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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 64x16 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)

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    \$\begingroup\$ You need to provide links to the datasheets for the battery and BMS. \$\endgroup\$ – Elliot Alderson Sep 25 '18 at 13:27
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    \$\begingroup\$ Are you quite certain this is a good idea? Please note that your matrix requires 5 volts x 10 amps, or 50 watts. A 3.7 volt, 3000 mAh cell will provide about 10 Wh, so the matrix will only run for about 12 minutes or so (10/50 x 60 minutes). Two cells will provide about 24 minutes, and this assumes a perfect voltage converter to get from the battery to 5 volts. Assume 80% system efficiency and your time drops to about 10 minutes/20 minutes. Are you sure this is adequate? \$\endgroup\$ – WhatRoughBeast Sep 25 '18 at 14:15
  • \$\begingroup\$ Fahgeddaboudit! \$\endgroup\$ – Olin Lathrop Sep 25 '18 at 14:22
  • \$\begingroup\$ any reason for choosing a 3.7V li-ion? why not just go for a single larger battery pack and step it down to 5V with a switching regulator? this is much easier than using BMS to parallel enough batteries to satisfy your power requirements \$\endgroup\$ – Zy Gan Sep 25 '18 at 15:03
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One way or another you want 5 V at 10 A. That's 50 W. Your batteries have to cough up that, plus some extra for the loss in the converter. That's basic physics. No amount of clever electronics can fix that.

Let's say you find a boost converter that is 85% efficient. It will then need 59 W as input. 50 W of that will go to the output, and the remaining 9 W will heat the converter.

59 W is a lot to ask of a 18650 cell. At 3.7 V, that would be 16 A. And, the current goes up as the battery voltage runs down. Check the datasheet of your battery. That's either past the reasonable limit, or high enough that you will get significantly reduced energy per charge, and probably reduced lifetime.

Pop up a couple of levels and think about the larger problem. You probably want a different way to address that altogether. Or, realize that what you want to do is not feasible.

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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.

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What about the min and max voltage ratings on the matrix display, whould it by any means be safe to run it on either 3.7v (single cell) or 7.4v (dual cell). If you're using standard LiPo cells the current draw is not going to be your problem when you do not need to scale the voltage (it's only 3C for the cell). (I'm drawing 120Amps out of a 4s LiPo pack everyday ;-) @80C).

One thing you have to keep in mind though is that LiPo batteries are unstable, so either too low or too high voltage can make them dangerous explosives. So use an existing protection circuit, or come up with some protection yourself.

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  • \$\begingroup\$ Thank you all for your input. I decided to discontinue the project. As it is too ambiguous for my current knowledge \$\endgroup\$ – Fran6 Nov 14 '18 at 9:15

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