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I found the IXGX400N30A3 at Digikey. The datasheet says the device is rated for 400A @ 25C, 1200A @ 25C for 1ms, with a voltage rating of 300V and PD of 1000W.
Really? This TO-264 package can control 400A of current all day long? I can short out my TIG welder with it in DC mode? How do those leads even carry 400A of current?
That device has a very low thermal resistance from junction to case, \$R_{thJC}\$=0.125 ºC/W (max), which means that, for every watt dissipated, the junction will only be 0.125 ºC (max) above the case temperature. So, for instance, for \$I_C\$=300 A, \$V_{GE}\$=15 V, and \$T_J\$=125 ºC (see Fig. 2) \$V_{CE}\$ will only be about 1.55 V. That's a power of P=300·1.55=465 W being dissipated (yes, more than some electric heaters). So, the junction will be 465·0.125=58.125 ºC (max) above the case temperature, which is a very low differential, for that massive dissipation.
However, in order for the junction temperature not to exceed its limit (of 150 ºC), the thermal resistance from case to ambient, \$R_{thCA}\$, which depends on the heat sink used, also has to be very low, because otherwise the case temperature would rise well above the ambient temperature (and the junction temperature is always above it). In other words, you need a very good heat sink (with a very low \$R_{th}\$), in order to be able to run this creature at 300 A.
The thermal equation is:
$$ T_J=P_D·(R_{thJC}+R_{thCA})+T_A $$
with
\$T_J\$ : Junction temperature [ºC]. Has to be < 150 ºC, according to the datasheet.
\$P_D\$ : Power dissipation [W].
\$R_{thJC}\$ : Thermal resistance from junction to case [ºC/W]. This is 0.125 ºC/W (max), according to the datasheet.
\$R_{thCA}\$ : Thermal resistance from case to ambient [ºC/W]. This depends on the heat sink used.
\$T_A\$ : Ambient temperature [ºC].
For instance, on an ambient temperature of 60 ºC, if you want to dissipate 465 W, then the heat sink has to be such that \$R_{thCA}\$ is at most 0.069 ºC/W, which implies a very large surface in contact with air, and/or forced cooling.
As far as the terminals, the approximate dimensions of their thinnest part are (L-L1)·b1·c. If they were made of copper (just an approximation), the resistance of each one would be:
\$R_{min}\$=16.78e-9*(19.79e-3-2.59e-3)/(2.59e-3*0.74e-3)=151 \$\mu\Omega\$
\$R_{max}\$=16.78e-9*(21.39e-3-2.21e-3)/(2.21e-3*0.43e-3)=339 \$\mu\Omega\$
At \$I_C\$=300 A, each one of them would dissipate between 13.6 and 30.5 W (!). That's a lot. Twice of it (for C and E) can be as high as 13% of the 465 W being dissipated (in this example) at the IGBT itself. But, usually, you will solder them so that that thin part is shorter than (L-L1).
Sure, it's possible. However, consider that the '400A@25°C' number is based on a \$T_C\$ of 25°C, not an air temperature. \$T_C\$ is the case temperature. At 400A, the voltage across the device, \$V_{CE(sat)}\$, can be 1.70 V. At 400A, that's a power dissipation of 680 W. You will need one hefty heat sink, which may not be physically possible, especially if the ambient temperature is 25°C.
As far as the leads carrying that current, the dimensioned drawing says that they're at least 2.21 mm wide and 0.43 mm thick. That's a cross-sectional area of about 1 square mm, equivalent to a 17-gauge wire. My reference chart says that 100A will cause a long segment of that thickness of (circular, uninsulated) wire to melt in 30 seconds. Of course, these leads will not be long segments, they'll be connected to heat-sinked copper planes. But even then, that's pushing it pretty tightly.
What have you learned from this analysis? Don't trust the first page of a datasheet! You can also happily ignore any table marked "Absolute Maximum". You're not guaranteed a functional device or an implementable design if you court these numbers. My professors always said that these pages are compiled by the marketing department, not the engineering department. In this case, the table you got that number from is marked "Maximum Ratings". Don't design your device to function near these numbers. Instead, scroll down to the characteristic graphs and standard operating parameters (the latter is not in this datasheet, but it will be in others) and design based on that. Determine how much current your PCB or wires can handle, and how much heatsink capacity you can add, and then decide whether this type of package is even feasible.
You mentioned that you were on Digikey; I'm guessing that you took a wrong turn and went looking for a high-current part in the 'Discrete Semiconductor Products' group, section IGBTS - single. This section is for PCB-mounted components. The realities of PCB manufacturing (soldering, copper thickness, heatsinking) will limit the practically achievable values here. If you want to get really high-current stuff, go to 'Semiconductor Modules', that's where the chassis-mounted parts connected to thick wires are located. The IGBTs section there has components like this beast, shown with a pencil for scale (borrowed from Wikipedia):
That device can actually handle 3300 and 1200 A; it's 190 by 140mm rather than a little PCB-mount device. There are plenty of smaller, more reasonable devices available as well.
A short answer: you don't do both 400A and 300V at the same time, at least not for very long.
The device passes almost no current when in the off state, and dissipates very little power when off. The device incurs very little voltage drop when conducting in the on state, and so dissipates a controllable amount of heat in that state.
The major burn comes when changing between the two conditions. Probably the worst case is turning on with a load like a large motor; the inrush current to spin a motor up can last significant fractions of a second, during which lots of heat can be developed.
Because You see things; and you say, ‘Why?’ But B. Jayant Baliga dreams of things that never were; and says, ‘Why not?’"
But seriously, the leads have a very low resistance, so they do not generate much heat. I think there are many bjt sections in parallel in the actual device to get the on resistance down very low also.
