I have a little project that is relying on a 12v battery pack for deployment. It powers the arduino through the vin pin, which I heard is fine so long as I'm not drawing too much power. I am only using it for switching on an H-Bridge motor controller, so it should still be OK for this application.
Anyways, I plugged in the Serial port to program the Uno when the V-in pin was still connected to the battery. I knew that you could not power the Arduino from the v-in pin and the firewire at the same time, but I did not realize that when one is plugged in, one power source is made directly available to the other. My battery pack has small ON/OFF switch and a small LED indicator, and the LED lit up making me think the battery was still on. In a panic, I flipped the switch on the battery pack, thinking I was damaging the electronics.
Whoooooops. :o The was an arc sound and I heard and smelled smoke. Immediately I switched the power back off and disconnected everything. I looked up the documentation on the v-in pin and later realized what had happened. Upon further diagnosing of the Uno board, I realized that the 5v power regulator must have failed, because now it was outputing 8.8v when powered through v-in, but the 3v3 pin seemed to work fine still.
So far, everything on the Uno seems to work when powered on using either sources. The only thing damaged seems to be the 5v pin. I am planning on continuing to use the controller for my project, but just switch the 5v pin over to a designated digital pin instead for certain modules.
This was a good learning experience for me. For the sake of research and knowledge, I'd like to understand what exactly happened, electrically speaking, when both power sources were turned on. You have two power sources-- one with higher voltage than the other and from what I understand, there is a diode attached to the serial port preventing it from damaging other things like my computer. Where does the excess current flow and what component(s) on the Uno were directly damaged because of it? Why were none of the other pins effected?
And would installing a diode on the output of the battery pack prevent this short from happening in the future?
Not a firewire cable. It kind of looks like one so I got them mixed up. It is just a standard USB A/B cable as included in many electronics kits.
Here is how I had it wired. It is exactly the same except now pin 7 is used as a replacement 5v for the now broken 5v pin. I had a hall encoder of the motor wired up to the 5v pin of the arduino at the time which I am now later discovering that it is busted. I am using a BTS7960 controller for polarity switching on the motor which is connected directly to it. When using the continuity mode on my multi meter, the 3.3/5v positive terminals of the encoder seem to be connected to the 12v output of the battery (Probing 5v+ of encoder to 12v+ of BTS7960 IN terminals). Somehow, somewhere, at sometime this bridge might have busted the 5v pin of the Arduino since that is where I had the encoder's 5v power connected to when all of this happened.
At the time of the actual smoking/short on the Arduino the USB was connected, sending current into the battery in the reversed direction. You can see the little "dot" LED indicator by the battery pack switch. This usually lights up when it's charging or when it is on. I did not have a power supply charging the input terminals of the battery pack while it was connected to Vin but the battery pack LED was on, leading to me think that the battery was flipped ON. When I flipped the switch, that is when the short and smoke happened. There was no indications of damage prior to this event and I know the Arduino board was good. The motor's encoder I am not so sure about... so it's possible that did not help the issue. Regardless, I am sure that having both power sources on at the same time is not good for the hardware.
I don't see why having the battery connected would affect the rest of the Uno. Take a look at the schematic diagram of the Uno. The USB feeds 5V to the output of the 5 Volt NCP1117 regulator while you supplied 12V to it's input via the Vin pin. That's functionally no different to having a capacitor charged to 5V on the output while running on batteries via Vin so shouldn't harm the regulator.
I suspect you've got something else connected incorrectly. Descriptions are hard to follow, can you draw out a schematic diagram of how you've got everything else connected? Hand drawn is fine.
