High amperage MOSFETs like the 511-STP200N3LL say they can handle 120 Amps of current. Is this possible? If yes, then what prevents the small wires coming out of the device from melting or getting too hot? The package for the specific device I'm referring to is TO-220-3.

Here is a datasheet link: https://www.mouser.com/datasheet/2/389/stp200n3ll-1063226.pdf

  • \$\begingroup\$ @Ale..chenski: The \$I^2\$ in \$I^2 R\$ means squared. \$(100A)^2(2\mathrm{m}\Omega) = 20W\$. \$\endgroup\$
    – TimWescott
    Jan 6, 2019 at 4:31
  • \$\begingroup\$ Probably it doesn't. IF you can hold the case to 25°C and the leads short (and presumably also at 25°C).. but holding the case and leads to 25°C or anything remotely close to that is impossible in most situations. \$\endgroup\$ Jan 6, 2019 at 4:56

2 Answers 2


The MOSFET is rated to handle that current when sufficiently cooled. The hole in the tab at the top of the chip facilitates use of a machine screw to attach the MOSFET to a heatsink. A MOSFET with a higher current rating will also typically be capable of a low ON resistance comparative to its physical size, and the ON resistance actually achieved will depend on temperature and gate voltage. You can find charts in the datasheet to see exactly what the ON resistance should be at the gate voltage you intend, and use Watt's Law (\$P=I^2R\$) to figure out how much power the MOSFET will dissipate at the current you intend. Your heatsink must be capable of dissipating this amount of power while keeping the MOSFET at your intended temperature. There will be a temperature gradient right from the semiconductors in the MOSFET to the extremities of the heatsink, which will be rated by this, in degrees C per Watt.

In the case of your mosfet, installed and cooled as intended, this amounts to \$120A^2*0.0024 \Omega=34.56W\$ used by the mosfet to switch 120A at 30V. In reality, switching losses may contribute considerably more to this depending on what you're doing.

The wires coming from the device are intentionally kept short, and usually reasonably chunky by electronics standards. They also benefit from the cooling of the MOSFET.


Well, the most immediate answer is, "Don't use TO220 packages". The rule I've seen is that anything over about 75 amps will run into problems with the leads getting too hot (as opposed to the semiconductor). Obviously, the manufacturer you've linked to feels otherwise. This may be because the Rds(on) is extremely low, and the package's heat sink may be adequate for the total heat load.

K.H. has a perfectly good answer. The two important issues are the lead sizes (they should have adequate cross-section and be short) and heat-sinking the package. Heat sinks are a subject unto themselves. Air-cooled heatsinks at these current and power levels tend to be either very large or employ forced cooling, either forced air or liquid. Heat pipes can be extraordinarily effective at pulling heat away from a source.

  • \$\begingroup\$ What package is the best for high amp usage? \$\endgroup\$
    – kuchi
    Mar 18 at 11:32
  • 1
    \$\begingroup\$ @kuchi - At these levels you might as well stick with TO220. Just be ruthless in keeping your lead lengths short with good thermal paths. If you look at the dimensions of the package of your FET, you'll see that the soldering part of the leads, about 1/2 inch long, have a minimum cross-section of 0.61 x 0.48 mm. If you use all of that and draw the rated 120 amps, the total power in each affected lead will be a hair under 10 watts, or about 20 watts for the package. Obviously you won't do that, but keep it in mind. \$\endgroup\$ Mar 19 at 13:31

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