Btw: 400 mA drop-out compared to 250 mA of drop-out means 60% higher power dissipation at a minimum.

That's what I would have thought myself, but none of the power dissipation equations I have seen in the datasheets or online appear to include drop out in the power dissipation calculations. And since the Iq for both parts is the same, the amount of power that needs to be dissipated for both is equal.

I think I found the difference between the two regulators though.

It's not that the two packages are mounted differently. The landing pad sizes are virtually identical.

The difference can be seen on page 5 note 3 in the first datasheet, and in the Absolute Maximum ratings on page 4 of the second datasheet. The junction-to-ambient thermal resistance: ?j-a

For the first part, ?j-a = 220°C/W

For the second, ?j-a = 206°C/W

And the equation for calculating the amount of power that can be dissipated at a specific ambient temperature is:

p(max) = ( Tj(max) - Ta ) / ?j-a

So, the power which can be dissipated is the maximum junction temperature, minus the ambient temperature, divided by the junction-to-ambient thermal resistance.

This power calculation isn't new to me, but yesterday when I was tired I could not for the life of me find those thermal impedance ratings and I saw something in one of the datasheets indicating what a typical value would be on FR4 pcb, and I thought that it must be the same for both parts because the two packages are the same size and neither has any kind of heat sinking.

This leaves me with a little problem though. A lower thermal impedance is better. And doing the calculations for am ambient temperature of 25 degrees, I get:

(125-25) / 220 = .455

(150 - 25) / 206 = .607

Which means the first part, the one which is 3x as expensive and has a higher current rating, can actually dissipate less power than the second part.

But that's for the same package. The first part comes in a second type of package with a heat sink on the bottom. And that package has a thermal resistance of 282 C/W on a 2 layer board with no heat sink (worse than the first two) and 62 C/W on a 4-layer board with 6 thermal vias.

Plugging that into my equation I get:

(125-25) / 62 = 1.61

Which is 2.65x better than the 50 cent regulator. So by all rights it ought to be rated for 400mA if the cheaper one is 150mA, but I don't know what criteria they used to decide to rate it... ie what amient temperature, what current, what input and output voltages. So that could explain the difference.

Anyway after all this, assuming I don't go with the third option, the part with the heat sink on the bottom:

http://search.digikey.com/us/en/products/LP2992ILD-3.3%2FNOPB/LP2992ILD-3.3CT-ND/567557

It would seem that the 50 cent part would be the better of the two choices.

Now I just have to figure out how much current it can really carry by doing the rest of those power calculatons.