# Charging circuit for a small capacity (13 mAh) li-ion battery

I got my hands on some of Panasonic's new "pin-type" li-ion batteries – the CG-320. However, I have never had to charge a battery with such a low capacity (13 mAh). Are there charge management ICs out there that can manage charge currents this low? The closest I've found is the MCP73831, which only goes as low as 15 mAh. That won't work, right? Any tips are greatly appreciated.

• In the "handling Guidelines" PDF from the link, it says "Please inquire about our recommended IC". So it would seem that Panasonic has an IC that works. Maybe you can email their tech support? Aug 25, 2015 at 21:50
• the Linear tech LTC4071 chip looks well suited for your application, Also the datasheet for the cell specifies a charge current of 0.5It, which would mean a max charge current of ~6.5mA. Sep 17, 2015 at 19:20

simulate this circuit – Schematic created using CircuitLab

I have to say, I haven't tried it in the real-world with small cells yet, and it depends on the trickle level the battery will "eat" versus the very low Vf leakage in the diodes + leakage in the transistor.

But basically, when the battery is 2.5V the collector of Q1 will be up at 2.4V, the base will be held near 1.4V by the diodes, so R2 will "see 1V", which pulls 2mA through the battery, allowing sufficient base current into Q1 while still forcing the diodes into forward drop.

This will cause the transistor to go into voltage following emitter, which puts approximately 0.7V across R1, which turns that into approximately 7mA. Making a total of 9mA, which is sufficiently less than 1C to allow for "production errors". (Tweaking can be done by changing R1 and R2 and the supply voltage, of course).

Then, when the battery would be 4.3V, that leaves a collector voltage of about 0.6V, which should be in the region where the two diodes leak in the range of single μA, contributing only to a few tens of mV across R2, which then makes the transistor's base the main contributor there. At 0.55V base voltage that would be 100μA, that's possible with some transistors, but with this one not very likely. Looking at the Datasheet on page 4, figure 4, we can see that at 0.55V on the base the transistor can be considered off, and is not likely to conduct more that 50μA. I'd expect the total leakage to be in the 50μA range to be honest. But this would require checking, as I have only used this trick with 500mAh or higher, where 0.1mA is easy enough to handle if the battery can't.

It comes to a balance between low-operating voltage leakages and the chemical build quality of the cell. A decent Li-Ion battery can have leakages in the 2% per month range, which would be about (30 * 24 =) 720hours, which for your cell would mean:

2% of 13mAh = 0.26mAh
over 720 hours: 0.26mAh / 720hours =~ 0.36μA; which I would find a very amazing feat of engineering on their part in such a small cell, but it is possible.

In which case this would need a lot of tweaking, but with a tiny transistor and 0603 diodes and 0402 resistors it would be tiny at least. If you have cells to spare, why not try and see with a decent μA meter what the residual trickle is.

You can force the cell to plateau off sooner and more securely by adding one more resistor, theoretically at least:

simulate this circuit

Where R3 would force the base to close, whenever the collector tries to leak when the battery gets close to 4.3V, in effect using the transistor's attempt at current amplification against it (although at 10μA collector currents amplification may be very, very low in this setup).

It also allows a higher voltage to begin with, as the extra resistor changes the balance a little there too.

If you haven't experimented enough yet, you can attempt the following for potentially lower leakages at 4.3V cell voltage:

simulate this circuit

This is a bit of a freaky one, but the transistor will try to keep pulling enough current to keep the balance such that Vbase =~ Vemitter + 0.7V. The diodes will cause an upper limit, keeping the balance a little, even if the cell is in bad shape.

if the battery is at 4.2V, that leaves 0V across R1, which would again equate to base leakage, which will now be a little higher (as such you take 4.9V again).