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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?

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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).

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  • \$\begingroup\$ At DC the current will use the entire cross sectional area of the leads. At AC it will use less. The resistance will be higher. Skin depth at 100 kHz more like 0.24 mm. As the leads are 0.6-ish mm thick, the effect may be important. Are you planning on PWMimg? Also, how's your gate drive? Slow Vgs transitions could increase the power dissipation. How much time will it take to get 560 nC into/out of the gate? \$\endgroup\$ – user9224 May 8 '12 at 4:26
  • \$\begingroup\$ Another way to look at electrical resistance is consider if solder bridged the skinny leads so only the stub length, L1 outside the case is considered. The lead frame resistance R=L1·b1·c and the ESR from Fig 3 is 1.5mΩ \$\endgroup\$ – Sunnyskyguy EE75 May 11 '12 at 20:30
  • \$\begingroup\$ See Fig3 ... Since ESR of the whole device is 1,500 μΩ ( @-40'C) to 2,500μΩ (+150'C) the lead size is adequate for the device current.. Amazing as it is hard to believe, now figure out why your car jumper cables are so fussy on connection.... ha \$\endgroup\$ – Sunnyskyguy EE75 May 11 '12 at 20:41
  • \$\begingroup\$ I remember testing a diffusion bonder (1979) using 10,000 amps via 6" copper electrode wheels bonding Steel-Zirc-Steel tubes for nuclear reactors. The EMI , sparks and water works were spectacular. Since the tube resistance drops as you weld around it the operator had to increase current around the joint to maintain power drop to weld 2 tubes together. My instrumentation gave him that data. \$\endgroup\$ – Sunnyskyguy EE75 May 11 '12 at 20:49
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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):

enter image description here

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.

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    \$\begingroup\$ Coincidently, I know a guy who designs the electronic systems for electric locomotives, who used exactly the shown IGBT (CM1200HC) to drive a 2MW electric motor on an HST loco. They had to get a custom-made heatsink to dissipate the heat. The test setup was amusing - a tiny pushbutton to switch the motor to 100% power, causing the entire loco chassis to tilt as the motor span up. It made a noise like a dragon undergoing a root canal. \$\endgroup\$ – Polynomial May 8 '12 at 0:45
  • \$\begingroup\$ +1 for correctly identifying my wrong turn. \$\endgroup\$ – Bryan Boettcher May 8 '12 at 14:11
  • \$\begingroup\$ IGBT == incredibly good to be true? ;) \$\endgroup\$ – Kaz May 9 '12 at 4:36
  • \$\begingroup\$ @Kaz - Insulated gate bipolar transistor, but I think I like your definition better :) \$\endgroup\$ – Kevin Vermeer May 9 '12 at 13:45
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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.

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  • \$\begingroup\$ if you're using IGBTs, motors generally don't have "inrush" current, because you control the current to whatever you like. \$\endgroup\$ – Jason S May 8 '12 at 11:16
  • \$\begingroup\$ @JasonS - yeah, you use the device and control the current, b/c without it, a modest size motor like 1/3 HP can look like a short circuit for a couple hundred msec when starting from stopped. \$\endgroup\$ – JustJeff May 9 '12 at 22:50
  • \$\begingroup\$ oh, it's worse than that. Ever looked at the current vs. time waveforms on three-phase induction or synchronous motors if they're slammed across the AC lines? Truly awful transients. \$\endgroup\$ – Jason S May 10 '12 at 0:29
  • \$\begingroup\$ heheheh and try looking at those transients with a cheap digital scope \$\endgroup\$ – JustJeff May 10 '12 at 2:27
  • \$\begingroup\$ I think it is the turn off state that current wants to continue from the inductive load and the switch voltage rises , which actually peaks closer to the SOA quadrant limits in most practical cases than the V or I max limits. ( My son-in-law Prof at U of T says his students blow these out all the time.. the really big IGBT's .. I think ignoring what I just said... \$\endgroup\$ – Sunnyskyguy EE75 May 11 '12 at 20:53
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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.

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  • \$\begingroup\$ They're probably solid copper leads, which do have a low resistance - but power is \$P=I^2*R\$; they'd still melt without active heatsinking. Also, you seem to be confused about BJTs, MOSFETs, and IGBTs: You can't parallel BJTs, only MOSFETs, BJTs are current-controlled devices and they don't have 'resistance' in the usual meaning of the term, and IGBTs are another device entirely. \$\endgroup\$ – Kevin Vermeer May 7 '12 at 22:11
  • \$\begingroup\$ No parallel BJT's, ever? Hmm, does the Wikipedia page on "Thermal runaway" need a fix? It claims that If multiple BJT transistors are connected in parallel (which is typical in high current applications), a current hogging problem can occur. Special measures must be taken to control this characteristic vulnerability of BJTs. \$\endgroup\$ – Kaz May 8 '12 at 1:01
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    \$\begingroup\$ @Kevin Vermeer Actually in the datasheet for ULN2803A transistor array it is explicitly said that the transistors can be connected in parallel. Under the key features:OUTPUT CAN BE PARALLELED. How do you comment on that? \$\endgroup\$ – AndrejaKo May 8 '12 at 6:27
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    \$\begingroup\$ @AndrejaKo - That's a special feature, not a common one. The part has Darlingtons with integrated current limit resistors and they're all on the same die, so they should be more closely matched. It's possible, but difficult, to parallel BJTs. However, my point still stands that the device in question does not have 'many BJT sections in parallel to get the on resistance down very low' \$\endgroup\$ – Kevin Vermeer May 8 '12 at 13:13
  • \$\begingroup\$ @KevinVermeer is right, that George Bernard Shaw quote just popped into my head and I felt compelled. Then I supposed at the answer without thinking enough about it. After a quick read of Wikipedia, I think they just parallel the entire IGBT many times. Although there are some reasons to parallel bjt's, they are not common and this is not one of them. The best of the group would tend to bogart all the current. They do, have a resistance...several in fact, that depend on their q-point. Again, forgive me. \$\endgroup\$ – Matt May 8 '12 at 16:21

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