Case 2

To start the analysis based on calculus formulas (considered by the specialists being more exact) must be said from the beginning that any leakage of current through a conductive track will determine a raise in the temperature of that track, so at the presence of an over-temperature than the temperature of the surrounding environment. Because of the fact that is known that the formula of power is RI², where R is the track’s resistance, it can be specified the fact that the relation between current and temperature won’t be one of linear dependence. Moreover, due to the complex mechanisms of heat transfer, in many cases it is possible that for conditions somehow identical to obtain different results. The simplest example is that of the tracks of equal areas but of different widths. It is obvious that the track more noose will evacuate by convection a quantity of heat larger than the narrower track.

The calculus of current intensity through track starts from a formula of the kind:

  

where: I- the intensity of the current [A];

            ΔT- (T track – T environment) – over-temperature of the interconnection track [K or °C];

            A- transversal area of the interconnection track [mil²]. “Mil” is an unit usually used in the electronic technology: 1 mil=25.4μm=0.001 inch);

             K, m, n – constants.