I'm looking at a dual diode, SBL3040. It's rated at 30A. The package is
TO-3P. The leads are no less than 1.9mm x 0.51mm. That's about the
density of smashed #20 solid wire, which makes me wonder; how can a
device with such tiny leads be rated at 30A?

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On Aug 27, 5:34 pm, clifto wrote:
I'm looking at a dual diode, SBL3040. It's rated at 30A. The package is
TO-3P. The leads are no less than 1.9mm x 0.51mm. That's about the
density of smashed #20 solid wire, which makes me wonder; how can a
device with such tiny leads be rated at 30A?

--
Tiger Direct forcibly checking receipts?
http://tinyurl.com/2hz3ht

(a) Heatsinking to the terminated end of the lead.
(b) I believe the 30A max average forward current is in the mode where
each rectifier conducts half the time, the average being for both
diodes, though note that the RMS current is commonly higher than the
average current.
(c) The case would normally be connected electrically to carry the
cathode current.
(d) Flat wire gets rid of heat to the air better than round wire.
(e) Fusing current for 20 AWG copper is 58 amps...expect a 20 AWG
copper wire in free air to melt at about 58 amps.
(f) 1.9mm x 0.51mm = 0.969mm^2; 20AWG at 0.032 inches diameter works
out to about 0.52mm^2, so your leads have almost the cross-section of
17 AWG. (though the Diodes Inc data sheet for that part says the leads
could be as small as 0.3mm x 0.51mm, more like 25 AWG which has a
fusing current for round copper conductors of about 24 amps).

If the 0.52mm^2 leads on your part get quite hot, they might be as
much as 0.6 milliohms per inch. At 31.6 amps RMS, that would be about
0.6 watts per inch. Compare that to a half-watt resistor, and it
doesn't sound so unreasonable.