Assume the power input to the bulb is 10 Watts.
Assume for now 100% efficiency from battery output to bulb input.
Efficiency of energy storage by the battery of energy supplied to it will vary with battery chemistry and how well the charger is designed. Best case using a Lithium battery of some sort, over 90% efficiency may be achievable. Lower or much lower efficiency is often achieved in practice.
Efficiency of energy provided at the battery terminals compared with energy out of the PV panel will depend on the interface design and will also vary with battery state of charge.
Power output from the panel at any moment (Wp) and compared to the maximum power the panel can make under ideal conditions (Wmpp) will vary with insolation level (sunshine level), panel conditions, atmospheric conditions and more.
SO overall, a say 100 Watt panel will produce 100 Watts in full sunshine when new and will produce the equivalent of 2 or 3 hours of equivalent sunshine in most continental US locations in winter and 5 to 6 hours of equivalent full sunshine.
ie you get 200 to 700 Watt-hours per day depending on season.
With the very best interface equipment (MPPT, intelligent battery sizing to minimise resistive losses, ... you may get 95% + of this energy at the battery terminals and, as above, 90%+ of this actually stored into the battery.
So PV Watts rating x 0.95 x 0.9 x hours_equivalent_per_day = Watt-hours available. Say 85%. Using 80% would be safer and still very optimistic in many cases.
At the start I assumed 100% battery out to bulb in power.
Regardless of load type (which is usually LED in this context), if you want constant brightness as battery varies or constant "bulb" input there will be some conversion losses. 90% from battery to bulb or LED would usually be excellent.
So overall PV "nameplate rating" watt-hours to 'bulb' input watt-hours is at best about 75%. Usually less.
When the sun is providing energy, some gains can be had by running the bulb from the panel without battery storage. This gain is useful but still a small part of the total energy needed via the battery. I'll ignore it in the following and it can be factored in later if needed.
From the above:
Watt hours available = (Panel Watts rated) x 75% x Sunshine hours.
Watt hours wanted = Load_Watts x 24.
Rearranging the above -
Panel Watts needed = Load Watts x 24 / (0.75 x Sunshine hours )
= Load_Watts x 32 / Sunshine_Hours
So eg 10 Watt load in winter with 2 hours/day sunshine hours /day (= equivalent full sunshine).
Panel Watts needed = 10 x 32 / 2 = 160 Watts !!!
10 Watt load in Summer with 6 sunshine hours/day.
Panel watts needed = 10 x 32 / 6 = 53 Watts.
In practice higher Watts will be needed.
Averge sunshine hours per day can be found at the wonderful Gaisma site here - this example is for Houston
Top line is insoltaion in kWh/m^2/day = sunshine hours/day = hours of equivalent full sunshine. I = January, II = February etc.
2.34 hours/day in January.
5.98 hours/day in July
These are means for many years and any year and any day in the montyh may vary widely from this. That's weather for you :-)
More later ...
There are two pieces that work together in order to light the LED.
The first is an IC such as the QX5251. This is a solar battery charger and LED driver in a TO-94 package. It rectifies the solar input as well as detects when it is dark in order to activate the LED.
The second is an inductor which the driver uses to boost the voltage from the battery high enough to light the LED. The driver uses the same mechanism as a boost regulator but without any of the filtering.
The result is an LED that is imperceptibly flickering. This flickering may interfere with your vision attempts, so do plenty of testing before deploying it in the wild.
Best Answer
I picked up a bunch of doorbell transformers (just for my junkbox) at $1 each, a while back. About the same as you describe. I'd probably use a bridge rectifier followed by a ripple capacitor to get the basic DC part done in preparation for an LM2596 buck converter. (You can get these for under a dollar on ebay, if you don't want a fancy voltage and current display, and they can easily handle your LED requirements. Most include potentiometers to use in setting the voltage and current limit, too.) Once you get the voltage into the right vicinity (say 5 V), you can then drop the rest of the voltage with a current limiting resistor in the usual way.
The first thing to do is to test your transformer under load. Just to make sure it works, mostly. I use one of those big 25 W resistors -- \$100\:\Omega\$ if you have one around? Load it down and see what it does. Chances are, it's fine though. But I like to know what the transformer does under load just to be sure.
But that's what I'd probably do (partly because I have a small box of buck converter boards laying around, too.)