Please I am new to electronics and new here. I'm doing a project which involves the use of a 18650 li-ion battery (or two as may be the case) to power a 64×16 led matrix display module which (according to seller info and datasheets) takes up a large 10A and 5V. I've read posts which exposed me to BMS modules. I have also researched on the bms modules but seem to find only those of 5A (considering the 18650 works with 3.7V).
Can I find a bms module that fits into this specification as a single module?
Does connecting two modules of 5A bms in parallel result in both working as one unit of 10A?
Will I still need a li-ion charger module to work with the BMS module(s)? (considering I'll have to recharge the batteries)
Electrical – How to get 5V 10A output from a 3.7V li-ion battery for 64×16 led matrix display module
battery-chargingbattery-operatedcircuit-protectioncurrent-sourceled-matrix
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Best Answer
What you want to achieve will be very challenging, even for someone with plenty of experience in electronics. For a newbee like yourself: forget it
Why?
You're asking too much from those 18650 cells. That LED panel wants 5 V * 10 A = 50 Watt. That's a ridiculous panel BTW but I digress.
A reasonable current that a decent 18650 cell can deliver is about 2 A at an average voltage of around 3.6 V: 2 A * 3.6 V = 7.2 Watt. At that current an 18650 cell of 2200 mAh will then be empty in about an hour.
For 50 Watt you need to use in the order of about (50 W / 7W ) = 7 cells. But since there will be conversion losses from converting 3.6 V to 5 V You might even need 8 or 9 cells to have some margin.
Also it is more efficient to use the cells not in parallel but in series and then use a buck converter to convert down the higher voltage to 5 V. That's for example how it is implemented in many laptops. Charging cells in series is more complex and definitely not something for a beginner. I would suggest that you use a ready-made battery pack which outputs 12 V for example.
The reason that upconverters that can output 5 V at 10 A is that the losses will be huge at such a current. If the upconverter was 100% efficient (which it is NOT) then at 3.6 V around 50 W/3.6 V = 14 A would need to flow. That's a lot of current, any extra series resistance will cause a massive voltage drop.
I would urge you to consider using a LED panel with a much lower power consumption as that will make things so much easier.