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Seemingly everything ok, but the Parkside 20V 2Ah battery is not charging

wujek_demon92 61215 30
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  • #1 19880846
    wujek_demon92
    Level 3  
    Hi. Firstly, I'd like to point out that I remember something from school, but I'm not much of an electronics technician. As in the subject Parkside PAP20 A1 battery. When connected to the charger, 2 red and green LEDs flash. The battery shows 1 dash by itself. I did a bit of reading so got on with it. The first thing I checked was the fuse - ok, nothing looks burnt - ok, cells - ok voltage on individual cells ok continuity of connections as such I checked ok. I lack further knowledge and ideas. I think the problem may be in the charging itself, maybe some transistor. Unfortunately, as I wrote, I have terrible deficiencies in electronics and I can't find a schematic of the whole board. Colleagues, please help.
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  • #2 19880942
    DiZMar
    Level 43  
    And what exactly is your problem? You have described the situation chaotically but you have not presented the problem. Write 2 short sentences about the problem. How long was the charger connected? Is the battery still discharged as it was before the charger was connected?
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  • #3 19880974
    wujek_demon92
    Level 3  
    I tried to add pictures but something is not going. The battery is not charging. On the outputs I have 12.8v and on each cell going from the beginning 3v another 6.2v further 8.18v and 10.37v

    Added after 1 [minute]: .

    The charger probably ran for an hour and no change. I read somewhere that the 2 blinking LEDs on the charger are a faulty battery

    Added after 4 [minutes]:

    I have now connected the screwdriver it does not work. In summary the battery is faulty (not charging and not passing power to the power tools)
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  • Helpful post
    #4 19881143
    DiZMar
    Level 43  
    wujek_demon92 wrote:
    and on each of the cells going from the beginning 3v another 6.2v then 8.18v and 10.37v
    .
    And a total of 12.8v. The first 2 cells have about 3 volts each and the last 3 cells only have about 2 volts each. The conclusion is that the battery should be replaced. The cells should have similar voltages regardless of their state of charge.
    It is possible that the battery is too far discharged and the charger (automation) does not "see" it. A 12 V of a 20 V battery is a faulty battery for the charger. You could try charging it directly with approx. 20-22 volts (approx. 15 minutes) and only then with the correct charger. Maybe it can be saved.
  • #5 19881214
    g107r
    Level 41  
    wujek_demon92 wrote:
    When connected to the charger, 2diids red and green flash.
    .
    Indicates a faulty battery, flashing does not charge.
    Measure the voltages of the individual cells without load, and loaded with e.g. a light bulb from the car.

    https://www.elektroda.pl/rtvforum/topic3691907.html
    You can recondition yourself.
  • #6 19881215
    DiZMar
    Level 43  
    g107r wrote:
    Measure the voltages of the individual cells at no load,
    .
    After all,
    wujek_demon92 wrote:
    On the outputs I have 12.8v and on each of the cells going from the beginning 3v another 6.2v further 8.18v and 10.37v
  • #7 19881255
    yogi009
    Level 43  
    With such a sparse description of the fault, one can bet on the failure of one section of the battery, which is hardly surprising in the case of this Lidl brand.
  • #8 19881296
    g107r
    Level 41  
    That was yesterday, besides, I have a desire to know the voltage drop on e.g. the directly loaded third, middle cell, not the drop on three cells (8.18V) to subtract the first two.
    Anyway, it depends on the author what he intends to do with the battery, as he has already answered the question himself.

    Intensive use of the screwdriver, will require replacing 5 cells, or buying a new battery.
    Occasional use, from time to time, then you could "glass" on the still remaining ones, but this will not guarantee the correct operation of the original charger.
    I use occasionally, and replaced the two dead cells with cells from a laptop battery, because I had and could, and for my requirements, it is enough for me....
    Also, if you want to play around, I think it would be simpler to measure specific cells, not two, three or four at a time (10.37V), but the third one (8.18V-6.2V= :?: ) to select suspicious pieces.
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    #9 19881336
    DiZMar
    Level 43  
    g107r wrote:
    in my opinion, it would be simpler to measure specific cells, not two, three or four at a time(10.37V), but a third one (8.18V-6.2V= ), in order to pick out suspicious pieces.
    .
    In post #4 13 Feb 2022 07:24 I drew my conclusions even after such imprecise measurements.
  • #10 19881341
    kortyleski
    Level 43  
    yogi009 wrote:
    , which I am not surprised about with this Lidl brand.
    .
    And it surprises me. I have three chargers, six Aku and quite a few power tools from them. Everything works perfectly. I dare say at this price it's top shelf quality. On one of the screwdrivers, such a powerful one with impact, even a 20mm drill bit in steel is not impressive.

    wujek_demon92 wrote:
    I have now connected the screwdriver does not work. In summary, the battery is faulty(not charging and not transferring power to the power tools)
    .
    Fair conclusion. But given that a new battery costs less than a hundred I wouldn't mess around with repairs.
  • #11 19881362
    DiZMar
    Level 43  
    kortyleski wrote:
    On one of the screwdrivers, such a powerful one with a hammer, even a 20mm drill bit in steel is not impressive.

    Impact in steel :?: :crazyeyes: :sm31: :sm31:
  • #12 19881402
    g107r
    Level 41  
    kortyleski wrote:
    And it surprises me. I have three chargers, six Aku and quite a few power tools from them. Everything works perfectly.
    I only have one battery, I boast, it was about two years old then.
    https://obrazki.elektroda.pl/9066460800_1618306726_org.jpg
    https://obrazki.elektroda.pl/7831796600_1616356103.jpg
    Same symptom as the author's, only the exploded cell had practically 0V already, and in a while the other one joined too.... Three are still alive, but I know that this may be temporary (in March I took a photo), as well as the life of cells from the laptop :) , but from 9 times I used it so far, of which from 3 times so seriously :wink: .
  • #13 19881409
    adamz74
    Level 32  
    Too uneven voltages on sections, perhaps damaged cells and balancer not allowing charging.

    You need to charge the individual sections with an external 1S charger, e.g.: on a TP4056 module. Only a battery charged in this way can be tested and see how it behaves.

    Attempting to charge lithium cells directly from the power supply without a suitable charger-controller is asking for problems.

    Regards!
  • #14 19881438
    LEDówki
    Level 43  
    The author really does not remember much from school. The battery in question is actually a battery of batteries connected in series (how do you connect one battery in series?). When connected in series, the current flowing through the individual cells of the battery is the same. The voltages on the batteries will depend on their condition - capacity and internal resistance.
    Lithium ion batteries are quite dangerous toys and are therefore used with protection circuits. Such a circuit disconnects the battery (electrochemical secondary cell) from the rest of the system (battery) when its voltage is too high, too low or when the current is too high. Sometimes the temperature of the battery is also checked and it is disconnected when it is too hot.
    If the previous analyses are correct,. then you can try to charge the batteries with a TP4056 charger (one charger per battery) or a stabilised power supply with current limitation. You will charge the batteries and perhaps the Parkside charger will start charging this battery. Further use of this battery will have similar problems if the colleague discharges it too hard and the weaker batteries are cut off.
    The colleague can replace the weaker batteries with new ones, but then they will work when the other (so far working) batteries will cause disconnections. The colleague will eventually come to the conclusion that all batteries should be replaced with new ones of equal capacity and equal condition, or the entire battery should be bought.
    You can verify any suspected damage to the charger by observing its behaviour when a working battery is connected.
  • #15 19881544
    gumisie
    Level 43  
    5S2P battery: 3.7V x 5 = 18.5V
    In this case: 12.8V : 5 = 2.56V
    This indicates something.
    The cells are "drained".

    The only sensible solution, would be to replace all the cells.
    You can "play around" with replacing only the most suspicious ones, but in my opinion this is "Sisyphean work".
    The next one will "fail" in a moment, then the next one, and so on.

    Regards.
  • #16 19881566
    DiZMar
    Level 43  
    gumisie wrote:
    5S2P battery: 3.7V x 5 = 18.5V
    In this case: 12.8V : 5 = 2.56V
    .
    There is nothing to average out because 2 each have about 3 V and 3 each have about 2 V.
    After all
    wujek_demon92 wrote:
    On the outputs I have 12.8v and on each of the cells going from the beginning 3v another 6.2v further 8.18v and 10.37v

    It must be taken into account that they are discharged. If they can be recharged they may yet come back to life. It was a mistake to discharge them beyond acceptable limits.
  • #17 19881599
    g107r
    Level 41  
    The most sensible option is to buy a new battery.
    The other option, replace the rechargeable batteries themselves. From £10 a piece, just what certainty to choose? Maybe it will close with half the price of a new battery. I see some ready-made package allegro.pl/offer/parkside-18v-18650-li-ion-battery-x0172lib-8855022548 for 75
    And the last option, other options :) because you already need and charging toys, and spare parts, and some grasp of the subject, although here you can learn fairly quickly.
    With just a battery and a meter, there is nothing about a soldering iron, I would choose the first option.

    DiZMar wrote:
    The mistake was to discharge them beyond the acceptable limits.
    .
    The protection is supposed to be in the device, the battery itself can be drained to 0.
    We don't know the history of the battery, maybe it was used with the adapter for other purposes as imagined by the manufacturer, or maybe it just wore out and needs to be ...
  • #18 19881660
    gumisie
    Level 43  
    DiZMar wrote:
    3 of about 2V each have.
    And they should not be below 2.7V.
    As far as I know, this battery is a cell arrangement: 5S2P, so not three but six cells to replace.
  • #19 19881769
    g107r
    Level 41  
    Well no, 2Ah is 5 cells, 4Ah has 10.
  • #20 19881788
    gumisie
    Level 43  
    g107r wrote:
    No, 2Ah is 5 cells, 4Ah has 10.
    "Sorry", You're right.
    I got the models mixed up. :cry: .
  • #21 19883129
    wujek_demon92
    Level 3  
    I'm from an electrical not electronics technical school. There was some dry theory about electronics and that was it. I have been involved in installation and installation design for a few years now. That is why I turned to much smarter people. Advice like "go buy a new battery" can be put in your pocket:d I started this topic for 2 reasons. I wanted to fix an old battery and to increase my knowledge of electronics. That's it collectively. Thank you for such a large response and I have additional questions.

    Added after 5 [minutes]: .

    DiZMar wrote:
    You could try charging it directly with about 20-22 volts (about 15 minutes) and only then with a proper charger. Maybe it can be salvaged.


    I understand that you unsolder etc the voltage directly to the + and - of the connected 5 cells?

    Added after 9 [minutes]:

    DiZMar wrote:
    g107r wrote:
    in my opinion it will be simpler to measure specific cells, not two, three, or four at a time(10.37V), but the third one (8.18V-6.2V= ), in order to select suspicious pieces.
    .
    In post #4 13 Feb 2022 07:24 I drew conclusions even after such imprecise measurements.


    I know I am an amateur, but I have tried to give you as much data as my knowledge allows. As my colleague dizmar underlined it is simple mathematics and I guess everyone knows how to subtract 6.2 from 8.18? It's just that my knowledge, or rather my lack of it, does not allow me to determine how big differences in voltage are acceptable on individual levers.

    Added after 4 [minutes]:

    g107r wrote:
    It's up to the author what he intends to do with the battery, as he has already answered the question himself.


    Colloquially speaking pi times door I know what works and what does not. My aim is :
    -to find out if I am right
    -whether it is possible and possibly how to rectify the fault

    Added after 10 [minutes]:

    LEDówki wrote:
    The author indeed remembers little from school. The battery in question is actually a battery bank of batteries connected in series (how do you connect one battery in series?). When connected in series, the current flowing through the individual cells of the battery is the same. The voltages on the batteries will depend on their condition - capacity and internal resistance.

    I don't know if a colleague read from the beginning perhaps the font is too small. I am almost a layman when it comes to electronics and my colleague instead of advising me is clinging to nomenclature.... If I didn't know that the battery was a set of batteries connected in series, I probably wouldn't be able to give the voltage drops. So all my colleague has done is to point out a lack of knowledge, which is probably not the point here....

    Added after 12 [minutes]:

    <span class="notranslate">
    g107r wrote:
    With just a battery and a meter, there is nothing about a soldering iron, I would choose the first option.

    I have the equipment I just lack the knowledge.

    Overall I have received some valuable tips for which I thank you very much. I would still be interested in the possible question of charging weaker batteries. Should I charge all of them together or only the weaker ones or simply replace them?
  • #22 19883199
    LEDówki
    Level 43  
    But I have already answered the question whether to replace the weaker batteries? It will help for a bit and you will have to replace the next ones. Some slight deviation in capacity differences may be there. Now the colleague should charge the batteries and check their capacity. The colleague may also believe that lithium ion batteries that are too deeply discharged are not usable.
    I give information not for clinging but for the knowledge of others. Vocabulary is used for communication and if it is not equal, it can become a serious communication barrier and lead to confusion. On the subject of lithium ion batteries and batteries composed of them, there is little to add.
  • #23 19883253
    g107r
    Level 41  
    Meaning you choose the options for now without buying a whole new battery? The other is still to buy a new pack, or good cells from a de-pack, with which you can probably repair the battery in a fairly simple way, because whether something has died in the charger or the battery PCB cannot be ruled out temporarily. And there is a good chance that the original charger will start charging such a battery, because with different rechargeable batteries it doesn't have to.
    And we don't know the history of the battery, mine was two years old, so quite "normally", after the warranty it stopped working :) in the morning it was gleefully flashing the charger :) which was already less funny....
    wujek_demon92 wrote:
    I understand that the unsolder etc voltage is applied directly to the + and - of the connected 5 cells?
    The battery outputs are connected to the terminals of the extreme cells, so theoretically nothing needs to be unsoldered, except for the sake of sanity about the battery electronics.
    wujek_demon92 wrote:
    Charge all together or just the weaker ones or just replace them?
    .
    And which are the weaker ones?
    In my case it was clearly the second cell that died, the ones next to it had lower voltages, i.e. the over-perishable cell must have already affected the charging of the adjacent ones, one of which survived and still works today, but I use it poorly....
    Altogether two to be disposed of, because the last normal use or charging damaged the second cell, and the first charge finished off the neighbouring one, probably the third middle one, because I can't remember anymore....

    Charge it as a whole, or individually, it depends what you want to charge with, but check the voltages on the individual cells so that none exceeds 4.2V, also it would be best to charge each battery individually, as they will all be charged to the same voltage straight away.
    I charge mine as a whole, 21V to the battery terminals, 0.8A current.
  • #24 19883314
    yogi009
    Level 43  
    DiZMar wrote:
    Under steel


    I have omitted the humorous bit.
  • #25 19883392
    gumisie
    Level 43  
    wujek_demon92 wrote:
    Advice like "go buy a new battery" can be put in your pocket:d I started this topic for 2 reasons. I wanted to repair an old battery and to increase my knowledge of electronics.
    Knowledge can always be increased by reading articles posted on the internet. In your case, in my opinion you should start by reading materials on charging and operating Li_ION cells.
    E.g.: Link .
    The advice that the best solution would be to buy new cells is by all means correct. The rationale has been given in several posts.
    As for the electronics in the battery itself, or the charger. You will not check if it is the culprit until you are sure that the cells are in working order.
    In your battery, they are definitely defective, so messing with the electronics, for the time being, makes no sense.
    Cells with a voltage below the acceptable level (e.g. 2.5V) are suitable for disposal.

    Kind regards.
  • #26 19883953
    kortyleski
    Level 43  
    yogi009 wrote:
    DiZMar wrote:
    Beat in steel
    .
    yogi009 wrote:
    I forgot this humorous passage.


    The excerpt is not so humorous. Because if my colleague DiZMar could read one sentence with comprehension instead of looking for a hole in the whole as usual, he would understand what I wanted to convey. However, you should have listened to this in Polish class in primary school, not just to make up a bunch of farts. Well, I will break this sentence down into two simple ones to clarify the ambiguity.

    kortyleski wrote:
    On one of the screwdrivers, such a powerful one with an impact
    I informed you that the screwdriver has considerable power and an impact function. I did this deliberately about which further on.

    kortyleski wrote:
    On one of the screwdrivers....... ....... even a 20mm drill bit in steel is not impressive
    .

    All the time talking about the same screwdriver.

    Nowhere did I write that I was drilling into steel using the impact function.

    So much for reading skills.

    Now clear as a peasant cow. This screwdriver uses a typical X20V series battery. And it has the performance of a mains drill. This is what I wanted to point out when informing you about both the impact function and the ability to drill into demanding material with a large drill diameter. This means so much that this typical battery has to have quite robust cells to power a powerful tool. I think 30-40 A are realistic currents. And here the cell replacement and its economic sense are not quite clear and simple. Firstly, all cells must be replaced. Secondly, high-current ones will not be bought for 5 zloty from a depakiet. Closer to £20 apiece. Thirdly, you still need to have what and know how to weld the new cells into a pack. And fourthly, at the end of this road, it may turn out that the battery monitoring system is faulty and the whole job will be ruined.... . And a new battery 99 zlotys. Sometimes it really isn't worth kicking down an open door
  • #27 19884322
    yogi009
    Level 43  
    kortyleski wrote:
    I think 30-40 amps are realistic currents.


    I think a cordless screwdriver has currents of 20-30A. Let alone a metal drill.
  • #28 19884377
    wujek_demon92
    Level 3  
    LEDówki wrote:
    Some slight deviation in capacity differences can be.

    What a relative term, and you yourself are talking about accuracy.With the voltage of each separate battery being 3.6v how much is that little? For me it may be 1v for others not much may be 0.1v

    Added after 3 [minutes]:

    LEDówki wrote:
    Now the mate should charge the batteries and check their capacity. You may also believe that lithium ion batteries that are too deeply discharged are not usable.
    .
    It is precisely how specifically I mean.... with what long what current to charge them to how many v. "Too deep" Well ok but how much is it? Because I don't know if 2v is already too deep or if I can try to charge them

    Added after 2 [minutes]:

    g107r wrote:
    wujek_demon92 wrote:
    I understand that the unsolder etc voltage is applied directly to the + and - of the connected 5 cells?
    The battery outputs are connected to the terminals of the extreme cells, so theoretically nothing needs to be unsoldered, except for the sake of the battery electronics.

    Well that's what I meant, wouldn't I damage the electronics by applying voltage directly to the terminals

    Added after 8 [minutes]:

    g107r wrote:
    wujek_demon92 wrote:
    Charge all of them together or just the weaker ones or just replace them?
    .
    And which are the weaker ones?

    Well...
    1aku-3v
    2-3.2v
    3-1.98v
    4-2.19v
    5-2.43v
    Which ones should I still try to resuscitate by charging them and which ones should I immediately howl...throw away? Trying to figure this out:D .

    Added after 5 [minutes]:

    g107r wrote:
    Charge, whole, or individually, depends what you want to charge with
    .
    Well, that's where it's all buried....
    For the whole thing the charger gives 21.5v and 2.4a so how do you charge the individual batteries? Individually they are 3.6v and 2000mah

    Added after 4 [minutes]:

    gumisie wrote:
    Cells whose voltage is lower than acceptable (e.g.; 2.5V) are suitable for disposal
    .
    Well, and tell me where or how to check the lowest acceptable voltage?
  • #29 19884479
    LEDówki
    Level 43  
    Where to check the lowest permissible voltage? The answer is - On the internet, in battery documentation, etc., etc. A colleague has given you a link to an article.
  • #30 19884636
    gumisie
    Level 43  
    wujek_demon92 wrote:
    Well...
    1aku-3v
    2-3.2v
    3-1.98v
    4-2.19v
    5-2.43v
    Which ones should I still try to resuscitate by charging them and which ones should I immediately howl...throw away? Trying to figure this out:D
    Dump them all.
    You could resuscitate: 1 and 2, but in my opinion, this is a game not worth a lollipop. In a moment these two will also "die".

Topic summary

The discussion revolves around a Parkside PAP20 A1 20V 2Ah battery that is not charging, indicated by flashing red and green LEDs on the charger. The user reports that the battery shows low voltage across its cells, with some cells significantly underperforming. Various contributors suggest that the battery is likely faulty due to uneven cell voltages and recommend replacing the battery or its individual cells. Some suggest attempting to charge the cells directly with a higher voltage to potentially revive them, while others caution against this due to the risks associated with lithium-ion batteries. The consensus leans towards replacing the entire battery for reliability, given the age and condition of the cells.
Summary generated by the language model.
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