I have purchased and received this battery from eBay. Upon opening it delivers 3v as oppose to 12v-10v and it has a petty output current. I am going to assume this is because it has not been charged much. I have a dc adapter similar to this and i am wondering if i can plug the adapter into the female plug of the battery to let it charge. Would this work and if so what voltage should i set the adapter to for charging?


closed as off-topic by pipe, Voltage Spike, Marcus Müller, Wesley Lee, Dmitry Grigoryev Mar 30 '17 at 8:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Questions on the use of electronic devices are off-topic as this site is intended specifically for questions on electronics design." – pipe, Voltage Spike, Marcus Müller, Wesley Lee, Dmitry Grigoryev
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    \$\begingroup\$ This is stated in the manual for the device. If you did not get a manual, tough luck, contact the seller because anyone else can only guess. Guessing means that it might break, possibly explode, so I hope no one tries. \$\endgroup\$ – pipe Mar 29 '17 at 18:45
  • \$\begingroup\$ ...Fortunately it says: Short circuit, leakage protective, safety, no need to worry. \$\endgroup\$ – pipe Mar 29 '17 at 18:46
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    \$\begingroup\$ @pipe fortunately, I really like your sarcasm :) \$\endgroup\$ – Marcus Müller Mar 29 '17 at 19:17
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    \$\begingroup\$ Assuming this battery had arrived fully charged - what was your plan to recharge it once you'd used it for a bit? \$\endgroup\$ – brhans Mar 29 '17 at 19:21
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    \$\begingroup\$ Unless you want to risk burning stuff down, get a Li-Ion charger. I've had a Li-Ion battery catch fire on my desk due to being improperly charged - its not much fun. \$\endgroup\$ – brhans Mar 29 '17 at 19:47

If your battery delivers only 3V (while being obviously 3-cell battery), it means it is fully discharged and possibly damaged because of overdischarge. It may or may not recover if properly charged.

Before asking questions, you should familiarize yourself with methods of charging Li-Ion batteries. If you do so, you would discover that Li-ion cells must be pre-charged with small current (100-200mA) first until they get to about 3V level (each, meaning ~9V for your battery). Then the charge should go at constant current (0.5C is considered as safe level if battery specifications are nor known). Then, when the battery voltage reaches certain level as defined by details of the particular battery chemistry (4.2V can be a safe assumption, giving 3x4.2=12.6V for your battery), the charge should proceed at constant voltage (12.6V, or 4.2V per cell) until the current drops below 100-200mA. This would conclude the charge cycle.

Now it is up to you to decide if your "suggested method" fits into above description. Keep in mind also that the cells in the battery pack are connected in series, so the voltage on inner cells might be distributed unevenly, causing imbalance, overvoltage, and undercharge. For optimal performance and longevity, batteries must have a more sophisticated, "balanced" charger, with more than two connected wires.

  • \$\begingroup\$ Great answer! Notice that 0.5C depends on accurate knowledge of C, the nominal battery capacity. It's sadly pretty common to see batteries be printed with capacities they don't have. By a factor >=5, sometimes. So yeah, go for 0.1C if you can't trust the battery \$\endgroup\$ – Marcus Müller Mar 29 '17 at 19:57
  • \$\begingroup\$ This answer assumes that the pack does not contain charging circuitry. Judging from the description "Over current, over charge, over temperature, out of date protective.", it does. \$\endgroup\$ – pipe Mar 29 '17 at 20:28
  • \$\begingroup\$ @pipe, do you read Chinese? Commendable... However, even if the cells include some silicon protection circuitry, it is not a charging circuit. The protection circuits usually are voltage-current monitors that shut off a pair of MOSFETs. I am not familiar with any simple battery pack that includes true charger, constant current to constant voltage. Again, after the charge is done, the charger must be bypassed by another low-impedance MOSFET. This is clearly cost-prohibitive. \$\endgroup\$ – Ale..chenski Mar 29 '17 at 20:44
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    \$\begingroup\$ @pipe I agree with Ali. The battery may have protection circuitry, but it is very unlikely that it contains circuitry to properly charge the battery. I would also like to echo the fact that the cells are most likely unusable if the battery is only 3V. Generally 2.5V per cell, maybe down to 2V is as far as you want to go. I believe at around 1.2V it starts plating lithium and once that happens the cell is useless. \$\endgroup\$ – Redja Mar 29 '17 at 21:31
  • \$\begingroup\$ ... and all of this is why I commented that I hope no one starts to guess what the heck that device contains. \$\endgroup\$ – pipe Mar 29 '17 at 21:33

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