A = .607 / (Vin - Vout)
Vout = 3.3v, so:
.607 / (5 - 3.3) = .357 (Piggybacking off my 5A 5v regulator)
.607 / (8.4 - 3.3) = .119 (7.2v NiMh -> 8.4v at full charge)
.606 / (12 - 3.3) = .070 (12v battery - probably more like 13v+ but whatever)
.606 / (16 - 3.3) = .048 (16v worst case)
Let me toss in a few ideas here. These last calculations are the most pertinent. They show
you are quite limited in how much current the v.regs can "realistically" supply, without melting
your pcb. Assuming the NiMH cells are the most likely to be used, you've got ~120 mA max.
HOWEVER, and it's a BIG however, you're allowing 150degC temp on the v.reg. How hot is that?
100degC is the temp of boiling water, and how long could you keep your hand in boiling water
for? 0.5 sec, that's it. IOW, you're really pushing it here.
PD = IOUT * (VIN - VOUT) + VIN*IQ
where VIN = input voltage; VOUT = output voltage; IOUT = output current, A; and IQ = quiescent current, A
I assume quiescent current is also known as ground current as that seems to match up with the minumum quiescent current listed in the absolute maximum ratings and there's no other quiescent current listed. In which case for both vregs, it's 850uA typical at 150mA output. Which means as far as the power dissipation equation goes, the two regulators are identical.
Secondly, you're initial ideas here were offbase. The PD eqn is simply how you calculate PD from
Ohm's Law and circuit theory, and has nothing to do with the characteristics of the v.regs. The 1st
term is the important one, and has nothing to do with what's "inside" the box, only with the
input-output characteristics. The 2nd term is basically insignificant, since IQ is > 150X smaller than
Thirdly, for my part [and I've designed a lot of pcbs], I wouldn't even think about using such a tiny
smt part at all, unless I was sure PD would always be maybe < 50% of your 0.6W value. Then the v.reg
would not overheat so much, and it would still be a 75degC = 167degF on the part. Still pretty
Fourthly, I don't even like the SOT223 parts [160degC/W], let alone the tiny ones you're using. I
go for the NCP1117 DPAKs [67 degC/W].
Fifthly, I think the answer to your initial question ["difference in current capacity"] mainly has to
do with the physical characteristics of the PNP output transistor in the parts, and not with the
size of the package, etc. Eg, the smaller the collector area of the transistor, the lower the