ronsimpson:
Supposedly if i follow equations 8,9,10 I should arrive at an approved value. Now, the big unknown in that is the Io ("output current"). In a dead short this is whatever the IC can provide without popping (I would guess this to be 5.3A) as the max output current limit? I tried to use some estimated actual application values for these initially, not thinking that the protection feature was dependent. Let's re-run with 5.3A as the Io
Ilp-p = Vout / Vin(max) * (Vin(max) - Vout) / Lo*fsw = 5/14 * (14-5)/(3.3e-6 * 650e3) = 1.50A
Ilpeak = Io+Ilp-p/2 = 5.3+1.5/2 = 6.8A
Ilo(rms) = sqrt(5.3^2+1.5^2/12) = 5.32A
This seems to be closer to the inductor you mentioned above. The probably simply went off of the 6.8A for the peak for this situation, but i would think they'd want to cover the rms as well.
As for my inductor being 1uH @ 5.3A, yeah, I didn't consider the inductor at the shorted current, I ran everything for 3A, and probably not correctly, then checked with ti's webench, which probably doesn't check for this sort of thing.
That all sounds fine and good, but it seems like we're sizing the inductor to run at the shorted current, when what we really want to do is have enough inductance to allow enough time for the IC to do the current limit right?
So for a single "ON" cycle, with the RMS current in the inductor already @ 3A, what happens in the 1/650kHz = 1.538uS when the output goes short? How fast does the current rise, in the time where the chip hasn't done anything yet. It seems like it can rise about up to about 5A in one cycle (crude ltspice attempt), so that would put our peak current around 8A.
And as the inductance drops, that rise happens faster, so really, I want the 3.3u Inductor, with an 8A+ peak current and an rms > greater than whatever my operating point is, or up to 4.7u, and i can get down in the 6-7A peak range.
Thanks for the discussion. I've got some stuff to try.