Charging 16 pieces of 3.7V lipo batteries together - Page 1
Forum section: electronics / projects

author: kian0079
author_rank: Contributor
timestamp: January 16, 2023, 06:06:44 am
content: Hi all,I am building a charging dock station to charge 16 pieces of 3.7V 1800mAh batteries together at the same time. I plan to use separate charging ICs for each battery (Part number: TP4056). According to the datasheet, the TP4056 has a maximum charging current of 1A.However it doesn't make sense for me to use the max charging current of 1A because that would mean drawing 16A from the power supply. I am thinking of setting the charging current to perhaps 400mA or 580mA, but that still works out to be a very high current demand for the power supply.I am hoping to get some advise or feedback on what are some of the potential problems that you would foresee with such a setup (to charge 16 batteries at the same time) and if there are other better solutions for this. I would like to use a standard AC to DC adapter with 5V, 10A output.Thanks in advance!

author: inse
author_rank: Super Contributor
timestamp: January 16, 2023, 06:52:47 am
content: You also have to consider the overall power dissipation: as those charging modules are fed from 5V and the average charging voltage is 3.6V, you have 1.4W per module and 22W of heat overall at 1A.The charger I have has a buck converter for each of it’s four cells. Of course you would need a controller to take care of them all.Are there switch mode LiPo charger ICs or modules out there?Those can be supplied by a higher voltage thus requiring less current.Why not line up commercial chargers?

author: enut11
author_rank: Super Contributor
timestamp: January 16, 2023, 06:54:16 am
content: Current is always the limiting factor with power supplies. Why not split the input load and use a second AC to DC adapter so each source only has to deal with 8 charge ports?

author: beanflying
author_rank: Super Contributor
timestamp: January 16, 2023, 07:13:09 am
content: You might get a few ideas from this rats nest of mine from a few years agohttps://www.eevblog.com/forum/projects/multi-cell-lithium-charger-from-a-single-5v-supply/msg1454259/#msg1454259The B0505S isn't essential but they do give a nice stable supply into the TP4056 boards regardless of what you might have across a busbar. I cut into the load setting resistors and hacked a DIP switch and some resistors to make them variable current.In theory the small board layout was scalable in banks of 4 so hitting your 16 would be 4 boards worth and because of the isolation from the B0505's it allowed me to change multi cell packs from 5V or singles as connected.Longer term I went a different route but it was a fun low cost plaything/learning.

author: dobsonr741
author_rank: Frequent Contributor
timestamp: January 16, 2023, 11:18:10 am
content: The tradeoffs of your chosen linear charger vs. switch mode:https://www.analog.com/en/technical-articles/switchmode-linear-and-pulse-charging-techniques-for-li-battery-in-mobile-phones-and-pdas.htmlIf you entertain switchmode, this explains how it will work:https://www.ti.com/lit/ml/slyp089/slyp089.pdfYou might not find switchmode PCB modules on eBay/AliExpress. The closest race to the bottom cost component can be something from LCSC:https://datasheet.lcsc.com/lcsc/1810010241_ShangHai-Consonance-Elec-CN3791_C154992.pdf

author: tunk
author_rank: Super Contributor
timestamp: January 16, 2023, 11:52:29 am
content: If you go for 580mA, then your board/TP4056s will dissipate up to 18.6W (2V*0.58A*16).

author: mariush
author_rank: Super Contributor
timestamp: January 16, 2023, 12:57:41 pm
content: I would recommend changing your design to (at least) 12v input - you can use recycled laptop adapters, wallwarts (adapters that plug directly into the wall).You would need to use a switching regulator to convert your input voltage (let's say 7.5v...18v but 12v recommended) to around 4.5v if you plan to use one linear charger IC for each battery. This reduces the losses on each regulator.For example AOZ2261 is cheap and can do up to 28v input, and up to 8A of output current :https://www.digikey.com/en/products/detail/alpha-omega-semiconductor-inc/AOZ2261AQI-15/10249957Use one of these for 8 batteries, and you'd be able to get your 1A per battery charge current.RT7258 can also do 8A if you want a more common package :https://www.digikey.com/en/products/detail/richtek-usa-inc/RT7258GSP/3503828There are charger ICs which have a maximum of 500mA charge, and with 16 batteries that would mean a maximum of 8A of current.But you could complicate your design with some microcontroller or something that detects how many batteries are inserted and maybe reduce charge current to stay within a budget - for example you use a 12v 2A/3A/5A adapter for a maximum of 24w/36w/60w and you use 1-2 dc-dc converters to convert that to ~4.5v, so you get around 5A/7.5A/12.5A (not 100% efficient due to conversion losses) so if you charge 16 at same time, then you could drop charge current to 250mA and get within that 4.5A budget.

author: kian0079
author_rank: Contributor
timestamp: January 17, 2023, 04:35:26 am
content: Hi all,Sorry, perhaps I would have been more detailed in my description. I actually have 16 devices, each having its own PCB with TP4056 charging IC on board. I plan to have pogo pin connectors on each device that makes contact with the charging connectors on the charging dock. The charging dock itself doesn't have any electronics. Its just has 16 sets of 2 pin connectors (5V and Gnd) in parallel. And I plan to get a off the shelf 5V AC-DC adaptor (somewhat like a laptop charger) that is capable of supplying up to 10A. I presume heating would not be a problem in this case since the devices are stacked on top of each other separated by a housing.

author: kian0079
author_rank: Contributor
timestamp: January 17, 2023, 04:45:06 am
content: Hi mariush,I don't quite understand your recommendation. Do you mean using a 12V supply and stepping down the 12V to 4.5V with a switching regulator before feeding it to the charging IC? If I can get a off the shelf 5V 10A supply, why would I want to use a 12V supply and have using extra switching regulators which will bring up my BOM cost?The AOZ2261 can supply up to 8A of current, so I presume I can tap the 4.5V 8A output and connect it to 8 charging ICs in parallel? And I would need 2 AOZ2261. In that case, what is the rating of the 12V laptop supply I should be looking at? Eg. how many Amps?Quote from: mariush on January 16, 2023, 12:57:41 pm

author: sleemanj
author_rank: Super Contributor
timestamp: January 17, 2023, 05:09:01 am
content: Since you already have the TP4056 in the devices, then it is no problem to buy a 5v supply which is capable of the maximum theoretical current you need, they are not even expensive. Example:Mean Well LRS-100-5 gets you 5V @ 18AThere really isn't any more simple or cheaper way.The TP4056 is not going to maintain 1A for long anyway as it will throttle down unless you have it heatsunk very well in my experience.

author: inse
author_rank: Super Contributor
timestamp: January 17, 2023, 05:20:53 am
content: Put a polyfuse on every dock in case of overload or short circuit.

author: mariush
author_rank: Super Contributor
timestamp: January 17, 2023, 07:27:14 am
content: If you want to make a one-off , a single item, just for your own needs, then by all means, use a 5v 10A power supply, or whatever power supply you find, because it doesn't matter.But, if you're thinking on making several "kits" like this, you may find that power adapters with higher voltage and lower current are cheaper than adapters with low voltage and high current - a 12v-24v 2-3A power adapters will be much cheaper than a 5v 10A power adapter.The higher the current, the higher the losses in the cable between the power supply and the actual connector on your product.For example, here's what I could buy from TME.eu today (you can change the language in top corner):wallwarts :6.03$ : 24W 12v x 2.0Ahttps://www.tme.eu/ro/details/posc12200a-h/alimentatoare-cu-stecar-integrat/pos/6.03$ : 24W 24v x 1.0Ahttps://www.tme.eu/ro/details/posc24100a-h/alimentatoare-cu-stecar-integrat/pos/6.55$ : 24W 15v x 1.6Ahttps://www.tme.eu/ro/details/posc15160a/alimentatoare-cu-stecar-integrat/pos/8.50$ : 36W 12v x 3.0Ahttps://www.tme.eu/ro/details/posc12300a-h/alimentatoare-cu-stecar-integrat/pos/8.50$ : 36W 24v x 1.5Ahttps://www.tme.eu/ro/details/posc24150a-h/alimentatoare-cu-stecar-integrat/pos/Now let's look for 5v wallwarts capable of same watts :26.8$ : 30w 5v x 6Ahttps://www.tme.eu/ro/details/sga60e05-p1j/alimentatoare-cu-stecar-integrat/mean-well/Same deal for laptop adapter style supplies - you won't find 5v laptop adapter style supplies:7.13$ 24w 12v x 2.00Ahttps://www.tme.eu/ro/details/posc12200d-c8-wh/alimentatoare-tip-desktop/pos/9.05$ 36w 12v x 3.00Ahttps://www.tme.eu/ro/details/posc12300d-c8-wh/alimentatoare-tip-desktop/pos/12.34$ 60w 24v x 2.50Ahttps://www.tme.eu/ro/details/posc24250d-c14/alimentatoare-tip-desktop/pos/11.16$ 45W 20v x 2.25Ahttps://www.tme.eu/ro/details/ak-nd-50/alimentatoare-pentru-laptopuri/akyga/cpsunotaky-07712/12.52$ 65W 19v x 3.42Ahttps://www.tme.eu/ro/details/ak-nd-01/alimentatoare-pentru-laptopuri/akyga/cpsunotaky-07051/A switching converter will convert the higher voltage to lower voltage with some efficiency, usually it's over 85% ... let's go with 90%. That means that 90% of the wattage will be available after conversion.So for example, if you go with 36w 12v power adapter and you convert to 4.5v ... 90% of 36w is 32.4 watts and if you configure the regulator to output 4.5v, that means you'll have up to 32.4w / 4.5v = ~ 7.2A of current.Instead of spending maybe 20-30$ on a 5v 10A adapter, you spend 8.5$ on higher voltage adapter, and maybe 3-4$ on the switching regulator and the other components it needs (inductor, mosfet maybe) and you also gain flexibility - you optimize the switching regulator to work with 12v but if you run out of 12v adapters, you could pair it with 15v or 24v adapter and it will still work, and your product will also work if the user accidentally uses the 7.5v or 9v adapter from another device with your product.You say you have those TP4056 modules already. That's a linear regulator charger IC ... it only needs an input voltage a bit higher than 4.2v (the maximum lithium battery voltage) in order to properly charge it. Whatever extra voltage you give it will be dissipated as heat on the chip. So, for example, if you charge an almost discharged battery (let's say it measures 3.8v) at 1A of current, and you power the module with 5v, you'll have at least (5v - 3.8v ) x 1A = 1.2 watts as heat produced by the charger chip. It's just wasted energy.If you have the ability to set the voltage a bit lower - 4.5v is a good value as it's around 0.3v higher than maximum needed - you'll get less losses in the form of heat and more modules would be able to charge at higher currents at same time.Each module will pull current as needed by the battery - it depends on how discharged the battery will be. It's not a constant 1A of current. When the battery is all discharged, there's a period where the charger will give it some low amount of current until the measured voltage goes above some threshold, then the charger IC will push up to how much you configure it, and as the battery gets close to being fully charged, the current amount will lower.You can see that graph in the bottom right corner of page 1 in the TP4056 datasheet :https://dlnmh9ip6v2uc.cloudfront.net/datasheets/Prototyping/TP4056.pdfSlow charge until the voltage on battery goes above around 3v , then straight to 800mA and higher, then as battery gets to around 4.1v the current starts to decrease until the battery reaches 4.2vIf you plug 16 almost discharged batteries at the same time, you'll have 16 modules each trying to pull 1A of current, but you don't have 16 x 1A = 16A available. if the modules pull too much current, the power supply may go into protection mode, over current protection or over power protection, and may shut down.For your own use, you could just let it be and just put a label on the box saying don't charge more than 10 fully empty batteries or something like that.For a proper device, you have several options.You could use a higher wattage supply (ex. a 60w adapter) and a strong switching regulator or several smaller switching regulators (ex 2 3$ ones, 1 for each group of 8 batteries or 4 $1 ones, 1 for each group of 4 batteries) to have the maximum current available for all modules.You could have some sort of microcontroller that monitors how much power each module consumes and adds all up and if it's too much power it could disconnect some modules for a few seconds - for example power modules 1-4 and 9-11 for 10 seconds, then turn off and power modules 5-8 and 12-16 for 30 seconds or power 3 out of 4 groups for 5 minutes at a time - you get less power consumption but longer charging times.You could have your microcontroller detect how many modules are active by measuring the power consumption each one has, and if the total power is too much, microcontroller could change the programmed maximum charge current set on each module (by connecting or disconnecting a resistor using a mechanical relay or other methods)

author: beanflying
author_rank: Super Contributor
timestamp: January 17, 2023, 07:49:02 am
content: Quote from: mariush on January 17, 2023, 07:27:14 amThis is now your second go at trying to over complicate a solution with a Microcontroller. There is just no justification for this level of switching and control over 16 different outputs. As someone who started playing with Electronics when 555 Timers were brand new the 'need' to add a Micro to a project 'always' is complete BS. KISS always where you can.Quote from: kian0079 on January 17, 2023, 04:35:26 amThe sensible option unless charge time is an issue is drop the current and size the now reduced power supply to suit a worst case.If you want to keep a little heat out of the packages then a lot of SMPS supplies and in particular the Meanwell ones they typically have a Pot on the board to drop the output voltage up or down a little.A Laptop type of Brick will see you running into Current limits and most of them are not 5V outputs so look elsewhere IMO.Don't overthink or over engineer you Solution

author: kian0079
author_rank: Contributor
timestamp: January 17, 2023, 09:36:39 am
content: Thanks mariush and beanflying for the insightful explanations.I actually need to make 100 sets of these charging docks. Possibly more in the future. So I need to get the design right the beginning. I think I understand where mariush is coming from and it does make sense to me now about using a high voltage lower current supply with a switching regulator. Charging time is not much of an important concern, good to have, but not a must.However, having a microcontroller seem over complicating, its probably something that I would have the knowledge implementing, but I need to find a justification for it. Perhaps its implementation is important from a safety point of view?I had chosen the TP4056 because if its price and also availability. I am not using off the shelf modules. The charging circuit is designed into the device PCB itself. Do you have other better recommendations? My considerations in the order of priority are: safety, cost, availability

author: NiHaoMike
author_rank: Super Contributor
timestamp: January 18, 2023, 04:43:19 am
content: Supply all the TP4056 with a current limited buck converter, then the voltage would be allowed to droop if a large number of outputs are in use. That can allow for lower cost but slower charging if a lot of outputs are in use at once. And by adjusting the maximum buck voltage to just enough to allow for the dropout of the TP4056, heat dissipation would be kept to a minimum.

author: Peabody
author_rank: Super Contributor
timestamp: January 18, 2023, 04:18:35 pm
content: It seems that sleemanj's suggestion to just get a bigger power supply would be the simplest solution. But if you underpower a bit, remember that the TP4056 won't draw any current if its input voltage is below the battery voltage. So if you have some variance in the discharge state of the batteries, and if the input voltage sags a bit, the batteries that are only a little discharged may shut down charging until the input voltage comes back up. So if there isn't enough current for all, the most deeply discharged batteries will be served first. That's assuming your power brick will tolerate being over-driven. Edit: And that ripple won't get out of hand.While you can set the charging rate on the TP4056 to 1A, that's just the maximum. It will use whatever lesser current is available so long as the input voltage is barely enough to make current flow into the battery. In fact, ideally that's where you want to be because it minimizes dissipation which can result in self-throttling from overheating. So if your power brick is 5V, it might make sense to insert a rectifier diode going to each TP4056 to drop the voltage, and to transfer some heat to those diodes.If you already have the TP4056 circuit on your board, I don't see any reason to change that. No other linear charger is going to behave much different, and a switching charger would require a completely different circuit - inductors and such - and probably not make much difference anyway.

author: kian0079
author_rank: Contributor
timestamp: January 19, 2023, 03:33:28 am
content: Hi NiHaoMike,I don't quite understand your suggestion. Are you able to provide more details? For example any specific current limited buck converter? What voltage and current should I be using for the buck converter? I also don't really understand what you mean by "adjusting the maximum buck voltage to just enough to allow for the dropout of the TP4056, heat dissipation would be kept to a minimum"Quote from: NiHaoMike on January 18, 2023, 04:43:19 am

author: kian0079
author_rank: Contributor
timestamp: January 19, 2023, 03:45:26 am
content: Hi Peabody,I am thinking of using something like this:https://www.amazon.com/LeTaoXing-100V-240V-Converter-Transformer-5-5x2-5mm/dp/B08HCS1X66/This is a 5V 10A supply. And I plan to charge all 16 batteries with 580mA. This would be a total current of about 9.28A which in within the rating of the power supply.What do you mean by the input voltage sagging abit? Wouldn't the 5V supply be providing a constant 5V to each of the 16 charging circuits? I also don't understand your suggestion to use a rectifier diode for each of the TP4056. Do you mean dropping the input voltage to something like 4.3V so that there is less heat dissipated from the TP4056? But wouldn't this put the input charging voltage too close to the fully battery charged voltage and cut of the charging prematurely? Or is that exactly what we want to achieve?Quote from: Peabody on January 18, 2023, 04:18:35 pm

author: Peabody
author_rank: Super Contributor
timestamp: January 19, 2023, 03:20:25 pm
content: If you set the charge current to 580mA, then there should be no sagging. But I was trying to think through what would happen if you set the charge current to 1A, which is the maximum the TP4056 can handle. If the batteries are not all equally discharged, then any sagging in voltage would shut down charging of the batteries with the highest voltage, but still allow the low voltage batteries to charge at 1A. But that would depend on your power supply being able to deal successfully with excess load, and you have no way of knowing that without a lot of testing. So it's probably not a good idea.The TP4056 is a linear charger, so any input voltage higher that what's needed to just barely supply 1A of charge current will result in heat dissipation. And the TP4056 is designed to reduce charging current if it gets too hot. So the optimum is to supply the lowest voltage that still gets the job done. The TP4056 will begin to charge if the input voltage is 30mV higher than the battery voltage, but I don't know if it will supply 1A at that voltage. You would have to experiment to see what voltage gives you 1A when the battery is at 4.2V, and then try to keep the input close to that. If your power supply is 5V, then a rectifier diode would nominally drop it to about 4.3V, which might work. But the drop might be too much at 1A, so you would just have to see what works. But no, you don't want to terminate charging early.Actually, the datasheet for the TP4056 has a Typical Application schematic that includes an input resistor of 0.2 to 0.5 ohms when used with a 5V supply. That might be an option. The problem is that if the supply voltage is 5V, heat dissipation will be much higher when supplying 1A to a battery that's at 3.5V than one at almost 4.2V. I just don't know of an easy way to have the input voltage follow the battery voltage. But dropping the input voltage by a fixed amount at the maximum charge current would provide some benefit. At 580mA, a 1-ohm 1W resistor might work.