91Ó°ÊÓ

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heat \(\sin k\) to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attractive option is to use a heat sink consisting of an array of square fins of width \(w\) on a side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fin width designated as the fin pitch \(S\). The method by which the heat sink is joined to the chip would determine the interfacial contact resistance, \(R_{t, c^{*}}^{n}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a dielectric liquid with \(T_{\infty}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat \(\operatorname{sink}\) is fabricated from copper \((k=400 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\), and its characteristic dimensions are \(w=0.25 \mathrm{~mm}\), \(S=0.50 \mathrm{~mm}, L_{f}=6 \mathrm{~mm}\), and \(L_{b}=3 \mathrm{~mm}\). The prescribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{t, c}^{\prime \prime}=5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q_{c} ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (subject to manufacturing constraints that \(L_{f} \leq 10 \mathrm{~mm}\) ). Assess the effect of such changes.

Step by step solution

01

1. Calculating contact resistance

The contact resistance \(R_{t, c}^{\prime \prime}\) is given as \(5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). To determine the total contact resistance between the heat sink and the chip, we'll multiply this value by the chip area, \(W_c^2\): \(R_{t, c} = R_{t, c}^{\prime \prime} W_c^2 = (5 \times 10^{-6} \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}) (16 \times 10^{-3} \mathrm{~m})^2\)

02

2. Heat transfer through the heat sink

We'll assume all of the heat generated by the chip is transferred through the heat sink. For this, we'll use the following formula: \(q_c = \frac{T_c - T_\infty}{R_{t, c} + R_h}\), where \(q_c\) is the heat transferred, \(T_c\) is the chip temperature, \(T_\infty\) is the coolant temperature, \(R_{t, c}\) is the contact resistance, and \(R_h\) is the heat transfer resistance of the heat sink. To find \(R_h\), we'll be using the relation: \(R_h = \frac{1}{hA_c}\), where \(h\) is the convection heat transfer coefficient, and \(A_c\) is the cooling area of the heat sink. For a fin-array heat sink, the total area for cooling \(A_c\) can be calculated by considering all fins to have a square cross-sectional area equal to \(w^2\), giving: \(A_c = N_{fins} [(2wL_f + w^2) + L_b w_c^2]\), where \(N_{fins}\) is the total number of fins, \(L_f\) is the fin length, and \(L_b\) is the base length of the heat sink. To find \(N_{fins}\), we need to use the fin pitch \(S\) and width \(w\): \(N_{fins} = \frac{W_c - w}{S}\)

03

3. Finding the maximum allowable chip power dissipation \(q_c\)

Now we can plug the values for \(R_h\), \(R_{t, c}\), and the temperatures into the formula for \(q_c\), and solve for it, knowing that the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\): \(q_{c, max} = \frac{(85 - 25)K}{R_{t, c} + R_h}\).

04

4. Assessing the effect of changing the width and length of the fins

We can now analyze how the chip's power dissipation capacity changes when we increase either the width of the fins (\(w\)) or their length (\(L_f\)). For this, we need to assess the impact of these variables on \(R_h\) and how that affects the overall heat transfer rate (\(q_c\)). For example, if changing \(w\) or \(L_f\) results in an increase or decrease of \(R_h\), that will have a direct impact on the amount of heat the chip can dissipate. The constraint given is that \((S - w) \geq 0.25 \mathrm{~mm}\) and \(L_{f} \leq 10 \mathrm{~mm}\), so we have to explore those ranges of values to make our assessment and draw conclusions about the optimal values for heat dissipation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermal Resistance

Thermal resistance is a concept used to describe a material's ability to resist the flow of heat. It's a critical factor in the design of heat sinks. Think of it like insulation, but for heat. We often use this concept to determine how effectively a heat sink can transfer heat away from electronic components like chips.

In a heat sink design, thermal resistance can be calculated by considering the material and its structure. For instance, with our copper heat sink, it's calculated by the formula for thermal resistance across an interface, which includes the contact resistance between surfaces. The total resistance is then affected by the size and area of the contact surface.

Lower thermal resistance means better heat transfer, which is essential for keeping chips cool under high performance. By reducing resistance, more heat can escape from the chip through the heat sink, thereby keeping the device operating safely and efficiently.

Convective Heat Transfer

Convective heat transfer is a form of heat transfer where heat is carried away by a fluid. In our heat sink scenario, the dielectric liquid circulating around the fins is responsible for this process. Convection depends on both the speed of the fluid and the heat transfer coefficient, which is given as 1500 W/m²·K in this exercise.

The effectiveness of convective heat transfer in a heat sink is crucial, as it defines how quickly heat can be moved away from a heat-generating component. This efficiency depends on factors like surface area and fin design. More surface area means more space for heat to transfer into the moving fluid, which enhances the cooling performance.

• It often involves enhancing surface area coverage, as seen with the fins on heat sinks.
• The fins increase the effective area for heat transfer, allowing more contact between the fluid and the surface.
• This increased contact area leads to better heat removal from the chip.

Convective heat transfer can thus be improved by modifying the heat sink design to ensure added surface area and efficient fluid flow.

Chip Cooling Efficiency

Chip cooling efficiency refers to how effectively a cooling system manages to keep a chip's temperature within safe operational limits. The maximum allowable chip power dissipation is a key metric for determining this efficiency. In our exercise, this was calculated by considering the maximum chip temperature and all resistive components that add to the thermal path.

Several factors contribute to cooling efficiency, including the heat sink's material, design, and the thermal management strategy employed. By optimizing parameters such as fin width ( w ) and length ( L_f ), cooling efficiency can be enhanced.

Here's how adjustments affect efficiency:

• Increasing the fin width reduces the space between fins, impacting airflow and convective efficiency.


• Extending fin length increases surface area, boosting convective heat transfer but also adds to material cost and weight.


• Balancing these factors is key to maximizing cooling efficiency without incurring excessive costs or design complexity.

The goal is to design a heat sink that efficiently dissipates heat at minimum material and manufacturing costs, achieving optimal chip cooling efficiency.