Heat transfer by conduction occurs between two solid masses in contact when they are at different temperatures, and also between different parts of a single mass if a temperature differential exists within it.
Two main mechanisms are involved, atomic vibrations and conduction by free electrons. Typically using free electrons to transfer kinetic energy is substantially the more effective of the two mechanisms, which explains why metals are much better conductors of heat than insulators such as ceramics and polymers.
Provided that we ignore the contribution from the other methods of heat transfer which will be discussed later, the conduction of heat is very similar to the conduction of electricity. The driving force becomes temperature difference rather than voltage, and heat flow instead of current, but otherwise the concepts of resistivity, resistance, conductivity and conductance are exactly the same.
The formula for calculating the heat transferred by conduction is:
Q is the quantity of heat transferred
S the area of the surface through which the heat is transferred
T the difference of temperature across the thickness of the material
d the thickness of material through which the heat has to travel
The specific conductivity k is the quantity of heat that can be transferred during unit time across a section of material with unit surface area and unit thickness, with a temperature difference of 1K. [Note that some writers use the symbol l (Greek letter lambda) for this parameter]
Working in SI units for all these parameters (watt; meter, square meter; kelvin), we can see that specific conductivity should be expressed in W/(m.K), although many non-SI units1 are also used.
1 If you have to make conversions, there is a wonderful collection of conversion calculators at http://directory.google.com/.../Online_Calculator.
Note from the equation that the heat transmitted increases linearly with k, S and T, and inversely with d. In other words, given the same material, thin sheets conduct heat better than thick blocks.
If we apply the formula to a piece of given size (so that T and d are constant), we can define conductance (reciprocal resistance) as:
so that the conduction formula becomes:
, whence T = Q x R, which is analogous to Ohm’s Law.
The useful concept of thermal resistance2 will be explored further in Design for Test and the Physical Environment. For the time being, just be aware that, when different materials are stuck together, you can calculate the thermal resistance of each and add them together to deduce the overall resistance.
2 Thermal resistance has units of K/W, not to be confused with kW!
The result of higher levels of silicon integration, combined with higher operating speeds, is that more heat is generated. As a result, substrates come under pressure, not only because they have to withstand higher temperatures, but also because they must be able to conduct heat away in order to keep the circuitry at a safe operating temperature.
As can be seen from Table 1, the only useful materials for heat removal are metal and ceramics. Ceramics such as alumina, beryllia and aluminium nitride are used as interconnection substrates in their own right, but have size and complexity limitations: for most purposes, reinforced polymers will continue to form the basis of most PCBs.
|material||thermal conductivityW.m−1 K−1|
3 Values of thermal conductivity vary widely between references, and any calculations you make will not be very accurate, hence our use of few decimal places in the table.
Metals are good conductors of heat, but also conduct electricity and need an insulating layer in order to make them into a useful circuit substrate. However, even a thin layer of insulation greatly reduces the thermal conductivity: for example, only 50µm of an epoxy coating reduces4 the conductivity of a 1.6mm thick sheet of copper from 390 W/(m.°C) to 5 – 8 W/(m.°C).
It is clear, therefore, that if significant amounts of heat have to be removed from a polymer-based circuit, then the best way is to cut access holes through the polymer, to get the heat source into direct contact with the metal heat sink. It is better to package integrated circuits with very high dissipation in a thermally efficient module mounted on a conventional board, and with a dissipating surface on the top of the package.
4 If you would like to check this, use the concept of adding together the thermal resistances which we discussed earlier.
Metal surfaces are actually quite rough at a microscopic level. This means that the thermal conduction between two solids is much less than you might expect, because they make contact only where the high spots on their surfaces touch, as illustrated in Figure 1.
When components are screwed to a heat sink, the clamping force applied tends to flatten the high spots and create additional contact areas, which reduces the thermal resistance of the interface.
However, much of the energy transfer still takes place across whatever medium fills the valleys between the many peaks. This is usually air, an excellent thermal insulator, which is why we try to find some way of ‘bridging the gap’, to improve intimacy of contact and provide a better conduction path between the two surfaces.
The earliest method was to use silicone greases (sometimes in combination with a mica washer). However, light silicone oils tend to migrate elsewhere in the circuit, to the detriment of both thermal effectiveness and solderability.
Current practice is to use either a thermally conductive gel to fill in the valleys, or thin shims of elastomer to provide both isolation and heat transfer. These shims are likely to have a carrier such as woven fibre-glass and be loaded with ceramic powder of high thermal conductivity.
Silicone is the most common elastomer used, and has the advantage that it exhibits cold flow, which excludes air from the interface as the elastomer conforms to the mating surfaces. Given that mechanical attachment is usual, to ensure that the joint has minimal interfacial resistance, care should be taken to ensure that the shim is not punctured by burrs on the parts.
The power semiconductor package on your circuit contains a silicon chip mounted on a 1.5 mm thick steel base of area 1 cm2. The package is insulated from its heat sink by a silicone rubber gasket that is 0.25 mm thick. Use the values of thermal conductivity for the materials given in Table 1 to calculate the thermal resistance between chip and heat sink.
What precautions should you take to ensure that this theoretical value is achieved as nearly as possible?
Apart from conduction, there are three main methods by which heat may be transferred from a hot body to a cooler environment.
Convection heat transfer occurs when a fluid (such as air, nitrogen, or water) passes over a solid (such as an SMT assembly). Heating or cooling requires contact between flow and the solid, and only the layer of the flow that is in contact with the part actually transfers heat.
Radiation: infra red (IR) radiation is emitted by all bodies at all temperatures. The amount of energy emitted, and the wavelength of that emission, both depend on the temperature of the object. IR energy is also absorbed by bodies, so that net transfer of energy occurs when two bodies at different temperatures are in sight of each other. The most obvious example of IR energy transfer is the heating of the earth by the sun. IR is a non contact heat transfer process.
Latent heat of evaporation is the heat which is absorbed by a fluid when it turns into vapour (for example, when a boiling kettle produces steam) and can be released when the vapour recondenses.
Of these, convection is commonly used for cooling, but most commonly in the form of ‘forced convection’, where a fan is used to increase the rate of transfer. Radiation plays only a minimal part in most cases, and latent heat evaporation appears only in the little-used heat pipe. However, both methods of heat transfer are used in soldering.
Three thermal fundamentals should be borne in mind:
With conduction and convection, the rate of heat transfer is a linear function of the difference in temperature; with infra red as the source temperature increases, the heat transfer output increases exponentially to the fourth power.
5 This is a very simple consequence of a very profound and fundamental physical principle. For an introduction to the subject, the first few pages of http://www.entropylaw.com are recommended.
When we are transferring heat, one of the values we need to know is the ‘heat capacity’ of the items involved. This term is used for the amount of heat needed to raise the temperature of an object by 1K. The ‘specific heat capacity’, usual symbol c, is the amount of heat needed for 1kg of material, derived from the equation:
and has the units J kg−1 K−1. Some typical values for specific heat capacity are given in Table 2. You will notice that, because the equation involves mass rather than volume, materials which are less dense (like aluminium) have higher specific heats than dense materials such as copper 6. When you are carrying out calculations, remember that parts of similar size may have substantially different weights, depending on the material used.
6 If you express specific heat as energy per mole, that st for the same number of atoms, the figures are very similar. For more details, search under ‘Law of Dulong and Petit’.
|material||specific heat capacityJ kg−1 K−1|
7 Thermal information on materials is hard to get, and often ill-defined. All figures quoted should be treated with caution.
Metals have much smaller heat capacities than plastics, so require less heat to reach a particular temperature than do plastics; they also conduct heat more readily. These are two good reasons why kitchen equipment traditionally uses plastic handles instead of metal ones.
What are the heat transfer processes that create solder joints in a reflow oven?
Your surface mount design contains both small resistor chips (made of alumina ceramic) and some large copper heat sinks. When both are exposed to forced convection in an oven, what differences would you expect to see in the rate at which they increase in temperature?