I am working on a 4 wheeled rover project. As the rover needs to be operated remotely, I will need some sort of batteries onboard. The major parts that need to be powered by the batteries are:
- Raspberry Pi 3
- Raspberry Pi Camera 5 MP
- Arduino Uno
- 6V Brushed DC Motors
The Pi Camera will draw its power from the Pi. The Arduino is to be connected to the via Pi USB. The Arduino is to control 4 x DC Motors for the wheels with an LM298 H-Bridge IC. The Pi will draw its power from the batteries through a 5V Step-Down converter. The Arduino, and the Pi Cam will draw their power from the Pi's USB port, and CSI interface respectively, and the motor voltage pin of the LM298 will be directly connected to the batteries.
I believe a battery with a 2A output will be sufficient. I am looking forward to using two protected 18650 3000mah batteries in series. This will give an output of 8.4 V at full charge, which is suitable for my application.
1) How do I charge two 18650 batteries with a single charger?
I found this Li-Ion charger module based on the TP4056.
With an output of 4.2 V and 1 A current, it can easily charge a single cell 18650 battery. However, I need to charge two cells simultaneously.
I came across this design on the Adafruit website for LiPo batteries and a different charging circuit:
Here, the two cells are connected to the charging circuit in parallel during charge mode. When the relay is switched, the charging circuit is disconnected and the batteries are connected in series.
I wish to use this design with a Lithium Ion 18650 and a TP4056 charging circuit. Is a TP4056 charging circuit capable of safely charging two batteries connected to its output terminals in parallel?
2) Are the batteries safe to be used in series configuration?
Since each battery has its own protection circuitry, will the protection circuitry work as it should when the two batteries are connected in series during usage?
3) Which step-down converter should I use for the Raspberry Pi?
I believe I need something which can take 7.4V input and give a 5V voltage and at least 2A current output.
I hope my questions are clear. Thank you for taking the time to read.