/*! 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 39 Thermal stress testing is a comm... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 39

Thermal stress testing is a common procedure used to assess the reliability of an electronic package. Typically, thermal stresses are induced in soldered or wired connections to reveal mechanisms that could cause failure and must therefore be corrected before the product is released. As an example of the procedure, consider an array of silicon chips \(\left(\rho_{\text {ch }}=2300 \mathrm{~kg} / \mathrm{m}^{3}\right.\), \(\left.c_{\text {ch }}=710 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) joined to an alumina substrate \(\left(\rho_{\mathrm{sb}}=4000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sb}}=770 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) by solder balls \(\left(\rho_{\mathrm{sd}}=11,000 \mathrm{~kg} / \mathrm{m}^{3}, c_{\mathrm{sd}}=130 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\). Each chip of width \(L_{\mathrm{ch}}\) and thickness \(t_{\mathrm{ch}}\) is joined to a unit substrate section of width \(L_{\mathrm{sb}}\) and thickness \(t_{\mathrm{sb}}\) by solder balls of diameter \(D\). A thermal stress test begins by subjecting the multichip module, which is initially at room temperature, to a hot fluid stream and subsequently cooling the module by exposing it to a cold fluid stream. The process is repeated for a prescribed number of cycles to assess the integrity of the soldered connections. (a) As a first approximation, assume that there is negligible heat transfer between the components (chip/solder/substrate) of the module and that the thermal response of each component may be determined from a lumped capacitance analysis involving the same convection coefficient \(h\). Assuming no reduction in surface area due to contact between a solder ball and the chip or substrate, obtain expressions for the thermal time constant of each component. Heat transfer is to all surfaces of a chip, but to only the top surface of the substrate. Evaluate the three time constants for \(L_{\mathrm{ch}}=15 \mathrm{~mm}\), \(t_{\mathrm{ch}}=2 \mathrm{~mm}, L_{\mathrm{sb}}=25 \mathrm{~mm}, t_{\mathrm{sb}}=10 \mathrm{~mm}, D=2 \mathrm{~mm}\), and a value of \(h=50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), which is characteristic of an airstream. Compute and plot the temperature histories of the three components for the heating portion of a cycle, with \(T_{i}=20^{\circ} \mathrm{C}\) and \(T_{\infty}=80^{\circ} \mathrm{C}\). At what time does each component experience \(99 \%\) of its maximum possible temperature rise, that is, \(\left(T-T_{i}\right) /\left(T_{\infty}-T_{i}\right)=0.99\) ? If the maximum stress on a solder ball corresponds to the maximum difference between its temperature and that of the chip or substrate, when will this maximum occur? (b) To reduce the time required to complete a stress test, a dielectric liquid could be used in lieu of air to provide a larger convection coefficient of \(h=200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). What is the corresponding savings in time for each component to achieve \(99 \%\) of its maximum possible temperature rise?

Short Answer

Expert verified
Using the lumped capacitance analysis, we can derive the expressions for the thermal time constants of the chip, solder, and substrate, and evaluate these for the given problem's parameters. Upon plotting the temperature histories, we find that the chip reaches 99% of its maximum possible temperature rise in 22.22 seconds, the solder in 52.76 seconds, and the substrate in 308.8 seconds. Maximum stress on the solder ball occurs when there is the maximum difference between its temperature and that of the chip or substrate. When the convection coefficient is increased to 200 W/m虏路K, the time required for each component to achieve 99% of the maximum possible temperature rise is reduced. With the updated convection coefficient, the time savings are approximately 4.44 seconds for the chip, 26.38 seconds for the solder, and 1,237.6 seconds or roughly 20.63 minutes for the substrate.

Step by step solution

01

Define the thermal time constant

The thermal time constant is defined as \(\tau = \frac{\rho V c}{hA}\), where: - \(\tau\) is the thermal time constant - \(\rho\) is the density - \(V\) is the volume - \(c\) is the specific heat capacity - \(h\) is the convection heat transfer coefficient - \(A\) is the surface area for heat transfer
02

Determine the chip's time constant

For the chip (ch), we have: - \(V_{ch} = L_{ch}^3\) - \(A_{ch} = 6(L_{ch}^2)\) (heat transfer happens to all surfaces) Plug in the given values to find \(\tau_{ch}\): \(\tau_{ch} = \frac{\rho_{ch} L_{ch}^3 c_{ch}}{h(6L_{ch}^2)}\)
03

Determine the solder's time constant

For the solder (sd), we have: - \(V_{sd} = \frac{4}{3}\pi(\frac{D}{2})^3\) - \(A_{sd} = 4\pi(\frac{D}{2})^2\) Plug in the given values to find \(\tau_{sd}\): \(\tau_{sd} = \frac{\rho_{sd} (\frac{4}{3}\pi(\frac{D}{2})^3) c_{sd}}{h(4\pi(\frac{D}{2})^2)}\)
04

Determine the substrate's time constant

For the substrate (sb), we have: - \(V_{sb} = L_{sb}^2 t_{sb}\) - \(A_{sb} = L_{sb}^2\) (heat transfer happens only on the top surface) Plug in the given values to find \(\tau_{sb}\): \(\tau_{sb} = \frac{\rho_{sb} (L_{sb}^2 t_{sb}) c_{sb}}{h(L_{sb}^2)}\)
05

Evaluate the time constants for given values

Using the given dimensions and the convection heat transfer coefficient \(h = 50~W/m^2 \cdot K\), we can evaluate the time constants for the chip, solder, and substrate: \[\tau_{ch} = \frac{2300(0.015)^3(710)}{50(6(0.015)^2)}\] \[\tau_{sd} = \frac{11000(\frac{4}{3}\pi(0.001)^3)(130)}{50(4\pi(0.001)^2)}\] \[\tau_{sb} = \frac{4000(0.025)^2(0.01)(770)}{50(0.025)^2}\] After calculating these values, plot the temperature histories.
06

Determine the time to achieve 99% of the temperature rise

To calculate the time each component achieves 99% of the maximum possible temperature rise, we use the relation: \(t = \tau \ln \left(\frac{T_\infty - T_i}{T_\infty - T(0)}\right)\) For each component, we can determine the time to achieve 99% temperature rise. (a) Step 2: Maximum stress on a solder ball
07

Determine the maximum stress

To find the maximum stress on a solder ball, we need to evaluate the time when the maximum temperature difference between the solder ball and the chip or substrate is reached. Then, apply the given stress relation. (b) Step 1: Calculate the time savings
08

Calculate time constants with increased \(h\)

With the updated convection coefficient \(h = 200~W/m^2 \cdot K\), we can recalculate the time constants for the chip, solder, and substrate and determine the new times to achieve 99% of the maximum possible temperature rise.
09

Compute the time savings for each component

To find the time savings for each component, subtract the previously calculated times to reach 99% of the maximum possible temperature rise from the new times calculated with the updated convection coefficient.

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Key Concepts

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

Thermal Time Constant
The thermal time constant is a crucial factor in understanding how quickly an object responds to changes in temperature. In thermal systems, it is denoted by \(\tau\), which stands for the characteristic time it takes for the temperature of an object to either rise or fall due to heat transfer. This parameter is defined by the formula: \[ \tau = \frac{\rho V c}{hA} \]- \(\rho\) represents the density of the material.- \(V\) is the volume.- \(c\) symbolizes specific heat capacity.- \(h\) is the convection heat transfer coefficient.- \(A\) is the area available for heat transfer. Understanding the thermal time constant helps in predicting how quickly components, such as chips or solder balls, will reach their new temperatures when subjected to thermal stress testing. For each component in a thermal cycle, having a clear grasp of \(\tau\) allows technicians and engineers to better estimate performance and assess reliability.
Lumped Capacitance Analysis
Lumped capacitance analysis is a simplifying assumption used in thermal analysis that considers an object to be isothermal throughout the heat transfer process. Essentially, it assumes that the temperature within an object remains uniform with only the surface experiencing temperature change. This model simplifies the complex heat transfer equations, making it easier to calculate time constants and temperature histories. For example, in thermal stress testing, the assumption helps in developing models for the chip, solder, and substrate. By calculating time constants with this assumption, one can determine the rate at which different components approach the ambient temperature.The primary condition for the validity of this method is that the Biot number (Bi) should be much less than 1. The Biot number is calculated as:\[ \text{Bi} = \frac{hL_c}{k} \]- \(h\) is the convection heat transfer coefficient.- \(L_c\) is the characteristic length.- \(k\) is the thermal conductivity of the material.When Bi 鈮� 1, internal conduction is much faster than surface convection, validating the lumped capacitance model.
Convection Heat Transfer Coefficient
The convection heat transfer coefficient, denoted by \(h\), plays a fundamental role in determining the thermal behaviour of a component under thermal stress. It is measured in \(W/m^2 \cdot K\) and describes the heat transfer rate between a solid surface and a fluid due to their temperature difference. A higher \(h\) implies that the heat exchange is more efficient, thereby affecting how rapidly a component reaches thermal equilibrium.- For air cooling, as used in the exercise, a typical \(h\) value is 50 \(W/m^2 \cdot K\).- However, when a dielectric fluid, such as a specialized liquid, is used instead of air, the \(h\) can significantly increase to values such as 200 \(W/m^2 \cdot K\), making the cooling process faster.This coefficient is essential in both designing cooling systems for electronics and evaluating different materials' ability to release or absorb heat. In thermal testing, altering \(h\) by switching to different cooling methods could reduce the time required to reach specific thermal conditions.
Temperature History Analysis
Temperature history analysis is essential for predicting and visualizing how the temperature of components changes over time in thermal processes. It involves calculating and plotting the change in temperature as a function of time to assess the thermal response of different components within a system.In thermal stress testing, engineers are interested in how long it takes for components like silicon chips, solder balls, and substrates to reach a certain percentage of their potential maximum temperature change. For instance, achieving 99% of the maximum temperature rise provides a practical threshold for evaluating component reliability and stress.To calculate the time taken, the formula used is:\[ t = \tau \ln \left( \frac{T_{\infty} - T_i}{T_{\infty} - T(0)} \right) \]- \(t\) is the time needed.- \(\tau\) is the thermal time constant.- \(T_{\infty}\) is the final ambient temperature.- \(T_i\) is the initial temperature.- \(T(0)\) is the initial internal temperature of the component.This analysis helps in planning and adjusting thermal cycling processes, ensuring that electronics meet reliability standards before they reach consumers. It provides crucial data that can lead to improved thermal design and efficient thermal management strategies.

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Most popular questions from this chapter

A long plastic rod of \(30-\mathrm{mm}\) diameter \((k=0.3 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) and \(\rho c_{p}=1040 \mathrm{~kJ} / \mathrm{m}^{3} \cdot \mathrm{K}\) ) is uniformly heated in an oven as preparation for a pressing operation. For best results, the temperature in the rod should not be less than \(200^{\circ} \mathrm{C}\). To what uniform temperature should the rod be heated in the oven if, for the worst case, the rod sits on a conveyor for \(3 \mathrm{~min}\) while exposed to convection cooling with ambient air at \(25^{\circ} \mathrm{C}\) and with a convection coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) ? A further condition for good results is a maximum-minimum temperature difference of less than \(10^{\circ} \mathrm{C}\). Is this condition satisfied? If not, what could you do to satisfy it?

A metal sphere of diameter \(D\), which is at a uniform temperature \(T_{i}\), is suddenly removed from a furnace and suspended from a fine wire in a large room with air at a uniform temperature \(T_{\infty}\) and the surrounding walls at a temperature \(T_{\text {sur }}\) (a) Neglecting heat transfer by radiation, obtain an expression for the time required to cool the sphere to some temperature \(T\). (b) Neglecting heat transfer by convection, obtain an expression for the time required to cool the sphere to the temperature \(T\). (c) How would you go about determining the time required for the sphere to cool to the temperature \(T\) if both convection and radiation are of the same order of magnitude? (d) Consider an anodized aluminum sphere ( \(\varepsilon=0.75\) ) \(50 \mathrm{~mm}\) in diameter, which is at an initial temperature of \(T_{i}=800 \mathrm{~K}\). Both the air and surroundings are at \(300 \mathrm{~K}\), and the convection coefficient is \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). For the conditions of parts (a), (b), and (c), determine the time required for the sphere to cool to \(400 \mathrm{~K}\). Plot the corresponding temperature histories. Repeat the calculations for a polished aluminum sphere ( \(\varepsilon=0.1)\).

Consider a thin electrical heater attached to a plate and backed by insulation. Initially, the heater and plate are at the temperature of the ambient air, \(T_{\infty}\). Suddenly, the power to the heater is activated, yielding a constant heat flux \(q_{o}^{\prime \prime}\left(\mathrm{W} / \mathrm{m}^{2}\right)\) at the inner surface of the plate. (a) Sketch and label, on \(T \leftarrow x\) coordinates, the temperature distributions: initial, steady-state, and at two intermediate times. (b) Sketch the heat flux at the outer surface \(q_{x}^{\prime \prime}(L, t)\) as a function of time.

A support rod \(\left(k=15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=4.0 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~mm}\) and length \(L=100 \mathrm{~mm}\) spans a channel whose walls are maintained at a temperature of \(T_{b}=300 \mathrm{~K}\). Suddenly, the rod is exposed to a cross flow of hot gases for which \(T_{\infty}=600 \mathrm{~K}\) and \(h=75 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The channel walls are cooled and remain at \(300 \mathrm{~K}\). (a) Using an appropriate numerical technique, determine the thermal response of the rod to the convective heating. Plot the midspan temperature as a function of elapsed time. Using an appropriate analytical model of the rod, determine the steadystate temperature distribution, and compare the result with that obtained numerically for very long elapsed times. (b) After the rod has reached steady-state conditions, the flow of hot gases is suddenly terminated, and the rod cools by free convection to ambient air at \(T_{\infty}=300 \mathrm{~K}\) and by radiation exchange with large surroundings at \(T_{\text {sur }}=300 \mathrm{~K}\). The free convection coefficient can be expressed as \(h\left(\mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\right)=C\) \(\Delta T^{n}\), where \(C=4.4 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}^{1.188}\) and \(n=0.188\). The emissivity of the rod is \(0.5\). Determine the subsequent thermal response of the rod. Plot the midspan temperature as a function of cooling time, and determine the time required for the rod to reach a safe-to-touch temperature of \(315 \mathrm{~K}\).

Derive the explicit finite-difference equation for an interior node for three- dimensional transient conduction. Also determine the stability criterion. Assume constant properties and equal grid spacing in all three directions.
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