Hi Dirk,
I have tried to provide an explanation below.
73, Dave
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My understanding is that the BMS can only effectively balance at relatively low currents and when the battery voltages are not too far apart.
In the example shown below, the balance resistors are 43?, so the maximum balance current is going to be around 100mA.
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An example circuit using a typical balance IC is shown below. I think most BMS modules use the N-MOSFET option.
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A simple charging model is shown below with two batteries at 4.2V and one battery at 4.18V. I have assumed an internal battery resistance of 5m? for B1 and B2 and 20m? for B3.
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With the charge voltage at 12.6V and no current limiting.
AM_CHG = 699 mA
AM1 = 98 mA ? ? ? ?SW1 closed
AM2 = 98 mA ? ? ? ?SW2 closed
AM3 = 0 mA? ? ? ? ? SW3 open
AM4 = 601 mA
VM1 = 4.2 V
VM2 = 4.2 V
VM3 = 4.19 V
Therefore there is approximately 600mA flowing through batteries B1 and B2 and 700mA flowing through battery B3.
The protection ICs will cut off charging if any of the battery voltages (measured at the BMS) exceeds approximately 4.28V.
The balance circuit will continue to operate and will discharge any overvoltage battery via the 43? shunt (100mA), until the battery voltage reduces to 4.19V.
The balance resistor will dissipate approximately 0.43W, when enabled.
The actual behaviour will also depend on the track resistances between the batteries.
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