1
\$\begingroup\$

I'm rather new to circuits & DIY electronics, but am thoroughly enjoying tinkering with my new RaspberryPi and assorted components. I'm trying to power the device using a few simple solar panels I stripped from cheap LED lawn lights from Menards.

I've combined three 2V solar units in series to output 6V total. The RasPi (B+ model) device needs 5V @ 800mA (including some components I've added). While I expect steady, consistent lighting (will be inside under relatively bright fluorescent lights, but I haven't gotten to measure their luminosity or anything yet), I believe the solar panels can only put out a couple hundred mA.

I have a couple old Samsung Galaxy batteries, which put out 3.7V, 1650mAh. My thought is that a charged battery (or perhaps two in parallel) could provide sufficient current, while the solar panel could keep the voltage in the circuit high enough. But my understanding of the intricacies of circuits (especially many elements in combination) is really limited, so I'm not sure this is possible.

My hope is this: Fully charged, two Samsung batteries in parallel could provide 3300mAh at 3.7V, which could supply the RasPi's needed 800mA for just over 4 hours. But the voltage is too low, so I hope I could "combine" the battery's current with the solar panel's 6V voltage supply "upstream" of the RaspberryPi.

I also know that using the solar to charge the battery is a whole other question, given the RasPi's power needs. Even without any load, at 1650mAh, each battery would take 11 hours to charge, or 22 for both! So I'm not too concerned about that just yet. But I'm curious for now to see what's possible with the pieces I've got.

Thanks in advance for any help or advice! Let me know if I can clear up or expound on anything I said.

Linking two posts I found that I think are related: 1) Multiple Power Sources Each For a Certain Purpose 2) Arduino Project Powered by Solar or Battery

I'm not entirely certain whether they address my questions, but I'm working on it.

\$\endgroup\$
  • \$\begingroup\$ Fluorescent lights are NOT a substitute for sunlight. You will get barely any current from the solar panels without direct sunlight. Plus, your theory on connecting the solar panels for the voltage and batteries for the current will not work, it invalidates some basic electronic principles. However, if you can get 8V of solar panel you can charge two 3.7V in series and regulate the 7.2V output to 5V. This can all be done with reasonable efficiency, but it's unlikely that you will be able to run solely on solar power, you'll just prolong the battery life. \$\endgroup\$ – CharlieHanson Jan 27 '15 at 18:53
  • \$\begingroup\$ Thanks for the note about fluorescent lights, that's something I hadn't even considered! As for the rest, you're right, bumping the solar to 8V and using the batteries in series (or otherwise just obtaining a 5V battery) may be the best way forward. But your comment about invalidating basic principles was what I was afraid of: details I don't know well. I'll ponder this, you may have Answered it. Thank you! \$\endgroup\$ – TCAllen07 Jan 27 '15 at 19:20
  • \$\begingroup\$ @chaaarlie2, what basic principles would be violated? You can put batteries in series for more voltage, you can put solar panels in series for more voltage... why can't you mix 'em? (Mind you I don't think it's a good idea either.) \$\endgroup\$ – George Herold Jan 27 '15 at 21:18
  • \$\begingroup\$ @GeorgrHerold you can't have the solar panels at 6V, the parallel batteries at 3.7V and the RasPi at 5V ... just by connecting them together. If you want the current to come from the batteries, their +ve terminals must be at the same potential as the solar panels, which are providing the voltage. You can raise the batteries' -ve terminal by 2.3V, but how? A potentially wasteful voltage reference? A resistor, hoping that the current through it (i.e. the battery current draw) is constant at all times to provide a static voltage? Then you still need to reduce the 6V to down to 5V for the RasPi. \$\endgroup\$ – CharlieHanson Jan 27 '15 at 21:33
  • \$\begingroup\$ A hint when you power the Pi from battery: It needs just a little current at 5V. Most current is drawn from 3.3V followed by 1.8V, and a little from 2.5V is also drawn. These voltages are generated by linear regulators which waste a substantial fraction of the total consumed power of the Pi. Consider replacing at least the 3.3V regulator by a switching regulator. Just search the web for this topic. \$\endgroup\$ – sweber Jan 29 '15 at 9:14
3
\$\begingroup\$

Don't even think about creating the setup you just described. It is bloody dangerous.

If you wire the "solar cell pack" and the two battery packs in parallel without connecting the Raspberry Pi, you'll get a loop. Kirchhoff's second law explains that the sum of voltages around a loop must be zero. In this case, if you start going around the loop in one direction, you'll encounter the two power sources with opposite directions, so now their difference must be zero - so they must be at an equal voltage. Will this be 6V (dictated by the solar panels) or 3.7V (dictated by the battery packs)? The following will happen:

  • Initially, without sunlight, the common voltage will be 3.7V. No current flows, since the solar panel does not let current flow backwards (its resistance goes near infinity). All is well for now.
  • Then you apply sunlight. The solar panels try to increase the voltage to 6V, but at this voltage the batteries would allow through much more current than they can supply. So the panels drop their voltage to 3.7V, but still begin to charge the batteries with the couple hundred milliamps they can supply, until the voltage in the battery packs reaches 6V. And there's a pretty good reason the batteries are rated at 3.7V.
  • If I learned anything about Li-ion and Li-polymer batteries is that they are very easy to upset. And they especially don't like being overcharged. If they are indoors, they will blow up your desk and burn your house down. If they are oudoors, they'll happily ignite the grass around them. Then burn your house down. Li-ion and Li-polymer batteries are not toys. Don't even think about putting them in a circuit where there's even a slight chance they'll get overcharged.

Connecting the Raspberry Pi before the detonation wouldn't work out well either. The 3.7V combined power supply is not enough for the Pi, which will then do one of the following (I'm not familiar with the Pi's power supply circuit):

  • Pull a lot of current overheating the battery, the solar panel and maybe even its own on-board voltage regulator. It will not boot, or even if it does, it will frequently crash and reboot because of the inadequate voltage. This goes on until one of the components fail: if it's the solar panel, you're safe. If it's the Pi, it's the time bomb scenario all over again. The battery packs also don't like being over-discharged, but as far as I know, they don't burst out in flames then. They just don't work anymore.
  • Don't pull any current at all. Then it's like you didn't connect it at all. Time bomb again.

Creating a circuit which safely combines solar and battery power requires advanced electronics skills and dedicated circuitry. In your case, I would follow S.J. Becker's advice (+1) and buy a (solar powered) power bank from eBay. The circuitry is there, pre-made for you and it can power your Raspberry Pi longer than your setup would have even if it worked. I know they are not as cheap as using things from your parts bin, but they are definitely cheaper than replacing your burnt furniture.

Additionally, does the thing have to be solar powered?

Edit: Some battery packs have built-in protection circuitry that shuts the power off if the battery is overcharged, so there's a chance your setup won't actually ignite but just not work at all.

\$\endgroup\$
1
\$\begingroup\$

On ebay you can buy power banks designed to charge phones. They use 3.7v batteries and a switching regulator. You can also buy them without a battery cheap (as low as $2.50). A perk is that it also allows you to charge your battery from a usb port or phone adapter. They are cheap, they convert the 3.7v to 5v, but not all of them will be capable of 800ma. Also with that big of a load you may want a bigger battery to run longer. Another option is to use a 12v sla battery and a switching buck regulator to make 5v or a car phone charger.

\$\endgroup\$

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.