Hi how long did you bump them for?

Cool

This worked for me as well! Thank you, It takes about an hour to go from 1.7v to 2.2v. Then charger can do normal charging.

1. Sure could. Dont have enough different types to look at neither enough info perhaps but some day. 2. Why do you want it on after sunset? It can be done either with a time controlled switch or a lightning sensitive switch.

Thanks. The 2x resistors or so is more if you got 1/4th or 1/8th W resistors you can still use those as long as you use a couple more of them :) For instance 2x 160Ohm in paralell is 80Ohm and will then spread the load :) Eya you get it :)

This is not difficult for me, learned this on school. 😉

Sweden so Close :)

Interesting

same happened to me was at like 2volts and jumped back up to 3.2v

@@Gamer-nc8qp mine got 0.7 v then after i push like in the video it back to 3.8 v.. wth is that!!!

The CID is most likely triggered or the cell is just very bad. I would recycle the battery and get a new one instead.

@@DIYTechRepairs OK, thank you very much for the very quick reply.

Thank you too!

If you have a hybrid inverter that one will do what you want for you in terms of using battery when not enough solar exist. You should be able to set in that inverter what order you should use as source. Ie Solar > Battery > grid. For the normal inverters always draw power from the battery and that will work out on its own.

hehe good on you. I'm still looking for laptop repair places willing to help. Some I called and they say they're busy and email them, so I did.. no response.. grr I think you'll need to cycle those cells a few times and see how well they work, probably just flattened the shit out of it and wasn't due to a problem causing them to go so low. hopefully if you charged them slowly, it wouldn't have damaged them internally, at least if not already from the 0.6v

Phizicks Yeah, but they are more than good, probably only the BMS stopped working. :) Anyway, I use to buy notebook batteries in a "exchange fair" (at least is how we call here in Brazil). You know? It's a fair where people sell new and used things, farm animals, even sell or give pets, and a lot of people sell parts of old and broken tools, DVDs and so on. It happens once a week, every Sunday. I already bought a lot of good stuff there, like dc motors and so on, that I use in my projects/videos. :)

so if the nonrechargable nickel reaches full charge off a lithium charger acting like an electrolytic just like an overcharged electrolytic it could blow up right? maybe you could take a dead nickle nonrechargable and use it in place of a electrolytic might hold more capacity.

You recycle Them. They have their cid triggered = crap

Depends on Charger. But 1.5v+

@@DIYTechRepairs any reason why I can't put two batteries in series, say they're both at 1v each, to fool the charger into thinking it's a single battery at 2v and get the charge started? How does battery resistance affect the charging?

@@areyouavinalaff You can but at same time you need to monitor so not one of them jumps to high and the other stays flat. Its better just to do it properly :)

Yes you can. Best is though ro bump each cell Group Up to 3v and then max 100mA per cell. Then when all 10 groups are done you're back in business. Bmping overall voltage is a bit riskier and you then need to monitor each cells voltgae

@@DIYTechRepairs it comes as a pack... Bumping each one means I would have to rip open the pack... If am to bump the pack... What is the suggested voltage though

@@stanleytan9717 You dont have access to the balance wires? If you dont have that you can only guess that it will work and that the internal BMS will do its job if one cell fails. Overall total voltage would be 30+ volt if its 10s.

@@DIYTechRepairs thank you very much for the information and really appreciate the help and quick response even though this video is out for so long!

U = I * R. If you get 500mA you have a resistor of = 4,2/0,5 = 8,4Ohm and not 82 ohm...

@@DIYTechRepairs Thank you for your reply. I did measure resistance on thr positive side of the bumper (with resistor) It gives me 82ohm. The other side (negative) gives 0 ohm. When I put two batteries in parallel mode( one is 4.2v another in 1.7v) It charges battery, but it happens too fast. When I measure current flowing on 1.7v I get around 500mA. I understand your math is right, I am just puzzled what could I be doing wrong. Maybe faulty resistor? It seems to be slowly dropping though. It dropped to 100mA. Theres is a possibility that it can't be measured properly while it is charging with the tools I have. Ok, finaly, I figured it out. If I have only one charged battery in the bumper (4.2v) in slot 1, and slot 2 is empty, it measures 50mA on the slot 2 leads. I guess it was drawing current from second battery as well, when I was measuring it first time.

Im not sure what you want to do? No you should not hook 2 in parallell like that unless they have proper bms atached

Nicklas Olsson för att driva elen i huset

DIY Tech & Repairs, ok 👍 riktigt intressant faktiskt.

No problem if voltage was det below 4.2

@@DIYTechRepairs hehehe guess i wont be messing with it any more i threw it away when i heard a hissing when i push it down with a very verryyy small pressure to it it scared me and also there was a fluid coming out of it. It does smell like acetone there is no way im messing the other battery btw thanks for this vid. And this method really does work but be very careful when doing this cause my battery was really soft when i push on its disc or something

Can be yes but dont have to.

First of all bin those cells. They are FAKE... You should have watched my video about fake cells :) But yes when you get real cells just put them in the charger

All cells taking charge is worth testing. I would say that many below 2V is not comming back good enough for my liking. But this depends on why they are on low voltage from the start. I test em all

You can but bumping a cell with resistor is as fine and easier than running above chargers in NiMh mode to get the voltage up. I would rather use a proper Bench PSU than above chargers to do that if so ;)

Did you think you just can wave a Magic wand?

The resistance of a new cell is in the 10s of milliohms, orders of magnitude less than the current-limiting resistor. A damaged cell will only have a higher internal resistance, limiting the current even more.

I kind of only went with the worst case scenario and made the calculation very very simple. There is much to be added there in you want the exact current flowing but for our use case i left that out. :)

To clarify point number one: the possible internal resistances of the cells is so small that it is masked by the manufacturing uncertainty of ... well, any physical resistor you could possibly attempt to produce economically. Your 1% 1k resistor don't give a crap about your puny 170 mOhm hissy-fit of worst-case-scenario.

The internal resistance is only interesting for the (dis)charge properties of the battery. When the value is to high you will experience a larger voltage drop at higher currents. The value of the internal resistance is in this application to small to limit the current, but it can create a problem with the voltage drop. For example that the battery can't be charged full at a higher current or that the the voltage drops to fast. Both situations can create an instability in the pack. AveRage Joe has videos about testing fuse wire where he explains why, if a cell is not equal to the others in the pack it can cause a current draw. Something you don't want to happen.