Device says dead battery but individual Li-ion cells are still over 4 V. - Page 1
Forum section: electronics / beginners

author: raptor
author_rank: Newbie
timestamp: February 26, 2024, 05:11:10 pm
content: Hi, I am about to attempt to replace cells in a vacuum cleaner battery pack (7S 18650 Li-ion 2600 mAh cells). The battery goes dead after 1 to 3 minutes (with full suction power though). When I pulled the battery pack out, I measured the individual cells and they all read over 4 V (mostly 4.03, some 4.02, one reads 4.01). I would expect them to be depleted to at least 3.7 V if the device indicates dead battery and refuses to run the motor. Or I would expect one cell to be bad (and read low voltage). Maybe the BMS doesn't like something else about the cells. Or maybe the cells are OK and the BMS has broken?As a beginner I might be missing something.Thank you for any ideas and advice.Philips PowerPro Aqua FC6409 vacuum cleanerHere you can see some of the electronics (not my video):https://youtu.be/CAh4RwBRqfk?t=136

author: RoGeorge
author_rank: Super Contributor
timestamp: February 26, 2024, 05:18:27 pm
content: Probably one or more cells can not deliver enough current, and their voltage drop under load.Measure the voltage with some load attached to the battery pack. Put a 10-20ohms/5W resistor in parallel with the voltmeter's probes to identify the weak cells.

author: Bud
author_rank: Super Contributor
timestamp: February 26, 2024, 06:03:33 pm
content: It may be a number of things, from a bad cell to unbalanced cells to how the battery management circuit calculates the capacity. You should probably start from googling your device/battery manufacturer and model to see if anyone has done troubleshooting or rebuild. Simply replacing cells may not work at all if the process of battery recalibration or management circuit reset is unknown. The battery may lock itself when you disconnect a cell and you may have hard time to even get to the state you had./

author: Infraviolet
author_rank: Super Contributor
timestamp: February 26, 2024, 06:59:29 pm
content: First check what form the wiring from the battery to the device takes, just two conductors for power, or something with extra wires for data signalling?In the former case, see if the device can be powered simply from a DC supply (of sufficient current driving capability) giving the nominal voltage of the battery pack. Does the device's indicator show this as being a mid-level, highly-charged or near-depleted battery level?In the latter you might need to probe the data wires with an oscilloscope, logic analyser or some sort of arduino based jig, to see what data is flowing, and see about decoding it. The device might only work if the battery says "I'm a battery and have this voltage available" or equivalent. If the manufacturers have been ****ers and use some sort of DRM cryptography on these signals, rather than just normal line encoding, then if you need to custom built a replacement battery, you'd probably find it easier modifying the device not to make these checks than you would trying to decrypt the data.

author: Bud
author_rank: Super Contributor
timestamp: February 26, 2024, 08:19:05 pm
content: The proper answer is to get a SMBus smart battery analyzer and software, connect it to the battery (smart batteries have i2c interface), read the data and look at it, why the device behaves as it does. This is going to go in the rabbit hole very fast though, this is why i recommended just google it first if anyone has done it.I am not an Arduino guy and do not know if a SMBus app exists for Arduino, but you can get away cheap using this guy's open source work and an inexpensive adapter from Aliexpress/Amazonhttp://www.karosium.com/2016/08/smbusb-hacking-smart-batteries.html

author: raptor
author_rank: Newbie
timestamp: February 27, 2024, 12:29:30 pm
content: Thank you, everyone, I am already learning a lot from your answers.Quote from: RoGeorge on February 26, 2024, 05:18:27 pmWould a 2W resistor be OK to use? If not, I will get the 5W one.Quote from: Bud on February 26, 2024, 06:03:33 pmUnfortunately I haven't found any project on the Internet dealing with this particular cell swap. But various vendors sell replacement cells for this pack, so that leads me to believe that there should not be any original cell lock or encryption in place.Quote from: Infraviolet on February 26, 2024, 06:59:29 pmThere are two extra wires, but they apper to be just from what I think is a thermal sensor that is placed between two cells.Unfortunately I don't have any bench DC power supply. Maybe I should eventually get one, it would be useful for troubleshooting.Just to be clear, I don't have the new cells yet. The behavior I descibed is happening with the old original cells.I will get the cells, make a temporary pack from them (without permanent connections) and test if the motor is running without the original BMS. If yes, I will connect them to the orig. BMS (replace the original cells).I will get the 18650 cells with welded tabs because I don't want to solder directly to the cells.But soldering to the tabs should still be done quickly, right? I hope my 70W iron and lead-free solder will be good enough for that. I will practice on a completely non-functioning cell first. Is it a good idea to use some sort of heatsink on the battery terminal or a section of the tab?

author: Zero999
author_rank: Super Contributor
timestamp: February 27, 2024, 12:59:12 pm
content: Quote from: raptor on February 27, 2024, 12:29:30 pmP = V2/RSo a 10R resistor will dissipate 1.6W.A 2W resistor will be fine, as long as you don't leave it for too long, otherwise whilst the resistor itself won't be damaged, it'll get too hot to handle and is a burn hazard.

author: RoGeorge
author_rank: Super Contributor
timestamp: February 27, 2024, 02:01:38 pm
content: 2W or even smaller should be OK in open air, and for only a couple of seconds while measuring.

author: Bud
author_rank: Super Contributor
timestamp: February 27, 2024, 02:43:47 pm
content: Communication is not encrypted, as it typically complies with SMBus specification. However "changing" some of the parameter values is protected with manufacturer's password or hash. You can read those values but you cant change them. There are paid tools that have a list of known passwords or use proprietary workarounds or brute force algorithms to unlock (unseal) batteries for updates.You may replace cells but that does not garantee the battery will start to perform to full charge capacity again until some specific values are updated, and that where the problem is.You can certainly try, please report your experience then.

author: raptor
author_rank: Newbie
timestamp: February 27, 2024, 05:41:27 pm
content: I think it should not be locked like that because there are manufacturers selling replacement kits for this very battery pack, for example here:https://allegro.pl/oferta/bateria-do-philips-powerpro-fc6409-2900mah-25-2v-14569880753. If the pack was somehow locked this would not be possible. At least I hope so.I won't be buying this kit though because it is too expensive. I will be using quality used Sony VTC5 cells from a local seller (tested for capacity and internal resistance, he will spot weld the tabs for me because I don't have a spot welder).

author: Bud
author_rank: Super Contributor
timestamp: February 28, 2024, 04:59:58 pm
content: Ask your local supplier if they can reset the battery for you if simple cell replacement wont work. They may have a software that can do it.

author: uli12us
author_rank: Contributor
timestamp: February 28, 2024, 07:21:17 pm
content: But, don't replace the batteries with higher capacity. Typically a battery with 2Ah can deliver 20Amps or more,but a 3Ah or bigger can only deliver 3-5Amps. Thats not enough for the Motor.

author: Bud
author_rank: Super Contributor
timestamp: February 29, 2024, 05:44:48 am
content: Not only that. You wont be able to utilize higher capacity cells without re-programming the controller.

author: raptor
author_rank: Newbie
timestamp: February 29, 2024, 10:30:56 pm
content: I put a load on each cell. I used 12 Ohm 2W resistor (12.2 Ohm was the measured resistance incl. the wires). The calculated (not measured) current was 0.33 A. I don't see any significant voltage drop that stands out. Should I try to draw more amps (get a resistor with less Ohms and more Watts)?at rest -> load (Volts)cell 14.00 -> 3.95cell 24.03 -> 4.00cell 34.02 -> 3.98cell 44.03 -> 4.00cell 54.02 -> 3.99cell 64.03 -> 4.00cell 74.03 -> 4.00