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I got a cheap powerbank, which didn't work at all, so decided to look inside it. There is a control IC AP5900, what I found about it is a chinese website: http://www.chipown.com.cn/en/case.asp?id=24 They included a reference design there with schematic: schematic Now I measured the battery voltage in turned off state, which was 1.7V...pretty unusual for a Li-ion battery, expected at least 3V. So I tried to charge it and measured the voltages during charging. The control IC gets 4,9V measured on pin 6 (referenced to GND pin 8), but the voltage on the pin 5 (so the charge voltage) was only 0,9V (referenced to pin 8). The battery voltage was still 1.7V, so I measured the voltage between pin 5 and battery positive and there was the 0,8V differnce, so its not connected on the PCB as on the schematic.

Can you give me any tips what could be wrong with it or how to try to find it out?

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  • \$\begingroup\$ The battery pack might be bad, the output of the boost converter could be shorted (caps or elsewhere) or the IC itself could be bad. You could temporarily remove D1 to see if the charge voltage comes up, or use an ohmmeter to see if the output is shorted. The link to the actual datasheet seems to be broken/not working so I can't tell much more. \$\endgroup\$ – John D Jun 3 '18 at 21:13
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    \$\begingroup\$ Battery at 1.7V is dangerously low. Some ICs refuse to try to charge at this level. TRY adding a say 1K resistor from 5V input to battery positive. This provides a few mA of trickle-up charge to the cell. Even 100 Ohms probably OK (I = V/ R = (5-1.7)/100 ~~= 33 mA. Disconnecting cell from cct at positive terminal better but probably not needed. IF cell does not ris in voltage to say 2.5V over time then probably dead. Charging cell at low V MAY cause "vent with flames" but usually not. YMMV. \$\endgroup\$ – Russell McMahon Jun 3 '18 at 22:05
  • \$\begingroup\$ Does the powerbank has a single cell, or two or more cells in parallel? \$\endgroup\$ – Ale..chenski Jun 4 '18 at 2:38
  • \$\begingroup\$ it was a single cell \$\endgroup\$ – U.L. Jun 5 '18 at 11:25
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Your Li-Ion cell is in a very discharged state. When the cell voltage is below 2 V, normal charger would use so-called "pre-charge" mode, trying to charge the cell very slowly, at 20-50-100 mA rate, until its voltage gets up to ~2.5 V. You might need to exercise more patience and wait, monitoring the actual cell voltage. If it doesn't creep up in 20-30 mins, your cell is likely dead.

Note that even if your power bank would recover eventually, don't expect any good capacity results - overdischarged Li-Ion cells usually suffer unrecoverable loss of capacity, down to nearly zero.

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    \$\begingroup\$ In a previous job I worked in consumer electronics design. Most of our products used a single lithium pouch cell for power. We always fully discharged the cell as part of our acceptance test. After a normal discharge, we would bypass the protection circuit and leave a resistor load connected for a long time (overnight). Then re-connect the protection circuit and attempt to recharge the battery. If we were not able to recharge it, the test result was fail and the battery was rejected. There was permanent capacity loss, but not "almost down to zero." The device was still usable. \$\endgroup\$ – mkeith Jun 3 '18 at 23:29
  • \$\begingroup\$ The charger IC in the system would use a "pre-charge" algorithm for recovering severely discharged cells. It is definitely not a good idea to charge a severely discharged cell at the normal charge rate. \$\endgroup\$ – mkeith Jun 3 '18 at 23:30
  • \$\begingroup\$ The OP states that the voltage on pin 5 is 0.9V when the charger is trying to charge- If the open circuit voltage is 1.7V and it drops to 0.9V when the charger starts something is really wrong. In recovery mode I would expect to see the cell voltage at or above the open circuit voltage, not nearly half of it. \$\endgroup\$ – John D Jun 3 '18 at 23:38
  • \$\begingroup\$ @mkeith, the loss of capacity of over-discharged Li-Ion cell depends on how long it was in the under-voltage state. If it was sitting in a subtropical warehouse for a couple of months in this state, the capacity might really go down. \$\endgroup\$ – Ale..chenski Jun 3 '18 at 23:52
  • \$\begingroup\$ @AliChen, that makes sense. When we did the test, we charged it back up within around 1 day or so of discharge. \$\endgroup\$ – mkeith Jun 4 '18 at 0:40
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The solution was in Russell McMahon's comment. The charger IC refused to charge it at that low voltage. So I used a 5V supply and a 100 ohm resistor in series (to limit the charging current to ~33mA) and charged the battery to 3.0 V (to avoid problems I disconnected it from the PCB). Then I connected back the battery in its original place and tried to charge it as normally. It charged the battery without problem and it's working as it should.

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  • \$\begingroup\$ Better documentation of the control IC would have helped me to solve it alone, but all I could find was the chinese site above and chinese documentation \$\endgroup\$ – U.L. Jun 5 '18 at 12:18

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