It seems at a fundamental level, the panel puts out less power than the thing you are trying to power wants. That means you can't "convert" your way out of this. You can convert some combination of volts and amps to a different one, but the volts x amps product of the output can't ever be higher than the input volts x amps product.
The first thing you need is proper specs for the thing you are trying to power. Apparently it expects around 5 V. So give it 5 V and measure the current. You can't design a circuit to meet a current requirement if you don't know what that current requirement is. That really should have been obvious.
It sounds like your switcher can deliver sufficient output current when given sufficient input voltage. The problem is that your panel can't supply the necessary power, so the switcher keeps trying to draw more current from the panel. That causes the panel voltage to drop, so the switcher draws even more current, which causes the panel voltage to drop even more, etc. Rather quickly, the panel voltage collapses. It then produces even less power than it could if managed properly.
So what to do? One possibility is to add another panel in parallel. With enough input current capability, the system will at least work in steady state in full sun. However, when the insolation goes away, the voltage will collapse again, and may not be able to recover when the insolation returns.
What you need is a circuit that disables the switcher altogether when the panel voltage is too low. If the switcher doesn't have a shutdown input, get one that does. That could be done with a external transistor, but at your apparent level it is better to drive a shutdown input that is meant for the purpose. Many buck switchers have shutdown (or enable) inputs, so this is not a onerous requirement.
Derive the shutdown signal from a comparator with hysteresis. Find what a reasonable maximum power voltage for the panel is under a bit less than optimum illumination, then set the comparator off threshold a bit below that. Set the compator on threshold a bit below the open-circuit output voltage at medium illumination.
Now connect a big capacitor across the panel. This should be 10s of mF at least, rated for 25 V or more.
What will happen now is that the panel will charge up the capacitor to the comparator on threshold. The buck switcher then makes 5 V and your device charges. The current drawn by the buck switcher will exceed what the panel is producing. That capacitor supplies the remaining current, but discharges in the process. After some time, the capacitor voltage drops to the comparator off threshold. The buck switcher turns off, stops charging the device, and stops drawing input current. The solar panel current now charges the capacitor, and the cycle repeats.
The larger the capacitor, the longer the device will be charged at a time. Depending on how the charger in the device works, it may need some minimum on time to do any useful charging. A larger capacitor will lengthen that on time. It will also lengthen the off time, but that shouldn't bother the device.
Overall, you still aren't getting more power out than in. However, the output power is now in burst of high power with gaps in between. The device charges during those bursts of high power in the way it is intended to work. Obviously the overall charging will take longer, but that's again due to basic physics limited by the available input power.
Some Lead Acid batteries and Nickel-Cadmium cells can be reconditioned.
I have never seen such claims for Lithium based batteries.
Lithium based batteries age even when they're not used (as you found out). Some batteries suffer more from this than others ! Their chemical structure changes in a way that cannot be restored. So in my opinion the marketing guys talked nonsense.
To be absolutely sure ask the battery manufacturer about the shelf life of their batteries. They have an interest in selling you new batteries of course so they might be inclined to say you need new ones. So I'd ask them how long their batteries last in storage and how they need to be stored.
Best Answer
How about:
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).