i am having a 6s 18650 battery pack, and used a usb-type c trigger to get 20V-4.5A and boosted the voltage to 25.2Vto charge the battery. when connected to bms the laptop charger stops supplying power, tried connecting directly to the battery positive to positive and negative to negative, the laptop charger produced power continuously and the voltage increased in battery by 0.1v

 

Welcome to AAC.

You haven’t really provided enough information to help. To get a useful answer instead of a hundred random questions and a thousand random guesses you‘ll need to tell us more.

First, you should provide a BoM (Bill of Materials), that is, a list of the parts involved with manufacturer (if available), model or part number), and relevant ratings if no datasheet is available. For modules like the boost converter, a link to the source is usually the best you can do.

Second, you should provide a schematic that shows how things are actually connected. (if you don’t know how to draw a schematic, make a wiring diagram) Doing the due diligence of tracing what you’ve done to ensure the schematic reflects reality rather than your intentions is often enough to solve the problem yourself as you spot an error you overlooked many times.

Third, you should measure relevant outputs. If something produces a voltage actually measure it and report the result. Same for things that must provide particular current, an AC signal (like audio), or a simple logic level. Also, if there are supposed to be particular resistances at various points, measure and report those. Make sure your meter is set to the proper range (e.g.: AC or DC for respective voltages so you don’t get confusing or misleading results).

Fourth, if there are different states involved (e.g.: switch on/off, timer running/elapsed, device working as expected or an anomaly, document the different settings at the different states.

And last, but absolutely no less important than anything above—and place it at the very beginning of your post: Explain the problem you are trying to solve with whatever you are building. Note that this is not the same as explaining the solution you have settled on—this is not your problem. The reason this is so important is that to the expert knowing the problem at hand implies a thousand little details that you fail to mention or might not even be aware of. This can lead to a completely unexpected solution that would otherwise never come to light because of the problem being kept in the dark.

Your original post covers some of these things, but please try to do all of it, from scratch. This in itself may lead to you solving your own problem, as is frequently the case, but if it doesn’t, it will certainly lead to better answers and a much quicker, more complete solution to your problem.

Again, welcome to AAC—it’s great to have you join us and I hope you stick around and help others when you know something they don’t.

 

Welcome to AAC.

You haven’t really provided enough information to help. To get a useful answer instead of a hundred random questions and a thousand random guesses you‘ll need to tell us more.

First, you should provide a BoM (Bill of Materials), that is, a list of the parts involved with manufacturer (if available), model or part number), and relevant ratings if no datasheet is available. For modules like the boost converter, a link to the source is usually the best you can do.

Second, you should provide a schematic that shows how things are actually connected. (if you don’t know how to draw a schematic, make a wiring diagram) Doing the due diligence of tracing what you’ve done to ensure the schematic reflects reality rather than your intentions is often enough to solve the problem yourself as you spot an error you overlooked many times.

Third, you should measure relevant outputs. If something produces a voltage actually measure it and report the result. Same for things that must provide particular current, an AC signal (like audio), or a simple logic level. Also, if there are supposed to be particular resistances at various points, measure and report those. Make sure your meter is set to the proper range (e.g.: AC or DC for respective voltages so you don’t get confusing or misleading results).

Fourth, if there are different states involved (e.g.: switch on/off, timer running/elapsed, device working as expected or an anomaly, document the different settings at the different states.

And last, but absolutely no less important than anything above—and place it at the very beginning of your post: Explain the problem you are trying to solve with whatever you are building. Note that this is not the same as explaining the solution you have settled on—this is not your problem. The reason this is so important is that to the expert knowing the problem at hand implies a thousand little details that you fail to mention or might not even be aware of. This can lead to a completely unexpected solution that would otherwise never come to light because of the problem being kept in the dark.

Your original post covers some of these things, but please try to do all of it, from scratch. This in itself may lead to you solving your own problem, as is frequently the case, but if it doesn’t, it will certainly lead to better answers and a much quicker, more complete solution to your problem.

Again, welcome to AAC—it’s great to have you join us and I hope you stick around and help others when you know something they don’t.

 

thanks for your reply Ya’akov,

my goal is to recharge the 6s battery pack, I went for a DIY method,
the configuration of the battery - three 18650 in parallel and 6 set in series with BMS - xcluma 6S 12A 24V PCB BMS Protection Board For 6 Pack 18650 Li-ion Lithium Battery Cell
used a laptop charger to get 19.5v with the help of USB QC/PD/AFC Trigger-Decoy Board module - DELL 90W 19.5V 4.62A Laptop Adapter
the voltage after the trigger board was 19.5v.
then connected to the in+ and in- boost converter and the adjusted the potentiometer to get 25.5v at out+ and out- .
this is the booster - Generic PZIN14013019 Imported 150W Adjustable 12 32V to 12 35V and directly connected out+ and out- to the bms positive and negative, that was the moment the laptop charger stopped producing power.

the purpose of this project to make a DIY power bank to that can charge the Samsung s22 at 25w and as well as charge the dell XPS 45w, 65w.
so made this battery pack to power this car charger INICIO 95W USB-C Car Charger PD 65W Type C + A QC3.0 30W Dual Port Fast Charging Car Adapter.
I wasn't able to get my hands on better modules that can charge and discharge and also do this PD functions.

the laptop charger -> USB trigger (19.5v) -> boost converter (25.5v), was able to power the car charger (max load).

when I was trying to charge the battery pack, the battery pack was at 20v approx.,
so I adjusted the potentiometer at boost converter to produce 23v approx., then battery pack was charging, 1.8A was flowing into the battery pack, after few hour, the battery was at 23v, again I increased the potentiometer to 25.5v, the battery pack was charging, 2A was flowing into the battery.
the battery pack was full at 25.2v after couple hours.
(didn't use any resistor)

it there any way to charge easily without adjusting the potentiometer. and is there anything to improve this design.

and I am basically an software + mechanical engineer, new to this electric circuits and modules (with basic knowledge).

 

I don't understand why you need the usb trigger and boost converter, why not modify the laptop charger to give out 25V?

 

You need a battery charger circuit. You cannot simply connect a voltage to the battery.

Your laptop supply (it is NOT a charger), shut down because too much current was being drawn. You are lucky you did, because otherwise the battery might have ignited.

Charging LiIon batteries is not for beginners. The likes of Apple and Samsung have created chargers that caused fires. Get a commercial charger for a 6S LiIon battery.

 

thanks for your reply Ya’akov,

my goal is to recharge the 6s battery pack, I went for a DIY method,
the configuration of the battery - three 18650 in parallel and 6 set in series with BMS - xcluma 6S 12A 24V PCB BMS Protection Board For 6 Pack 18650 Li-ion Lithium Battery Cell
used a laptop charger to get 19.5v with the help of USB QC/PD/AFC Trigger-Decoy Board module - DELL 90W 19.5V 4.62A Laptop Adapter
the voltage after the trigger board was 19.5v.
then connected to the in+ and in- boost converter and the adjusted the potentiometer to get 25.5v at out+ and out- .
this is the booster - Generic PZIN14013019 Imported 150W Adjustable 12 32V to 12 35V and directly connected out+ and out- to the bms positive and negative, that was the moment the laptop charger stopped producing power.

the purpose of this project to make a DIY power bank to that can charge the Samsung s22 at 25w and as well as charge the dell XPS 45w, 65w.
so made this battery pack to power this car charger INICIO 95W USB-C Car Charger PD 65W Type C + A QC3.0 30W Dual Port Fast Charging Car Adapter.
I wasn't able to get my hands on better modules that can charge and discharge and also do this PD functions.

the laptop charger -> USB trigger (19.5v) -> boost converter (25.5v), was able to power the car charger (max load).

when I was trying to charge the battery pack, the battery pack was at 20v approx.,
so I adjusted the potentiometer at boost converter to produce 23v approx., then battery pack was charging, 1.8A was flowing into the battery pack, after few hour, the battery was at 23v, again I increased the potentiometer to 25.5v, the battery pack was charging, 2A was flowing into the battery.
the battery pack was full at 25.2v after couple hours.
(didn't use any resistor)

it there any way to charge easily without adjusting the potentiometer. and is there anything to improve this design.

and I am basically an software + mechanical engineer, new to this electric circuits and modules (with basic knowledge).View attachment 303414

(This is part one of two in my response because I write far too much and am short time this morning. But I should get to the rest later today, or failing that, tomorrow. As it stands it doesn’t answer your question, but I promise part two will tackle that so don’t despair. I do believe that the safety issue is important enough to spend substantial time on, so here it is…)

Nice write up. Very helpful.

So, first—

As @BobTPH said, it is very important to consider safety first when it comes to Lithium batteries. They can and do burst into nasty flames when provoked. However, the nature of this behavior is often misunderstood so let me review a few points to bring it into perspective.

Lithium battery fires are about heat, in a process called thermal runaway—and the highly flammable electrolyte in the cell. The electrolyte is the connection between the positive and negative terminals of the cell. This can be a liquid, a solid, or even a gel (like a “gel cell” SLA Sealed Lead Acid battery) Which of these is used depends on the chemistry and desired performance characteristics.

It is worthwhile noting that another sort of Li cell, Lithium Iron Phosphate (or LiFePO⁴ or LFP) doesn’t use a flammable electrolyte and so is not subject to the problems that plague its Cobalt-using fellows. The downsides of LFP are higher cost and lower energy density. The former will probably be mitigated by increased production but the latter is inherent in the chemistry and the demand for smaller, thinner devices with infinite run times make the use of LFP chemistry a non-starter for many manufacturers.

So, for example, a typical cylindrical 18650¹ cell² will have Lithium Cobalt Oxide³ (LiCoO₂ or LCO) cathode. The cathode is the component of the battery that comprises the positive terminal. LiCoO₂ cells use a highly flammable organic electrolyte in a liquid form. The LiCoO₂ provides the ions that transport the charge through the electrolyte to the anode which in this case is usually graphite.

The typical LiPo (Lithium Polymer) prismatic (non-cylindrical) cell doesn’t use a liquid electrolyte. It uses LiCoO₂ cathode but the polymer in the name refers to a solid electrolyte that allows for different physical construction which, in some applications, has distinct advantages. One is the arbitrary shape they can be made allowing for high powered batteries in spaces where cylindrical cells couldn’t possibly work (e.g. mobile phone, tablets, wearables, &c.). They also have much larger surface area and so can potentially provide much more current than a cylindrical cell.

Unfortunately, the internal components of a LiPo cell are also flammable and from a phenomenological perspective, the dangers and outcomes are the same as the liquid electrolyte versions.

There is almost no Lithium (Li) metal in a Li cell⁴. While there are fears of metal fires (a nasty sort of fire indeed) this is just not the issue with LiCoO₂ and related chemistry batteries It is that organic electrolyte. The confluence of attributes of the cells (cylinders or prismatic “pouches”) and the chemical reactions, create a uniquely scary and now ubiquitous device.

Here’s the problem: there are at least four times that a Li cell might find itself getting excessively hot. The first is in a fire. Well, this is obvious, keep everything not specifically meant to be placed in a fire out of one. In this case, it’s a matter of the cells just being cooked until the effects we see from proper thermal runaway are created by the heat of the fire.

The others have to do with electrical current in one way or any other. Specifically the charging or discharging of a cell in a way or condition that should not exist. Let’s look at these three, ending on the one that is most relevant to your charging circuit (or lack thereof).

If a cell suffers a physical insult (not to anthropomorphize but I already have the cell “finding itself, so…), and the damage leads to a short circuit by connecting internal layers of the cell so current can flow inside the cell, the result will be heat. How much heat with depend on the nature of the damage and the SoC (State of Charge) of the cell.

This is a case where the stored chemical energy is relevant. The fire itself is not necessarily connected to the SoC, but the propensity to start the thermal runaway process certainly is. This is a subtle distinction but while ignition is tied to electrical discharge the fire itself is not. It will affect the speed of the thermal runaway but not the temperature of the resulting flames.

The process looks like this: a short circuit is formed inside the cell which leads to an electrical discharge and heat proporational to the energy involved in the discharge. The power can be characterized in Watts (W) by taking the Voltage (V) and multiplying it by the current in Amps (A). So, in an arbitrary example of a fully charged cell, let’s say we are dealing with 4V and the short circuit causes a current flow of 20A inside the cell.

The power, in W would be W = VA, or W = 4V x 20A, or 80W! This is a lot of power, and the temperature rise in the concentrated area where this is being discharged will be fast and furious. At first, the electrolyte will boil, then an 18650 or similar cylindrical cell will vent from the built-in pressure relief valve under the positive cap of the cell (a LiPo is even more problematic, which we’ll see in a moment).

When the very, very hot flammable electrolyte exits the cell as a jet of fluid it is ignited by the spark of the discharge and beginning a very scary flamethrower, literally. It spews a flaming jet of hot fluid at high pressure, burning people unfortunate enough to be in the path and igniting flammable things nearby.

In the case of a LiPo “pouch” cell, if the damage includes a hole in the exterior, it acts like a vent, and the hot liquid fire will be eject from it. If the problem does not include having been pierced, the outer pouch will inflate and eventually explode. The rest is the same. An 18650 cell can explode if the pressure relief valve fails to operate. This can add shrapnel to the other dangers, and can also mean a much larger fireball and not just the nasty jets which are bad enough.

In failure mode three, these same things can happen if the cell is discharge by a short circuit across its terminals. It will take longer, and the cell must be in a high SoC, but shorting a Li cell is not a good idea no matter how you look at it.

These failure modes depend on the cell being in a SoC that is high enough to produce the heat required. A discharged cell does not present this problem. But if you introduce power to a discharged cell, you bypass this safe state, and experience failure mode four.

But, in your case and in a large number of the cases where serious problems have occurred, it is during charging that disaster strikes. This can happen even if the charger is doing everything “right”, but just doesn’t have any way to sense there is something going terribly wrong.

As Li cells discharge to lower levels, something called dendritic growth can occur. This is the formation of conductive ”whiskers“ which penetrate the separator keeping the layers that store the chemical energy from coming in to contact and discharging. These are generally very small, high resistance connections so they have the effect of diminishing cell capacity and increasing self discharge but are generally not a real danger.

However, when the cell SoC falls far enough, generally considered ≤1V, the danger of low resistance, large dendritic shorts becomes something needing attention. For this reason, if a Li cell is allowed to discharge this far, protection circuits are supposed to prevent charging. This is because if you try to charge the cell with a short in it, it will heat up the same way it does when it is shorted by damage—but not from the chemically stored energy in the cell, because of the chargers current being innocently supplied to charge it.

Once the cell gets hot enough, the same game plays out as above. But probably in your bedroom or living room! This is also where we can look at the Li battery not just the cell. Thermal runaway includes the heating of adjacent cells. And even if they are perfectly good with no damage and no dendritic short, they are now “cells in a fire”, and recall the rule is to keep your cells out of fires.

This leads to a chain reaction, which comprises one cell heating its neighbor to thermal runaway and that cell paying forward by donating heat to another. A process that results in a cascade of flamethrowers until there are none left or something else stops the process.

One way proper chargers manage this possibility is by monitoring the temperature of the cell or cells and shutting down the charging process if the heat rises too far. But, not all chargers use a temperature sensor.

Is a BMS a charger? No, strictly speaking a BMS is not a charger. But it incorporates a lot of the same safety mechanisms that should be in a charger. A good BMS, in addition to providing balancing* also provides protection against:
(*a closely related topic for various reasons but I am not going to cover it)

• over-voltage / overcharge
• under-voltage /over-discharge
• over-current
• over-temperature

What a BMS doesn’t provide is management of the charge curve. All cell chemistries have a characteristic curve of applied b voltage relative to the cell’s SoC. In the case of Li cells, this takes the form of a strategy called constant current/constant voltage. In which the first the current is fixed at the ideal value (generally, the lower the better for the cell life, as charging produces heat which degrades capacity) for the time to completion desired. Most cells will be specified at 1A for a “normal” charge, fast charging will use more current (always check the datasheet to see what the ideal and maximum charge current for the particular cell is).

As long as the current being drawn by the cell or battery is greater than the current limit, constant current mode continues. At a certain point, the battery voltage will be such that it will no longer draw as much as the current limit with the maximum applied voltage (generally 4.2V for Li cells). At this point the switch to constant voltage happens. The cell is allowed to draw whatever current the 4.2V produces.

Eventually, the cell reaches the 4.2V of the charger and the charging is stopped. This can happen (mostly) naturally since the two voltages are the same and nothing will flow. But the BMS, or the cells own protection circuit will shot off the input from the charger when the maximum voltage is reached (over-voltage).

One of the problems is, that while over-voltage protection works great, the over-current protection has to be designed to allow the maximum discharge current (as much as 50A in the case of high capacity cells!) because it can’t distinguish between charge and discharge. So the charging circuit must provide this protection.

It is this gap in the BMS/cell protection regime that can lead to charging related fires.

I will cover more of how this affects your project, and how you can fix it, in part two.
—
1. 18650 is the nomenclature of the most common type of cylindrical Li-Ion cell. The name is derived from the size of the cell casing. which is 18mm in diameter and 65mm tall. Other cells follow that pattern such as 16350 with Φ16mm and a 350mm height. Primastic cells follow a similar convention but there are many more ppossibilities.

2. A cell is a single element of a battery. So an “18650“ is a cell, while an 18650 battery is a collection of 18650 cells. This terminology is due to Ben Franklin, who was a very productive and celebrated researcher into electricity in the early days (1700s) and thought that the collection of cells used to get increased voltages were like an artillery battery used to increase fire power. Franklin made prodigious contributions to the field of both theory and practice in the then nascent field of electricity though he seems most remembered for the coin toss he lost in naming the positive and negative terminals of a cell leading to all sorts of confusion when it comes to how electricity flows. But I will leave that for another time.

3. There are other Li compounds used, such as Lithium Nickel Cobalt Aluminum Oxide (NCA), and Lithium Nickel Manganese Cobalt Oxide (NMC). In general, the thermal instability of the cell can be judged by the fact and extent of the Nickel in the chemistry.

4. There are Lithium Metal cells, and these do add the danger of a Li metal fire. Fortunately, you are exceedingly unlikely to encounter any.

 

Here's some chips that might do the job:
https://www.analog.com/en/parametri...=chem|Li-Ion_vmax|25.2_icharge|1&p5380=Li-Ion
Or buy a 25.2V lithium-ion charger from ebay.
https://www.ebay.com/itm/263767410279

 

thanks for all your responses, I got an constant voltage - constant current module
boosted the 20v to 25.5v, using the above cc-cv module adjusted the current to 3A, the battery can accept 1.5A each cell(0.7C - 1C max) two cell are in parallel, so 3A. and charged fine with the help of 6S bms (balanced charging )