Electronics Cooling Toolset Formulation

Chip—Resistor Networks Formulation

For all network types that are available within the Electronics Cooling Toolset, the junction node temperature $T_{J}$ is obtained by solving the following equation:

Figure 1. EQUATION_DISPLAY

\sum_{i} (\frac{T_{J} - T_{i}}{R_{J i}}) = S_{J}

(560)

where:

$T_{i}$ is the temperature of the thermal network node i that is connected to the junction node.
$R_{J i}$ is the thermal resistance between the junction node and the connected node i.
$S_{J}$ is the heat source at the junction node.

At interfaces between thermal networks and Quick Parts, face temperatures $T_{f}$ are calculated as:

Figure 2. EQUATION_DISPLAY

- q_{f}^{''} \cdot ⅆ a_{f} + \sum_{i} (\frac{T_{f} - T_{i}}{R_{f i}}) = 0

(561)

where:

$q_{f}^{''}$ is the heat flux out of surface face f.
$a_{f}$ is the area vector of surface face f.
$T_{i}$ is the temperature of the thermal network node i that is connected to the respective interface node.
$R_{f i}$ is the thermal resistance between the interface node and the connected node i.

If a chip surface is not represented by a network node, then the faces of the contacting Quick Part are modeled as adiabatic:

Figure 3. EQUATION_DISPLAY

q_{f}^{''} = 0

(562)

PCB—Thermal Conductivity Formulation

Basic

For a basic PCB with specified layers, the thermal conductivity of the PCB is modeled as anisotropic.

The thermal conductivity tensor is calculated as:

Figure 4. EQUATION_DISPLAY

K_{PCB} = (\begin{matrix} {\begin{matrix} k \end{matrix}}_{XX} & 0 & 0 \\ 0 & k_{YY} & 0 \\ 0 & 0 & k_{ZZ} \end{matrix})

(563)

with:

Figure 5. EQUATION_DISPLAY

k_{XX} = k_{YY} = \frac{1}{D} \sum_{i = 1}^{n} d_{i} k_{i}

(564)

Figure 6. EQUATION_DISPLAY

k_{ZZ} = \frac{D}{\sum_{i = 1}^{n} \frac{d_{i}}{k_{i}}}

(565)

where:

Figure 7. EQUATION_DISPLAY

k_{i} = f_{m, i} k_{m} + (1 - f_{m, i}) k_{d}

(566)

and:

$n$ is the number of layers.
$d_{i}$ is the thickness of layer $i$ .
$D$ is the overall thickness of the PCB.
$f_{m, i}$ is the metal fraction of layer $i$ .
$k_{m}$ and $k_{d}$ are the thermal conductivity of the metal and the dielectric material, respectively.

Detailed

For a layer of a detailed PCB with specified metal fraction, the thermal conductivity is isotropic throughout the layer and calculated as:

Figure 8. EQUATION_DISPLAY

k_{i} = f_{m, i} k_{m} + (1 - f_{m, i}) k_{d}

(567)

For a layer with a specified trace image, the thermal conductivity is a function of the image grayscale (0 - 255, where 0 = all dielectric and 255 = all metal) and calculated as:

Figure 9. EQUATION_DISPLAY

k_{i} (x, y) = k_{d} + \frac{(k_{m} - k_{d}) {grey}_{i} (x, y)}{255}

(568)

where ${grey}_{i} (x, y)$ is the greyscale value of the trace image for layer $i$ at position $(x, y)$ .