I have built a Raspberry Pi NAS and it is powered by a lithium-ion battery through a 5 V step-up module - the one used in powerbanks. The battery is a 18650 Li-Ion battery which is kept continuously at 3.7 V by an adjusted LM2596 module.

I am not directly powering the Pi through a 5 V mobile charger because of two reasons - brownouts and blackouts will cause the NAS to reboot. Second, no cheap-ish mobile charger's output is free from AC ripples and Y-cap leakages, which might shorten the life of the Pi. And it'll be overkill to buy a well-constructed Apple charger that's fairly foolproof.

Now, I was wondering, is it harmful for the lithium-ion battery to be kept at 3.7 V? Does it cause continuous trickle charging, that is harmful for lithium-ion batteries? Or does the battery take zero current once it's at 3.7 V?

The circuit is like this:

circuit diagram!

  • \$\begingroup\$ What purpose does the battery serve if the Pi is getting a step-down 12V voltage input? \$\endgroup\$
    – Jay
    Commented Feb 28, 2020 at 21:13
  • \$\begingroup\$ Powercuts, brownouts protection. \$\endgroup\$
    – Bagho
    Commented Feb 28, 2020 at 21:19
  • \$\begingroup\$ If you put a diode in front of the battery cell to protect against reverse voltage from the 12V supply, then this may be okay. I would consider looking into a UPS charging circuit where you are able to charge the battery when you have an external supply plugged in. \$\endgroup\$
    – Jay
    Commented Feb 28, 2020 at 21:24
  • \$\begingroup\$ There are inexpensive "smart" charger IC boards that might be safer, e.g.amazon.com/Anmbest-Charger-Protection-Lithium-Battery/dp/… , though 3.7 V is not 100% charged and theoretically should be safe. \$\endgroup\$ Commented Feb 28, 2020 at 21:27
  • 1
    \$\begingroup\$ It's to protect the battery and the LM2596. You may not always get a perfect 3.7V. So current will backflow into either the LM or battery depending on which one has the lower voltage potential. \$\endgroup\$
    – Jay
    Commented Feb 28, 2020 at 21:30

1 Answer 1


does the battery take zero current once it's at 3.7 V?

Yes. Or to be more precise, once the battery voltage reaches the regulated power supply voltage the charge current will reduce exponentially until it reaches the battery's leakage current (which is very low).

Can I keep Li-Ion battery 'charging' continuously at 3.7 V?

Yes, you can do that. But the battery will only be charged to ~10% of its rated capacity.

Apart from the obvious capacity issue, this could be a big problem if the power supply input is removed and the battery is not switched out of circuit soon after. If the battery drops below 2.5 V it will be permanently damaged and won't charge. When you reconnect power the LM2596 will try to charge it with up to 3 A of current, which could damage it if the voltage is below 3.0 V. Trying to charge a damaged Li-ion battery at high current may cause it to explode!

Bottom line: for safety and to prevent damaging the battery you should use a proper li-ion charger which is designed to 'float' the battery while powering the load. Here's an example:-

2.5A Single Cell Switch Mode Battery Charger with Power Path Management

enter image description here

  • \$\begingroup\$ Thank you so much for your answer. That is a very important thing you addressed. As for the undervoltage, the power bank boost module automatically cuts off once the battery reaches 3.2V. And that happens after 30 -45 mins when the battery is at 3.7 V initally. I only need at most 5 mins until the generator kicks in. \$\endgroup\$
    – Bagho
    Commented Feb 28, 2020 at 21:55
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    \$\begingroup\$ Even though your booster automatically cuts, the LM2596 may draw some current when not powered. So long as you keep it powered continuously (apart from short blackouts) it should be OK. But I would set the LM2596 to a higher voltage (eg. 3.85V for 50% charge) so the battery will have enough capacity to handle longer blackouts. Remember that the battery voltage under load is lower than at rest or 'floating', so 3.7V under load may be equivalent to ~4V unloaded. Higher input voltage also makes the buck and boost regulators' jobs easier because the current is lower. \$\endgroup\$ Commented Feb 28, 2020 at 22:23
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    \$\begingroup\$ @Bagho: Please take care to verify the recommended charging voltage of your battery. Whereas 3.9V would correspond to a moderate 70%-ish charge for NMC, with LFP cells you would be risking a thermal event. \$\endgroup\$
    – doynax
    Commented Feb 29, 2020 at 2:10
  • 1
    \$\begingroup\$ LFP stands for LiFePO4 = Lithium-Iron-Phosphate. It reaches full charge at ~3.65V, but quickly drops back to ~3.3-3.2V on discharge (nominal voltage is 3.2V). If your battery follows this voltage curve then it is LFP, not Li-ion. But 18650 size LFP cells have capacity of 1100-1600mAh, so to get 2100mAh you would need 2 cells in parallel. Is your battery a single round cell? How did you charge it for the discharge test? \$\endgroup\$ Commented Feb 29, 2020 at 22:38
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    \$\begingroup\$ OK, so it's definitely Li-ion. \$\endgroup\$ Commented Feb 29, 2020 at 22:58

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