Material thickness, thermal conductivity and electrical resistivity

Question

I have a plastic housing (for a printed circuit board- PCB) in the shape of a box (a *b *c) with L (thickness). Its material properties are:

*thermal conductivity:                              $k_h~[Wm^{-1}K{-1}]$

*specific electrical resistance:                    $sigma_h [Omega m]$

(elastic modulus:                                   $lambda_h [2600N/mm^2])$

thermal contact 1:  area of plastic and copper:          $M = 50*20 mm^2$

thermal contact 2:  area of plastic and PCB:             $N = 10*30 mm^2$

temperature inside housing minus temperature ambiental:  $dT = (60 - 27)K$

I would like to know how heat conduction and electrical resistivity are changing with thickness $L$. So I can find the optimal $L$.

Floris · Answer

The flow of both current and heat depend on the gradient (for current: gradient of voltage over distance, i.e. electric field; for heat: temperature difference over distance). If you increase the thickness, you increase the thermal resistance perpendicular to the walls.

Note - both thermal and electrical conductivity are functions of temperature. You indicate rather low temperatures: it might matter to make sure that you are using the correct values of conductivity (and not, for example, values at room temperature).

Material thickness, thermal conductivity and electrical resistivity

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