Thermal resistance

The diagram shows an equivalent thermal circuit for a semiconductor device with a heat sink:
${\displaystyle Q}$ is the power dissipated by the device.
${\displaystyle T_{\rm {J}}}$ is the junction temperature in the device.
${\displaystyle T_{\rm {C}}}$ is the temperature at its case.
${\displaystyle T_{\rm {H}}}$ is the temperature where the heat sink is attached.
${\displaystyle T_{\rm {amb}}}$ is the ambient air temperature.
${\displaystyle R_{\theta {\rm {JC}}}}$ is the device's absolute thermal resistance from junction to case.
${\displaystyle R_{\theta {\rm {CH}}}}$ is the absolute thermal resistance from the case to the heatsink.
${\displaystyle R_{\theta {\rm {HA}}}}$ is the absolute thermal resistance of the heat sink.

The heat flow can be modelled by analogy to an electrical circuit where heat flow is represented by current, temperatures are represented by voltages, heat sources are represented by constant current sources, absolute thermal resistances are represented by resistors and thermal capacitances by capacitors.

The diagram shows an equivalent thermal circuit for a semiconductor device with a heat sink.

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 absolute thermal resistance from junction to case (symbol: ${\displaystyle R_{\theta {\rm {JC}}}}$), and the maximum allowable temperature of the semiconductor junction (symbol: ${\displaystyle T_{J{\rm {max}}}}$). 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 ${\displaystyle \Delta T_{\rm {HS}}}$ above the ambient temperature. Note: T_HS appears to be undefined.

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 ${\displaystyle \Delta T_{\rm {HS}}}$ above ambient: this is all we need to know.

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

Total absolute thermal resistance from junction to ambient = ${\displaystyle R_{\theta {\rm {JC}}}+R_{\theta {\rm {B}}}}$

where ${\displaystyle R_{\theta {\rm {B}}}}$ is the absolute 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 absolute thermal resistance.

Maximum temperature drop from junction to ambient = ${\displaystyle T_{J{\rm {max}}}-(T_{\rm {amb}}+\Delta T_{\rm {HS}})}$.

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

${\displaystyle \Delta T=Q\times R_{\theta }\,}$.

Substituting our own symbols into this formula gives:

${\displaystyle T_{J{\rm {max}}}-(T_{\rm {amb}}+\Delta T_{\rm {HS}})=Q_{\rm {max}}\times (R_{\theta {\rm {JC}}}+R_{\theta {\rm {B}}}+R_{\theta {\rm {HA}}})\,}$,

and, rearranging,

${\displaystyle Q_{\rm {max}}={{T_{J{\rm {max}}}-(T_{\rm {amb}}+\Delta T_{\rm {HS}})} \over {R_{\theta {\rm {JC}}}+R_{\theta {\rm {B}}}+R_{\theta {\rm {HA}}}}}}$

The designer now knows ${\displaystyle Q_{\rm {max}}}$, the maximum power that the transistor can be allowed to dissipate, so they can design the circuit to limit the temperature of the transistor to a safe level.

Let us substitute some sample numbers:

${\displaystyle T_{J{\rm {max}}}=125\ ^{\circ }{\text{C}}}$ (typical for a silicon transistor)

${\displaystyle T_{\rm {amb}}=21\ ^{\circ }{\text{C}}}$ (a typical specification for commercial equipment)

${\displaystyle R_{\theta {\rm {JC}}}=1.5\ ^{\circ }{\text{C}}/{\text{W}}\,}$ (for a typical TO-220 package)

${\displaystyle R_{\theta {\rm {B}}}=0.1\ ^{\circ }{\text{C}}/{\text{W}}\,}$ (a typical value for an elastomer heat-transfer pad for a TO-220 package)

${\displaystyle R_{\theta {\rm {HA}}}=4\ ^{\circ }{\text{C}}/{\text{W}}\,}$ (a typical value for a heatsink for a TO-220 package)

The result is then:

${\displaystyle Q={{125\ ^{\circ }{\text{C}}-(21\ ^{\circ }{\text{C}})} \over {1.5\ ^{\circ }{\text{C}}/{\text{W}}+0.1\ ^{\circ }{\text{C}}/{\text{W}}+4\ ^{\circ }{\text{C}}/{\text{W}}}}=18.6\ {\text{W}}}$

This means that the transistor can dissipate about 18 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 absolute thermal resistances of the layers and the temperature drops across the layers.

From Fourier's Law for heat conduction, the following equation can be derived, and is valid as long as all of the parameters (x and k) are constant throughout the sample.

${\displaystyle R_{\theta }={\frac {\Delta x}{A\times k}}={\frac {\Delta x\times r}{A}}}$

where:

• ${\displaystyle R_{\theta }}$ is the absolute thermal resistance (K/W) across the thickness of the sample
• ${\displaystyle \Delta x}$ is the thickness (m) of the sample (measured on a path parallel to the heat flow)
• ${\displaystyle k}$ is the thermal conductivity (W/(K·m)) of the sample
• ${\displaystyle r}$ is the thermal resistivity (K·m/W) of the sample
• ${\displaystyle A}$ is the cross-sectional area (m²) perpendicular to the path of heat flow.

In terms of the temperature gradient across the sample and heat flux through the sample, the relationship is:

${\displaystyle R_{\theta }={\frac {\Delta x}{A\times \phi _{q}}}{\frac {\Delta T}{\Delta x}}={\frac {\Delta T}{q}}}$

where:

• ${\displaystyle R_{\theta }}$ is the absolute thermal resistance (K/W) across the thickness of the sample,
• ${\displaystyle \Delta x}$ is the thickness (m) of the sample (measured on a path parallel to the heat flow),
• ${\displaystyle \phi _{q}}$ is the heat flux through the sample (W·m⁻²),
• ${\displaystyle {\frac {\Delta T}{\Delta x}}}$ is the temperature gradient (K·m⁻¹) across the sample,
• ${\displaystyle A}$ is the cross-sectional area (m²) perpendicular to the path of heat flow through the sample,
• ${\displaystyle \Delta T}$ is the temperature difference (K) across the sample,
• ${\displaystyle q}$ is the rate of heat flow (W) through the sample.

A 2008 review paper written by Philips researcher Clemens J. M. Lasance notes that: "Although there is an analogy between heat flow by conduction (Fourier’s law) and the flow of an electric current (Ohm’s law), the corresponding physical properties of thermal conductivity and electrical conductivity conspire to make the behavior of heat flow quite unlike the flow of electricity in normal situations. [...] Unfortunately, although the electrical and thermal differential equations are analogous, it is erroneous to conclude that there is any practical analogy between electrical and thermal resistance. This is because a material that is considered an insulator in electrical terms is about 20 orders of magnitude less conductive than a material that is considered a conductor, while, in thermal terms, the difference between an "insulator" and a "conductor" is only about three orders of magnitude. The entire range of thermal conductivity is then equivalent to the difference in electrical conductivity of high-doped and low-doped silicon."[3]

Similarly to electrical circuits, the total thermal resistance for steady state conditions can be calculated as follows.

Parallel Thermal Resistance in composite walls

The total thermal resistance

${\displaystyle {{1 \over R_{\rm {tot}}}={1 \over R_{B}}+{1 \over R_{C}}}}$          (1)

Simplifying the equation, we get

${\displaystyle {R_{\rm {tot}}={R_{B}R_{C} \over R_{B}+R_{C}}}}$          (2)

With terms for the thermal resistance for conduction, we get

${\displaystyle {R_{t,{\rm {cond}}}={L \over (k_{b}+k_{c})A}}}$          (3)

It is often suitable to assume one-dimensional conditions, although the heat flow is multidimensional. Now, two different circuits may be used for this case. For case (a) (shown in picture), we presume isothermal surfaces for those normal to the x- direction, whereas for case (b) we presume adiabatic surfaces parallel to the x- direction. We may obtain different results for the total resistance ${\displaystyle {R_{tot}}}$ and the actual corresponding values of the heat transfer are bracketed by ${\displaystyle {q}}$. When the multidimensional effects becomes more significant, these differences are increased with increasing ${\displaystyle {|k_{f}-k_{g}|}}$.[9]

Equivalent thermal circuits for series-parallel composite wall

Spherical and cylindrical systems may be treated as one-dimensional, due to the temperature gradients in the radial direction. The standard method can be used for analyzing radial systems under steady state conditions, starting with the appropriate form of the heat equation, or the alternative method, starting with the appropriate form of Fourier's law. For a hollow cylinder in steady state conditions with no heat generation, the appropriate form of heat equation is [9]

${\displaystyle {{1 \over r}{d \over dr}\left(kr{dT \over dr}\right)=0}}$          (4)

Where ${\displaystyle {k}}$ is treated as a variable. Considering the appropriate form of Fourier's law, the physical significance of treating ${\displaystyle {k}}$ as a variable becomes evident when the rate at which energy is conducted across a cylindrical surface, this is represented as

${\displaystyle {q_{r}=-kA{dT \over dr}=-k(2\pi rL){dT \over dr}}}$          (5)

Where ${\displaystyle {A=2\pi rL}}$ is the area that is normal to the direction of where the heat transfer occurs. Equation 1 implies that the quantity ${\displaystyle {kr(dT/dr)}}$ is not dependent of the radius ${\displaystyle {r}}$, it follows from equation 5 that the heat transfer rate, ${\displaystyle {q_{r}}}$ is a constant in the radial direction.

Hollow cylinder with convective surface conditions in thermal conduction

In order to determine the temperature distribution in the cylinder, equation 4 can be solved applying the appropriate boundary conditions. With the assumption that ${\displaystyle {k}}$ is constant

${\displaystyle {T(r)=C_{1}\ln r+C_{2}}}$          (6)

Using the following boundary conditions, the constants ${\displaystyle {C_{1}}}$ and ${\displaystyle {C_{2}}}$ can be computed

${\displaystyle {T(r_{1})=T_{s,1}}}$          and          ${\displaystyle {T(r_{2})=T_{s,2}}}$

The general solution gives us

${\displaystyle {T_{s,1}=C_{1}\ln r_{1}+C_{2}}}$          and          ${\displaystyle {T_{s,2}=C_{1}\ln r_{2}+C_{2}}}$

Solving for ${\displaystyle {C_{1}}}$ and ${\displaystyle {C_{2}}}$ and substituting into the general solution, we obtain

${\displaystyle {T(r)={T_{s,1}-T_{s,2} \over {\ln(r_{1}/r_{2})}}\ln \left({r \over r_{2}}\right)+T_{s,2}}}$          (7)

The logarithmic distribution of the temperature is sketched in the inset of the thumbnail figure. Assuming that the temperature distribution, equation 7, is used with Fourier's law in equation 5, the heat transfer rate can be expressed in the following form

${\displaystyle {Q_{r}={2\pi Lk(T_{s,1}-T_{s,2}) \over \ln(r_{2}/r_{1})}}}$

Finally, for radial conduction in a cylindrical wall, the thermal resistance is of the form

${\displaystyle {R_{t,\mathrm {cond} }={\ln(r_{2}/r_{1}) \over 2\pi Lk}}}$ such that ${\displaystyle {r_{2}>r_{1}}}$