Question

Two-Dimensional Steady and Transient Conduction - Cooling a very large scale microelectronic chip,

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 provide for uniform volumetric heating at a rate of q. However, 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. Coefficient of chip is k_c = 50 W/m.K, and q = 10^7 W I m^3 , the heat conduction coefficient for the substrate is k_s = 5 W/m.K.

For the condition shown on the sketch, will the maximum temperature in the 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 (α = 89.2 ×10^-6)

Plot the temperatures of these three successive nodes versus time within 5 min of chip's operation.

Assumption:

Constant Properties

Thermal heat transfer coefficients are constant

Radiation heat transfer negligible

Coolant T = 20°C h=500 W/m2.K. Chip k=50 W/m•K q=107W/m3 00 -13 Substrate, * 5 Wim H = 12 mn -0.27 mm Irx,0) = T; = 600 K ---- = 300 K -Po h = 500 W/m? K

Answer 1

chip Energy balance on Tman à (1/4) q (+14) + be To 2 k ci b lotx (0.003) => + 10 X 0.003 500 + 20°c 2 x 50 Trmase = 80.9°C - Since Tman <85°c onctual 20 Situation with the Conducting Substrate the Maximeem temperature should be less than 80° c di electric ) haa 500 X 00003 Ks 2 0,3 La 2 - 2 Кs kei + 1 = 1.818 Кс +1 a 2 2 The shtu - 50 0-182 Ks + 5 0.0910 ks +1 aloi Scanned by TapScanner

4 х Духду — 1-30, Node 1 :- Кs xpxx Ts-т, + ks ay T,-т ДУ + дX hDL (то-т, ) = 0 27 T, + Т. + Те = - hДхте с) - 2:35, +Т, + Те — 6. 1 Node 2: Т. – 3-37, + Ta+T1 = -6 2 No4 е 3: T, -Tз тч - Tз Кs x548 + Дm (*) | Kay + (A) JAGAY + Kg An Ts-T3 Ах hДх (тас -3)=0 t -Т, - (2+ d + be</us x3) + LT4 - Tg = - har toe Scanned by TapScanner

T, -Ч. 12)2 +1-82T + 3 = -6 - 3 Node 4: Та-тч + Вс Ду 15-ти (*) БИ ksayt (A) / KcXAY + Ta - Tч бу (р»үру) 2) [иқоя» 225 / bx + box (те:) -> Рт, -( 1+2 (3+[һди [k]) 4+ тя + в та = - (he) те - thе: ) те – буд» для КС г) o •182 Та - 1-39 ту+Ts + 0-182T = - 2.9 (Ч Node 5: Кс Xду x Tч Т. То - , + (e){ux lov)+ bf%) Кс вку. ) Scanned by TapScanner

thCox) (Tcc -75) + (Tcc-75) + gay (Ah) -> 274 -2.2175 +0+182710 = -2-4 5 odes 6 and in Kg x D xx6, -76) KS XDy x(77-76) t sy Dn КsxДx (т-те) DY T-3T6 +Tq + T1 = 0 6 To -3711.+ Ti2 + T16 = 0 Node 7,8,12,13,14 Interior points Tz + T6-4T7+Tg + T12 = 0 Tz + Tt - 4 Tg + Tq & T 13-0 7 8 T7 + 71 - 4712 +213 +15= 0 2. Tg + Ti2-4713 + T4+Tig=0 (13 Tq + Ti3-4 T4+T15 +7,90 Scanned by TapScanner

Node 9 :- Кѕх дух Та-та Ksxay toto + 1 Ду ! an + ks x Д У Тета тч - Та (9) (As/..) Ay + e-с до Kg An - Ч Node Jo;- 1. 42T4 +Ts – V•. 14 + Tio + тіч = 0 - symmetry. |:33 T5 H 2 Та – Ч. 82To + + Ts «О О, Node 15 5) Тio + 2Try – Ч Т5 + To Node 66 Inspection - 4 % о 2T16 +Tia коле 11,13, 14,0 Т. 1 16 - 3та +Т8 = 0 (? Tз + Tra - 3T18 +1 = 0 (18 19 Tiy.t 18 -3 3 *14 + Тро = 0 ho s + 2T1 - 3 To - 0 Scanned by TapScanner

matrix inversion method, finite-difference equations is written in matrix notation, [A][T] =[C] 1 -6 -6 -6 -2.4 -2.4 0 0 0 1 1 1 0 -2.3 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 -3.31 0 0 0 1 0 0 OOOOOOOOOOO 0 1 -4.12 1.82 0 0 0 0 00000000000 0 0.182 -1.39 1 0 0 0.182 00000000000 0 0 0 2 -2.21 0 0 0 0 .182 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -3 1 0 0 0 1000000000 0 1 0 0 0 1-4 0 00100000000 0 0 1 0 0 0 1 -4 0 0010000000 0 0 0 1.82 0 0 1 -4.82 00010000 0 0 0 0 0 1.82 0 0 0 2 -4.820 000 100 000 [A] = 0 0 0 0 0 1 0 0 0 0 -3 1000 10000 0 0 0 0 0 0 1 0 0 0 1 -4 1000 1000 0 0 0 0 0 0 0 1 0 0 0 1-4 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0001-4100010 0 0 0 0 0 0 0 0 0 1 0 0 0 2 -4 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -2 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 10001 - 100 0 0 0 0 0 0 0 0 0 0001000 1-3 1 0 0 0 0 0 0 0 0 0 0 0000 10001 -3 1 0 0 0 0 0 0 0 0 0 0 0000 1000 2-3 [C] = 0 0 0 0 0 0 0 0 and the temperature distribution (°C), in geometrical representation, is 34.46 36.13 40.41 45.88 46.23 37.13 38.37 40.85 43.80 44.51 38.56 39.38 40.81 42.72 42.78 39.16 39.77 40.76 41.70 42.06 The maximum temperature is Ts - 46.23°C less than 85°C.