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I want to create 5V power supply for a portable lighting project, using 18650 batteries.

The power supply needs to provide power to an arduino (which can take 3.3V to 13V input) and some addressable LED strips, which would take 5 volts input ideally, but a somewhat lower voltage is acceptable.

The project would draw a maximum of 2 Amps (10 watts) with all LEDs on.

I’d like to use 18650 batteries as the power source, which are nominally 3.7V but which I believe would provide an actual voltage rannging between around 4.1V (when fully charged) down to maybe 2.5V (at which point they would be considered discharged).

The options I’m considering are:

  • 2 18650s in series, with a buck converter to step the voltage down to 5v.
  • a single 18650, with a buck booster to bring the voltage up to 5v.

EDIT

I’m aiming for around 3 hours of battery life. I think the average draw of the project would be maybe 5 watts, so I need ~15Wh of capacity. A good quality panasonic 18650 provides around 12Wh, so if I went with the boost option I’d consider adding a second 18650 in parallel. Perhaps this means I’d also need a BMS?

END EDIT

I don’t believe that an LDO voltage regulator is a good approach - it would be pretty inefficient and would need a lot of heat dissapation support, given the wattage of the LEDs.

Are there any majors pros or cons between these options? Are there other options I haven’t considered?

My main decision factors are efficiency (for longer battery life) and cost.

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    \$\begingroup\$ You can buy boards which will do the boost conversion for you - designed for use in 5V "USB" battery packs. \$\endgroup\$
    – pjc50
    Commented Sep 13, 2019 at 16:59
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    \$\begingroup\$ You are probably best just buying a power bank and realizing that such a scheme won't run very long before draining even a large one. \$\endgroup\$
    – Chris Stratton
    Commented Sep 13, 2019 at 17:02
  • \$\begingroup\$ I don't know but it's likely that an Arduino can take 3.3 to 13 volts because it has a linear regulator which isn't very efficient then. Maybe it doesn't matter if you're going to drive LEDs too. \$\endgroup\$
    – pipe
    Commented Sep 13, 2019 at 17:20
  • \$\begingroup\$ How long do you want this setup to run on one charge? I guess that with 2 off 18650s you will get about 1 - 1.5 hours. \$\endgroup\$
    – Peter Jennings
    Commented Sep 13, 2019 at 17:22
  • \$\begingroup\$ regarding using an off-the-shelf usb battery pack, I’m a bit worried about the current draw. From a bit of research, the USB 1.0 and 2.0 spec provides a max of 500mA of current. I guess I’d have to check that whatever battery pack I bought was rated to provide a higher current. \$\endgroup\$
    – Pete Hodgson
    Commented Sep 13, 2019 at 17:35

1 Answer 1

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2 cells in series provide twice the energy of a single cell, and current is halved so capacity will be a bit higher. Buck converters are also usually more efficient than boost converters, so in practice it will probably be more than double. 2 cells in parallel will probably provide a bit less than 2 in series.

One possible downside of cells in series is charging. If they are charged in situ then the charger must put out up to 8.4V, and a balancing circuit is required. This means you won't be able to use those cheap TP4056 modules, and the charger may need its own power supply rather than working off USB.

Balancing is not required for cells in parallel. However 'BMS' functions to prevent over-charge and over-discharge are always required. For long cycle life the discharge voltage should not be allowed to go below 3.0V.

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