I need to build for a hobby project, a active balancing circuit for large number of LiIon series cells (90 cells). So it needs to be cheap but efficient :) (ie. not resistive dup). After lot of research, I found this circuit on internet

enter image description here

with the schematic as follows

enter image description here

Now the explanation of the author is quite poor, saying that the buck will "track" the VBAT/2 which is VCOM. I really don't fully understand the principle, so I can scale for my needs. Also, I'm puzzled about efficiency.

If buck's FB is feed by VBAT/2, then the buck will (try to) regulate VCOM (middle) to VBAT/2 which can happen only when (after) batteries will have the same voltage.

This can work only if bottom BATT has lower voltage, so BUCK, absorbing energy from both, will charge it until become equal.

But what if this bottom BATT has higher voltage? The BUCK will try to equalize the same, but as far as I see, will discharge it through buck SW itself (synchronous, lower side mosfet). So BUCK needs to be synchronous one for sure.

Am I missing something?
What does the 2'nd OPAMP denoted LPF filter?

Thank you in advance for further clarifications.

  • \$\begingroup\$ can you link to where you found this? Quite clearly, adding the context will help us. \$\endgroup\$ – Marcus Müller Oct 7 '20 at 9:32
  • \$\begingroup\$ Yes it is possible. But for 90 cells in series (that's around 360 V DC, I hope you know what you're doing but I'm afraid that you lack enough experience) that will be a challenge even for an experienced engineer. I guess that the circuit could work but I fear that it might have issues like the circuits working against each other. I would expect a "proper" cell balancing system (like in an electric car) to be much more complex than this circuit. Also: why do you need 90 cells in series? \$\endgroup\$ – Bimpelrekkie Oct 7 '20 at 9:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.