The following isn't complete, but this is what I just remember 

The typical voltage/current [5.5V, 540mA or ~3Wp] is for standard conditions, which are:

* Light intensity: 1000 W/m² (!)
* Temperature: 25°C
* Angle of light beam: 90°
* Light-Spectrum: AM 1.5

You can look these up here: http://en.wikipedia.org/wiki/Solar_panel (section 'Module Performance and Lifetime')


## Things not to forget about: ##

* Assume worst condition for season (=1.18 kWh/m²/day, winter)
* You worst case (winter) is 1.18 kWh/m²/day with ~10 hours sun
* Average Energy: 0.118 kWh/m² = 118 Wh/m²; o: ~11.8% of 'typical' energy
* 3 W * 11.8% = 0.354 W (avg); or 0.354 Wh each hour during daylight in winter (1day: 3.54 Wh) 
* Know that panels are aging: At ~80% of original efficiency it should still work.
* At ~80% efficiency your panel will give you ~2.832 Wh per winter day
* Note: at night it's dark, your battery should be able to hold enough energy that period
* Also: These are averages. There may be some longer period of rainy or cloudy days..
* You will not be able to store all energy into your battery, even with enough capacity.
* Battery-Efficiency (charge/discharge): about 80-90% for LiIon
* Storing 2.832 Wh into the battery gives you only 2.27 Wh back discharging
* This still does not take the efficiency of the charging circuit into account!
* The up-converter circuit that brings 3.7 V from battery up to 5V: ~90% efficiency.
* 2.043 Wh per day left...
* We have still not taken care of suboptimal angle of the light...


Ok, let’s assume we have 2 Wh per day (which is higher than to be expected). This is how you could check if this is enough:

* Check your circuit’s energy consumption. Measure it..
* Or: Take a fully charged battery of known capacity, check how long it lasts (e.g. 1 Wh capacity for 20 hours)
* The test battery should have a lower or equal capacity
* Now you can estimate how long your circuit will survive with 2 Wh