..... in a processor a 0.02 mm thick layer of glue separates the Si layer from a

Question

.....

in a processor a 0.02 mm thick layer of glue separates the Si layer from an A1 (240 W/mK) substrate. The sides of the processor are insulated and the Si and A1 layers are 1 mm and 10 mm thick respectively. The processor is cooled on both the surfaces by ambient air (T alpha = 25degreeC, h = 100 W/m2K). The chip dissipates heat at 104 W/m2 in steady state. Estimate the operating temperature of chip. Assume negligible thermal resistance in Si layer (isothermal chip), negligible radiation. Not that the contact resistance of the glue is 0.9 times 104 m2K/W (Hint: hence the conductivity of the glue is 1.1Times104 W/m2K).

Explanation / Answer

Let the temperature of the chip be T degree celcius ...

so. heat radiated directly to air = h*A*( T_chip - T_ambient ) = 100*A*(T - 25 )

for the heat radiated through Aluminium

Total resistance = L_glue / ( K_glue * A ) +   L_Al / ( K_Al * A )   + 1 / ( h * A )

= 0.00002/(1.1*10^4 * A ) + 0.01/(240 * A ) + 1 / (A*100 )

= (1/A) * (1.8 * 10^-5 + 4.1667 * 10^-5 + 1000 * 10^-5 )

= ( 1/A) * ( 1005.9667 * 10^-5)

so... heat radiated from aluminum side = temp diff / total resistance = ( T - 25 ) * A * 10^5 / ( 1005.9667)

                                                                = 99.40687 * A * ( T - 25 )

so... total heat rodiated = 100*A* ( T - 25 ) + 99.40687 *A* ( T - 25 ) = 199.40687*A*(T - 25 )

total heat generated by chip = 10^4 * A

so... 199.40687 ( T - 25 ) * A = 10^4 * A

so... T = 75.1487 C ...

so ... operating temp of chip = 75.1487 Celcius

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