91Ó°ÊÓ

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heot sink 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 attmctive option is to use a heat sink consisting of an atmy of square fins of width w on s side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fan width designated as the fin pitch \(S\). The method by which the hent sink is joined to the chip would detesmine the interfacial contact resistance, \(R_{\mathcal{U}^{*}}^{*}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a diclectric liquid with \(T_{w}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat sink is fahricated from copper \((k=400\) Whe \(\mathrm{K})\), and its characteristic dimensions are \(w=0.25\) \(\operatorname{mmn}_{,} S=0.50 \mathrm{~mm}, L_{y}=6 \mathrm{~mm}\), and \(L_{1}=3 \mathrm{~mm}\). The preccribed 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 \(R_{\mathrm{W}}^{*}=5 \times 10^{-6} \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q, ?\) 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 \(L_{f}\) (suhject to manufacturing constraints that \(L_{y} \leq 10 \mathrm{~mm}\). Assess the effect of such changes.

Step by step solution

01

Calculate Thermal Resistance Components

First, calculate the thermal resistance through the heat sink, which includes conduction through the fins and convection from the fins to the liquid. Use the formula for fin efficiency and the convective thermal resistance \( R_{th} = \frac{1}{{n * \eta_f * A_f * h}} \), where \( n \) is the number of fins, \( \eta_f \) is the fin efficiency, \( A_f \) is the fin area, and \( h \) is the convection coefficient.

02

Determine Fin Efficiency

Calculate the efficiency of the fins using\( \eta_f = \tanh(mL_f)/(mL_f) \),where \( m = \sqrt{{2h}/{kAf}} \), \( Af \) is the fin cross-sectional area, and \( k \) is the conductivity of copper. Substitute the known values to find \( \eta_f \).

03

Calculate Overall Thermal Resistance

Sum the individual thermal resistances including the interfacial contact resistance \( R_{total} = R_{cu} + R_{conv} + R_{\text{contact}} \). Use the values from the previous steps to find \( R_{total} \).

04

Determine Heat Dissipation

Using the calculated total thermal resistance, compute the power dissipation using the formula \( q = \frac{(T_{max} - T_w)}{R_{total}} \),where \( T_{max} = 85^{\circ} \text{C} \) and \( T_w = 25^{\circ} \text{C} \).

05

Evaluate Effects of Changing Parameters

Consider changes in \( w \) and \( L_f \) to increase heat dissipation according to \( q = \frac{(T_{max} - T_w)}{R_{total}(w, L_f)} \).Recalculate \( q \) with these adjusted parameters while considering the constraints on spacing \((S-w)\) and fin length \(L_f \).

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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 crucial concept in heat transfer, especially when dealing with systems like heat sinks. It represents the system’s resistance to heat flow through different materials or interfaces, akin to how electrical resistance operates in a circuit. In the case of a heat sink attached to a high-performance chip, understanding thermal resistance allows us to measure how effectively heat can dissipate from the chip’s surface into the surrounding medium, which is often a liquid or air. Thermal resistance is cumulative and consists of several components:

• Conduction Resistance: This is the resistance against heat flow through solid materials, such as the copper fins in the heat sink. It depends on the material's properties and geometric dimensions.
• Convection Resistance: This resistance represents the heat transfer to a fluid, like air or a dielectric liquid, surrounding the heat sink. It's influenced by factors like the convection coefficient, surface area, and the effectiveness of fin design.
• Interfacial Resistance: The thermal resistance at the contact point between the chip and the heat sink, which can be minimized using high-quality joints.

Each component adds up to create the total thermal resistance for the system, which should be minimized to enhance cooling.

Fin Efficiency

Fin efficiency is a measurement of how effectively a fin transfers heat from its base into the surrounding fluid. Ideally, a fin would have an efficiency of 100%, indicating perfect heat transfer, but real fins often fall short due to several heat losses.The efficiency of a fin depends on various factors:

• Material: Conductivity plays a major role. High conductivity materials like copper are preferred for manufacturing fins.
• Geometry: The shape and size of a fin affect how heat spreads throughout the material. For instance, longer or thinner fins usually aid in better heat distribution.
• Environmental Conditions: The convection coefficient of the surrounding environment impacts how effectively heat is exchanged between the fin and the surrounding fluid.

To calculate fin efficiency, we use the formula: \[ \eta_f = \frac{\tanh(mL_f)}{mL_f} \]Where \( m = \sqrt{\frac{2h}{kA_f}} \), and \( A_f \) is the cross-sectional area of the fin. Here, \( m \) represents how swiftly heat can travel within the fin material. Understanding fin efficiency helps in evaluating a heat sink's performance and suggests ways to improve heat dissipation by tweaking fin parameters.

Heat Dissipation

Heat dissipation refers to the process by which heat is expelled from an area, such as a computer chip, preventing overheating and maintaining optimal functionality. In heat sinks, this process predominantly relies on enhancing the thermal pathway from the chip to the surrounding cooling medium via convection and conduction.The rate of heat dissipation is influenced by elements such as:

• Overall Thermal Resistance: Lower thermal resistance facilitates better heat flow through the heat sink.
• Heat Sink Design: Factors like the number of fins, their size, and their placement affect the overall heat transfer rate.
• Ambient Conditions: The temperature difference between the chip and the cooling fluid (\( T_{max} - T_w \)) drives the rate of dissipation.

The formula used for calculating the power dissipation is: \[ q = \frac{(T_{max} - T_w)}{R_{total}} \]By minimizing \( R_{total} \), and optimizing factors like fin size and material, engineers can maximize \( q \), increasing the chip's power capacity without risking overheating. Adjustments in fin dimensions or material properties can significantly impact \( q \), making the understanding of heat dissipation critical in design and performance assessment.

Interfacial Contact Resistance

Interfacial Contact Resistance is the restriction to heat flow at the interface where two bodies meet, such as the chip and the heat sink. Despite seeming minor, this resistance can significantly impact the effectiveness of heat dissipation in electronic systems, as it acts as a bottleneck for heat transfer.Contact resistance is typically influenced by:

• Material Mismatch: Different thermal properties of contact materials can create inefficiencies.
• Surface Roughness: The smoother the contact surfaces, the better the thermal interface between them. Rough or uneven surfaces trap air pockets that act as insulators.
• Bonding Quality: Better bonds, such as those made through metallurgical joining, can offer lower contact resistance compared to adhesive or mechanical joints.

For the given exercise, the interfacial contact resistance \( R_{\text{contact}} \) was specified and needs to be taken into account when calculating the total resistance: \[ R_{total} = R_{cu} + R_{conv} + R_{\text{contact}} \]By aiming to minimize this resistance, the heat transfer from the chip to the heat sink interface can be optimized, thus maintaining lower chip temperatures and enhancing performance.