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I have a circuit of a battery pack of 4 18650 and a 4S BMS. See the picture below.
The question is: is my understanding correct that this circuit means that the batteries are connected in a series, so the output from BMS should have 6.7 A current (current of 1 battery) and voltage is 4*3.7=14.8 V (4 times the voltage of one battery). I think it should be so but I want to be sure to avoid a dangerous situation. I'm not sure because of these wires that go from the "middle" of the battery pack, i.e. not from the head and the tail.
The circuit and BMS are standard. I say that circuit is standard because I don't know how to visualize the circuit without wires intersection.

enter image description here

pascalamin
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    Wires from the "middle" of the pack probably allow the BMS to monitor the voltages of individual cells and maybe balance cell voltages. But, what does the data sheet say? – Solomon Slow Nov 15 '21 at 15:02
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    I'm only going to buy. For example this one: https://www.aliexpress.com/item/1005001916022803.html?spm=a2g0o.productlist.0.0.5e486d6fP4ITjE&algo_pvid=8050e8d3-43ce-4ca8-8fbd-c5511b20aff1&algo_exp_id=8050e8d3-43ce-4ca8-8fbd-c5511b20aff1-1&pdp_ext_f=%7B%22sku_id%22%3A%2212000018105332513%22%7D – pascalamin Nov 15 '21 at 15:07

1 Answers1

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All four cells are connected in series between B1- and B+. These will be the main connections for carrying the current.

So the total voltage will be the sum of all the cells (14.8V). The maximum current will be that of any one cell (6.7A), unless the BMS applies a lower limit of its own.

There are three extra sense wires B1+, B2+ and B3+. This allows the BMS to measure the voltage on each of the cells, while the pack is charging or discharging.

The BMS can make sure that none of the cells is ever discharged too much. When charging, it can check that all four cells are charging evenly. If not, then it can adjust the voltages across the cells to balance out the charge.

Simon B
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    So on average, the output of the BMS will be 6.7 A current (current of 1 battery) and voltage is 4*3.7=14.8 V (4 times the voltage of one battery)? I understand that both will change depending on how much the batteries are charged, I just need a simple answer without these details – pascalamin Nov 15 '21 at 15:29
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    @pascalamin Yes. – Simon B Nov 15 '21 at 20:19
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    @pascalamin "6.7 A current (current of 1 battery)" - should be "current of 1 cell". The same goes for "4 times the voltage of one cell" and "how much the cells are charged". It is important to use correct terms so we know you understand the answer. – Maple Nov 15 '21 at 21:54
  • @Maple thanks for correction – pascalamin Nov 16 '21 at 06:33