Question

A simplified representation for cooling in very large- scale integration (VLSI) of microelectronics is shown in the sketch below. A silicon chip is mounted in a dielectric substrate, and one surface of the system is convectively cooled, while the reminding surfaces are well insulated from the surrounding. The problem is rendered two dimensional by assuming the system to be very large in the direction perpendicular to the paper. Under steady state (1) and transient (2) operation electronic power dissipation in the chip provides for uniform volumetric heating at a rate of q. However, the heating rate is limited by the restrictions on the maximum temperature that the chip is allowed to achieve. Cooling information is T-20 °C and h=500 W/m2.K the conductive heat transfer of chip is kc 50 W/m.K, and q = 107 WIms, the heat conduction coefficient for the substrate is ks 5 W/m.K For the condition shown on the sketch, will the maximum temperature inthe chips exceed 85 °C, the maximum allowable operating temperature set by industry standards? A grid spacing of 1.5 mm is suggested. Find the temperature distribution in the chips and substrate (1) Steady state two dimensions (2) Transient only for three successive nodes from the center of chip in two dimensions (a = 89.2 x 10-6 Plot the temperatures of this three successive nodes versus time within 5 min of chip's operation. Assumption: Constant Properties Thermal heat transfer coefficients are constant .Radiation heat trans fer negligible Coolant = 20°c T h 500 W/m2 K Chip ke=50 W/m K q=107 W/m3 Coolant Chip k-50W/m.K q=10 W/m2 T 20°c oo h=500 W/m2 K HI4 3 Substrate, k5.Wm-K H= 12 mm L-27 mm

Answer 1

SOLUTION:

Perform an energy balance on the chip q(H/4) (H4) T h Tmax = 2kc 107 W/m3 (0.003 m)10' W/m2 (0.003 m) ,C = 8o 9'C 2 x 50 W/m K 500 W/m K Since Tmax < 85°C for the assumed situation, for the actual two-dimensional situation with the conducting dielectric substrate, the maximum temperature should be less than 80°C ax3mm 15 ?j4y-5mm 1 12 13 14 16 17 18 19 20 X Using the suggested grid spacing of 3 mm, construct the nodal network and write the finite-difference equation for each of the nodes taking advantage of symmetry of the system. 2 500 W/m Kx 0.003 m 500 W/m K x 0.003 m hAx hAx = 0.30 -0.030 5 W/m K 5 W/m K ks 2 2 1 =1.818 =0.0910 5/50 1 (kg /ke)+1 kc/kg +1 2 2 -0.182 (kc/ks)+1 50/51 QAXAY1.800 kc find the nodal equations: Node 1 T2-T T6-T +hAx (T-Ti)0 AX +kgAy Ay

hAx 2 + hAx T (1) Ti+T T6 - 2.30T1 T2 + T6 = -6.00 Node 2 (2) TI 3.3T2 T3 + T7 = -6.00 Node 3 Ts - T3 kgAx Тз — Тз kgAy Та - Тз +hAx (T -T3 ) = 0 (Ax/2)/keAy(Ax/2)/kgAy Ay Ax T, - (2+а+(hдхk, )T;) + «Т, + Ts%3D- (Һaxk)T. (3) Т, - 4.12T; +1.82T4 + Tg 3D -6.00 Node 4 Тз - Та (Ax/2)/k Ay (Ax/2)/kgAy Ts - Т4 k Ay Ax Tо - Т4 (Ay/2)/k2Ax+(Ay/2)kAx +q(AxAy)hAx (T -T4)= 0 FT3-(1+2p+hAx/k)T4 + T5 + T9-(hAx/ke)T -AxAy/k (4) 0.182T3-1.39T4 T5 +0.182T9 =-2.40 Node 5 Та - Ts kçAy Ax Ti0 T5 +h(Дх/2) (Tp -T;) + qдy(Дх/2) 3D0 (Ay/2)/k (Ax/2)+(Ay/2)/k (Ax/2) (5) 2T4 2.21T5 0.182T10 = -2.40 Nodes 6 and 11 k Ax (T-T6)/Ay kAy(T7 - T6)/ Ax + kAx(T1-T6)/Ay 0 (6,11) T6 3T11T12 + T16 = 0 T1-3T6 T7T11 = 0 Nodes 7, 8, 12, 13, 14 Treat as interior points, (7,8) T3 T7 -4Ts+ T9 + T13 = 0 T2 T6-4T7 Tg T12 0 (12,13) T + Ti1- 4T12 + Tiз + Tiz %3D0 Tg T12 4T13 + Tj4 + T18 = 0 (14) T9 T13 4T14 +Ti5 + T19 = 0 Node 9 ТА - То Tg - То kgAy Ax Tio - То T14-To +kgAx Ay kgAy = 0 (Ay/2)/kcAx (Ay/2)/kgAx Ax (9) 1.82T4 Ts 4.82T9 T10 +T14 0 Node 10 Using the result ofNode 9 and considering symmetry, (10) 1.82T5 2T9 4.82T10 T15 0 Node 15 Interior point considering symmetry (15) T1o 2T14 -4T15 +T20 = 0 Node 16 By inspection .(16) T11 2T16 T17 0 Nodes 17, 18, 19, 20 T12 T6-T17T18 = 0 T4+Ti8-3T19+T20 = 0 T13 T17-3T18T19 = 0 Tis+2T19-3T200 - (17,18) .(19,20)

matrix inversion method, finite-difference equations is written in matrix notation, [A][T] = [C] 10 0 0 0 0 0 0 0 0 0 0 0 0 -6 -2.3 1 0 0 0 0 -6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1-3.3 0 0 0 1 0 0 10 0 1 -4.12 1.82 0 0.182 -1.39 1 0 0 0 0 0 -6 1 0 0 0 0 0 0 0 0 .182 0 0 0 0 0 0 0 0 0 -2.4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10 0 0 0 0 -2.4 0 2 -2.21 0 0 0 0 .182 0 0 0 0 1 0 0 0 10 -3 1 0 0 0 1 0 1 -4 0 0 0 1 0 1 0 0 C 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 10 0 0 0 0 0 1 0 0 0 -3 0 1-4 1 0 0 0 1 0 0 0 1 -4 1 0 1-4 1 0 0 0 1 0 0 0 2-4 0 0 0 0 1 0 0 0 0-2 1 0 0 0 1 0 0 01-3 0 1 0 0 0 1 0 0 1 0 0 0 0 0 0 1 0 0 0 2-3 0 0 0 1-4 1 0 0 0 1 0 C 0 0 1.82 0 1 0 0 0 0 0 0 0 0 0 0 0 10 -4.82 0 0 0 0 0 0 0 1.82 0 2 -4.82 0 [C] [A] 0 0 0 1 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 10 0 1 0 0 0 0 10 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 10 0 0 0 0 0 0 0 3 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1-3 1 0 0 10 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 and the temperature distribution (°C), in geometrical representation, is 40.41 45.88 34.46 36.13 46.23 37.13 38.37 40.85 43.80 44.51 38.56 39.38 40.81 42.72 42.78 39.16 42.06 39.77 40.76 41.70 The maximum temperature is T5 = 46.23°c less than 85°C