So I went over the data sheet again (specifically https://www.meanwell-web.com/content/files/pdfs/productPdfs/MW/DKM10/SKM10,DKM10-spec.pdf ), and I’ll try and articulate the reasoning that they’re specifying a 4A time delay fuse.
- Nowhere does it mention polyfuses. As I mentioned, polyfuses are a different beast to a simple wire fuse, and in most cases you cannot simply just replace one with another and expect things to be the same. In some ways they are similar, but in others, they can be very different. This is why I brought it up initially, as it’s quite easy to get tripped up. That said, it does recommend delay type fuses (which polyfuses are more similar to than a say a fast blow wire fuse), which means they expect it to draw more current in certain “worst case” conditions. They don’t seem to mention any specific fuses (specs or part numbers), and certain information isn’t spelled out in the data sheet (see next points) which would make the details to figure this out more apparent.
- The datasheet mentions specifically under Protection → Overload that it will supply up to 140% of the rated output power. This means that if there’s a higher than normal current draw on the output, the unit will supply at least 140% of the rated power at that point. As such, it needs to be able to draw more input current to do this. Using a higher rated fuse will cover that extra current draw. The fact it doesn’t mention any specific max values is a lack of detail issue in the data sheet, particularly the lack on how long the module will supply an output when in an overloaded condition, and therefore how long it would need to draw more current.
- The input voltage range on the modules is 9-18V. The values for max current are based on a 12V input with a conversion efficiency of 87% typical (for the +/-12V module). If you look at the table in the datasheet, point 1 specifically states “All parameters are specified at normal input (E:5Vdc, A:12Vdc, B:24Vdc, C:48Vdc), rated load, 25°C 70% RH ambient”. In essence, they are the typical values (maximums and minimums) when the above parameters are met. At values outside those parameters, the values will change. With lower input voltages (eg: 9V input), that max input current has to go up, otherwise you would not be able to draw the max rated output current. See more below, but once again, I think the datasheet is somewhat lacking here in not explicitly pointing this out (even at the very least by providing input voltage vs input current or even input voltage vs conversion efficiency graphs somewhere so the ratings can be estimated for various conditions).
So, lets work through all this. Remember that we’re considering all the worst case conditions, as the datasheet suggests that the module will perform (at least for some duration) under those conditions. Anything beyond those worst case conditions is what would be classed as a fault, and would be when a fuse needs to kick in, at least from the module designers perspective. There will be overhead allowed as well, simply to cover component tolerances.
As the module is converting voltages, what matters is the overall power draw (we’ll convert this back to current in the end). The max output current is 416mA per output supply rail. That means the max power required is about twice the current times the output supply voltage. So that’s 2 x 0.416A x 12V, which gives you 9.984W. As the module is only 87% efficient typically (more on this later), this means that under full load it will typically draw 11.476W from the supply. This roughly matches the input current of 0.957A when the input voltage is at 12V.
Now, the module, in a overload condition can handle up to 140% of it’s output power. That takes the input max power up to 16.066W at peak before the modules overload protection cuts in.
As per the data sheet, those values are typical values. The real values may be larger. Unfortunately, as mentioned, it’s not quite laid out in the datasheet how things change when you go outside those values, so you usually need to expect a worst case. One example is that the 12V module shutdown input voltage is 8V. This means that if it’s pulling 16.066W at 8V (just before the module wants to shut down), the matching current could be as high as 2A (2.008A from the calculations I did).
All the above also assumes that the module is operating at 87% efficiency, which definitely won’t be the case in extreme operating conditions. If we instead assume 80% (I’m estimating, based on the lowest typical value in the table being 81% - that is just a guess and I suspect it’s even lower than that), then this pushes everything up. Going though the numbers again, that puts the absolute max power at 17.472W, which at 8V is 2.184A. If the efficiency drops to as low as 60%, then that takes the power to 23.296W, which at 8V is 2.912A. This is quite close to 3A, so it may even be worse than that in certain conditions.
I will also note that the input current values we used are for the +/-12V module. The fuse values in the datasheet are independent of the output voltage, so they obviously allow for all the module configurations. Specifically, the +/-5V module actually draws slightly more input current (0.985A), so this would push some of the values higher still, possibly just over the 3A mark at a peak. The next convenient value is 4A, so this is probably where the fuse value comes from and gives some further leeway for any component tolerances.
The module is supposed to be able to recover from this sort of situation as long as those values aren’t exceeded. This means that short bursts of higher input current could be expected when such a condition occurs very close to all the recommended values (ie: just short of reaching them). While this is obviously where the need for a time delay fuse comes from, the fact it’s not stated how long it can continue to work in this mode, nor is a recommended fuse type listed (ie: so we could estimate things ourselves based on it’s specs), is quite annoying, as it leaves a lot of uncertainty.
In addition, the fuse also provides protection for the power supply (and the tracks on your circuit board) if the module fails with an internal short, so long as the supply can produce the current needed for the appropriate time to trip the fuse. In the case of a reversed input voltage, a diode across the module input (as per the circuit you show in the original post) provides close to a short circuit across the module, allowing the fuse to work to protect things in the same manner. In the reversed voltage case, you need to consider how long the diode will pass the needed current, before the diode might fail. If it fails before the fuse trips, then you most likely need to reconsider what diode you are using.
FWIW: A 1N400x series diode should handle a peak current spike of 30A, but only for a very short time (8.3ms, since it’s designed really for rectifying AC). The normal max current rating is 1A. As such, with a time delay fuse, this may not be long enough for the fuse to trip. That said, it’s likely that if the diode fails, it will fail short (though not guaranteed). A 1N5819 is fairly similar, though I suspect it might fare slightly better.
If that’s not possible (for whatever reason, be it cost, availability, size constraints, etc), then you need to reconsider if the extremes that are assumed above are worth handing, or if you need to adjust things accordingly to suit your application. You may decide that handling the reverse voltage protection scenario is much more necessary than handling output overloads and low voltages, in which case you may not need as high a fuse value on the input.
All this said, this is just me taking the values from the data sheet, assuming the worst possible conditions (at least, the ones I can see) that the module is supposed to be able to handle before it shuts down, making some basic assumptions, and then working with them. Only you can determine if you need to take them into consideration.
If there’s an application data sheet (ie: a guide on how to implement the module in a design, rather than just it’s specifications), then it might answer some of these questions. I did look for one but did not find it on the MeanWell site. Beyond this, I suspect the only way you’d get a bulletproof answer on this would be to reach out to an engineer at MeanWell. That said, they may just recycle the rote answer from the data sheet.
