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I am facing an issue with my common ground for my system. The system below has a voltage of the battery equal to 16 V and the power supply is supplying my MCU with 3.3V and 0.20A.

What is the difference between option 1 and option 2 knowing that option 1 is the recommended and option 2 lead to burning my MCU and having a short circuit between vin and GND is there any difference between two scenarios. It was mentioned to me that there is a high current path that might path through MCU and might lead to burning it can any one further explain it please. Knowing that B- is connected to GND layer of the BMS. enter image description here

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  • \$\begingroup\$ Usually in battery-control devices the negative battery is not GND. Low-side switches and sense resistors for example, are in between. When something bad happens, they disconnect the B- from GND (but the B+ stays connected). \$\endgroup\$ Commented Jul 16 at 10:24
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    \$\begingroup\$ This question is similar to: BMS-Battery System. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. \$\endgroup\$
    – winny
    Commented Jul 16 at 10:59
  • \$\begingroup\$ Mohamed Smaili - Hi, Please don't materially change a question, after it has been answered (a.k.a. "chameleon question"). While clarifying a question after receiving answers can sometimes be appropriate (subject to review), editing-in new interpretations isn't appropriate. || If you want to ask an answer's author whether your interpretation of their answer is correct, then ask in a comment on that answer. If you want to add image links in a comment, you can use this technique. Note that the draft answer must not be submitted. (Edit reversed) \$\endgroup\$
    – SamGibson
    Commented Jul 16 at 12:20
  • \$\begingroup\$ @SamBison Thank you for mentioning I'll do. \$\endgroup\$ Commented Jul 16 at 12:28

1 Answer 1

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First off, your drawing is a bit mislabeled: you have a BMS board, not a BMS chip alone (see your previous question); and what you label "VDD" is not what VDD should be, namely the positive supply voltage of a digital chip, but is probably "GND" on the MCU board.

Because we really had a bad time understanding each other on your previous question, please go through every single sentence that follows and check whether you understand it. If you don't, or have any other questions, ask about that first, before asking anything else.

Now to the multiple problems:

Ground shift due to abusing a battery cable contact

In a battery management system, large currents flow.

In your case, large current flows on the B- cable.

A large current means that even with small cable and resistance, there can be a potential difference: Ohm's law.

So, even if there is a cable or trace between the point where you connect and your BMS's ground potential, there can be a substantial potential between these two points.

So you connect the ground of your MCU board to something that is above the voltage of the ground of your BMS board.

That means that, for example, the voltage that the BMS pulls the SDA line to to signal a "0" will be below the ground voltage of the MCU. The MCU's protection diodes will conduct. If the potential difference is high, too much current flows from MCU to BMS, and your MCU burns.

Externally forced potential difference

The external supply attached to the MCU makes matters worse: If that is not isolated, it will pull ground of your MCU board to some specific potential.

Your charger, if not isolated, will set the negative charger rail to a potential, which might be the same as the GND from your "Power Supply" block.

Now, between that CH- contact on your BMS board and the B- contact, there's a current sense shunt: So, by design, the B- pin is above the potential of CH-. If CH- and power supply GND are connected externally through power supply and charger supply, then the same current as through the current sense resistor can flow, one way or another, from MCU to BMS, and something burns.

Solutions

Clearly, don't use B-. It's the least sensible connector for ground in your system! Simply don't.

Then, instead of using an external power supply for your MCU, simply use the voltage supplied by the MCU for controller things. That's what it's for! If that seems impossible, use an external supply that you know is well-isolated, or simply use a separate, small battery, to power the MCU.

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  • \$\begingroup\$ Thank you for your very detailed explanation. Based on What I have understand from your explanation I have added a picture in the question above showing the flow of current in both scenarios is it true what I have understood ? \$\endgroup\$ Commented Jul 16 at 12:12
  • \$\begingroup\$ @MohamedSmaili - You just commented: "I have added a picture in the question" I have removed your change to the question and I added a comment below the question explaining why, and what to do instead. You can add a link to that image in a comment here. Please don't start a discussion about an answer, by changing the question. Stack Exchange doesn't work that way. TY \$\endgroup\$
    – SamGibson
    Commented Jul 16 at 12:26
  • \$\begingroup\$ @SamGibson Thank you for mentioning \$\endgroup\$ Commented Jul 16 at 12:27
  • \$\begingroup\$ @MarcusMüller Find attached the link showing what I have understood from your explanation: i.sstatic.net/XIplWOgc.png \$\endgroup\$ Commented Jul 16 at 12:39
  • \$\begingroup\$ You're missing the path where current I5 returns; current can't just be flowing into something and never "come out". But yes, the fact that there was somehow current flowing into or out of your MCU board was clear. Your drawing shows one way current could be flowing into your board, but it's not related to what I wrote in the answer, at all. \$\endgroup\$ Commented Jul 16 at 13:04

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