I'm thinking that if the comparator has a 3. If nothing is connected to the VIN pin, then the VIN pin voltage stays at 0V, which results in the comparator outputting 0V, so the 0V gate voltage at the FET helps to route any USB voltage to the 5V pin. ![]() So if the VIN pin has an applied voltage on it, then the comparator is meant to output a high voltage, which stops the FET from routing any USB voltage over to the +5V pin. With this comparator scheme, it looks like the output of the comparator drives a p-channel FET. 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). Arduino Mega 2560 Rev3 Pinout, Atmega2560 Pin Mapping, EAGLE Files, Schematics, and More Here you will find all of the technical documentation and support files for the Arduino Mega 2560 Revision 3. 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. ![]() The Mega uses around 50ma of that leaving less than 90mA (max) for everything else. With 12V into the regulator the max current is about 140 mA (1W / (12V - 5V)). ![]() The recommend max power dissipation for the regulator is 1 Watt. I've been reading information about this board and currently focusing on this piece of information: The on board 5V regulator is not heat sinked so will supply limited current before it overheats and shuts down. Hi there! I'm fairly new to using arduinos and my first ever arduino is the mega 2560 R3.
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