0
\$\begingroup\$

For powering my portable circuit, I need a bipolar power supply of ±2.5 V. The circuit is low-power, but requires very precise power.

For its stabilization, I am using ADM7154 regulators for positive and ADP7183 for negative power, respectively. They have a slight difference in load current (using ultralow noise regs with the same load for me is not possible), hence the same current won't be drawn from both batteries.

Two 18650 batteries are connected in series, and I need to simultaneously charge them both from USB without replacement (two TP4056-based modules with isolated ground via a transformer? DC-DC isolator? I need to avoid using switching electrical components).

Also, I am unsure about the BMS connection scheme in this case. Individually for each battery, connecting the plus of one with the minus of the other to simply have bipolar supply? Or perhaps this is incorrect, as my assumption about isolated charging might be wrong? I am just getting into using batteries, so some questions remain unclear to me.

In summary: how do I properly connect the BMS and charge bipolar 2s battery supply?

EDIT: my solution at the moment is as followsenter image description here

\$\endgroup\$
3
  • \$\begingroup\$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. \$\endgroup\$
    – Community Bot
    Commented Apr 19 at 21:43
  • \$\begingroup\$ "Employ separate BMIs for each cell". Body Mass Index? ;-) \$\endgroup\$ Commented Apr 20 at 0:34
  • \$\begingroup\$ "Perhaps the solution could be...:" You're sinking deeper and deeper into a monster. Keep it simple. Single 7.2 V Li-ion battery. Regulate to 5 V. Regulate to 2.5 V. Invert the 5 V to -5V. Regulate that to 2.5 V. \$\endgroup\$ Commented Apr 20 at 0:36

2 Answers 2

0
\$\begingroup\$

What you have there is not a single battery with two 18650 cells. Instead, what you have is two batteries, each battery having a single cell. Therefore, you will need to treat them as two separate batteries. Each battery has:

  • Its own BMS (a single-cell BMS) with its own protection switch
  • Its own charger (a single-cell charger, isolated from the other one)

A simpler design is to use a single battery with a single cell. Then, convert the positive 3.7 V voltage from that battery into a negative voltage -3.7 V with an inverting switching regulator. Then you can charge that single battery with a TP4056-based module.

\$\endgroup\$
12
  • \$\begingroup\$ Hey Davide, thanks for the quick response! Unfortunately, a switching regulator won't work for me due to the specifics of the circuit. Additionally, I also plan to use a common 5V level in addition to ±2.5V (I apologize for not mentioning this earlier). As a result, I will still need to combine the cells into one battery. In such a case, how can protection and charging be competently implemented? Leave individual BMS and physically disconnect the load during charging? This seems like a rather suboptimal solution. \$\endgroup\$
    – nighma
    Commented Apr 19 at 23:03
  • \$\begingroup\$ I didn't give this much thought earlier as I got caught up in figuring out the proper charging, and it slipped my mind! It's important that this 5V shares the same ground as the 2.5V. In that case, it probably necessitates using an additional cell. Or should I just increase the voltage of one of the two? Then the load will be quite different. \$\endgroup\$
    – nighma
    Commented Apr 19 at 23:10
  • \$\begingroup\$ But, you are considering a switching regulator! You said: " with isolated ground via a transformer? DC-DC isolator?" That is a switching regulator. \$\endgroup\$ Commented Apr 20 at 0:32
  • \$\begingroup\$ " It's important that this 5V shares the same ground as the 2.5V." Sure. My suggested solution (a single battery, but using 2 cells in series) does meet that requirement. \$\endgroup\$ Commented Apr 20 at 0:32
  • \$\begingroup\$ Yes, exactly, the method I described for isolated ground involves a switching regulator. The use of which I absolutely have to avoid. That's why I don't like the solution I have at the moment. \$\endgroup\$
    – nighma
    Commented Apr 21 at 17:51
0
\$\begingroup\$

I bet you $ 1 that the reason you want a +/- 2.5 V supply is because you have a sensor that produces a bipolar signal and you need to feed it into an AD converter. If so, you don't need a 2.5 V supply. What you need instead is an op-amp to convert that bipolar signal into a unipolar signal centered on +2.5 V. Better yet, power that sensor at 0 and 5 V and it will already produce a unipolar signal centered on +2.5 V; no need to convert.

\$\endgroup\$
1
  • \$\begingroup\$ No, I have a different purpose for using a bipolar signal. It's related to the use of an ADC, but discussing further circuitry deviates too much from the topic of the request. I did use unipolar power previously, but due to some peculiarities of my circuit, it brings additional issues. I simply need to have clean +/- 2.5V, +5V; the ADC board is already designed. \$\endgroup\$
    – nighma
    Commented Apr 21 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.