I see a lot of MOSFET transistors with suprisingly high continuos currents in TO-220 packages. To make matters more magic open ressistance is very low.
For example I want to use IRFB7545PbF N-MOSFET transistor. In datasheet, there's stated that maximum continuos current is 67A and RdON is lower than 5,9mOhm.
I don't believe, that this is possible.
Tiny legs of TO-220 package will have far bigger resistance that few mOhms at this conditions, it'll heat up and unsolder.
Where am I wrong?
I believe that semicunductor inside is capable of such currents, but I don't believe, that TO-220 package can deliver...
Answer
Lead resistance is not a problem. Look at the IRFB7545PBF data sheet. Lead cross-section is ~0.38 mm x 1.14 mm (minimum). Cross-section area is ~ 0.5 mm\$^2\$. This is equivalent to 20 ga wire, and has a resistance of ~10 mΩ/ft. A lead length of 0.3 inch will give a total resistance of 0.25 mΩ. Two leads, of course, will have a total resistance of 0.5 mΩ (less than 10% of the stated 5.9 mΩ). Total power dissipation for the package is 26 W (67 x 67 x .0059), which is well within limits for a TO220.
Under this load, each lead will dissipate 1 W. For a long run in free air, this is enough to melt the lead, but that is not how a TO220 is used. The leads will be soldered into a pc board with very heavy traces (to avoid delamination of the board), and these traces will serve to conduct heat away from the leads.
And for what it's worth, the data-sheet specifies the 67 A figure for a case temperature of 100°C.
ETA - According to a FAQ note at irf.com, the TO220 package has a current limit of 75 A, so the original suspicion that a TO220 can't handle high currents is almost correct.
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