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So, I understand that a 3Cell LiPo battery can output 12.6V when fully charged and that the cells must remain balanced to guard against Fire/Explosions/and general bad stuff happening.

However, each cell can only store a maximum of 4.6V. These Cells are then connected in Series to produce the 12.6V.

Is it possible to Charge a 12.6V LiPo from a 5V (or 4.6V) supply using a circuit that resembles something like the one below: enter image description here

I understand that the ground plane on the charging circuit would have to isolated from the ground plane of the battery.

The MicroController would just cycle through the 3 cells charging one of them at a time. Some feedback loop to 3 Analogue inputs on the same microcontroller could monitor the current voltage of each cell and ensure that more time in the cycle is provided to the cells that are lagging behind until they are again balanced.

Would this work?

Any thoughts and suggestions would be MUCH appreciated. If any circuits like these exist already it would also be super helpful if you could point me at them.

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    \$\begingroup\$ Can't you get rid of all the dots? They make the schematic look very messy, and hardly readable. \$\endgroup\$
    – stevenvh
    Commented Jun 26, 2012 at 14:12
  • \$\begingroup\$ Sorry, I just used circuitlab.com to quickly draw up the schematic. \$\endgroup\$
    – Gineer
    Commented Jun 26, 2012 at 15:52
  • \$\begingroup\$ The circuit from my answer is also made with CircuitLab, but doesn't have all these dots. I don't know how you did it, so can't advise you how to get rid off them. I only see the dots while I'm placing a part, or connecting a wire, but then just that one part, never all of them. \$\endgroup\$
    – stevenvh
    Commented Jun 26, 2012 at 16:29

2 Answers 2

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No that doesn't work. If you want to charge the top cell switching Q1 on (without current limit!), your Q2 (also no current limit) will pull Q1's base below the cell's negative voltage.

enter image description here

This is your circuit for one cell. The signal coming from the microcontroller switches on Q2, but also is pulled down to 0.7 V. So the base of Q1 is also at 0.7 V, while it needs at least 5.1 V (0.2 V saturation voltage of Q2 + 4.2 V battery + 0.7 V base-emitter junction).

A solution could be to have a current output boost converter. This can generate a higher voltage from a lower input voltage. Constant current boost converters are often used as LED drivers.

enter image description here

Replace the LEDs with your battery, and set the charge current by choosing R1 = 95 mV/I.

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  • \$\begingroup\$ Would I replace the LED's with my battery or with one of the cells in the battery? If this circuit could produce voltages above say 18V I could feed this into a basic LiPo Balancing Charger circuit, right? \$\endgroup\$
    – Gineer
    Commented Jun 26, 2012 at 16:01
  • \$\begingroup\$ @Gineer - The complete battery. The datasheet has an application example driving 7 white LEDs from 5V input, which typically will be more than 20V. What do you mean by balancing? Aren't you using all cells in series, so that they always have the same current, either charging or discharging? \$\endgroup\$
    – stevenvh
    Commented Jun 26, 2012 at 16:08
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Stevenvh, you underestimate need for balancing in LiPo circuits. Yes, usually cells are equal but LiPo is known to catch fire in case or abuse. So in real life sometime single cell is broken/short and if you charge pack to 12.6V, remaining two cells will be overcharged.

I would recommend going to chip makers websites to look what specialized chips they have. Meanwhile yes, Boost converter is keyword. I love Texas Instruments online tool called Power Architect on their website where you just enter parameters you need and it provides you with optimized schematic. One reference schematic for LiPo balanced charger is available http://www.ti.com/tool/PMP5214 where some ideas can be taken about balancing.

In case balancing is not used, temperature sensor should be next to cells for automatic stop in case of temperature getting high.

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