I am working on a concept idea at the moment that I would like to take forward later in the form of wearable computers. At the moment, I am using a Raspberry Pi as the computer itself.

It presently is fairly simple. There are WiFi and Bluetooth Adapters connected, as well as a small shield connected to the GPIO for some status LEDs, the whole lot using very little power. I am unsure of specifics, but a 1A 5v power supply runs it pretty well.

But that requires it to be plugged into the mains, which is not so good for wearable computers.

So, I am presently designing the power system. The concept is simple. 4 parallel sets of power source modules. Each power source module is comprised of 3 AAA (and rechargeable) batteries, as well as an internal circuit to determine how much current each module has available, the operating temperature, battery lifetime, and provide some control functionality.

Ideally, the system will determine which module is running low, throw a warning for a user to replace the battery, and disable it (allowing it to be removed and placed on charge).


question 1) Will 3 AAA (each at 1.5v, totaling 4.5v out) provide enough voltage to power up the Pi and its Peripherals? Or will I need to beef it up a bit further? (I have gotten a Pi with a few peripherals to run on 3 batteries, but they felt a little warm after about 15 minutes)

question 2) Can a power system that is designed with a "hot swap" like power module function be safe to operate (very low voltages), and allow for effective continuous operation of a pi module?

  • I'd go with a higher voltage batteries, and add an DC-to-DC convertor. That way you'd have a very constant voltage. As battery voltage drop as it's getting empty. I'm also unsure how batteries react if you put empty ones (low voltage) in parallel to full ones (high voltage), though some schlocky diodes should prevent current passing between battery packs. Secondly I'd suggest using some rechargeable battery packs. Preferably lithium, as they store a lot more power for a lot less weight.
    – Gerben
    Nov 25 '13 at 16:33
  • Well, I have been looking online, and I have found nicer battery packs in the form of NiMH or LION that can create 6 volts, which would be better. I think at first, I will just build it with one power module (with diagnostic circuits), and when it is running low, it will start bleeping at me/the wearer. Ideally, having multiple power packs for redundancy would be great, but I don't want damage to be incurred because of running differently charged packs in parallel and loosing charge. With each module being surrounded with diodes and isolated with relays, and a power merger that should be good
    – topherg
    Nov 25 '13 at 16:44
  • I don't think there are DC-DC converters that can convert 6V to 5V. Most need at least a few volts of difference. You could place 2 packs in series to generate 12 volts. Or get some 9.6V packs, which are very common in the RC word.
    – Gerben
    Nov 25 '13 at 19:04
  • 1
    FYI, I've discovered with my Pi that I can remove the micro-USB power connection while my Pi is receiving power from the main USB-A connection maintaining continuous operation of my Pi. Therefore if you have two USB battery power sources you can disconnect one while the second provides power. Then reconnect the first one after it's been charged and disconnect the second to charge it. In other words flip-flop supplying power from the micro-USB and the full USB. Dec 4 '13 at 22:07
  • @HeatfanJohn Ooh, thats a really nice idea, and have a little current feedback circuit on the box, so when it detects it gets low (or, the act of it cutting out releases a relay that activates the second one, maybe with a small capacitor to try to keep the pi online in that time)
    – topherg
    Dec 5 '13 at 9:50

answer to question 1)

the simplest solution I think is usage of a cheap USB Backup External Battery Pack like this:

Anker® Astro 3E 10000mAh Dual USB Backup External Battery Pack

it provides proper voltage and shows current charge status. This pack provides 37Wh but there are much bigger packs out there.

answer to question 2)

maybe an

USB 2.0 Micro 5 Pin to A Male Data Power PC HDD Y-Cable

commonly used for USB disks could help to swap 2 battery packs. Prior to testing this make sure it's allowed to connect 2 batteries of the above type shortly for doing the swap.

  • Wow, how long would a 37Wh pack keep a Raspberry Pi (no usb devices or ethernet, just linux running a simple looped test) running? (would I be right in assuming about 7 Hours, assuming the pi is drawing 1A consistently). Also, would having several of those battery packs in parallel at different stages of charge (and the sudden removal) cause a problem (even if each battery is surrounded by diodes)?
    – topherg
    Nov 26 '13 at 18:34
  • 1
    some interesting facts about uptime with Astro 3E 10000mAh can be found here: (raspberrypi.org/phpBB3/viewtopic.php?t=55216&p=421276) Raspberry Pi • View topic - RPI + 10,000mAh Battery Running Benchmark
    – sparkie
    Nov 26 '13 at 21:48

I've build an independent Raspi. A 12V 4.5Ah Battery is suitable for 18h. But, no matter how big the Supply is ... at some point you have to halt the Raspi. I used a microcontroller and a voltage divider to determin what's left of the battery, since there was also a Solarpannel involved which recharged the battery ... and I needed something to turn the Raspi on again. You could monitor the voltage directly with the Raspi in order to shut it down and prevent dataloss or worse. As someone else already mentioned, a DC to DC Voltage converter is a very good idea - get a stepping DC to DC Voltage converter, since every Watt counts.

  • are you willing to share your schematics on how you measure 12 V DC on the pi and how you power it from this source? I'm having some issues trying to make the same. Thanks.
    – otmezger
    Feb 17 '14 at 2:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.