Question

(8 marks) The heat generated from power dissipation of a silicon chip is transferred by convection heat transfer from top and bottom of the chip. Top surface of the chip is covered by a 4-mm thick aluminum cover (k = 240 W/m.k) which is subjected to convection heat transfer on its top surface. Bottom surface of the chip is directly subjected to convection. Convection heat transfer coefficient of fluid is h = 1000 W/m²K and To = 25 °C. The contact resistance of Aluminum/chip interface is 0.5x10-4 m²K/W. If the chip surface area is 100 mm and its maximum allowable temperature is 85 °C, what is the maximum allowable power dissipation in the chip? (Note that chip has a uniform, constant temperature) Fluid Toh Aluminum cover Chip Fluid Toh

Answer 1

Given data :-

• Thickness of Aluminium Cover, t = 4 mm = 0.004 m
• Thermal conductivity of Aluminium, K = 240 W/mk
• Convective heat ttansfer coefficient, h = 1000
  W/mk
  
• Temperature of surrounding Fluid,
  T
  = 25 °C
• Temperature of Chip, T = 85 °C
• Thermal contact resistance of chip/Aluminium interface, R =
  0.5 x 10-4mºk/W
  
• Surface area of chip A = 100
  mm:
  = $100 \times 10^{-6}m^{2}= 10^{-4}m^{2}$

Fluid 3 h, To q , T 4 mm x Al cover Chip/ Aluminium Interface chip e 'T=85 °C Fluid > h, To 4/2

Let the heat generated due to power dissipation per unit area in chip is q $W/m^2$ . From that heat generated, some amount of heat will be transfered by the top surface ( Let say $q_1$   $W/m^{2}$ ) and remaining amount of heat will be transfered from the bottom surface of the chip ( Let say $q_2$ $W/m^{2}$ ).

So, $q=q_1+q_2$

(1) Heat transfer from the top surface,   $q_1$  

Now heat transfer from the top surface of the chip per unit area to the surrounding fluid can be found by,

$q_1 = \frac{\Delta T}{\sum R_{th}} = \frac{T -T_\infty}{\sum R_{th}} W/m^{2}$   

Where , $\sum R_{th}$ = Total Thermal resistance between the top surface of chip and surrounding

So, $\sum R_{th}$ = Thermal contact resistance of chip/Aluminium interface + Conductive resistance of Aluminium cover + Concective resistance of fluid

So, $\sum R_{th} = R + \frac{t}{K} + \frac{1}{h}$

So, $\sum R_{th} = (0.5 \times 10^{-4})+ \frac{0.004}{240} + \frac{1}{1000}$

So, $\sum R_{th} = 1.066666667 \times 10^{-3}$   $m^{2}k/W$

Now,

$q_1 = \frac{T -T_\infty}{\sum R_{th}} W/m^{2}$

So,  $q_1 = \frac{85 -25}{1.06666666\times 10^{-3}} = 56250W/m^{2}$

(2) Heat transfer from the bottom surface of the chip ,  $q_2$

Heat transfer from the bottom surface of the chip per unit area to the surrounding fluid can be found by,

  $q_2 = \frac{T -T_\infty}{R_{convective}} W/m^{2}$

So,  $q_2 = \frac{T -T_\infty}{\frac{1}{h}} W/m^{2}$

So,  $q_2 = h ( T - T_\infty)= 1000 ( 85 - 25 ) W/m^{2}$

So,  

Now, Total heat generation in the chip per unit surface area , q

$q=q_1+q_2$

So,  

Now, Total Power Dissipation in the chip can be given by,

Q = q × Surface area of the chip

So, Q = 116250 × $10^{-4}$ = 11.625 watt

So, maximum power dissipation allowed in the chip is 11.625 Watt.