91Ó°ÊÓ

Consider a surface-mount type transistor on a circuit board whose temperature is muintained at \(35^{\circ} \mathrm{C}\). Air at \(207 \mathrm{C}\) flows over the upper surface of dimensions \(4 \mathrm{~mm}\) by \(8 \mathrm{~mm}\) with a convection coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Three wire leads, each of cross section \(1 \mathrm{~mm}\) by \(0.25 \mathrm{~mm}\) and length 4 nim, conduct heat from the case to the circuit board. The gap between the case and the board is \(0.2 \mathrm{~mm}\). (a) Assuming the case is isothermal and neglecting rudiation, eximate the case temperature uhen \(150 \mathrm{~mW}\) is dissipoted by the thansistor and (i) stagnant air or (ii) a conductive paste fills the gap. The thermal condoctivitios of the wire leadk, air, and cunductive paste are 25, \(0.0263\), and \(0.12\) W/m - \(\mathrm{K}\), respectively. (b) Using the conductive paste to fill the gap, we wish to determine the extent to which increased heat dissipation may be accommodated, subject to the constraint that the case lemperature not exceed \(40^{\circ} \mathrm{C}\). Options include increasing the air speed to achieve a lager convection coefficient \(h\) and/or changing the lead wire material to one of larger thermal conductivity. Independently considering leads fabricated from materials with thermal conductivities of 200 and \(400 \mathrm{~W} / \mathrm{m}+\mathrm{K}\), compute and plot the maximum allowable hean dissipation for variations in \(h\) over the range \(50 \leq h \leq 250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

Step by step solution

01

Calculate Thermal Resistances

First, calculate the thermal resistance due to convection: \[ R_{conv} = \frac{1}{hA} \] where the convection area \( A = 4 \times 8 = 32 \times 10^{-6} \text{ m}^2 \) and \( h = 50 \text{ W/m}^2 \cdot \text{K} \). \[ R_{conv} = \frac{1}{50 \times 32 \times 10^{-6}} = 625 \text{ K/W} \] Next, calculate the conduction resistance for air and paste (gap) using \[ R_{cond} = \frac{L}{kA} \] where \( A = 4 \times 8 = 32 \times 10^{-6} \text{ m}^2 \) and \( L = 0.2 \times 10^{-3} \text{ m} \). \[ R_{air} = \frac{0.2 \times 10^{-3}}{0.0263 \times 32 \times 10^{-6}} = 236.16 \text{ K/W} \] \[ R_{paste} = \frac{0.2 \times 10^{-3}}{0.12 \times 32 \times 10^{-6}} = 52.08 \text{ K/W} \] Finally, calculate the lead thermal resistance with \( A = 0.25 \times 1 = 0.25 \text{ mm}^2 = 0.25 \times 10^{-6} \text{ m}^2 \) \[ R_{lead} = \frac{4 \times 10^{-3}}{25 \times 0.25 \times 10^{-6} \times 3} = 213.33 \text{ K/W} \] (there are three leads, hence the factor of 3.)

02

Calculate Case Temperature for Scenario (i)

For stagnant air filling the gap, the total thermal resistance, \( R_{total} = R_{conv} + R_{air} + R_{lead} \), is: \[ R_{total} = 625 + 236.16 + 213.33 = 1074.49 \text{ K/W} \] The power dissipation \( P = 150 \text{ mW} = 0.15 \text{ W} \). The temperature rise \( \Delta T = P \times R_{total} \), \[ \Delta T = 0.15 \times 1074.49 = 161.17 \text{ K} \] The case temperature \( T_{case} \) is then \[ T_{case} = T_{ambient} + \Delta T = 35 + 161.17 = 196.17^{\circ}\text{C} \]

03

Calculate Case Temperature for Scenario (ii)

For conductive paste in the gap, the total thermal resistance becomes \[ R_{total} = R_{conv} + R_{paste} + R_{lead} = 625 + 52.08 + 213.33 = 890.41 \text{ K/W} \] Thus, the temperature rise is \[ \Delta T = 0.15 \times 890.41 = 133.56 \text{ K} \] So, the case temperature for this configuration is \[ T_{case} = 35 + 133.56 = 168.56^{\circ}\text{C} \]

04

Analyze Heat Dissipation with Conductive Paste

Determine the maximum power \( P \) with conductive paste such that \( T_{case} \leq 40^{\circ} \). We know \[ 40 = 35 + P \times 890.41 \Rightarrow P = \frac{5}{890.41} = 0.00561 \text{ W} \] Recalculate to allow more power dissipation by adjusting convection coefficient \( h \) and wire material with increased thermal conductivities. Each wire's thermal conductivity \( k = 200 \text{ or } 400 \text{ W/m}K \) is considered, and the impact on \( R_{lead} \) and subsequently \( R_{total} \) needs to be reassessed and plotted.

05

Iteratively Adjust and Compute Maximum Power Dissipation

As \( h \) varies, so does \( R_{conv} \), iterating from \( h = 50 \) to \( h = 250 \). Recompute \( R_{total} \) for each \( h \) and maximum \( k \). Each calculation helps set up data points to plot maximum permissible power dissipation while ensuring the case temperature remains below \( 40^{\circ}\text{C} \). Outcomes are graphed showing combinations of \( h \) and improved lead material conductivity.

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Key Concepts

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

Heat Transfer

Heat transfer is an essential process in which thermal energy moves from one place to another. In our given exercise, heat moves from a transistor into the surrounding air. This specific process involves a few mechanisms which include conduction, convection, and sometimes radiation. However, in this scenario, radiation is neglected for simplification.

Heat transfer helps maintain desired temperatures of electronic components like transistors. When these devices work, they tend to heat up. Without effective heat management, their performance can be compromised. It is crucial especially in circuits to continuously disperse heat to protect sensitive functions.

To understand how heat flows, we consider:

• Conduction: Heat moves through a solid material, like the wire leads or a conductive paste.
• Convection: Heat transfers between a solid and a fluid, such as air.

Knowing how these elements interact helps engineers design systems that efficiently handle heat, keeping temperatures within operational limits.

Thermal Resistance

In the study of heat transfer, thermal resistance plays a vital role. It represents how much an object resists the flow of heat. Imagine it like an electrical resistance, but for heat. It's a crucial factor when evaluating how warm or cool a system will get based on its design.

Thermal resistance can be calculated for both conduction and convection processes. The formula for conduction consists of the material's length, thermal conductivity, and area, whereas convection depends on the heat transfer coefficient and the contact area.

In our exercise:

• Conduction Resistance (R_cond): It incorporates the gap filled with air or conductive paste, where both have different thermal conductivities.
• Convection Resistance (R_conv): It incorporates the air movement over the surface of the transistor, depending on the convection coefficient.

Reducing the total thermal resistance in a system can lower the temperature of a component, resulting in better performance and increased lifespan. Adjusting factors like materials and surface configuration can help achieve this.

Convection Coefficient

The convection coefficient, denoted by the symbol "h," describes how effectively heat can be transferred between a solid surface and a fluid - in this case, air.

The convection coefficient is vital when calculating the convective heat transfer resistance. It is influenced by the speed and properties of the fluid, such as air.

In the exercise, as air flows over the surface of the transistor:

• Higher airflow rates increase the convection coefficient, enhancing heat dissipation.
• This relationship allows for increased power dissipation without overheating.
• Engineers often adjust the convection coefficient by controlling air flow dynamics.

Thus, by increasing airspeed and improving convective capabilities, the system can handle more thermal stress and maintain components at usable temperatures.

Conductive Paste

Conductive paste can bridge the gap between surfaces, providing a path for heat transfer where an air interface exists. In electronics, this material is extensively used to enhance heat flow efficiency.

In this scenario, the transistor uses conductive paste to more effectively transfer heat than air:

• The paste has higher thermal conductivity compared to air, reducing the overall thermal resistance.
• Introducing it to the gap decreases the temperature rise in the system significantly.
• It ensures better contact between surfaces, promoting efficient heat transfer.

The adaptation to use conductive paste results in lowering the case temperature which is crucial to keeping electronic components safe from heat damage. By using conductive paste, systems are more resilient to heat, letting them operate safely under higher power conditions.