/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 178 Gold is commonly used in semicon... [FREE SOLUTION] | 91Ó°ÊÓ

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Gold is commonly used in semiconductor packaging to form interconnections (also known as interconnects) that carry electrical signals between different devices in the package. In addition to being a good electrical conductor, gold interconnects are also effective at protecting the heat-generating devices to which they are attached by conducting thermal energy away from the devices to surrounding, cooler regions. Consider a thin film of gold that has a cross section of \(60 \mathrm{~nm} \times 250 \mathrm{~nm}\). (a) For an applied temperature difference of \(20^{\circ} \mathrm{C}\), determine the energy conducted along a \(1-\mu \mathrm{m}\) long, thin-film interconnect. Evaluate properties at \(300 \mathrm{~K}\). (b) Plot the lengthwise (in the \(1-\mu \mathrm{m}\) direction) and spanwise (in the thinnest direction) thermal conductivities of the gold film as a function of the film thickness \(L\) for \(30 \leq L \leq 140 \mathrm{~nm}\).

Short Answer

Expert verified
In this problem, we determine the energy conducted along a gold thin-film interconnect and plot the lengthwise and spanwise thermal conductivities as a function of film thickness. a) The energy conducted along the \(1-\mu\mathrm{m}\) long gold thin-film interconnect is approximately \(9.45 \times 10^{-9} \mathrm{W}\). b) To plot the lengthwise and spanwise thermal conductivities as a function of film thickness, use software like Excel, Python, or MATLAB to calculate the corresponding thermal conductivities for film thicknesses \(L\) ranging from \(30\mathrm{nm}\) to \(140\mathrm{nm}\).

Step by step solution

01

Write down the formula for energy conducted

We use the Fourier's Law formula for heat conduction to determine the energy conducted along the thin-film interconnect. The formula is given by: \[Q = kA \frac{\Delta T}{L}\] Where: - \(Q\) = energy conducted, - \(k\) = thermal conductivity of gold, - \(A\) = cross-sectional area of the thin film, - \(\Delta T\) = applied temperature difference, and - \(L\) = length of the thin-film interconnect.
02

Find the values of the parameters

We have the following values given in the problem: - \(k \approx 315 \frac{\mathrm{W}}{\mathrm{mK}}\) (thermal conductivity of gold at \(300\mathrm{K}\)), - Cross-section of the film: \(60\mathrm{nm} \times 250\mathrm{nm}\), - Applied temperature difference, \(\Delta T = 20^{\circ} \mathrm{C}\), and - Length of the thin-film interconnect, \(L = 1-\mu \mathrm{m}\).
03

Calculate the cross-sectional area

The cross-sectional area of the thin film, \(A\), can be found by multiplying the width and height of the film. Make sure to convert the units to meters: \[A = (60 \times 10^{-9}\mathrm{m})(250 \times 10^{-9}\mathrm{m}) = 1.5 \times 10^{-14} \mathrm{m^2}\]
04

Calculate the energy conducted

Now we can substitute the given values into the Fourier's Law formula: \[Q = (315 \frac{\mathrm{W}}{\mathrm{mK}})(1.5 \times 10^{-14} \mathrm{m^2}) \frac{(20^{\circ} \mathrm{C})}{(1 \times 10^{-6}\mathrm{m})}\] \[Q \approx 9.45 \times 10^{-9} \mathrm{W}\] The energy conducted along the \(1-\mu \mathrm{m}\) long gold thin-film interconnect is approximately \(9.45 \times 10^{-9} \mathrm{W}\). #b) Plotting the lengthwise and spanwise thermal conductivities as a function of thickness# For this part, we will simply need to recall that thermal conductivity is dependent on the thickness of the gold film. To plot the lengthwise and spanwise thermal conductivities, use the values of \(L\) ranging from \(30 \mathrm{nm}\) to \(140 \mathrm{nm}\) and calculate the corresponding thermal conductivities for each \(L\). Note that solving this part of the problem requires a graphical output. As a text-based AI, I can't generate a plot directly. However, you can use software like Excel, Python, or MATLAB to plot the lengthwise and spanwise thermal conductivity values for different film thicknesses (\(30 \leq L \leq 140 \mathrm{~nm}\)).

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Key Concepts

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

Fourier's Law of Heat Conduction
When learning about conducting heat through materials such as gold thin-film interconnects in semiconductors, one must become acquainted with Fourier's Law of heat conduction. This fundamental principle governs how thermal energy is transferred in materials due to a temperature difference. It states that the rate at which heat is conducted through a material is proportional to the negative gradient of the temperature and the area through which the heat is traveling.

Mathematically, Fourier's Law is expressed as:
\[Q = -kA\frac{d\ T}{d\ x}\]
where:
  • \(Q\) is the rate of heat transfer (energy per time),
  • \(k\) is the thermal conductivity of the material,
  • \(A\) is the area perpendicular to the direction of heat transfer, and
  • \(\frac{d\ T}{d\ x}\) is the temperature gradient along the direction of heat transfer.
In simpler terms, it shows how quickly heat will move through a material from a hot region to a cooler one. Different materials have varying values of thermal conductivity \(k\), with gold being a particularly good conductor of heat, hence its use in semiconductor packaging. Understanding this law helps in calculating the amount of thermal energy transferred under specific conditions, such as those presented in exercises concerning semiconductor interconnects.
Thermal Energy Transfer
Thermal energy transfer is a process that plays a crucial role in various applications, especially in electronics where it can affect the performance and longevity of semiconductor devices. The transfer of thermal energy involves moving heat from one place to another and can occur through different modes including conduction, convection, and radiation. In the context of semiconductor technology and especially in exercises involving gold thin-film interconnects, conduction is often the primary focus.

During conduction, thermal energy is passed from molecule to molecule without any overall transfer of matter. In semiconductors, managing this thermal energy transfer is vital because excessive heat can damage electronic components. This is where materials with high thermal conductivity, like gold, come into play. They conduct the heat away from sensitive areas to prevent overheating. The effectiveness of this process can be measured using Fourier's Law, thus allowing for precise calculations and designs that optimize heat dissipation.
Furthermore, in exercises such as computing energy conducted along a gold thin-film interconnect, it's important to understand that the amount of energy transferred will depend on the cross-sectional area, the material's thermal conductivity, and the temperature difference across the material. By mastering the concept of thermal energy transfer, students can better evaluate and predict the thermal behavior of electronic devices.
Gold Thin-Film Interconnects
In the realm of semiconductor packaging, gold thin-film interconnects are particularly significant due to their exceptional ability to carry electrical signals and conduct thermal energy. These interconnects serve as bridges between different devices within the package, ensuring both electrical connectivity and thermal management. Their thin-film nature allows for miniaturization, which is essential in modern electronics where space is at a premium.

Gold's high electrical and thermal conductivity makes it a superior choice for these interconnects. The thermal aspect is especially important in high-density circuits where heat generation could compromise device integrity. In exercises focused on thin-film interconnects, it is often required to assess the conductive properties under varying conditions, such as changes in the film's thickness. Students might be asked to plot the thermal conductivities lengthwise and spanwise to understand how the film's dimensions affect its ability to transfer heat. This is critical for the design of efficient cooling strategies in microelectronic devices.
Moreover, as film thickness decreases, quantum effects can become more pronounced, leading to a deviation from bulk material properties—another important consideration for advanced technology and materials science courses. Through such exercises, students gain insights into the practical application of concepts like thermal conductivity in cutting-edge technologies.

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Most popular questions from this chapter

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.

The wall of a spherical tank of \(1-m\) diameter contains an exothermic chemical reaction and is at \(200^{\circ} \mathrm{C}\) when the ambient air temperature is \(25^{\circ} \mathrm{C}\). What thickness of urethane foam is required to reduce the exterior temperature to \(40^{\circ} \mathrm{C}\), assuming the convection coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) for both situations? What is the percentage reduction in heat rate achieved by using the insulation?

One method that is used to grow nanowires (nanotubes with solid cores) is to initially deposit a small droplet of a liquid catalyst onto a flat surface. The surface and catalyst are heated and simultaneously exposed to a higher- temperature, low-pressure gas that contains a mixture of chemical species from which the nanowire is to be formed. The catalytic liquid slowly absorbs the species from the gas through its top surface and converts these to a solid material that is deposited onto the underlying liquid-solid interface, resulting in construction of the nanowire. The liquid catalyst remains suspended at the tip of the nanowire. Consider the growth of a 15 -nm-diameter silicon carbide nanowire onto a silicon carbide surface. The surface is maintained at a temperature of \(T_{s}=2400 \mathrm{~K}\), and the particular liquid catalyst that is used must be maintained in the range \(2400 \mathrm{~K} \leq T_{c} \leq 3000 \mathrm{~K}\) to perform its function. Determine the maximum length of a nanowire that may be grown for conditions characterized by \(h=10^{5} \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(T_{\infty}=8000 \mathrm{~K}\). Assume properties of the nanowire are the same as for bulk silicon carbide.

A spherical tank of \(3-\mathrm{m}\) diameter contains a liquifiedpetroleum gas at \(-60^{\circ} \mathrm{C}\). Insulation with a thermal conductivity of \(0.06 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and thickness \(250 \mathrm{~mm}\) is applied to the tank to reduce the heat gain. (a) Determine the radial position in the insulation layer at which the temperature is \(0^{\circ} \mathrm{C}\) when the ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection coefficient on the outer surface is \(6 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (b) If the insulation is pervious to moisture from the atmospheric air, what conclusions can you reach about the formation of ice in the insulation? What effect will ice formation have on heat gain to the LP gas? How could this situation be avoided?

Determine the percentage increase in heat transfer associated with attaching aluminum fins of rectangular profile to a plane wall. The fins are \(50 \mathrm{~mm}\) long, \(0.5 \mathrm{~mm}\) thick, and are equally spaced at a distance of \(4 \mathrm{~mm}\) ( 250 fins \(/ \mathrm{m})\). The convection coefficient associated with the bare wall is \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), while that resulting from attachment of the fins is \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

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