Hello, We have broken in 2 weeks 2 mega boards by having short problems or led draining too much current from pin out with 1k2 resistor.
The problems is when too much current is drained the usb chip breaks. The Windows does not recon Arduino anymore but software on Mega still works.
The Mega board needs protection for usb CHIP.
We had 12v input and usb connector connected while led was on A7 with 1k2 resistor so that 4mA should go, how ever there may have had short since cap was removed because led was too fast with the code and we did try to slow with cap and there fore might left short to the ground which caused hard amperage drain. What breaks before anything is USB chip and the board is not able to download anything anymore.
Please for future boards, please add some protection for the usb circuit.
So, you tried to modify the circuit of a running Arduino, and you fried it. And now you want to be protected from your own stupidity. I get it. You need to be.
Nope Mega is original, we made shield for it. If shield causes too much current usage in any pin, the usb chip is broken for some reason, and Windows cannot connect to board anymore. Fuse is working, program works on mega and all pins work but USB chip is broken. For what I can say is that smoke comes from if you take a look mega while usb connector is left side, the USB chip right side causes smoke to appear. The weard thing is that we actually do not use anything other than Vin and 3.3V and 5V pins. It appears to be 5V that causes the situation. If suddenly 5V is draining too much current something at the USB chip breaks before fuse. This is circuit layout bug. Fuse SHOULD break if too much current drained.
And to be honest, my Instructor in electronics engineering school said that Arduino Power source is "not well defined" which kinda translates to "Amateur". Based on the schematic and what Teacher said it looks like there might be problems.
Jeroi:
If suddenly 5V is draining too much current something at the USB chip breaks before fuse. This is circuit layout bug. Fuse SHOULD break if too much current drained.
The point of the fuse is to protect you from a fire. From what you've described it served its purpose.
I think you are probably overstressing the 5V regulator, it is leading to a high voltage on the 5V line and damaging things.
I don't know why the 16U2 is more susceptible to damage than the 2560 when that occurs.
The fuse only prevents more than 500mA from being drawn from PC USB power. It doesn't do anything to protect parts powered from the 5V regulator.
Why are you putting caps on the IO pins?
led was on A7 with 1k2 resistor so that 4mA should go, how ever there may have had short since cap was removed because led was too fast with the code and we did try to slow with cap
That doesn't sound like you understand what you're doing. Fix your code in this case. Leave the output on longer or something.
There a common saying, attributed to Carl Sagan: "Extraordinary claims require extraordinary evidence".
Here, you're claiming Arduino Mega has some sort of design flaw where the USB chip burns up. That's a pretty extraordinary claim, for a circuit board that's been very widely sold and used for many years by many, many thousands of people.
Apparently we're supposed to accept is your own conclusions, which you've described with inconclusive language like "for some reason", "the weard thing" and "something at the USB chip breaks" which demonstrate a lack of through investigation and detailed understanding. You've presented almost no evidence at all, only a vague agreement by your academic instructor.
Unlikely as it seems, there many indeed be some design flaw with Arduino Mega. But extraordinary evidence needs to be presented to back up such an extraordinary claim. At a bare minimum, complete technical details of the custom shield you've designed (schematic, layout, high-res photos) would be needed. The exact code you're running on Mega, and the exact sequence of steps taken that cause a Mega to fail would also be needed.
I should also mention you've misunderstood the "fuse". You're not alone. Many people don't really understand how it works. The fact that it's called a "fuse" or "self resetting fuse" and the published specs are rating similar to conventional fuses causes a LOT of very common misunderstanding.
It's actually a resistor, with a steep positive temperature coefficient (PTC). It doesn't "blow" like a normal fuse. Instead, the current flow through its resistance causes self heating. Internally, it's made of a high thermal expansion ceramic with suspended metal particles. When cold, the particles tend to touch and form a low resistance. When it gets hot, the ceramic's expansion separates the particles, causing the resistance to increase rapidly.
In an over-current condition, the PTC resistor gets hot. When the power comes from a constant voltage source, increasing resistance decreases the current. But as the current decreases, the PTC's temperature drops, allowing more current. Eventually, an equilibrium is reached, where the PTC's resistance is in series with whatever circuit (or short) is causing the fault condition. The current keeps flowing indefinitely, but at a lower level that the fault alone, because the PTC resistance increased dramatically. It never "blows" and stops the current. It merely adds more resistance in series, which has the net effect of limiting the current, at least in the common case where the energy is coming from a constant voltage source.
PTC "fuses" offer little protection for unusual energy sources, like constant power or constant current. Even with constant voltage sources, the protective effect is merely limiting the current to a maximum amount which can flow indefinitely. It never "blows" like a real fuse. Shortly after the fault condition ends, the part cools down and returns to normal low resistance mode.