Electrical – How to get 5V 10A output from a 3.7V li-ion battery for 64×16 led matrix display module

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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)

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.