In the IKW75N65ET7 simulation model, the formula for on-state loss is: max(lookup(‘e_rg_on’, Rgon) * lookup(‘e_tj_on’, T) * lookup(‘e_i_on’, i) * lookup(‘e_v_on’, v) / (lookup(‘e_i_on’, 75)^3) * 1e-3, 0). Why is it divided by (lookup(‘e_i_on’, 75)^3) * 1e-3, 0) in this formula? How did the ^3 come about? What do 1e-3, 0) mean?
If you look at the datasheet, Infineon provides 4 separate 2D curves for E = f(IC), E = f(RG), E = f(Tvj) and E = f(VCE), where for each one they provide the fixed values of the other three dependencies (e.g., E = f(IC) at VCC = 400 V, Tvj = 175 °C, and RG = 4.7 Ω)
While it would be helpful (and more accurate) to have a 4D lookup table with E = f(IC,VCE,Tvj,RG), this would require that each curve was provided for at least 2 values of the other dependencies. However, when behavior is linear for a given variable we can make some simplifications to “fill in” the missing data.
So basically the formula is:
E(IC) * [E(VCE)/E(IC@75)] * [E(Tvj)/E(IC@75)] * [E(RG)/E(IC@75)] , where each of the energy lookups for VCE, Tjv, and RG are normalized by what the losses are from the IC table at 75A.
As for why they chose E(IC) to normalize against, the other three curves for Eon are quite linear, whereas the Eon vs. Ic behavior is not.
The 1e-3 is to scale for units given in the plots (mJ).
And the zero at the end of the formula is related to the max function- they are saying here to provide a max of the calculated losses and 0: max(all of the E lookup table arithmetic, 0).
Let me know if this is clear or you have further questions.