[cFyrute]

It's not a terrible mistake. A terrible mistake would have been having such power available on ports that even a reasonable person might short out by innocently connecting a USB C cable between them.

A couple 5.1k resistors add about $0.00001 to the BOM cost. The terrible mistake is on the designers of devices who try to forego these.

[willis936]

It's really not the BOM cost that drives these decisions but the assembly cost of adding placements. Time on the assembly line is very valuable and doesn't have a simple / clean representation on a spreadsheet. It's dependent on the market and right now assembly line time is worth a lot.

[f1shy]

That is exactly the reality. I work in a place where we build HW. The resistor costs almost nothing. But installing it, having it in stock, making sure the quality of yet another component is correct, eventually managing another vendor, all costs. So much, that the cost of a resistor we put a value of some cents (up to ten) even when the part itself cost so little that the software has problems tracking it.

[derkades]

Except that connecting 5V to 5V does not cause a short circuit. No current will flow without a voltage difference. If there is a difference, the capacitors in one of the power supplies will charge up to the same voltage and then current stops flowing again.

[zettabomb]

That would be true if both sides were exactly 5.0V, but they're not. There's a 5% tolerance, from 4.75V to 5.25V, and in practice you will see chargers often run "5V" at 5.1V intentionally, to account for resistive loss in the charging cable. If you accidentally connect your "5V" device to the host's "5V" you may find that the host simply disables the port, which has happened to me more than once. So no, you can't just blindly connect them together.

[f1shy]

Yes you can. They will not blow. Nothing will happen.

[zettabomb]

What do you mean by "blow"? There's often a polyfuse which will trip, and needs to cool down to reset. I haven't seen a normal fuse but I believe it's possible. Efuses are also common, to allow the system to automatically reset as soon as the fault condition is removed.

It's unlikely that anything will be damaged, but the device likely will not work until the issue is resolved.

[f1shy]

No it will not. Have done it thousands of times. No no no. You can downvote all that you want, but you will still be wrong. It will happen nothing. Period. If you do not know about it, educate yourself before downvoting and commenting about fuses.

Because of all these supplies work with transistors they do not act as a load to the other. Is like if the 2 had a diode in the output (in fact they do have one, but not directly in the output).

This is my typical experience in HN lately, is getting full of people with absolutely no idea what they talk about, and are constantly downvoting good comments.

[zettabomb]

>Because of all these supplies work with transistors they do not act as a load to the other.

This is utter nonsense, Ohm's law doesn't magically stop working with a transistor. I do know about this, I've designed power supplies and USB devices, and I've destroyed more than a few components accidentally by connecting two switching supplies together. Yes, there will be current flowing, and yes, sometimes a fuse or breaker will trip, I have experienced this many times, and just because you haven't doesn't mean it doesn't happen.

>Is like if the 2 had a diode in the output (in fact they do have one, but not directly in the output)

Sounds like you're referring to either ESD protection diodes, or flyback diodes, neither of which do anything in the case of two similar but unmatched power supplies.

I'd advise you to get a degree in engineering (as I have), or do some serious studying, as this kind of uninformed discussion is not productive or helpful to anyone, it's just noise.