5

Your proposed balancer is similar to the one you posted in your own answer. Your understanding of its function seems to be correct - the charge current is distributed to the cells depending on the discharge state of each cell. A cell with lower voltage will get more current than a cell with higher voltage. In that sense it will work. The folks who wrote the ...


5

This is a bit like asking if you can fill a 42 foot high tank through a spout 12 feet off the ground. No. Your source has to be higher than the destination or you need a pump (a voltage booster in your case).


4

It is presumably to avoid power from the cell discharging into the USB port. Particularly of the host/charger is powered off. The 0.4 ohm resistor will help reduce the power dissipated within the IC at high charging currents.


3

for a beginner how do you explain this? Assume a tank is full of water, and there's a faucet attached to the tank. You can adjust the flow rate of the faucet: If you open it more, the tank will drain more quickly. Now the Amp-Hour (i.e. charge) is the amount of water in the tank, the flow rate of the water is Amperes. 7 Amp-Hours means the source can ...


2

Your 3s9p cells @ 3Ah /cell may supply 36Ah in 12h to a pack and with 27Ah capacity pack but additional soak time is required to achieve this capacity during CV mode. This is a tradeoff between battery life cycles and maximum capacity and may be defined in the charger design specs as 2 to 5% of CC rate with different CV thresholds from 4.1 to 4.2. This ...


2

tl; dr: consider a 4s RC hobby pack+charger solution. All the problems are solved for you. I can't recommend 3s + boost converter for running a high-current motor load as this adds losses and increases the current demand on the battery pack, shortening its life. Using 4 cells (4s) is better, and as it so happens this is a popular pack for RC cars anyway. ...


2

And would it work without it too? Probably, depending on what is inside the 'TP4056 charger' block. The resistor won't be needed if the USB cable has sufficient resistance, and even if it doesn't all that will happen is the TP4056 charger IC gets hotter and has to reduce the charging current so the battery takes longer to charge. The diode stops current ...


2

The problem is that you don't have a lithium battery charger, and you are directly connecting a 12V power supply to your batteries. Don't do that, as it will damage the power supply and the lithium batteries, and they may explode or burst into flames. Only charge the lithium batteries with a lithium battery charger.


1

Lenovo has this algorithm for planned efficient usage @ 60% SoC. Ref


1

A Lithium ion battery pack must be charged with a circuit specifically designed to do so, and should be discharged with a protection circuit as well. Batteries with multiple cells in series(like yours) should be charged with an appropriate BMS(Battery management system). If you're using Lithium ion for a project, do your research. You might want to check ...


1

Yeah, this will work, however: Better get a good quality power supply (brands like Meanwell) than no-name junk. These individually addressable LEDs draw current pulses. Some cheap power supplies have trouble regulating the voltage, or they will emit audible whine, which is annoying. Some junk brands are quite dangerous and omit stuff you really want to have, ...


1

T is the terminal for the battery charging device to measure battery temperature with a temperature sensor. The blue part is the temperature sensor. The sensor resistance changes according to the temperature.


1

Laptop power supply is a power supply. It is not a charger for lithium batteries. It can not and must not be directly connected to batteries. Most likely what happens is that the 19V output will drop when it is connected to batteries with lower than 19V, and the power supply tries to give as much as current possible to try rising voltage back to 19V, but it ...


1

There are three main reasons why a 'fully charged' cell does not stay at 4.2 V after charging. Firstly the cell has internal resistance, so any current going into it will raise the terminal voltage, while any current drawn from it will lower the voltage. Secondly, most chargers do not put in a full charge. Thirdly, voltage will gradually drop after charging ...


1

You must take into account the charging (and recharging) profile and charging cycle. For instance in the following profile given on Digi-Key's site: the hysteresis is clear where you can see the lower threshold. So it will wait to discharge till 3.9v (for this particular profile) before recharging.


1

Voltage is not a very accurate way to determine how full a battery is, especially a flat discharge curve battery like a Li-Ion. The most accurate ways are super inconvenient though. Chargers also don't use voltage as the sole method of determining when a battery is full. They use a combination of methods.


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