The halved voltage is compared with a 3.3V value. This particular voltage divider appears to be set up to halve the incoming VIN voltage. When the VIN pin is used to power the arduino, it can be seen that the VIN pin leads to a voltage divider (as mentioned earlier by bud). Or if not using the barrel jack (and not using USB power), then we could hook up an external voltage to the VIN pin - maybe in the range between 7V to 12V. I can see barrel jack, and according to the arduino 2560 document, it is best to supply this jack (if using it) between 7V and 12V. I went to take a look at the schematic of the 2560 and see what was discussed above. I was just wondering whether or not there's some mixed up information in those details. Seems kind of contradictory - as the first instance speaks of regulated supply, then the next instance speaks of 'bypassing' a regulator, and advising not to use the 5V pin as a voltage supply. Now, if the information is specifically indicating that the pin OUTPUTS regulated 5V, then may I ask why the information follows up by saying that using the 5V pin as a SUPPLY (5V voltage source) 'bypasses' the regulator. The details above indicates that the 5V pin is an output pin, and is basically a regulated 5V supply rail. Supplying voltage via the 5V or 3.3V pins bypasses the regulator, and can damage your board. The board can be supplied with power either from the DC power jack (7 - 12V), the USB connector (5V), or the VIN pin of the board (7-12V). This pin outputs a regulated 5V from the regulator on the board. I've been reading information about this board and currently focusing on this piece of information: Hi there! I'm fairly new to using arduinos and my first ever arduino is the mega 2560 R3.

Arduino mega 2560 rev3 pinout