4
\$\begingroup\$

I've got two LIR2450 rechargeable 3.6v coin cells. These need to be connected in series to form a 7.2 volts which is being regulated down to 5v to power an ATMEGA328P. I need to charge the two batteries using a USB power supply. As USB can only supply 5v, I can't use this to recharge the two batteries when they are connected in series, as this would require a 7.2v source.

I've found a charging circuit which works well on the LIR2450. If I can have the two batteries connected to this in parallel, it should work, but I can't figure out how to change the connection from series to parallel when the USB is plugged in.

Here's a picture of what I've got so far: Circuit Diagram

Basically, when the switch is off, and the USB is plugged in, the connection between the batteries needs to become parallel, and remain this way until the USB is unplugged. Any suggestions?

\$\endgroup\$
8
  • \$\begingroup\$ MOSFETs. Several of them. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Jun 18, 2014 at 4:57
  • \$\begingroup\$ How would I attach the mofsets? \$\endgroup\$
    – Ben Ramcharan
    Commented Jun 18, 2014 at 5:59
  • \$\begingroup\$ Changeover switch or relay or MOSFETs between batteries. When in series mode B- top connects to B+ bottom. When in charge mode disconnect link between batteries and ground bottom of btop. Apply V+charge to top of both. \$\endgroup\$
    – Russell McMahon ♦
    Commented Jun 18, 2014 at 11:01
  • \$\begingroup\$ Would someone be able to post a quick sketch of what I would need to do with the transistors? I'm not too skilled in this area, and any help would be much appreciated. \$\endgroup\$
    – Ben Ramcharan
    Commented Jun 18, 2014 at 14:49
  • 2
    \$\begingroup\$ Probably better and simpler to leave them in series all the time, and charge them via a boost converter. \$\endgroup\$
    – user16324
    Commented Jun 18, 2014 at 17:36

1 Answer 1

Reset to default
1
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Here's one circuit you could use to switch from series to parallel. Check that both switches are break-before-make or there could be sparks. The diode protects the charger if the batteries are switched into series. This will waste a little power, but it should still leave about 4.3V which is plenty for charging the cells in parallel. You could get a bit fancier with a 5V relay if you want, which would mean no external switch is required at all. I wouldn't bother with mosfets because they would increase the component count unnecessarily and require a higher gate voltage than you have at hand for high-side switching.

\$\endgroup\$

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