Electronic – When IGBT is off what is the leakage current between collector and emiter

They give you two points on the datasheet: 580W with case temp of 25C, and 400W with a case temp of 80C. You can assume an approximate third point of 0W with a case temp of 150C, which is the maximum junction temperature. Throwing those points into Excel and getting a scatterplot shows those three points are colinear to R^2 of .98. So as a first-order approximation, you could say your maximum acceptable pulse watt-loss is 725C, minus 4.7 times the temperature above zero in Celsius.

This function can not be modified! It's a function of the internal structure of the device. Temperature reduction is the only way that value is going to improve, short of picking a different IGBT. If you have any sort of budget, or plan to buy new parts in the future, I might recommend changing to a different IGBT. That device isn't even listed on Infineon's site any more, and word here is that the entire BSM series is dead; we're having to update our UL listings that include that series. Sorry I don't have a source for that.

Now, you can mitigate the problem by reducing the actual watt-loss caused by the pulse. You can calculate your watt-loss if you know your current by looking at the output characteristics plot in the datasheet. Voltage times current is power. You'll note that Vce goes down for a given current if Vge increases. So by hitting the gate as hard as you can without exceeding its spec (20V), you may buy yourself a few amps. But not much.

As a general rule, if you're really operating on the raggedy edge like that, you need to reconsider what you're doing.

Yeah, that explanation is fine. When BE junction is forward biased, majority carriers in Emitter(electrons) are pushed towards base where those electrons become minority carriers(As base is p - type). As CB junction is reverse biased, the minority carriers (electrons in base) in base has no problem to reach the collector. This is only in NPN transistor.