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I want to build a battery for my robot, and I don't understand how to choose components to use after BMS to charge/discharge the battery.
I have 4 pieces of 18650 batteries: 3.7 V, 6.7 A maximum.
This is my scheme:
enter image description here
I think I will buy BMS like this.
In short, the main characteristics of this BMS:

  • 4S, 40 A.
  • Charging voltage: 16.8V - 18.1V
  • Continuous charge current (upper limit): 20A
  • Continuous discharge current (upper limit): 40A

I don't understand what voltage will be in the output of BMS. I guess the maximum current will be 4*6.7 A. I need this information to properly choose step-up or step-down DC-DC converter for charging and discharging. This is the main question. After that, I wrote some additional information that shows my goals globally.

First, discharge:
I need to source:

  • Jetson Nano (USB-C, 5V)
  • Motor (link) which inputs 7-35V.

I'm going to split a wire and then use step-up or step-down DC-DC converter to make appropriate voltage for each part.
Second, charge. From the description of BMS, I think I will use 5V 3A power adapter, then step-up DC-DC converter to make it about 17 V.
I'm going to use USB-C port to both charge and discharge the battery for Jetson Nano. I'm going to use XT60 to discharge it for the motor.
Do I do it right?
Anticipating the question, the normal power bank doesn't work for my case: it doesn't provide enough current for the motor. I think so because the car shuts down sometimes and these shutdowns correlate with the speed I choose. Also, a normal power (Xiaomi Mi Power Bank 3 Ultra compact) bank turns off automatically because the current is not enough for it, even if I use the low-current charging.
Thank you for making it this far.

UPD: Currently, I'm using 2 power banks. First powers Jetson Nano, second powers the motor. Power banks are normal, intended for smartphones and whatnot. If I use one power bank for both, it doesn't work for some reason.
UPD: Also, I just realized that the motor inputs maximum 2A so I'm confused because I don't understand why it shuts down.

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2 Answers 2

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what voltage will be in the output of BMS.

4 * 3.7 V = 14.8 V, nominally. The range is 12 V (nearly empty) to 16.8 V (full).

A BMS doesn't change the voltage. It it either on (same voltage as the string of cells) or off (0 V).

I guess the maximum current will be 4*6.7 A.

No, the maximum current "might" be 6.7 A because the cells are in series. I say "might" because battery vendors lie and exaggerate the capabilities of cells. Yes, you can draw 6.7 A from this battery, but doing so will reduce its cycle life. My recommendation is not to exceed 3 A.

Charging voltage: 16.8V - 18.1V

18.1 V is too high. The maximum voltage of the charger should be 16.8 V. If you use that charger, it will not charge properly: the BMS will shut down before the battery is full.

Motor (link) which inputs 7-35V.

As you know, that's not a motor. That's an ESC (Electronic Speed Control).

the normal power bank doesn't work for my case: it doesn't provide enough current for the motor.

Well, this battery you're designing will be even worse. Based on the kinds of questions you are asking, and how dangerous a poorly designed Li-ion battery can be, my strong recommendation is that you reconsider. Buy a larger power bank instead, one that can power your load.

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    \$\begingroup\$ Good info. Don't agree with the last part though. Pre-built battery with BMS would be the obvious substitute, not a power bank. \$\endgroup\$
    – Drew
    Commented Oct 18 at 16:24
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I think you may be confused about what a BMS actually does. It does 3 things only.

  1. It monitors the voltage of the cells, and has a solid state switch which can disconnect the battery output if the cells go above or below pre-set limits (to avoid damage to the battery).

  2. Most BMS's monitor current and can also disconnect the battery if that exceeds preset limits.

  3. The BMS monitors the cells of the battery if one is higher than the others it can discharge that cell with a very small current to bring it in line.

Note that the BMS output switch is on, or off. Except in fault conditions you can basically pretend it's not there. It doesn't change the voltage or current put out by the batteries.

Building a lithium ion battery pack is fairly simple, but be careful. There's a lot of energy stored in it and if you make a mistake it CAN burst into flames. An alternative route would be to buy a premade battery pack with a BMS built in. A powertool battery for example might suite you.

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