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Author Topic: DUE losing program while on 3.3V battery  (Read 934 times)
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Michigan
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I am powering my arduino DUE with a 4.2V charged lithium ion polymer battery which feeds a 3.3V low voltage drop out regulator.  There is a 470 uF capacitor on the leads, and I tie it directly to the 3.3V pin.  I get about 3.279V with the multimeter.  After cycling power about 5 times, the arduino DUE loses its program, and I must re-flash the chip.  This happens every time, but does not do that when attached to USB 5V input.  Can the DUE be powered off 3.3V?  I see nothing in the schematic that says otherwise...
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IR Combat laser tag developer/inventor

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Without +5V the 16u2 chip will not be powered, but the 16u2 controls the SAM3X erase line. So when the voltage on the line goes low enough the SAM is erased.
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Michigan
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Hmm, so if I make my own SAM3X board I won't have this problem right?  I'm planning to program w/ an FTDI USB to serial chip w/ my own board and am using the DUE for prototyping purposes.  

So apparently I MUST power on 5V and can't use LiIon now, boo.

Any simple board modification that can work around this so I can use a 4.2V battery?
« Last Edit: April 03, 2013, 07:59:37 pm by zachtos » Logged

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Looking quickly through the datasheet it may be possible to disable the auto-erase functionality in software. If you understand low-level coding the idea would be to enable PIO on port C pin 0, set it as an output and drive it low. However that will make it impossible to upload new code to the board. If I have read the data sheet correctly you should be able to erase the chip by unplugging the board and holding down the erase button whilst plugging the board back in. Try it at your own risk smiley-cool
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