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This my third question about a 3S Li-ion charger circuit.

In the first one, I learned about the functions of a BMS: BMS adjusting charge current

In the second one, I learned that a charger circuit is needed to limit the charging current: 3S 18650 battery charge current limiter

I couldn't find a low cost 3S charger IC system.

I came across some examples using three TP4056 ICs on the internet. It has 2 limits:

  1. When I connected three TP4056s in series, I saw that there is not an equal 5 V drop across each.
  2. It also needs 15 V; I need to use a 12 V adapter.

I decided to use a step-down converter to supply 5 V to every TP4056 and also isolate their negative terminals using DC-DC converters.

Like this one.

enter image description here

Or with my drawing:

enter image description here

Is this diagram logical and suitable for my purpose? Are there things I missed? Specifically using DC/DC converters for isolating.

I need validation and criticism.

EDIT-1
Probably I need to use a TP4056 model with integrated BMS:

enter image description here

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  • \$\begingroup\$ Your drawing is quite different from what's shown in the video you linked. Why are there only two wires going from your TP4056's to the battery pack? (The two red ones.) Each TP4056 must be connected with two individual wires to a single cell of your pack. \$\endgroup\$ Dec 28, 2021 at 15:09
  • \$\begingroup\$ You're right. I'm a little confused there. I thought of using this bms, but it probably won't be suitable tr.banggood.com/…. \$\endgroup\$
    – berker
    Dec 28, 2021 at 15:16
  • \$\begingroup\$ I guess I have to use the batteries individually in this circuit. In this case, how the balance will be, that is another problem. \$\endgroup\$
    – berker
    Dec 28, 2021 at 15:17

1 Answer 1

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The protection PCB you linked in the comments will work. You just have to wire the TP4056s to the cells individually, bypassing the protection PCB. This is acceptable since the TP4056 itself limits the battery voltage to 4.2V while charging. The load of course has to be connected to the output terminals of the protection PCB to prevent over-discharge of the battery pack. You will get balancing for free this way as each battery is charged individually.

This will work just fine as long as you wire the TP4056s exactly like in the video you linked. A BMS is not needed; a simple protection PCB (for discharging) is enough.

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  • \$\begingroup\$ These models include protection circuits (same modules in the video) alibaba.com/product-detail/…. Do i need big green BMS indeed? Using induvidually how to balance? One of the 3 batteries may be high and the other voltage. \$\endgroup\$
    – berker
    Dec 28, 2021 at 15:37
  • \$\begingroup\$ Also, is there another cheaper option than DCDC converters at the input side for isolating? \$\endgroup\$
    – berker
    Dec 28, 2021 at 15:39
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    \$\begingroup\$ You do need the "big green BMS" (which is not actually a BMS, it's just a protection circuit). It makes sure that the battery doesn't get damaged by over-discharging. You don't need a TP4056 with an integrated protection circuit, however, since you're not going to discharge the battery through these. And no, there's no cheaper option than three isolated DC/DC converters. The TP4056s will charge each cell to 4.2V individually so the pack will be automatically balanced. \$\endgroup\$ Dec 28, 2021 at 15:41
  • \$\begingroup\$ BMS provides balance between batteries, is this right? Since indvidual charging, this circuit can't do balancing? Also, TP4056 with protection can detect over-discharge, it has output pins. I want to to get rid of big green one. \$\endgroup\$
    – berker
    Dec 28, 2021 at 15:55
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    \$\begingroup\$ The problem arises when only one of your protection PCBs disconnects its battery: The other two (still connected) batteries then force a reverse current through the tripped PCB, destroying it in the process. The reverse polarity protection diode diverts this current. Given that your batteries aren't exactly equal, it's guaranteed that one of the three PCBs will disconnect first (and break without the diode). You also have to connect only the output terminals in series, not the battery terminals (these must each be connected to exactly one cell). \$\endgroup\$ Dec 28, 2021 at 16:33

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