1
\$\begingroup\$

I'm currently working on a project where I need to design a device to power a 24 V, 350 W motor using 18 V tool batteries. The 18 V output from a single battery is sufficient in terms of voltage, but unfortunately, it doesn't provide enough current to power the motor. Therefore, I'm considering using two 18V batteries in parallel to meet the current requirements.

The challenge I'm facing is how to safely parallel these batteries without running into issues like one battery discharging into the other. I've considered a few options:

Diodes: Adding diodes to each battery line could work, but this approach generates a lot of heat and would require large heatsinks, which is not ideal and one battery takes the load if the voltage is different between them.

Ideal diodes: These create an OR-ing condition. Some have a sharing condition if the voltages are close enough.

Powermuxes: Just a switching mechanism between sources.

I've also thought about more complex and potentially expensive solutions like using two stackable boost converters. Another idea was to connect a buck/boost converter to one battery, with a load sharing IC that could sense the other battery and match its output. However, these options seem overly complicated for what I'm trying to achieve.

I would greatly appreciate any insights or suggestions on how to tackle this problem effectively. Are there any simpler or more efficient methods to safely parallel two 18 V batteries for my application?

Thank you in advance for your valuable input!

\$\endgroup\$
1
  • \$\begingroup\$ Depending of the type of the motor and the device it is cranking, it might be a bad idea to use it at lower voltages than it is designed for. If you demand the full 350 W power in terms of n • M your motor will draw a significant higher current than expected. I'd double check that or go for @GodJihyo solution. \$\endgroup\$
    – Ariser
    Aug 30, 2023 at 7:49

2 Answers 2

2
\$\begingroup\$

You might look at putting them in series instead of parallel to get 36V and then using a buck converter to bring the voltage down to 24 V.

You would need to look at the efficiency of the converter and how much current your batteries can deliver to see if it’s feasible. You might need to go with three packs in series, but that might be harder to find a converter and BMS for.

\$\endgroup\$
2
  • \$\begingroup\$ +1, I would also try to run them in series. One comment here is that OP would want to listen to each BMS as one might indicate undervoltage before the other and stop operating. Basically ORing the BMS fault or power good signals. \$\endgroup\$
    – winny
    Aug 30, 2023 at 7:16
  • \$\begingroup\$ In general, the problem with putting two Li-ion batteries in series is that, when either one opens under load, the voltage across its protector switch is twice its rating, and a negative voltage to boot. The protector switch may not be rated for twice the voltage, and negative. If so. it may blow up. However, if the two "batteries" are just "modules" (no BMS), then you can add a single BMS for twice the cells in series, in which case all is well. \$\endgroup\$ Aug 30, 2023 at 11:39
0
\$\begingroup\$

DO NOT connect Li-ion batteries with a BMS directly in parallel. When one of them turns back on, there will be huge inrush current from the more charged one to the least charged one, which will at best degrade them, and worst start a fire.

Do not connect Li-ion battery modules without a BMS directly in parallel. A single BMS cannot manage both battery modules . It can only sense the cell voltages of one module, allowing the cells in the other module to become unbalanced and possibly reverse, which can start a fire.

Avoid using diodes: they do allow you to parallel batteries safely, but at these low voltages they waste significant power.

MOSFET switches do let you parallel batteries safely and with a lower power loss, but it's a complicated circuit.

Therefore, all things considered, if you need a bigger battery, buy a single bigger battery.

\$\endgroup\$
2
  • 1
    \$\begingroup\$ When using Schottky diodes to parallel the batteries, you will waste only 2.7% of the energy. I think this might be an acceptable loss for such a makeshift solution. \$\endgroup\$
    – Ariser
    Aug 30, 2023 at 7:45
  • 1
    \$\begingroup\$ I was calculating 12W of power dissipation through diodes (350W / 18V * 0.6 forward voltage). It means a large heatsink. \$\endgroup\$ Aug 30, 2023 at 13:25

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.