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So I've built this circuit on my breadboard with an arduino uno and largely have it working. http://makezine.com/projects/make-34/the-dryer-messenger/

Now I'm trying to miniaturize it by using a pro mini; however, I keep blowing them up. I've confirmed that the soldering is fine and there are no shorts on the board since I've powered it up fine with a FTDI friend.

I have 9.2 DCV coming from the wall wart, when I connect the positive to the raw pin and the negative to the ground pin then I get a red flashing light on the arduino, and the wall wart also flashes red. If I leave it on for about 10 seconds, then it blows up and I see sparks.

A picture of the setup: breadboard with arduino pro mini

The arduino pro mini : http://www.sparkfun.com/tutorials/244

The arduino pro mini schematic : http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Dev/Arduino/Boards/Arduino-Pro-Mini-v13.pdf

The dc plug: http://dlnmh9ip6v2uc.cloudfront.net/datasheets/Prototyping/18742.pdf

The voltage regulator stage of the arduino pro mini board voltage regulator stage

Any idea as to what I'm doing wrong?

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    \$\begingroup\$ Can you connect your wall wart supply to an oscilloscope and check what actual voltage and how much ripple it delivers at its DC output? Possibilities that come to mind are either peaks greater than 12 Volts, wall jack providing AC instead of DC, and a polarity reversal. The Pro Mini is rated for input of 3.35 -12 V on the raw pin. Also: Using a lower voltage wall wart, such as an universal 5 Volt phone charger might be a far better idea than a 9 Volt supply - it will reduce thermal stress on the on-board regulator, which needs a mere 300 mV of voltage headroom to operate. \$\endgroup\$ Jun 19, 2013 at 5:42
  • \$\begingroup\$ A picture of your setup may help. And a link to the product page / datasheet of the pro mini of course. Do you have the original Arduino or one of its clones? \$\endgroup\$
    – jippie
    Jun 19, 2013 at 6:10
  • \$\begingroup\$ Here is a picture of the setup. twitter.com/samdelagarza/status/347334221951541250/photo/1 \$\endgroup\$
    – sam
    Jun 19, 2013 at 12:46
  • \$\begingroup\$ I confirmed with a multimeter that the voltage is 9.2v of DC \$\endgroup\$
    – sam
    Jun 19, 2013 at 12:46
  • \$\begingroup\$ I'm plugging red to the raw and negative to the ground right next to it. \$\endgroup\$
    – sam
    Jun 19, 2013 at 12:47

2 Answers 2

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I think you're killing the tantalum capacitor across the input. In fact, it looks like it's bulged up in your photo.

enter image description here

Although the capacitor may be rated at 16V (I don't actually see a rating anywhere, but I'm inferring it from the size and input voltage spec) seasoned engineers will never use tantalum capacitors in this sort of situation. The input current needs to be limited by several ohms and/or the rated voltage should be something like three times the operating voltage. Your wall wart can probably supply several amperes.

enter image description here

enter image description here Years ago, this information was anecdotal, but today you can find it buried deep within the manuals for such parts. They bury it, because it means you very seldom should use tantalum capacitors, and almost never across power supply rails unless the current is limited to much less than an ampere.

Many have been burned by this (perhaps literally). Often failures take place long after product has been shipped, and can cause localized charring, meaning (if your standards are high, as in aerospace or medical) the entire board must be scrapped.

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Since there is a regulator IC in the board, you can feed 9.2V to the board. But the filtering circuit inside the wall wart might not be good. Even though it is DC, there will be lots of ripples which causes unwanted issues in your system. To avoid this, connect a 1000MFD/16V capacitor to the DC jack. This capacitor assures the filtering of ripples from the DC source.

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