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What is the formula to find Ef for an IGBT? Can I calculate it from datasheet parameters?
E is for energy dissipation and I found the graph from here: https://www.researchgate.net/post/How-can-I-calculate-the-losses-of-an-IGBT-using-datasheet-parameters
That's not energy, that is power, and it's simply the multiplication of forward voltage and collector-emitter current.
An IGBT stores only a negligible amount of energy (because the insulated gate acts as a capacitor), and that energy is not usable -- quite the opposite: turning the IGBT on or off requires changing the charge of this capacitor, and the energy required to do so is lost as heat. The datasheet should mention the capacitance, as this is an important performance characteristic; but this is not related to what is labeled "E" in your graph.
That depends on what you mean by "calculate", and how close of an answer you want. In general, a datasheet doesn't go into enough detail, and those energy dissipations depend circuit operation. While it's theoretically humanly possible to do the calculations by hand, in practical terms it takes seconds to do on a simulator what it would take months to do by hand.
Most manufacturers have SPICE models of their parts -- I would start there. If you want to use a part that doesn't have a SPICE model, then your best bet is to find a similar part, and tweak the SPICE model for the differences. This will not be trivial for an IGBT model.
For all that -- you can get pretty close for the $E_f$ -- that's just the $V_{CE}$ of the device when it's on times the current flowing through it. Assuming that what's on the transistor is inductor current, then for a typical circuit it has a starting current and an ending current, and the waveform is triangular. $E_f$ is just the integral of $V_{CE}$ and the collector current.
$E_{on}$ and $E_{off}$ are more difficult, because they're the losses that happen when the IGBT are partially on during the turn on/turn off transisents. I have little experience with IGBTs, but with MOSFETs you can come do a rough estimate by assuming that the $V_{SD}$ varies on a straight line while the drain current stays constant (because it's being driven by an inductor). Just replace $V_{SD}$ with $V_{CE}$, and $i_{D}$ with $i_C$, and you're golden -- except that the turn-on and turn-off times are very heavily dependent on how the gate is driven.
To really know the turn-on and turn-off times you need to take into account the gate drive voltage, the output resistance (or nonlinear current-drive capabilities) of the gate driver, whether you have an RC or resistor-diode-capacitor network between the gate driver and the gate, whether you have any impedance between the emitter and the gate driver's common (either intentional resistance or wiring/PCB resistance).
It's this dealing with the complex interplay of the gate drive and the switching characteristics of the part that drives the community to use simulation. You only get to the point of knowing what the turn-on and turn-off time will actually be through experience, and you only get that experience through working with circuits.
