Oh, something I forgot to mention earlier..
Very high power converters will often use a quasi resonant topology (i.e. soft switching, ZVS/ZCS) to reduce semiconductor switching losses and perhaps any PN junction reverse recovery times such as those of non-schottky rectifier diodes and the IGBT's internal collector-emitter flyback diode. They're not as simple to design as hard switcing converters though. Ideally it'd be a full bridge converter, but with only 2 switches and running off-line, it will be a half bridge.
Here's a detailed application/datasheet for a ZVS half bridge controller IC:
L6591
No idea if it's appropriate or not, just googled "ZVS half bridge" and linked it for reference.
Conduction loss is relatively simple to calculate as it's just ohms law multiplied by the duty cycle.
Looking at the curves on the
Mitsubishi CM800DZ-34H datasheet, I think for the purpose of calculating conduction loss, you can model the collector-emitter path through an IGBT as a near ideal diode in series with the drain of an ideal MOSFET and one resistor in series with the diode, one in parallel with it. Roughly 20mR in parallel, 1.25mR in series by my calculation and 1V forward drop in the diode. Note the diode won't conduct until it's passing 50A (it seems the device is intended to be used between 50-800A). You have to be careful looking at just the data in the table as most of it comes from full voltage or current and I don't think you're likely to be putting 800A through it. Once you know the equivalent resistance of the conductive path, it's D*I^2*R or D*V^2/R.
Switching loss isn't as simple to calculate. You might be fooled into calcualting the energy that goes into charging and discharging the gate capacitance rather than that which is dissipated in the conductive path. If you were to assume the switching loss is the product of maximum current, maximum voltage, switch transition time and switch frequency will probably be a vast overestimation. E.g. if the device was switching its full voltage 1700V and 800A at 10kHz with a turn on time of 2.9us and turn off time of 3.3us, that's 83kW dissipation. I don't think that's particularly accurate. I don't have much experience with IGBTs, but knowing the
switching waveforms in a FET I know that there is some time where there is both voltage across and current through the channel and its dissipation will be high, but most of the time they're both high one will be ramping up and the other ramping down, so the dissipation is less than if they were both at maximum for the full switch transition.
I can't comment on rikkitikki's suggestion which quotes a known amount of energy for switching a certain current and voltage and then using the ratio of current and voltage in your circuit to this standard value. I haven't seen it before but it doesn't seem to account for the actual switch transition time which will be different in different devices so I can't see how there can be a fixed amount of energy to switch a fixed voltage and current, unless I've missed something. About his conduction loss equation, it's right but if you just go by the datasheet value for Vce(sat), you won't get the right answer as that value was measured at Ic=800A
