0
\$\begingroup\$

I plan to run a circuit with two 18650s in parallel to get 7Ah of capacity and a 3.3V rail (for running the ESP32 MCU). I'll use an LDO to get my 3.3V until the battery voltage drops to 3.3V. At that point, I'd like to use MOSFETs to rearrange them into series configuration for charging via 2-cell balancing and charging ICs (inspired by this Adafruit product but in reverse).

I'm wondering how to set up my MOSFETs to achieve this. I could use P-channel enhancement-mode MOSFETs that are by default ON if the gate voltage from the batteries is >3.3V and then shut off when the voltage drops below 3.3V to open up a new branch. I've modelled this below using SPDT switches for readability.

Battery Charge Switching

Would it be better to turn the MOSFETs on/off via a digital signal from the MCU? I could monitor the battery voltage using a comparator and some code and have it all controlled in software. That way I don't have the output of my pack voltage driving the switches that are reconnecting the output of my pack voltage.

\$\endgroup\$
  • \$\begingroup\$ If you're not discharging fast, and you're running at close to room temperature, a parallel-connected battery is going to be almost out of charge at 3.3V anyway -- you could just shut down at that point. \$\endgroup\$ – TimWescott Dec 31 '18 at 0:43
  • \$\begingroup\$ Or you could keep the cells in series always, and use a buck (switching) converter to 3.3V. \$\endgroup\$ – TimWescott Dec 31 '18 at 0:45
  • \$\begingroup\$ Hi @TimWescott, thanks for your comments. Yes, the point is that at 3.3V they would be out of charge. Why would I shut down at 3.3V instead of charging the batteries? Also, the reason I have them in parallel is to get me more running time (hence the 7Ah capacity). Bucking them down from 7~8.4V would only give me 3.5Ah. Perhaps I don't understand your comment? \$\endgroup\$ – YNGVV Dec 31 '18 at 1:17
  • \$\begingroup\$ If you choose the right one, a switching converter (buck/buck-boost/cuk, etc) chopping your voltage down to 3.3V could potentially give you more efficiency than an LDO. The LDO is guaranteed to waste the power represented by the excess voltage and no more, so if you start at 3.7V, an LDO will have a loss of (3.7V-3.3V)/3.7V about 10% and when your battery is just above output voltage, losses will be negligible, but the battery doesn't have much power left to provide at this voltage. Switching converters can reach as high as 98% efficiency, with an exceptional design. \$\endgroup\$ – K H Dec 31 '18 at 3:10
  • \$\begingroup\$ Also when you're talking about capacity, you must consider Wh, not Ah. To get Wh, multiply capacity in Ah by voltage. You'll see your capacity is the same regardless of how you arrange the batteries. This means there may be considerable advantage in having the batteries in series both for charge and run mode. \$\endgroup\$ – K H Dec 31 '18 at 3:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.