I'm assuming that you tested the panel voltage into an open circuit (just test leads across the panel) and the panel current into a short circuit (again, just test leads across the panel). This is not a valid test. When you put your ammeter across the panel, it acts as a short circuit. So the panel is putting out 130ma at almost no voltage!
As soon as you connected the regulator to the panel, it drew enough current to pull the voltage below what it could handle.
Look at it this way: a 1W 6V panel will put out 1/6 amps (167ma) at 6V in full sunlight. There's no way it's going to put out anywhere near that in dim light.
1) This should be fine with a low-power solar panel, but it will charge very slowly.
First a bit of theory. Charging LiPo batteries needs to follow two rules:
- the end voltage (called float voltage) must be lower than or equal to 4.2 V
- the charge current must never be higher than a certain value (typically 1C, meaning 100 mA for a 100 mAh battery, 1 A for a 1000 mAh battery, etc.)
If you have a 4.2 V regulator then the battery voltage can't go above 4.2 V. The charge current will decrease to zero as the battery voltage approaches the regulator target voltage.
When using a wall adapter, you need something to limit the charge current, even if the adapter has a low power rating. Otherwise you might damage both the battery and the power adapter. This is because a LiPo battery has a very low resistance and there will be a rush of current when you connect the power.
But with a low-power solar panel you could rely on the fact that your particular panel is incapable of supplying more than 1C of current to the load. Then your solution might work but it will be very, very slow to charge. That's because whenever you try to draw too much power from a solar panel, its voltage collapses abruptly and you get no power at all from it.
For a 5 V, 1 W solar panel in good light, this is what happens to the voltage and power when you increase the current draw by decreasing the resistance of the load:
- 20 mA: 6.0 V (0.12 W)
- 100 mA: 5.5 V (0.55 W)
- 200 mA: 5.0 V (1 W)
- 250 mA: 0.1 V (0.025 W !!)
So you would have to get an oversized panel and use a resistor in series with the battery after the voltage regulator to limit the charging current. This will work but it is wasteful.
For good efficiency you need a circuit that reduces the charge current when the solar panel gets less light. See for instance this one (tutorial). What this board does is reduce the charge current whenever the panel voltage goes below 4.75 V, and increase it when it goes above. It also handles the charge current limit for the battery. For further reading google "Maximum Power Point Tracking".
2) You only need to interrupt one wire to open a circuit. One diode is enough. You don't need two switches for your lights.
3) This would work with the reserves outlined above. But there is another danger: LiPos should NOT be discharged too deep, otherwise they will be damaged. So you really should use a protection circuit to cut power to the load if the battery voltage is too low, and also to make sure that the load current isn't too high. Solar charger boards like the one I mentioned should do all that.
I recommend reading the tutorial above, and especially the "design notes" section if you want to understand more about solar panels.
Best Answer
It is a little confusing since you jump from watts to amps and back and forth, but I believe that you do know the difference between the two.
I suspect that even though you are outside, you are not in the full sunlight that the panel requires. Broad daylight is not enough, it needs bright blistering sunlight.
Remember that a solar panel will output full voltage at light intensity levels as low as 10% of maximum, while amperage is linear. So 10% light level = approx 10% max amperage (for a given voltage)
You are correct, the short circuit amperage reading should be close to the actual current output.
Also, the smaller panels may not have bypass diodes and so even the smallest little speck of shade on one part of one solar cell will dramatically decrease amperage (I.e, your thumb provides a small amount of shade on the panel while doing the measurement and the amps will drop by 50% or more!)
If you want to map your own I-V curve, you just hook up a number of different resistors one at a time in series from the positive to the negative of the panel and measure the voltage drop across the resistor and using ohm's law calculate the current.
Good luck.