Indeed there has been. Best way to explain it is with a handy dandy photo!
[img]https://lh3.googleusercontent.com/-2osm ... 20Flow.png
So in the above example let\'s say your PSU has the following specs
Output: 5V DC, 1A
So from the PSU it goes straight into the R-pi via a DC power jack using the micro usb form factor.
Once it hits the r-pi it goes through 3 protections features (I make no claims that this is the order it goes through internally, I\'m just listing them):
1) Polarity Protection Diode - Keeps the voltage traveling from PSU to R-pi and not the other way around. While damage to the PSU from the r-pi wouldn\'t technically be the RPF\'s (raspberry pi foundation) fault, they\'re really focused on their mission here and are looking out for the end user
2) Voltage Clamp - Basically a surge protector, it makes sure the voltage coming in isn\'t greater than the r-pi can handle
3) Self-Resetting Fuse - Protects from too much current going into device. Technically, I think the purpose for this is more of a feature to protect your PSU than the r-pi. In theory, under normal circumstances, you shouldn\'t receive more current than absolutely needed. This fuse then is to protect the PSU in the event that the attached devices request too much
current. Of course, the r-pi team can\'t know the capabilities of each of our PSUs, so they\'re choosing a limit they think is reasonable and safe.
Okay, so once it leaves the protection features it splits, 5V going to the Soc (where the internal switching regulator steps it down to 1v2 and, I believe, the 3v3 used by the GPIOs) and to the USB devices.
I realize this is a long, roundabout way to answer your question novelty, but it\'s important that we clear the air about this confusion.
So let\'s assume that the self-resetting fuse is 1A as well and we can use every last mA the PSU has to offer. For the sake of argument and just because this is an illustration, let\'s also assume that this is a perfect system and we loose nothing through the entire process
So the power comes in and we immediately use 350 mA to power the r-pi itself, from the moment it turns on it\'s working at 100% power draw and stays that way the whole time. Let\'s also assume that the HDMI port is using it\'s full 50 mA as well (which for the record, I should have drawn one more line coming off the 5V \"rail\" right after the protection features going to the HDMI port meaning that the 5V \"Rail\" goes three places (HDMI, USB, and SoC) rather than the two places I show in my photo).
This will leave 600 mA available for the USB ports to use. Maybe enough to power some 2.5 drives, but it will have a slow spin-up. Of course all of this is just an example! If you use a 2A PSU then obviously you\'d have more mA available by the time you got to the USB ports provided that the fuse allows up to 2A!!!
That\'s really the million dollar question for me at this point. What are the specs of that fuse???
Of course for the more intrepid, you could always try replacing the fuse
[img]https://lh3.googleusercontent.com/-39z5 ... images.jpg