First, I am no expert. I am sharing where I think you erred hoping/expecting that some real expert will either confirm me or correct my error. We can both learn something.
I think you are calculating power wrong here.
...
Knowing all of that and doing the calculations tells me that i can't use the MAX17561 even if i supply it with the absolute minimum operating voltage of 4.5V noted in the datasheet let alone 6V -12V as i was planning on using, and 1A (Even though the datasheet says the programmable current limit is anywhere between 0.7A to 4.2A)
1A x 4.5V = 4.5W
4.5W x 39C/W = 175.5C
...
Here, you are calculating the wattage you are supplying, but not the wattage the MAX17561 has to dissipate.
It is the wattage the chip has to dissipate that heats the chip. So we have to evaluate
how many watts the chip is consuming, so we must find the voltage drop across the chip. I can't find the Voltage Drop or Voltage Forward listed in the electrical characteristics, but I did find the resistance.
Obviously V
drop (V
forward) will be missing because it depends on your V
max limit. When you are supplying a voltage below between selected V
max and V
min, I take it you expect the same voltage out (or as much as it can).
On page 1 of the spec under "Benefits and Features", listed is "
Low RON 100mΩ (typ)", so when ON, the thing looks like a 100mΩ resister, typically. You are supplying
at allowable min, but say it turns on just fine (instead of locking out as I suspect it may do as well with the smallest of voltage sag)...
P=I*V, and V=I*R, so P=I*I*R
P = 1A * 1A * 100mΩ = 1A * 1A * 0.1Ω =
0.1 WattThe 4.5W you calculated is what you supplied and expect the chip passed along to your load, 0.1 Watt is what the chip consumed.
When you are lock-out, then you use the undervolt/overvolt leakage current and (100nA in both cases) and do the P=VI to see how many watts it has to dissipate. Since current is in mere 100 nano Amp, you would be talking nano-watts to micro-watts, not even milli-watts.
EDIT:
added and then removed a situation when Voltage in > selected max pretending it would be a V
drop (heat dissipation had it regulate the voltage down) calculation -- it is pointless since the chip is suppose to lock-out instead of regulate it down. The paragraph was replaced with lock-out current calculation instead.
Additional EDIT - initial reply I had said "below allowable limit" with your 4.5V supply, wrong wording, you are at limit, not below, so reworded accordingly.