0
\$\begingroup\$

I'm investigating options to power a small trolling motor, which is designed to be used with a 12V lead-acid battery.

My main constraint is weight, and from the research I have done so far it looks like Li-ion batteries have the highest energy/mass ratio, so this would potentially be the best option for getting the most power in the smallest package.

So I am comparing to a commercial LiFePO4 battery with the following specs:

  • Voltage: 12.8V
  • Continuous Discharge: 40A
  • Capacity: 40Ah
  • Weight: 5.9kg

So for instance, if I look at a Panasonic NCR18650B, with the unit specs:

  • Voltage: 3.6V
  • Continuous Discharge: 4.9A
  • Capacity: 3.4Ah
  • Weight: 47.5g

Then I have calculated that I should be able to configure these with 4 in series, and 12 in parallel to achieve:

  • Voltage: 14.4V
  • Continuous Discharge: 58.8A
  • Capacity: 40.8Ah
  • Weight: 2.28g

So it looks as if this battery would achieve more power than the commercial version, with the same capacity at less than half the weight.

So my questions would be:

  1. What will happen if I connect a 12V motor to a 14.4V power source?

  2. Is there some way to regulate the voltage coming out of the battery? I.e. can I have the battery output at 12V somehow?

  3. If it is possible to convert the voltage, then is there a drawback to this approach? I.e. am I going to lose efficiency? Or what effects does this have on the system?

edit: The motor takes max 17A (204W) at top speed.

\$\endgroup\$
6
  • \$\begingroup\$ The 3.6 V of a Lithium based battery is the voltage at which it is considered depleted (empty). A fully charged Li-Ion cell can have as much as 4.2 V. If you're unfamiliar on how to treat Lithium based batteries then educate yourself because using them in the wrong way can damage them and is dangerous as these cells can catch fire. A lead-acid battery is much more robust and user friendly in that respect. Also you omit one very important detail: how much current does the motor take? \$\endgroup\$
    – Bimpelrekkie
    Commented May 12, 2020 at 10:14
  • \$\begingroup\$ Thank you for the warning, I am aware of the dangers of the dangers of Lithium based batteries, and I am taking it seriously. I'm in the initial feasibility assessment phase of the project, so at the moment I'm mostly interested in figuring out what is possible. I will add the motor details to the post. \$\endgroup\$
    – sak
    Commented May 12, 2020 at 10:35
  • 1
    \$\begingroup\$ As noted - LiIon = 3V- 4.2V cell so 4S = 12V - 16.8V. || For 4S you also need a BMS(battery management system). What you are looking at is very commonly done. If you want to DIY there are many groups that will provide much practical advice. If you are on Facebook search for groups relating to liion, powerwalls, RC cars, ... many more. If you cannot find anything suitable advise and I can suggest a group. \$\endgroup\$
    – Russell McMahon ♦
    Commented May 12, 2020 at 10:35
  • \$\begingroup\$ How a 12V motor runs on 16VDC varies. Is it BLDC or brushed. What wattage. You can get buck converters on ebay etc at good prices that give say 13-16V in and 12V out. Or even <12V and so variable speed if a brushed motor. || Max specs and brand and model and links helps us help you. || The LiFePO4 you mention is lower energy density that LiIon but longer cycle life. \$\endgroup\$
    – Russell McMahon ♦
    Commented May 12, 2020 at 10:38
  • \$\begingroup\$ @RussellMcMahon thanks - yes I have read that I will need a BMS. Do you know if the BMS will also handle voltage regulation, or would that be a separate component? I have read that the BMS is important for making sure the battery charges within safe limits, but I am not sure if I also would need something like a separate buck converter to achieve 12V output (or if I even need to). \$\endgroup\$
    – sak
    Commented May 12, 2020 at 10:39

0

Reset to default

Your Answer

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

Browse other questions tagged or ask your own question.