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 $L$.

  • 1
    $\begingroup$ Electrical resistivity is a material property. It does not change with the dimensions of the object. $\endgroup$
    – nasu
    Oct 20, 2021 at 13:53

1 Answer 1


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).


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