Charging 2x18650 batteries via BMS - Page 1
Forum section: electronics / beginners

author: mike519
author_rank: Contributor
timestamp: March 31, 2025, 08:05:29 pm
content: So, I have 2 18650 cells (3.7Vx2=7.4V) 9in series running through a 2S BMS. The output of this battery pack first has an on/off switch then a buck converter to get it down to 5V before heading to the load.I am trying to figure out what the min and max charging voltage should be. I had been thinking to get a 9V 2A power supply (higher A if I can find one) to charge but now am finding that the max is probably 8.4V since the max charge voltage for these cells is 4.2V each. This is my first time planning such a setup.1) Can I use the 9V power supply since its going through the BMS (doubt it now)?2) Could I use a 5V power supply since they are common (also doubt it...read that the minimum needs to be 6.4V (3.2V minimum per cell). Is that right?3) If I cannot do the above, I guess it means I need to find a PS between 6.4 and 8.4V. I don't really want to use another buck converter after the PS before BMS as space is now getting tight in the spacce this all has to go.Thoughts?

author: microwhat
author_rank: Contributor
timestamp: March 31, 2025, 08:11:20 pm
content: A BMS doesn't handle charging. It only exists to protect the batteries from insults like overcharging, overdischarging, overtemperature, excessive voltage imbalance, etc.A constant voltage power supply should never be used to directly charge a lithium cell string. You must put something between your power supply and the BMS to provide the proper charge profile. Generally for lithium cells, the charger supplies constant current until the string hits some voltage limit, and then a constant voltage until some other time limit.

author: tooki
author_rank: Super Contributor
timestamp: March 31, 2025, 08:12:03 pm
content: Exactly what microwhat said: don't do any of the options you're looking at. You need an actual charger.

author: tooki
author_rank: Super Contributor
timestamp: March 31, 2025, 08:18:00 pm
content: Quote from: microwhat on March 31, 2025, 08:11:20 pmAlmost.The first* charging phase is the fast-charging constant-current (CC) phase, which ends when thetermination voltageis reached (4.1-4.3V for most Li-ion chemistries, normally to 50mV accuracy). This is followed by the constant-voltage (CV) phase, which ends when thetermination current(typically C/20 to C/10) is reached.Any time-based limits are merely safeguards in case something has gone wrong with the primary termination method.*Many chargers actually have a "pre-charge" phase before this, at very low current (C/20 or so), if the battery voltage is below a threshold voltage (usually around 2.8-3V per cell). This is to gently bring a battery back into the voltage range where fast charging is safe to do. They will also often do pre-charge if the cell is too cold.

author: Peabody
author_rank: Super Contributor
timestamp: March 31, 2025, 09:08:30 pm
content: Something like a TP5100 module could be used to charge a 2S pack, but it doesn't do balancing.Another option would be to switch to a single cell, or two in parallel, and a boost converter to 5V. Then a simple charger like the TP4056 could be used, and you could use a regular 5V USB supply.

author: timeandfrequency
author_rank: Frequent Contributor
timestamp: March 31, 2025, 09:35:16 pm
content: Hello mike519,You may use theIP2326which is a boost charger for 2 Li-ion batteries that you can source fromLCSC.Vin = 5 VDC, provided by any vanilla USB wall wand.

author: mike519
author_rank: Contributor
timestamp: April 01, 2025, 08:30:19 pm
content: OK, slightly confused. I had originally thought that I needed a charger and BMS but then found this board and that it did both as it says charger in the description…https://www.amazon.ca/gp/aw/d/B09YYHFMV8?psc=1&ref=ppx_pop_mob_b_asin_titleSo if it does not charge, can I use this before the BMS?18650 battery charger 2S 3S 4S C type A 8.4 V 12.6 V 16.8 V Boost Lithium battery 7.4 V 11.1 V 14.8 Vhttps://a.aliexpress.com/_mNdpAhv

author: tunk
author_rank: Super Contributor
timestamp: April 01, 2025, 08:54:19 pm
content: The title says charger, the description only mentions BMS functionality.Did you consider what Peabody suggested - using one cell and a step-up converter?

author: mike519
author_rank: Contributor
timestamp: April 02, 2025, 01:49:10 am
content: I’m already in on this as a 2S. I’m running a number of LEDs, a mini amp board, etc off of this. If I go to one cell and boost it, do I not lose current?

author: daisizhou
author_rank: Super Contributor
timestamp: April 02, 2025, 05:45:43 am
content: Considering durability,2S connection needs to consider balanceWhy not step up the 3.6v-4.2v to 5V?Then use common TP4056 ...... to control charging.

author: tunk
author_rank: Super Contributor
timestamp: April 02, 2025, 10:22:33 am
content: How much current do you need?You could use two (or more) cells in parallel.

author: Peabody
author_rank: Super Contributor
timestamp: April 02, 2025, 02:37:27 pm
content: Quote from: mike519 on April 02, 2025, 01:49:10 amYes, one cell has half as much energy as two cells. But 2S and 2P are about the same. 2P and a boost converter is a simpler charging situation (no problem finding a 5V supply), without the need to balance the cells. Are your 18650s unprotected? I've always thought they should be unprotected if connected in parallel, with protection somewhere else in the circuit.Attached is a sample circuit that uses two modules - a charger with protection, and a boost converter - and a load sharing (aka power path) circuit in between, with an on/off switch placed so you can continue to charge the batteries when the device is switched off.

author: mike519
author_rank: Contributor
timestamp: April 03, 2025, 03:25:50 am
content: Sorry, what I meant was using a boost converter...if I use one cell only then a boost converter to bring it up to 5V, the process of boosting do I not lose current?I have a proper 7.4-8.4 charger board now so my new thought is running the output to the 2S bank with BMS.I appreciate all of the input and suggestions, however, consider me at the kindergarten stage compared to you folks and some of this is over my head. I'm hoping the above will work fine.

author: daisizhou
author_rank: Super Contributor
timestamp: April 03, 2025, 04:25:22 am
content: The current loss depends on which solution you choose.Usually 75%-95%

author: daisizhou
author_rank: Super Contributor
timestamp: April 03, 2025, 04:35:17 am
content: This is a recent project I worked on to convert 4.2v to 1.5v or 3.0v.It is suitable for 1.5v or 3.0v digital clock, changed to lithium battery power supply circuitYou can make a PCB to suit your liking, it's very cheap

author: Peabody
author_rank: Super Contributor
timestamp: April 03, 2025, 01:55:36 pm
content: Quote from: mike519 on April 03, 2025, 03:25:50 amI don't know what you mean by "lose current". Switching converters, both buck and boost, operate on the principle that power in equals power out, ignoring the losses involved in doing the conversion. Power is volts times amps. So if you're boosting, you have input of low voltage and high current, and output of high voltage and low current.The same happens using a buck converter, with high voltage low current input, low voltage high current output. But the power is the same either way.So a boost converter using one cell will of course give you half the battery life of a buck converter using two cells. But two cells in parallel feeding a boost converter is about the same as two cells in series feeding a buck converter.

author: daisizhou
author_rank: Super Contributor
timestamp: April 03, 2025, 10:31:02 pm
content: Losses during conversion。For example, my MC34063 booster has an efficiency of about 65-75 percent.

author: tooki
author_rank: Super Contributor
timestamp: April 04, 2025, 07:25:22 am
content: Quote from: mike519 on April 03, 2025, 03:25:50 amAssuming the boost converter is adequately sized for the load current you need: you do realize that lithium ion batteries are capable of delivering enormous current, so even if the maximum current is reduced by the voltage boosting, you should still be perfectly fine. And of course, by switching to a 2P configuration, your input current availability doubles, so even with conversion losses this won’t be your bottleneck.So if your load current could be supplied by a bucked 2S configuration, then it can be supported by a boosted 2P configuration.Standard 18650 cells usually have a specified maximum current draw of 8-10 amps or so, while high-current cells (used for things like power tools) specify 20-30A. In both cases, this is the maximum current you can draw continuously without causing damage. The cell is capable of delivering far more current pulses, or in a damaging short-circuit event. (The short-circuit current is easily 50-100A!!)

author: xvr
author_rank: Frequent Contributor
timestamp: April 04, 2025, 10:15:47 am
content: Quote34063 celebrated its 40th anniversary. Let the old man die in peace. Use something more modern, with an efficiency of 90-95%LTC3111, LTC3112,https://www.ti.com/download/aap/pdf/13634tirev1.pdfhttps://www.amazon.co.uk/Converter-Efficiency-Lightweight-Working-Frequency/dp/B0DLZVT615?th=1

author: daisizhou
author_rank: Super Contributor
timestamp: April 04, 2025, 10:25:54 am
content: Quote from: xvr on April 04, 2025, 10:15:47 amIndeed, 34063 is very old.But it is very cheap。The most interesting thing about this circuit is that the output has an 8550 transistor.The quiescent current is very small and no cut-off switch is required

author: tooki
author_rank: Super Contributor
timestamp: April 04, 2025, 10:39:22 am
content: Quote from: daisizhou on April 04, 2025, 10:25:54 amWhat are you talking about?!?The MC34063 does not use an S8550 transistor (MC34063 is an integrated circuit: it doesn’t containanydiscrete transistors), and its quiescent current is 4mA, absolutely huge compared to the <50μA of modern chips.

author: MrAl
author_rank: Super Contributor
timestamp: April 04, 2025, 11:45:35 am
content: Quote from: mike519 on March 31, 2025, 08:05:29 pmHi,There are a lot of ways to charge Li-ion batteries but the best way is to use a charger made for what you are charging, exactly. If you can't do that then you should understand the specifications as well as you possibly can. Here are a few notes.First the maximum voltage for each cell alone is 4.200 volts and that is considered a 1 percent tolerance measurement. It's better to call that 4.150 volts though as that gives you some safety margin and also prolongs the life of the cells.Second is the maximum charge current. You get this from the data sheet for the cells. Long time ago the max for most cells was 700ma, but you can probably go to 1 amp without a problem if the cells are not really old. If the data sheet says more, then you can go higher.Then there is the nature of the max current which is said to be a constant current (CC), but it really does not have to be 'constant' it just has to be 'limited'. That means that if the max current was 2 amps, then the current has to be limited at 2 amps, but it does not have to be 2 amps the total charge time and in fact it can not be that way anyway because there is a taper to it where the current gradually decreases. So the current could be 2 amps, 1.9 amps, 1.8 amps, 1.7 amps, ..., 0.5 amps, etc., and the minimum current should be C/20 and that is where the charging should be terminated.If you want to use a constant voltage source like 9v to charge two 4.200v cells at 1 amp, then you need to add some resistance between the source and the cells. That resistance depends highly on the max current AND the voltage of the two cells at the time they are being charged.If the max current was 1 amp and the cells voltage was 8.000v, then you would need a 1 Ohm resistor to limit the top charge current to 1 amp. The power rating of the resistor should be at least twice the actual power dissipation, which would be 2 watts in this case.But let's say the cells voltage was 7.000v instead, then we would need a 2 Ohm resistor:R=(9.000v-7.000v)/1.000a=2 OhmsNow that works out pretty well, but when the cells get to 8.000v then we only have 0.5 amps flowing into the cells, so they are no only charging at 0.5 amps. Here we can either let it go at that, or decrease the resistance to 1 Ohm so that we get the full charge current again.Also note that the 2 Ohm resistor would have to be rated at 4 watts because the required resistor power rating is:P=i^2*R*2If the voltage of the two cells started out even lower, like 6.000v, then we would need a different resistor:R=(9-6)/1.000=3 ohms, and the power rating would have to be 6 watts.We might change that to 2 ohms later when the voltage rises to 7v, then again later when it rises to 8v, etc.So this can work, but you need to monitor the voltage manually unless you don't care if it takes a lot longer to fully charge the two cells.If you have a variable power supply then you can set the voltage to accommodate the cells using just one resistor, and if that power supply has a current limit setting, you can just set it to 1 amp, and with the max voltage set to 8.4volts or better 8.3 volts.So as you can see, there are other options, but the simplest is to get a charger that does it automatically where you do not have to set anything. It would then be like "Plug and Play"

author: daisizhou
author_rank: Super Contributor
timestamp: April 04, 2025, 12:12:43 pm
content: Quote from: tooki on April 04, 2025, 10:39:22 am

author: tooki
author_rank: Super Contributor
timestamp: April 04, 2025, 06:54:02 pm
content: Quote from: daisizhou on April 04, 2025, 12:12:43 pmThe text "this circuit" in your reply that I was responding to makes no reference to the board you showed, so it is to be assumed that "this circuit" means the onein the post you were replying to.And of course the thing I said about the quiescent current remains true: the MC34063 is an absolute pig compared to modern ICs.

author: daisizhou
author_rank: Super Contributor
timestamp: April 04, 2025, 08:00:05 pm
content: Quote from: tooki on April 04, 2025, 06:54:02 pmI usually apply this circuit to the 9V battery of a multimeter，Boost from 4.2V lithium battery to 9V。Simply replace the 9V voltage regulator diode to boost any voltage

author: tooki
author_rank: Super Contributor
timestamp: April 07, 2025, 06:37:31 pm
content: Quote from: daisizhou on April 04, 2025, 08:00:05 pmSo what? It's not optimal for that application.Quote from: daisizhou on April 04, 2025, 08:00:05 pmHuh? What do you mean?