:UI;1indhoven " -,,,

3 ~ THERMISTOR CRACKING i Marco van der Veen March 1989

4 1 O. SUMMARY A small, eletrical current limiting device called a thermistor is considered in this paper. These thermistors sometimes crack in practical applications. A suggestion has come from industry that cracking is caused by large temperature gradients inside the thermistor, resulting in large thermal stresses. In this paper, a simple, one-dimensional mathematical model is developed, yielding a one-dimensional heat-equa.tion for the temperature inside the thermistor. Numerical calculations are done on the temperature distribution inside the thermistor. No large gradients occur, however. H large temperature gradients ca.use the thermistor to crack, then this one-dimensional model cannot account for it.

5 2 1. INTRODUCTION A thermistor is a small device used in electrical circuits to protect vital components against large currents. The thermistor is a small cylinder, made of ceramic material, to which on both sides an electrical contact has been attached. The ceramic material has an electrical conductivity that varies with temperature. For small temperatures it has a high conductivity and for high temperatures it has a low conductivity. In an intermediate temperature range the conductivity sharply decreases from its maximum value to its minimum value. Suppose we build a circuit consisting of an external voltage, a resistance, the thermistor and a switch, yielding the circuit of Fig. 1. Once the switch is closed, a current starts to flow. This results in an increasing temperature in the thermistor, which causes its conductivity to decrease and its resistance to increase. As a result of the increased resistance, the current falls. Eventually, an equilibrium is reached, where all the heat generated in the thermistor is lost to the outside. Thus we have a small, but steady current in the circuit. A problem that occurs in practical applications, is that of the thermistor cracking, while it heats up. A cause of this problem might be large variations of the temperature inside the thermistor, yielding large temperature gradients. The large temperature gradients then could give rise to large thermal stresses, causing the thermistor to crack. The aim of this paper is the calculation of the heat distribution and the resulting temperature over the thermistor. This way we can check whether or not the assumption of the occurrence of large temperature gradients inside the thermistor is right. As will be seen, the occurring temperature gradients are quite small, though. From this we will conclude that if large temperature gradients cause thermistors to crack, a one-dimensional model as posed here, does.not account for it... ~ I Fig. 1: An exemplary circuit in which a thermistor is applied. Once the switch is closed, a current flows through the thermistor. As a result, the thermistor heats up and the conductivity decreases. Therefore, the resistance increases, whereby the current decreases. In the end an equilibrium is reached, where all the heat generated in the thermistor is lost to the outside.

6 3 Fig. 2: A thermistor.....;.. --, j _.--..,j I- I -1, I -, -., - I , '-, i,- - l, '7.,;; -'-II T.. Fig. 3: A typical example of a. curve of (1 against T. Typically, (1" = 2(nm)-1, TA = 100 C, Ts = 200 C.

7 4 2. MATHEMATICAL MODELLING (i) DERIVATION OF DIFFERENTIAL EQUATION The thermistor, introduced to us above, will be modelled one-dimensionally in space (Fig. 4). A justification for using such a simple one-dimensional model seems to lie in the following fact. Most of the generated heat is transported by the wires at the top and the bottom. The sides of the thermistor are nearly insulated. This fact leads to the expectation that the heat distribution will be homogeneous across each cross-section of the thermistor. We will put an z-axis along the side of the thermistor, as shown in Fig. 4. The thermistor is supposed to have thickness 2L, where the top lies at % = - L, the centre at % = 0 and the bottom side at % = + L...._ ;--~.. I,- -'" ~ '1"01' C :1- _L _.. _w 0 +L lrllttom X. Fig. 4. We denote the time by t and the temperature at % = %, t = t by T(%, t). Sometimes we will omit the % and the t. The heat, represented by T(%, t), will be generated by the current flow and will be conducted away to the edges % = ±L. There it goes into the wires and into the air. We will model the heat How by the well-established heat equation. Since we have assumed there is actually only one space dimension, this equation yields (2.1) where p c k Q(%,t) - the density of ceramic material, = the specific heat of the ceramic, = the heat conductivity of the ceramic and - the volumetric heat source from the current.

8 5 To this equation we add the following boundary conditions, the so-called radiation conditions: (2.2.a) (2.2.b) 8T -k - ax = -h(t - T. ) atx=-l, el for all t 8T -k ax = h(t - Ta) at x = L, for all t. Because of symmetry we can replace the first of the two equations by (2.2.c) at _ 0 0 ~ all ax - at x =,lor t, and solve the equation (2.1) for 0 S x S L. By putting T(x,t) = T(-x,t) for -L S x S 0, we again obtain the solution for the whole region. In these equations we have h - the heat transfer coefficient and T" - the exterior temperature. For solving the equation (2.1) we still lack the initial condition. For this, we take (2.3) T(x,t = 0) = Tel, -L:5 x :5 L, which simply states that for t = 0 the temperature inside the thermistor equals the exterior temperature. We will now derive an explicit formula for the volumetric heat source, Q(x,t). Therefore, we will look at the current flow inside the thermistor. The current flows from x = +L or x = -L according to the electrical potential +(x, t), so that the current I, crossing an area. A parallel to the top and the bottom, equals 8+ I = - Au(T) ax. This is simply stated by Ohm's law. The electrical conductivity, which depends on T(x,t), is denoted by u(t). I ~~---...". - I Now Q, the rate of heat generation per unit volume, is, simply said in words 'current density' times 'voltage gradient', or in formula,

9 6 Q =! = -O'(T) ( 0+)2. A ax ax Yo We have let) = Ro + Rth' where Vo Ro Rth the external voltage applied, = the resistance in the circuit serially connected to the thermistor and = the total resistence of the thermistor. Since the temperature distribution across the thermistor is not uniform and thus neither is 0"8, we cannot use 2L R t = Au ' but instead we should use jl dx' R t = -L AO'(T(xl,t»). This results in (2.4) 1 Q(x,t)= ( ) u T(x, t) A2 vel Altogether, we have the following equations: (2.5) with the boundary conditions (2.6.a) (2.6.b) at ax (x = 0, t) = 0, at -Ie ax = h(t - Tel) at x = L, for all t, and one initial condition, namely (2.7) T(x,t = 0) = Tel

10 7 (li) NON-DIMENSIONALIZATION We will scale the equations (2.5)-(2.7) to decrease the number of parameters in the problem. The scaling will be such that we balance the left-hand side of (2.5) with the first term on the right-hand side. This is done in this way because of the fact that we scale with the heat conduction time and we expect to have to consider conduction in order to find temperature gradients. We now introduce the dimensionless quantities T*, x*, (1* and t* such that T* is the scaled temperature, T - Til = AT T* with AT a typical value from the (1- T-curve (Fig. 3): AT=TB-TA, x* is the scaled length-coordinate, x = Lx*, (1* is the scaled conductivity, (1 = (1u(1*, where (Tv. is the conductivity at T = 0, and t* is the scaled time, t = L: t*. The equations (2.5)-(2.7) now transform into (2.8) (2.9.30) (2.9.b) 8T 8 2 T 'Y 1 8t = 8x 2 + 0: (I-' + 2 j hi ) 2 ' o (T ( T(x',t») 8T 8x (x=o,t)=o, 8T 8x (x = 1,t) = -fjt, (2.10) T(x,t = 0) = 0, where we immediately omitted the superscripts '*'. In these equations, three dimensionless parameters appear, (1v.vJ 'Y = kat' p, = Ro (1u A &nd L 'Y can be interpreted as the factor determining the size of the electrical heat generation, relative to the size of the heat conduction.

11 8 The parameter p. is the ratio of resistance Ro and the initial resistance, that is, resistance at T = 0, of the thermistor. The third parameter, {j, determines the heat transfer at the edges ofthe thermistor. We will now try to solve the equations (2.8)-(2.10). Since we want to solve them for values of the parameters where none is small compared to the others, we will solve them numerically. The non-linear electrical heat generation term prohibits any profound analytical work.

12 9 3. NUMERICAL SOLUTION We will solve for T in the equations (2.8)-(2.10) by discretization in both space and time. We will use the values of t at the current time step (and probably former time steps), to find the values of T at a next step. We will achieve this as follows. Instead of solving (2.8)-(2.10) for T(x, t), x E [0,1], t ~ 0, for all x E [0,1] and for all t ~ 0, we will solve them for T in so-called gridpoints, only. These gridpoints we establish as follows. We define a griddistance ~x and a number M such that the interval [0,1] is subdivided into M subintervals and we get M + 1 gridpoints Xi = i~x, i = 0,..., M, where M ~x = 1. We do this at t = 0, ~t, 2~t... k~t... where ~t is the socalled timestep. Next we define the quantities Tl' = T(i~x,k~t), i = O,...,M, k E IN. The points (i~x,k~t), i = O,...,M, Ie E 'Ii, are the gridpoints where we calculate the values of T _ _.-..._. _._.-..'.-...-~ ,_._.-.-._ o o ~ ~ ~--+---~----~x 4" "'Il JlI1t... _ Hu.ad. (II..,)4/( Fig. 6: The pidpoints. Since we agreed to calculate T in the gridpoints only, we can replace the equations (2.8)-(2.10) by their discretisized counterparts. This is done as shown below. We will write down a Taylor-expansion of T(x, t) around (i~x, k~t). That is, II 8Tl ~t2 8 2 Th Tl+1 = T(i~x,(k + l)~x) = Ti + ~t -nt + 2! at; + O(~t3) T!<=-l, - T!<=,

13 10 From these relations we can. derive several formulae: 1'.1+1 T" i!:t.; i + O(!:t.t), or, as is similarly shown {JT~+1 I _ 8t - T"+1 1'." i - i + O(!:t.t)!:t.t. All three quantities enable us to replace the left-hand side of (2.8) by a linear relation between the quantities 1'l:+1, 1'i" and Ti"-I. Similarly, we have (j2t~ T " - 2T" + T~ We see that we can. approximate 8x~ by It 1!:t.~2.-1, where we introduce an error O(!:t.x2). It is obvious that by replacing the derivatives by their discretisized versions and neglecting the remaining O(!:t.x2) and O(!:t.t) terms, we introduce errors into our calculations. Nevertheless, when decreasing!:t.t and!:t.x we hope that the errors will decrease as well. Before actu~y saying which descretization we are going to take, let us dwell a bit more on the possible ways to discretisize and their respective pros and cons. Clearly, we have to deal not only with the two discretisized partial derivatives, but with the second term on the right-hand side of (2.8) as well. This term involves a ~ factor 1, and in the denominator, a squared integral J (dx ). Both may give rise to a o (1 1'(x',t) nonlinear term. Suppose, for example, that we discretisize with respect to (k + i)!:t.t, making a second 8T~t i Tlct 1 _ Tic order approximation of in by I!:t.t I. Consequently, if we do not want to lose our O(!:t.t 2 ) accuracy, we have to approximate the heat generation term at t = (k+ i )!:t.t, which we will shortly denote by Q~t! = Q(k + i)!:t.t,i!:t.x), also by

14 11 When knowing the values of T for t = kat, this specific discretization will give rise to nonlinear equations in the Tl'+l, i = 0,..., M. Nonlinear equations are far less easy to solve than linear equation, and therefore we would like to choose a discretization such that the resulting equations are linear in the Tl'+l, i = 0,..., M, supposing the T;", i = 0,..., M are known. A way to achieve this is the following. We discretize with respect to (k + 1 )At in the following way: (3.1) T~+l - T~ T~+l - 2T~+l + T~+l 'V, =,t1 I 1-1 +_' At Ax 2 (7~ I 1 i = 0,..., M, k 2:: 0. This is a totally implicit discretization with an error O(At, Ax2), that is, first order in time but second order in space- coordinates. The integral in the second term on the right-hand side can be evaluated using some well-known integration-rule like the two point method or the Simpson-method. Equation (3.1) can be rewritten as follows: (3.2) T.. 1e +,At l --n- I (7~ (1 ) 2 ', f.l + 2 f dr' o (T(T(.,k~t») i=o,...,m, k2::0. Supposing we know Tl for i = 0,..., M, (3.2) gives a system of linear equation of the Tk+l' i, I = 0,..., M. The boundary conditions (2.9) can be discretized by introducing two virtual points (-Ax,(k + I)At) and (M + l)ax,(k + l)at). Doing so, (2.9.a) gives (3.3.a) T ktl _ Tk+l , (2.9.b) gives (3.3.b) (3.3.a) gives us T~tl = Tt+l, which we use in (3.2) for i = O. (3.3.b) yields T;'~l = Tkl..\ - 2/3AxT!tl, which we will substitute into (3.2) when i= M. Lastly, the initial condition (2.10) is now being restated as

15 12 (3.4) TP=O, i=o,...,m. Summarizing, we have 1 H1 ~t -To --(TI-To) 2 ~x2 1, kejn; (3.5) _ T~+ "1~t - S (1. k S 1, i = 1,..., M - 1, k e IN ;, kein. These are M + 1 linear equations for the M + 1 unknowns Tik+!, i = 0,..., M, supposing Tf, i = 0,.'" M are known. We therefore can solve these equations easily, step after step, starting from k = where we have the initial conditions (3.4). A minor adjustment to these equations has to be made if we introduce a sliding grid along the x-axis, instead of an evenly spaced grid. A sliding grid is a grid where the grid distance hj = Xj - Xj-1 is steadily decreasing or increasing with increasing Xj,,f} I hi 2T i b n t s case, we approximate f}x2 y (3.6) hi-it,+! - (hi, + ht-i)tt + hitt_1! hihi-l(h, + hi-i) where the error is or order O(hi - hi-i). S'mil 1 8Tt b 'd b 1 ar y, 8x can e approximate y (3.7) Ti+! - Tt-l hi + hi-l ' where the error is O(hi - hi-1).

16 13 4. RESULTS For practical purposes we chose a slicling grid with the first x-interval 10 times as small as the last interval. A time step was chosen that was made 4 times as small once the temperature T rises to values over 0.8. The results can be viewed on the following pages.

17 14 TeDl rature 111 thernustor LiD r. '1') ;J.L'" o.:j T - :: t=o. (loci) : ~~~... ~~... *...,.... _...,..~... ~.. ~ -""'... "... "t"'...,.... "\...,....,.....,......,,.,, ':' I,,! '..."...,...,.,...,....,J..."..._...,,1'..._ ~... _..., I I' : 'l f, I,,I,, I,,I, }... ~... ~... ~... ~...:... ~... ~...-t... ~... I 'l I I! I ~! I I', 1 I'.,...,...,...,... E ~~'~ ~:. : : :. ~. I I ~~~~:, "..-...~-~,... ~... ~... ~...:...""""--:---=...::.::"-~~... +,., J.,,I,,, -... ~ ~ ~ * ~~... ~~ ~... ~ ~... ~ ~. ~... ~ ~... "... ~... "... "... ~... *...,... ~... " ~.. ~... ~ ~ , '''''' ~... ~... "'... ~... ~... "..., ;... 1;.... "... :... ~... ~... ~"" I I ",. I, t,,!! I : :, I..."t......"i..."...,... 1'..."f.....",... ~..."... r"... r _...- I I l I I I I., I It.,, "!, '"... ~..._-....,.."...-_... ~ "r"...,. _... ~~ ~ ~.. _ ~. - - ~ - _. _..., _., ~ _, t... _...,...."'...,.. 0.", '1' I " I !],7D x Fig. 7: The temperature T plotted against x. The temperatures were plotted at t = 0.00, t = 0.05, t = 0.10,... till t = 0.40.

18 Temrerature in therml~or LiD 0.B r'; "':,r, ~'.'--.. ~ T,.... ~... :,. -:: t::[l.ooo~,.o.400~.. ~ ~ ~ ~ ".....,... ~ ~, ~ ~ "... ~, I I : t I " I,, ",, " I 1 t,, ~ ~..._ " /0... I!,, ; 1, I l,, I, t I I I I, I I I ~ I,. I j,.; \,,, ~ "'.. "' t ~... ;"..., :-.....,,,, It., I' I \, :... ' ~ ~... ; J ), I, I.,... ~"... " "-"...,...,....,., I, I I I I I 1 I I, I.;, 1 t.! I, l,, I i!.: I :. I ~...,.... _... ~... ~" ~... " "'...,... a... _... _._......,!! t, 1 ", \ I..., I....,, I ; I,...,I' I,..._~~...;..."... ~...,...,.. "... ~... ~...,... ~... a'".. '"a~... ~..,..."'.. _"'~.*.. ~,.*,....:... ~... ~..._... ~..._ j $...,... ~..-..."'.."""...,...,...;... ~ ~.."' r r.....* "';,...,...,...,...,.,...,;.....,...;..... r..."..... j, I,, \ '7 ; t, ~...,.... *... ~ "?'.~"...""'.. ~... ~...,,~...,,~... ~... "t"... ~...,. _.oo,ooo.oo...,o..,-....o....o,.... ~... ~,,,. x ,7Q 0.90 Fig. 8: Again, the temperature T as function of both :t and t. The temperature was plotted at t = 0.00, t = 0.05, t = 0.10,... till t = 0.40.

19 16 Ten}" rature in thel1ilbior G.BS 0.66 n 44 w. ; 1 r.!')r; \,oti..i.... ~ T ~--~*~*...,.,-:; t=d.ooo~,.o.1qo' I "'1"... ~ ~ ~... ~... _...'"... "t"...,......,...,.... *~..... I,,, 1 I, If, I:,, ',,.,.,......,... # ~.... I.,!' ~~~';';':':-' '.,., I ~ I : :: -:_ ~_... _..... ~.. I~... ~..,..... ~.. _.. ~~.. ;O'...,.. ; ~... '". ~;... -.~ ~... ; ~ "'... ~"':':' I I ",... ~... ~...,...,.~...,... ~...,... f... ~ ~-.. ~ --.., ~ ~ r.. ~.. =~,, ~ r, I \ T. l ; 1 "l I' _... _....,...,.. ~... ~... ~... _..,.... _... ~ ~ ~... ",;-"... _... -'t".' -... r "" ( I 1 ) C.4D I ' x Fig. 9: Again, the temperature T as function of both z and t. The temperature was plotted at t = 0.00, t = 0.05, t = 0.10,... till t = 0.40.

20 17 (li) CONVERGENCE Suppose T is the exact solution of the partial differential equation (2.8) with boundaryand initial conditions (2.9)-(2.10). Suppose we denote the solution achieved using step lengths 6.t and 6.z, by T(~z,6.t). Similarly, we denote the solutions achieved using (26.z,26.t) and (46.z,46.t) as step lengths, by T(26.z,2~t) and T(46.z,46.t), respectively. Since we use a discretization O(~t), 0(6.z 2 ), the global errors are of these orders as well. That is, I ~ Suppose ~t and 6.z are of same order. Then, (~z)2 -T(4~z,46.t) = 4~t: + (46.Z)20 + h.o. (4.1) T - T(26.z,26.t) = 26.t/ + (26.z)20 + h.o. T - T(6.z,6.t) = 6.t j + (6.z)2 0 + h.o. < 6.t and therefore the second term on the right-hand side can be negledted as long as i and 0 are of the same order of magnitude. "h.o." denotes higher order terms. Although T is unknown, some manipulation with (4.1) shows that, supposing OU) = 0(0), (4.2) T = 2T( 6.z, 6.t) - T(26.z,26.t) + h.o.. That is, from T(6.z,6.t) and T(26.z,26.t) we can derive a better guess for T. Also from (4.1), it follows that (4.3) T(26.z,2~t) - T(4~z,46.t) _ 2~ti _ 2 T(6.z,6.t)-T(26.z,26.t) - 6.tj -. In practical. equations Equation (4.3) is a good method to compute the order of convergence of a numerical scheme. A first order scheme will yield a value 2, stating that by halving step lengths, the errors will be halved as well. A second order scheme will yield a value 4, implying that when halving the step lengths, the errors will be 4 times as small. NUMERIC RESULTS FOR THE THERMISTOR Results with Beta = 0.10, Gamma = 500, Mu = 7.5. The average interval size dz = This is achieved by introducing 100 intervals and requiring the first interval to be 10 times as small as the 100th interval.. The time step dt = (1/200). When T(X, t) increases to values over 0.8, the critical value corresponding with 100 degrees Celcius, the time step is made 4 times as small.

21 18 Time = X T(4dx,4dt) T(2dx,2dt) T(dx,dt) (T2 - T4)/(Tl- T2) Time = X T(4dx,4dt) T(2dx,2dt) T(dx,dt) (T2 - T4)/(T1- T2) Time = X T(4dx,4dt) T(2dx,2dt) T(dx,dt) (T2 - T4)/(Tl - T2)

22 19 s. CONCLUSIONS AND RECOMMENDATIONS From our numerical results we conclude that the one-dimensional model of the thermistor as we chose it here, does not account for thermistor cra.cldng because of high temperature gradients. H high gradients do cause the thermistor to crack, a more sophisticated model of the thermistor has to be developed, in order to find these high gradients. One can think, for example, of a two-dimensional model. It is possible that the heat conduction at the side of the thermistor does play a role of importance, too. This would result in a lower tempera.ture at the sides and therefore, larger differences in the temperature inside the thermistor. On the other hand, we need to know more about the conditions under which the thermistors crack. H all thermistors sooner or later crack, the cause could be material fatigue as well, requiring a different modelling approach than we took. If the thermistor does not always crack in the same wa.y, fracture mechanics might be involved in the process. As seen, thermistor cracking is still an interesting subject of study.