2
\$\begingroup\$

I want to make a battery pack with 7.4 volt output and I want to charge them with 1A each so I connect them parallelly and input them with 5A 6A DC adapter. (No charging and using at the same time) Is this a good way?

My first circuit diagram that is shorted

Update: I have change my circuit to this diagram, is it still short circuit?

My new circuit diagram

\$\endgroup\$
4
  • \$\begingroup\$ It may be possible while not advisable, but the modules don't look isolating: the "green connections" may be a problem, same for each battery's connection to either + or - IN or OUT. \$\endgroup\$
    – greybeard
    Commented Jun 10, 2023 at 5:19
  • \$\begingroup\$ If I replace green line with diode like 1N4007, is it solve the problem? Also with evey red line at Vout. \$\endgroup\$ Commented Jun 10, 2023 at 5:26
  • \$\begingroup\$ Replacing the green line with a diode will result the battery to be shorted through the diode. \$\endgroup\$
    – Justme
    Commented Jun 10, 2023 at 12:24
  • \$\begingroup\$ It is still not a good way. You do realize the six chargers and six isolated converters cost more than a proper charger, and your pack could still be 3P 2S if you want a single 7.2V pack and charged with a single 7.2V charger. \$\endgroup\$
    – Justme
    Commented Jun 10, 2023 at 15:53

4 Answers 4

3
\$\begingroup\$

No, that's not a good way, you have just shorted the batteries together through the charging modules.

\$\endgroup\$
1
  • \$\begingroup\$ If I replace green line with diode, is it still shorted the circuit? \$\endgroup\$ Commented Jun 10, 2023 at 12:18
2
\$\begingroup\$

This will not work in the general case.

The charger board, however it looks like, is in fact a short between the two of its terminals (usually the negative ones) and a variable resistance between the others (the positive ones).

So your drawing depicts a dead short circuit.

Cells (or any other electronic elements for that matter) cannot be connected in series and in parallel at the same time. You have to actively switch between the two states. A single DPDT switch with an adequate current rating will do. As a bonus, you will need a single charging board.

In fact, this is done in a lot of (older) designs. The newer approach usually utilizes some kind of switching-mode voltage converter.


Another viable approach is to use a galvanically-separated chargers. E.g. leave off the left side of the circuit and use a separate USB charger for each battery. Or at least two for each of the two battery banks. Messy, isn't it?

And third, you can simply get a 7.2V (2S) charger instead of 1S ones you are planning to use.

\$\endgroup\$
2
  • \$\begingroup\$ For other methods that you told me that if I do a galvanically-separated on left side of the circuit (power supply side for charge) by having a DC to DC (5v to 5v) isolator such as the wrb0505s-3wr2. Will it still short circuit like before? \$\endgroup\$ Commented Jun 10, 2023 at 12:16
  • \$\begingroup\$ Sure, this will work. But wrb0505s-3wr2 is somewhat weak for your purposes, you won't get a whole ampere from it. The whole thing will get somewhat expensive as well. \$\endgroup\$
    – fraxinus
    Commented Jun 10, 2023 at 18:40
2
\$\begingroup\$

Your second approach is smashing a square peg in a round hole:

  • The DC-DC converters are a solution to a bad approach
  • The cells and the charger/protection boards will be damaged whenever one of the BMSs turns its switch back on and the other cells are at a different voltage

This is the only effective way to achieve what you need:

  • Connect 3 cells directly in parallel to form a block
  • Connect 2 blocks in series to form a string
  • Get a protector BMS for a 2S Li-ion battery
  • Get a charger for a 2S Li-ion battery; for example
  • Install the 2S BMS on the string of cells; for example
  • Charge the battery with the 2S charger
\$\endgroup\$
2
  • \$\begingroup\$ If doing this, each BMS must have its own charger or can I find an adapter such as 9V 6A to connect in parallel? \$\endgroup\$ Commented Jun 10, 2023 at 16:37
  • \$\begingroup\$ What do you mean "each BMS"? I said "Get a protector BMS for a 2S Li-ion battery". "a' = 1. There is only one BMS. That's the only safe solution. The moment you have more than 1 BMS, things get nasty. \$\endgroup\$ Commented Jun 10, 2023 at 17:55
0
\$\begingroup\$

The DC/DC converter is unnecessary. Use an LM2596/LM2576 adjustable IC and just connect the batteries in parallel. Or use your IC without the converter and fix the circuit.

If you decide to use LM2596, remember it has no current limitation. You need to add an external current limiting circuit.

The 18650 is charged at 4.2V max, 300mA.

https://www.aliexpress.com/item/32464248769.html

\$\endgroup\$
4
  • 3
    \$\begingroup\$ CFCBazar - Hi, Your shortened Aliexpress URLs (here and your previous answer) all contain tracking parameters i.e. they are hidden affiliate links. Please post either plain URLs or risk them being treated as spam (which has severe consequences for your account here). Thanks. \$\endgroup\$
    – SamGibson
    Commented Jun 10, 2023 at 14:02
  • \$\begingroup\$ But the problem is that what I want is to have 3 2s batteries in parallel so that each pair will be in series first and then paralleled. \$\endgroup\$ Commented Jun 10, 2023 at 14:14
  • \$\begingroup\$ If you feel like the links are spam, remove them, I will post the long ones as soon as I can. \$\endgroup\$
    – CFCBazar
    Commented Jun 11, 2023 at 15:58
  • \$\begingroup\$ You can't charge 18650 in series with a 5V power supply, 2x4.2 is 8.4. They are not charged in series parallel, which is why you have a short on your circuit. \$\endgroup\$
    – CFCBazar
    Commented Jun 11, 2023 at 15:59

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.