I have updated the original post with the revised formula.
It seems extremely low, so I'm not sure if its correct.
What are you using for Vin and Vout?
Notice that your reference is discussing synchronous rectification: this is when the transistor is switching on and off, opposite another one -- by averaging the "up" and "down" voltages, a new (lower) output voltage is obtained, without wasting power in the process. (Or put another way, you get more current output than input -- a sort of DC transformer. Or, equally well, it works in reverse, boosting a low voltage and high current, to a higher voltage at lower current.) They aren't very specific about this I think, hence the confusion (and, if you don't know what a synchronous rectifier is, well, that too). Well, for that application, Vin and Vout differ, and the duty cycle is given by the ratio of voltages. That's why they put it here. The switch is only carrying current while it's on, and off the rest of the time, dissipating zero. So P has to be less than I^2 * R, and how much less is given by the duty cycle, given by the voltage ratio (and that one-minus part). So that's what the extra stuff on the right is for.
Your OV/RV FETs will simply be at 100% duty, so the whole right side term goes away (=1) and you get P = I^2 R.
This is the equation for H from the document.
RDS(ON)HOT = RDS(ON)SPEC [1 + 0.005 × (TJ(HOT) - TSPEC)]
For some reason I replaced 0.005 with R. I think it may have been that I didn't understand what the 0.005 was, and it being a value fairly close to R I ended up making a stupid assumption that it was an example for 5mΩ.
Also, I guess they give as justification, the tempco: 0.5%/C being the upper limit, so they put that in the equation. This must have units of 1/°C, which they kindly omit for... clarity, I'm sure.
And as I mentioned, it's worse for higher voltage MOSFETs; this figure isn't true in general -- the Rds(T)/Rds(25C) curve goes up to 2.5-3x for 600V+ parts near Tjmax (150-175°C). And, not that you're working with OV/RV protection at those kinds of voltages... just for flavor.
Always check the datasheet (RTFDS, Read The "friendly" Data Sheet)!
And yeah, for low voltage parts like this, where the curve is linear, this linear fit works fine.
If I just take the voltage drop (0.26V) and apply ohm's law using 5A I get 1.3W.
I originally was going to use TO-252 part, but if it is really this low I think I may actually use a SOT part.
Bingo, you're doing DC (at least, I assume so), so this is the correct result.
You might consider getting a lower Rds(on) part. SOT-23 aren't good for much over say 500mW, and even that may be pushing it.
Note that supplier/datasheet headline "drain current" ratings are optimistic at best; indeed D(2)PAKs they outright lie to you and assume you have the capability to immerse the part in boiling freon at 25°C -- it'll keep the tab nice and cool sure, but how in the &*$! am I supposed to do that on a PCB? Simple, you don't. SOTs at least they tend to use thermal figures for typical scenarios (minimal footprint, 2-layer, 4-layer, etc.), and DPAKs they usually give a few too (just not the headline figures).
So, to put it another way -- you could very well have identical chips in SOT-23 and DPAK packages, and one is rated a measly ampere and the other ten. Because the DPAK can ideally shed off 50 goddamn watts or something like that. But it won't ever in practice, so you need to read below the fold and see what it's
actually good for.
All that really matters is chip temperature, and all of these typical figures assume, well, typical conditions -- if your ambient isn't 25.0°C, maybe it's still air inside a box, maybe it's got extra cooling (heatsinking, thermal pads, etc.), whatever -- all you can do is estimate the thermal resistance for your application, add the part's to it (RthJA or RthJC as appropriate) and that's your actual temp rise above ambient. And then, whatever current corresponds to the power that can be handled at Tjmax, in that setting, that's what you can run.
So yeh, rather roundabout, and in our defense as engineers, heat transfer and convection are notoriously difficult topics. Heat transfer not so much really, in the grand scheme of things, but it's maybe so seemingly trivial that it's easy to overlook -- until you have half a million gadgets getting warranty returns due to a design oversight.
Comes to mind, it's like statics class. That is, the study of objects and structures just resting in place. Basically Newtonian physics, but no motion, just force balancing; weight, tension and compression, rotational moment. Such an easy class, but deceptively so; most students did rather poorly in it as I recall.
I hope to one day have math skills such as yours.
Math used to be my best subject in school, and I used to be able to do multiplication and division quickly in my head.
Then 7 or so years ago I ended up in a coma, and ever since then my math skills have been lacking.
Remembering 2 numbers at once is tough for me now.
Weirdly enough my memory is still great, writing programs I can remember where and what every function does and its name, its just if I have to remember more than 1 number at a time.
Can't blame that for being bad at algebra though, that's just because I haven't really done any in 15 years.
Only really started to again when I started working with FPGAs and battery operated circuits, when I needed to start worrying about power consumption.
Ah yuck, coma is definitely no good! But you made it through, that's great.
On the upside, computing is just applied mathematics -- more on the logical side (program flow, conditional statements, etc.) than numeric, but numbers derive in turn from logic, so it's all the same in the end. Maybe give yourself some exercises, practice symbolic representation of program statements, data flows, expressions, etc. Turn algebraic expressions into program code, and, that can lead into numerous quirks of our machines, like integer overflow, limitations of floating point numbers, etc. (For real numbers, a*(b+c) = a*b + a*c, and (a*b)*c = a*(b*c); but for floating point, these are only true in very particular cases -- understanding how floats are aligned and rounded off, makes a huge difference to the numerical accuracy of a chain of operations.)
Might also take a look at little refresher problems, or riddles or what have you. This comes to mind,
https://www.youtube.com/c/MindYourDecisionsalso, 3Blue1Brown and other math-education channels have FANTASTIC content, ranging from simple (algebra, geometry?) to ludicrous (say, a surprisingly simple explanation of a topic deep in analytical calculus).
I have no idea what kinds of things might capture your interest, or prove helpful, if you're even looking for that sort of thing -- or if you aren't aware of them already! -- but there's lots of good stuff out there to be found, in any case.
A couple days ago I was looking at a SOT-23 that could dissipate up to 3 watts, but while reading the datasheet I found out it required a full square inch of copper to be able to handle 3 watts, gave me a chuckle.
I assume they were trying to abuse the parametric search, who knows.
Not sure why anyone wouldn't just upgrade to a larger package to save a ton of space at that point.
Keep in mind, dissipation is a surface thing. You don't magically get more dissipation from a DPAK, if the pours are the same size as on the SOT-23.
Well, not more beyond the basic effect of heat spreading: you get a lower Tj for the bigger device, because its tab is thicker and wider -- spreads out heat further before it even flows into the board. Whereas a SOT-23 in the middle of a board can't do any more than heat the tiny bit of board around it, which heats a little more around that, which... Each little ring of board-heating comes with a small temperature drop, and you add up all the drops from all the rings to get the total. Well, that temp drop is a heck of a lot higher for tiny rings with little circumference, so you do have some advantage using a bigger part there.
Beyond that though, if you're talking whole watts -- you simply need board area, or heatsinking. A square inch is good for a watt or so, if you don't mind it getting fairly toasty in the process. That's a square inch of board area, component surface, heatsink fins, enclosure, whatever, doesn't matter -- air doesn't know the difference.
Tim
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