6

I've tested the Pi with a single 3.7V 18650 battery and this < 5$ power boost/power bank.

It works, and this single battery runs for at least 1.5 or 2 hours.
(Sidenote: during my tests, the output of the powerboost circuit was more like 4.2V than the expected 5V, but still, it worked!)

Is it possible now to power the Raspberry Pi with 2 rechargeable AA batteries?


(source: ictradenet.com)

Are there some power boost than can go 2*1.2V = 2.4V ------> 5 V ?

4

3 Answers 3

4

The Pi itself only needs 3V3 and less for its circuitry. It's supplied with 5V because USB peripherals and HDMI expect 5V.

If you don't need anything connected to the USB or HDMI then powering from 3V3 seems quite reasonable.

I think the chap who sent a Pi up into the upper atmosphere hoisted by a weather balloon cut out all the 5V circuitry and powered from 3V3.

9
  • What batteries did that bloke use? Most likely LiPo, which supply a a nice constant voltage during 85% of its capacity. Alkaline(non rechargable) or NiMh (recahrgable) batteries drop as its power depletes, which is a big problem and hence not used for professional battery systems, like RC Racing or Smart Phones. (But NiMh is becoming popular again because of the new atomic level cell structuring though, which can provide 3-6 times more capacity on the same size and weight of battery - more power output and much faster charging)
    – Piotr Kula
    Commented Apr 24, 2015 at 20:47
  • @ppumkin I'm not sure. This is the chap daveakerman.com/?p=1154
    – joan
    Commented Apr 24, 2015 at 21:00
  • Yea. 4xaa lithium batteries (lipo) with some custom regulator. Not much details but lipo was mentioned.
    – Piotr Kula
    Commented Apr 24, 2015 at 22:01
  • Hello @joan, I am curious about the 5V situation. In answering this I tried to read up a little more - and wonder now about the BCM's internal SMPS for the core voltage (feed in via the VDD_BATx pins). You happen to know anything about that?
    – Ghanima
    Commented Jan 1, 2016 at 15:24
  • 1
    @Ghanima This thread probably contains the definitive information for powering the Pi from less than 5V. raspberrypi.org/forums/viewtopic.php?p=863978#p863978. Of course this is a pretty specialised usage.
    – joan
    Commented Jan 1, 2016 at 15:35
3

You can use a DC to DC circuit, but bear in mind.

Taking Ohms law into consideration though, it will cut your tested time in half because it needs more amps to produce a higher voltage.

So this means that if the Pi requires 500mA, then the DC-DC on the battery converter actually needs 1000mA from the batteries, more or less. Since 2.4 to 5v is roughly half.

The next problem with powering 3v3 is that you need it be 3v3 constantly, because once the batteries get to about 3v it will start to brown out and rebooting your Pi, cutting your actual time into something more like 1/4 of what you'd expect.

The voltage regulator on the first Pi's was a NCP1117-3v3 which claims it can take up to 20v maximum input, I wouldn't try this because allot of changes are required to make that work properly. It also claims that it can handle a 1.2V maximum drop out voltage, but that comes with a price too, read further.

It is true that the input 5v is generally used for USB and HDMI... but that was not a design choice by Pi, but a world wide standard that is used in electrical designs. The 3v3 regulator works best when its powered by 5volts, not by 3v3 (unless its constant which a battery wont provide)

Taking that in to consideration, it would be best to connect 4*1.2v (rechargeable batteries) and connect it to 5v rail (USB), but you shouldn't use HDMI or USB devices to help prolong the battery life.

The 3v3 regulator will work at its best then, giving you a much better result in the areas of 75-85% efficiency.

If you connect 3 x 1,2 you will only yield about 40-50% efficiency on the 3v3.

This is obviously taking into account you are using 2xAA batteries with limited mah - If on the other hand, you batteries with 10 times the capacity, you could use 3.6volts because the batteries will be able to sustain an acceptable voltage for much longer... but still only towards the high 50% efficiency.

Once the voltage starts to drop, more power is required by the regulator to sustain the output voltage (ohm law again), and plotting a graph you would produce a parabolic curve over time, with an extremely sharp downward trend under 3.3v and a complete cut out near 3v.

If you interested in a true battery operated Pi that wont cause you headaches, look at my other answer. The initial investment surpasses the return value, since my Pi can run for many days if you adjust accordingly. You dont need to buy all the equipment, you can replace my solution with 8xAA reachable batteries instead, but investing in that will only return a fraction of power up time because of smaller capacities.

2
  • 1
    Thanks for your helpful answer. What kind of DC DC converter would you suggest? I really need +5V as well as +3.3V, because my embedded device will need to interact with USB devices.
    – Basj
    Commented Apr 25, 2015 at 12:51
  • UBEC is great. Like on my linked answer but that is DCDC step down. So high voltage to regulated. To go up search for DCDC step up.
    – Piotr Kula
    Commented Apr 26, 2015 at 9:50
2

Yes you can power the Raspberry Pi from 2 AA batteries. I don't know if your boost converter could do it.

Adafruit's Minty boost should be able to.

How long it will run depends the batteries you use and the efficiency of the boost converter you are using. This page goes into the details of how to calculate it.

The rough result from that page is that for 2 alkaline AA's you could get about 1000mAh at 5 volts. My Raspberry Pi (Model B rev.2) draws about 430mA when idle so I would guess you might get about 2 hours.

1
  • More like 40 minutes being optimistic.
    – Piotr Kula
    Commented Apr 24, 2015 at 20:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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