I have arduino mega and Lipo batterry such this and Lipo battery charger 5V USB Boost. I am powering arduino normally from barrel Jack with an 9 Volt adaptor. But if there is power cut occur, I dont want to stop, I want powering arduino with Lipo battery. After power come back, arduino contunie to powering from barrel jack again at the same time battery start to charge. How can I succes this. Do you have any sample circuit?

After Majenko♦ comment, I've tried circuit but there is heating problem occur. More then 70 celsius. Is that normal?


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Termal Circuit

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    Plug the (boosted) battery into the USB socket. Job done. – Majenko Aug 7 at 16:18
  • ;) if i try that, boost is heating much more. Is that normal – mehmet Aug 7 at 17:27
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    No, that's not normal. If the barrel jack power is greater than 7V then the USB power should be isolated. – Majenko Aug 7 at 18:34
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    Majenko, you have it backwards. That boost circuit is designed to have 5V OUT at the USB and <5V in at the other end, so you can use a 1S LiPo battery to charge a cell phone. So, the battery goes to the input, while the power can be tapped from the USB connector output to the 5V pin on the Arduino to power the Arduino with 5V. A Schottky diode should probably go from the 5V out on the boost converter to the 5V pin on the Arduino to prevent backflow as well. Charging the LiPo battery from the barrel jack is another issue. That will require a LiPo charge controller IC. – Gabriel Staples Aug 7 at 22:45
  • @Majenko I've edited question with your solution but termal problem seen – mehmet Aug 8 at 9:15

As is described in this post from the arduino forum, There is a MOSFET switch in series with the USB connector +5V input line. If the battery booster 5V output is connected to the USB input jack, then it will automatically power the UNO whenever power is disconnected from the barrel jack, and will be automatically disconnected whenever at least 7.0V is available at the barrel jack.

If you are experiencing over-heating problems, then you are probably overheating the linear regulator that steps down the barrel jack voltage to 5V. You need about 7-7.5V at the barrel jack to overcome the 2-2.5V loss through the linear regulator, but anything higher than about 7.5V will just cause the linear regulator to get hotter.

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  • thanks for technic answer. If so I have something on my mind, only powering from barrel jack (9v) should make heating problem has to do it but it doesn't work like that with only powering from barrel dont make any heating problem – mehmet Aug 12 at 18:13

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