I have a project in mind with Arduino but I wanted to use a battery to power all of it. I have to power a Wi-Fi module and a small servo/stepper for a short period of time, maybe a RTC circuit and a few leds/buttons.

It's something that I will use extensively, so changing battery everytime would be expensive and boring. For those reasons I want to implement a charging solution for my device.

To power it, I've discovered this 9V Li-ion battery (http://anrieff.net/batterytest/details/others/e3-LiIon-9V-en.html), which seems great. The writer claims:

Protection circuit: This battery is not a "pure" 2-cell one; it has a protection circuit. It switches off the battery out of the way when the internal voltage drops below about 6.1V. There's an upper limit as well (or so they claim), but I haven't tested it, it should be around 8.5V.


Li-ion are charged at 8.4V (2 × 4.2V).

So, if I understand well, I need to charge it at 8.4V. My questions are:

  1. I have one of those multi-voltage power supply, which can power 5V or 9V. Can I use a step-down circuit to transform 9V to 8.4V? Charging it with 5V will extend the charging time but charge it in the end?
  2. 1A (max) power supply will work well?
  3. Do I need extra special measures, like some protection circuit I'm unaware of?.
  4. How do I know when to stop charging?

Thank you

  • 5
    \$\begingroup\$ do not charge it with 8.4v. Charge it with a 2S Li charger. Does the battery have a built-in balance circuit? \$\endgroup\$
    – Neil_UK
    Jul 27, 2018 at 12:45
  • 1
    \$\begingroup\$ "Charging it with 5V will [...] charge it in the end?" - No. You must provide a voltage that's higher than the batteries current voltage to charge. And for the 2S-LiPo: Never allow more than 8.4V to be applied to the battery. Never allow more charging current than the battery is specified to handle. In other words, as long as you limit the voltage applied to 8.4V and the charging current to the specified value at all times and you should be good. If the battery already incorporates a balancing circuit that is. \$\endgroup\$
    – JimmyB
    Jul 27, 2018 at 13:21
  • \$\begingroup\$ What a strange LiIon Cell. I was expecting the voltage being a multiple of 3.6 or 3.7... \$\endgroup\$ Jul 27, 2018 at 13:23
  • \$\begingroup\$ "How do I know when to stop charging?" - In fact, you don't. As long as you limit the voltage to 8.4V max, you can continue 'charging' "indefinitely". The question is more, when do you want to stop charging? Common practice is to declare a LiPo as 'full' when the current at maximum charging voltage drops below some threashold like 0.01C. \$\endgroup\$
    – JimmyB
    Jul 27, 2018 at 13:26
  • 1
    \$\begingroup\$ @JimmyB The intersting thing is the nominal voltage, not the charging voltage. But even at 4.2V this doesn't add up to a 9V battery \$\endgroup\$ Jul 27, 2018 at 13:27

1 Answer 1


Since the Battery is Chinese then its description has no details. Does its "protection circuit" limit the charging current to prevent an explosion? Maybe not then kaboom! The person who wrote about this battery showed that he used a fairly complicated circuit to charge it with only 55mA and it limited the voltage accurately. DealExtreme says this battery cannot be shipped to many countries because it is very dangerous (and cheap). Why do you think it can be charged with only 5V? It says it is disconnected when the voltage is less than 6.1V.

  • \$\begingroup\$ Maybe the protection circuit protects only against explosions. It is dangerous because Li-ion batteries are always seen as dangerous goods, they can explode badly. And about 5V I don't understand what are you referring to, I think it can be charged with 8.4V. It stops supplying current when it reaches 6.1V. \$\endgroup\$
    – Cris
    Jul 27, 2018 at 15:01

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