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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...

The Photon
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Viliam
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  • I can't find the link right now, but I've read an article about what you're asking. Basically, a researcher proved that the package itself has no chance of sustaining the currents from the datasheets. I think that the article too referenced IRF datasheets. – AndrejaKo Apr 30 '14 at 14:48
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    @AndrejaKo Would you be referring to [this article](http://www.mcmanis.com/chuck/robotics/projects/esc2/FET-power.html)? "*Discussions with the International Rectifier FE and later with a rep from Motorola confirmed that the limits on the TO-220AB were approximately 75 amps. This limit was due to the heating of the lead frame to the point where the **legs would melt**.*". Point to note, though, is that **solder would melt** much before this. – Anindo Ghosh May 01 '14 at 04:35
  • @Anindo Ghosh That's the article I was thinking of. – AndrejaKo May 01 '14 at 04:48

2 Answers2

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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.

Andy aka
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WhatRoughBeast
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  • Now I have missing part of datasheet from FAQ note. They assume you have infinite heatsink. :D I'll order one from IRF ASAP. :D:D – Viliam Apr 30 '14 at 21:02
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Usually if you read carefully, those currents are specified when the case temperature is (magically?) held at 25C. So if you had some way to cool the package and maintain the case at that temperature you would be able to pass those kinds of currents. You could argue that the leads are part of the case as well.

Absolutely not a real-world spec, but I think IR started it and other companies followed so that their FETs didn't look inferior. Nobody pays much attention to that number these days.

John D
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  • I did read carefully and I'm sure there are some info missing. Let's assume I can hold whole package at 25°C. I don't think RdON will be 5,9mOhm at 67A because of increased resistence of pins. – Viliam Apr 30 '14 at 15:08
  • Well if the FET measures 5.9mOhm with no current flowing at 25C, what's causing the increased resistance of the pins with 67A flowing and everything at 25C? They don't specifically state it, but the leads can be considered part of the case, so they will remain at 25C as well in order for the spec to be valid. – John D Apr 30 '14 at 15:12
  • The 67 A figure is for 100 C case temperature. For 25 C case temperature they spec 100 A(!). – The Photon Apr 30 '14 at 16:51
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    IRF specifically states that the TO220 package has a 75 A limit. – WhatRoughBeast Apr 30 '14 at 19:37
  • I think any resistance shift due to current density would happen at much higher currents. In pulse applications these FETs are fine, you don't see a shift in RDSon + package for very high pulse currents that don't cause significant heating. – John D Apr 30 '14 at 20:34