Comsol Laboration: Heat Conduction in a Chip

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1 Comsol Laboration: Heat Conduction in a Chip JO, CSC January 11, Physical configuration A chip on a circuit board is heated inside and cooled by convection by the surrounding fluid. We consider two cases 1. The stand-alone chip, with dimensions H W L as indicated in Figure The same chip but with aluminum cooling flanges added to improve cooling, see Figure 2. We wish to investigate What is the steady state temperature distribution, and how long does it take to reach steady state? How do the chip dimensions influence the max. temperature? How does the temperature change when cooling flanges are added? Figure 1: Chip geometry

2 2 Model 2 Figure 2: Chip with cooling fins 2 Model The mathematical model is an initial-boundary value problem for the heat equation, to determine T (x, y, z, t) for t > 0 when ρc T t = (k T ) + Q, x Ω, (1) k T n = h (T ext T ) x Ω, with initial data T (x, y, z, 0) = f(x, y, z). In particular, the steady state temperature obtained after long time is of interest, 0 = (k T ) + Q, x Ω, k T n = h (T ext T ), x Ω. The parameters here are the mass density ρ [kg/m 3 ], the specific heat C [J/(kg K)], the thermal conductivity k [W/(mK)] and the heat source Q [W/m 3 ]. Moreover, T ext is the ambient temperature and h [W/(m 2 K)] the convection heat transfer coefficient between the chip surface and surroundings. We make some simplifying assumptions: 1. No heat transfer to the card (h = 0). 2. Constant heat transfer coefficient h on all the other sides. 3. Uniform heat source Q in the whole chip. 4. The chip is thin, H W, H L, and we shall study the impact of thinness. We use symmetry to compute on a quarter chip, H (W/2) (L/2) part of the chip. Effective material data for the silicon chip is ρ Si = 2300 [kg/m 3 ], C Si = 750 [J/(Kg K)], k Si = 3.6 [W/mK], and for aluminum ρ Al = 1600 [kg/m 3 ], C Al = 900 [J/(kg K)], k Al = 60 [W/mK]. Total power is P [W], ie. Q = P V [W/m3 ] where V is the volume of the chip. The heat transfer coefficient h when the chip is cooled by a weak fan is h coef (e.g. 100) [W/m 2 K] and T ext = 20[ o C].

3 3 Preparatory analysis 3 3 Preparatory analysis An important characteristic quantity is the Biot number Bi = h k/d, the ratio of thermal resistance 1/h at the surface and d/k inside the chip itself. Here d is a characteristic length for heat transport through the chip which in our case would be H, the smallest dimension. When Bi is large, the surface temperature is close to T ext, and when Bi is small, the temperature will be nearly uniform in the body. 1. What is the steady temperature when Bi (ie h) is small? Give a formula. Hint: All the power produced must be removed by the cooling, and the temperature is almost uniform. 2. Derive for large Bi the approximate value T max = H 2 Q/(2k) + T ext. Is this perhaps an upper or lower limit? Hint: When H/L 1 and L W the differential equation can be simplified to 0 = k d2 T dz 2 + Q, 4 Lab session work Task A 4.1 Chip dt dz z=0 = 0, T (H) = T ext Model: Use 3D, COMSOL Multiphysics/PDE Modes/PDE, Coefficient Form / Time-dependent analysis. This will also do the steady calculation. It is worthwhile to define the material parameters as Constants under Options. Then you can easily change values etc, and the equations get to be in understandable terms. Geometry: A brick! Give the right dimensions under Properties PDE: Set the value of k and Q (named c and f) in the menu. Boundary conditions: Neumann conditions insulation on the bottom and symmetry-boundaries is obtained by q = g = 0. On the cooled surfaces, with q = hcoef and source term g = hcoef*text, where hcoef and Text may first be defined as Constant under Options. Solver: For steady case use Stationary linear, for transient runs : Time Dependent. Choose a finish time based on the calculated time-scale (see below). Tasks: 1. Calculate the maximum temperature when P = 30 [W], h coef = 20 [W/(m 2 K)], W = L = 20 mm and H = 2 mm 2. Check your result by choosing hcoef to make Bi = 10 and 100 and comparing to the formula derived in the preparatory analysis (2.) above. 3. Do the mesh refinement study for the temperature of an upper corner. Try linear and quadratic elements. 4. Plot the temperature profiles and find maximum temperatures for L = 20 mm and W = 10, 1, and 0.2 mm. Keep the power per unit volume Q constant. Explain why the max. temperature drops when W decreases.

4 5 Lab session work Task B Chip with cooling flanges The temperature can be reduced by better cooling. Let the chip-octant be mm 3. Place a 1 mm thick aluminum cap on the chip, then 5 cooling fins 1 mm thick and 1 cm high, essentially as shown in Figure 2 (but higher fins!). Geometry: First one creates the top as a brick and places it by the filling the right geometry data under Draw/Object Properties. Then make an outer flange, place it, then copy it and give the right translation in the dialog box (four times). Create the new Composite Object from the union of all the flanges and the lid without Interior borders. Finally, make an object composed of top, flanges, and chip, with retained Interior borders. That s it! PDE: Fill in coefficients c and f in the two Subdomains (Al and Si). Boundary conditions: Select Neumann for all boundaries, with heat transfer coefficient q and source term g like above. Then change the (few) symmetry-boundaries to insulation. Tasks: 1. How much is the maximum temperature reduced by the cooling fins? 2. Is it worthwhile to make the flanges higher and / or thinner and more numerous? A thought: Total surface area exposed to air is an important parameter in the model (see 1. in the preparatory analysis above). Then, maybe we could do with just a little Al, by making many densely packed thin flanges. But that is not how the actual chip cooler looks. Why? What is wrong with our model? 5 Lab session work Task B Now we return to the chip without cooling flanges and investigate transient warm-up and cooldown so you must use the time-dependent solver. Assume that the chip, after being on for a long time, is turned off at t = 0 so Q = 0 for t > 0. How does it cool off? 5.1 Preparatory analysis The Newton cooling law, often used in crime fiction to determine time of death, is when Bi is small. τ dt dt = T ext T, What is the time-scale τ for the chip? Give a formula. Hint: Start from the PDE (1), integrate over the entire chip and use Gauss s theorem. Use that T is almost constant through the chip. Derive a similar model for heating when the chip is turned back on. Hint: It needs a source term. 5.2 Lab tasks The question "Is it as fast to heat up as to cool down?" is imprecise. Please specify, observe (by computing) and justify/motivate. Select Time dependent under Solver/Solver Parameters. Modify the coefficient da in the Subdomain window and add initial data under the tab Init in the same

5 5 Lab session work Task B 5 window. It is easiest to run a simulation starting with T = T ext, turn on Q at t = 0 and shut off after so long time that the temperature has become (almost) stationary: Q = (t < t end ) Q 0. Compare your observed warm-up / cool-down times to the time-scale τ determined above.

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