Will a Duracell 9V battery power my Raspberry Pi (plus camera module) for 24 hours?

I've tried power banks and the 50,000mAh rating seems a bit exaggerated, considering I'm getting just 5/6 hours from my Raspberry Pi on the 1.0A USB charging port.


5 Answers 5


The short answer is no, a 9V battery will not power a Raspberry Pi for 24 hours.

Firstly, it is the wrong voltage. RPi's require 5V (see Link). It's possible to build a regulator that will step the 9V battery down to 5V, but that probably won't be much use as explained below.

Secondly, it doesn't provide sufficient charge. A typical commercial 9V battery has a 400-600mAH charge. Raspberry Pi B+ current draw will depend on your use and the appliances attached (such as your camera). Some sample draws are shown at What's the current draw and supply voltage tolerance?. These show that a conservative draw would be 350mA or so, but can go much higher.

You don't get the benefit of the full charge. Duracell provides a nice performance spec for their MN1604 9V battery (search: Duracell coppertop nm1604 datasheet). The voltage regulator will also not be 100% efficient.

Divide you supply by your usage, multiple by .5 (give or take), and you'll get a good estimate of how long you can run your RPi. 500mhA/300ma * .5 = .83 hours, or about 50 mins. If you're taking lots of pictures, I'd expect draw to be much higher. if it's cold or hot, you'll get less of the charge from the battery.

I think you'd be much better off buying a commercial battery pack that meets the power requirements of the references above.


Not sure how long it will last, I'm guessing it will have a VERY short life span... If you want to get the most out of it, you should be looking into a DC-DC / buck converter. You can buy them from ebay for a dollar or so. These converters will actually convert the 9V to alternating current, transform them down then make them DC again. Efficiency is usually north of 90%.

If you use something like a LM7805 it will work, but the 7805 will basically "dissipate all extra volts into heat" which you do not want.

I'd also look into underclocking the Pi to save power.


Haven't tried this myself yet, so this is a little bit theoretical.

The raspberry runs on 5 volts, not 9. Not really sure about autonomy of a battery powered pi, but first you would need to reduce the input voltage. Then you need to be able to supply the pi's required milliamps (check http://www.raspberrypi.org/help/faqs/#powerReqs for your particular model), and last but not least, would be the matter of how long it would run (which is related to the mAh of your supply).

The math is very simple, check this online calculator. Basically it's:

(mAh / mA) * 0.7 = hours of battery life

where mAh is the battery's mAh, mA is the current consumption, and 0.7 is a security factor.

I'm currently trying to achieve this by using a 6V battery and a voltage regulator to take it down to 5V. If I succeed I will share it here!

Also check for example this pi battery pack from adafruit: http://www.adafruit.com/product/962

  • Edited the answer to add the math for the battery life calculation and also corrected the pi's current consumption data. Commented Jan 13, 2015 at 19:09

You don't specify which kind of battery, this could range from the 9V used in smoke alarms (which definitely won't work).

As others have pointed out 9V will not work. If you are serious you should consider purchasing a UBEC - you will find lots on eBay.

Depending on exactly which Pi model and what you have connected you will need ~500mA, so you need a battery with at least 24*500 i.e. 12,000 mAH


You would need a regulator to drop the voltage to 5V, and then some way to connect it to microusb and I also doubt it would last very long. I've used a anker astro power bank (either 15,000 or 20,000) before and it lasted overnight but that was without a camera module or any extra peripherals. Maybe someone out there knows the math to work out what would be best?

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