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

Question

......

in a processor a 0.02 mm thick layer of glue separates the Si layer from an Al (240 W/mK) substrate. The sides of the processor are insulated and the Si and Al layers are 1 mm and 10 mm thick respectively. The processor is cooled on both the surfaces by ambient air (Talpha = 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 0.9Times10-4 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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