Thermal resistance

From Academic Kids

Thermal resistance has two different meanings:

1) the temperature difference across the structure when a unit of heat energy flows through it in unit time or
2) the temperature difference across a unit area of a material of unit thickness when a unit of heat energy flows through it in unit time.

It is the reciprocal of thermal conductance.

The SI units of thermal resistance are:

  • K/W for the first meaning
  • K·m²/W for the second meaning (see closer two different definitions general/buildings in thermal conductance).

In textiles, a tog value may be quoted instead. In the building trade in the USA, the traditional non-metric R-value is sometimes used instead, with units ft² °F·h/Btu.

Worked example from electronic engineering

The thermal resistance of materials is of great interest to electronic engineers, because most electrical components generate heat and need to be cooled. Some electronic components malfunction when they overheat, while others are permanently damaged.

Consider a component such as a silicon transistor that is bolted to the metal frame of a piece of equipment. The transistor's manufacturer will specify parameters in the datasheet called the thermal resistance from junction to case (symbol: <math>R_{\theta JC}<math>), and the maximum allowable temperature of the semiconductor junction (symbol: <math>T_{JMAX}<math>). The specification for the design should include a maximum temperature at which the circuit should function correctly. Finally, the designer should consider how the heat from the transistor will escape to the environment: this might be by convection into the air, with or without the aid of a heat sink, or by conduction through the printed circuit board. For simplicity, let us assume that the designer decides to bolt the transistor to a metal surface (or heat sink) that is guaranteed to be less than <math>\Delta T_{HS}<math> above the ambient temperature.

Given all this information, the designer can construct a model of the heat flow from the semiconductor junction, where the heat is generated, to the outside world. In our example, the heat has to flow from the junction to the case of the transistor, then from the case to the metalwork. We do not need to consider where the heat goes after that, because we are told that the metalwork will conduct heat fast enough to keep the temperature less than <math>\Delta T_{HS}<math> above ambient: this is all we need to know.

Suppose the engineer wishes to know how much power he can put into the transistor before it overheats. The calculations are as follows.

Total thermal resistance from junction to ambient = <math>R_{\theta JC}+R_{\theta B}<math>

where <math>R_{\theta B}<math> is the thermal resistance of the bond between the transistor's case and the metalwork. This figure depends on the nature of the bond - for example, a thermal bonding pad or thermal transfer grease might be used to reduce the thermal resistance.

Maximum temperature drop from junction to ambient = <math>T_{JMAX}-(T_{AMB}+\Delta T_{HS})<math>.

We use the general principle that the temperature drop <math>\Delta T<math> across a given thermal resistance <math>R_{\theta}<math> with a given heat flow <math>Q<math> through it is:

<math>\Delta T = Q \times R_{\theta}\,<math>.

Substituting our own symbols into this formula gives:

<math>T_{JMAX}-(T_{AMB}+\Delta T_{HS})=Q_{MAX} \times (R_{\theta JC}+R_{\theta B})\,<math>,

and, rearranging,

  Q_{MAX} = 
    { T_{JMAX}-(T_{AMB}+\Delta T_{HS}) } \over { R_{\theta JC}+R_{\theta B} }


The designer now knows <math>Q_{MAX}<math>, the maximum power that the transistor can be allowed to dissipate, so he can design the circuit to limit the temperature of the transistor to a safe level.

Let us plug in some sample numbers:

<math>T_{JMAX} = 125 \ ^{\circ}\mbox{C}<math> (typical for a silicon transistor)
<math>T_{AMB} = 70 \ ^{\circ}\mbox{C}<math> (a typical specification for commercial equipment)
<math>\Delta T_{HS} = 10 \ ^{\circ}\mbox{C}<math> (arbitrary figure)
<math>R_{\theta JC}<math> = 1.5 K/W (for a typical TO-220 package)
<math>R_{\theta B}<math> = 0.1 K/W (a typical value for an elastomer heat-transfer pad for a TO-220 package)

The result is then:

<math>Q = {{125-(70+10)} \over {1.5+0.1}} <math> watts
<math>= 28.125 <math> watts

This means that the transistor can dissipate about 28 watts before it overheats. A cautious designer would operate the transistor at a lower power level to increase its reliability.

This method can be generalised to include any number of layers of heat-conducting materials, simply by adding together the thermal resistances of the layers and the temperature drops across theämpöresistanssi


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