Hi John;
Here are some “guess-timates” (emphasis on the “guess”) of what I think you can do and the performance you will get. These estimates would be a lot more accurate once the current draw is known for your light with the DC to DC adapter in place (i.e. current actually being drawn from the motorcycle battery).
Here is a guess of the current draw of your light:
The AAA battery has a 0.55 amp hour rating. Let’s assume that the LED light you purchased will last say two hours with fresh batteries installed. Then you can get the amount of current draw by 0.55 / 2 = 0.275 amps. Let’s round this off to a guess of a quarter of an amp (0.250).
You will need an adapter go from 12 volts to 4.5 volts as you stated in your first post. You can get a cheap converter from WayneW’s links above or maybe go to Rat Shack and get
this one.
This one has a “self-resetting” fuse so you will not need to purchase an in-line fuse assembly as I previously suggested. You will have to somehow clip/solder/wrap wiring from the battery to the cigarette adapter's studs. Also, make sure you know which wire from the adapter is positive and which is negative as you will need this for connecting to your light.
Of course this adapter will use current and I’m going to throw out a number of 30 percent efficiency or 0.075 amps. This means your total current draw guess is 0.250 plus 0.075 or 0.325 amps (remember this is a guess-timate). For 12 hours of use the light will draw from the battery at a rate of 0.325 * 12 = 3.9 Amp Hours (let’s round this off to 4 amp hours).
Now let’s take an estimate on your charging system:
I looked at motorcycle batteries and most are around the 20 amp hour rating. To be safe and not go under 50% of its capability, let’s use 10 amp hours as the battery supply capability.
This means your battery can supply power to your light/adapter assembly for 10 / 0.325 or 30 hours, which is good because even if I’m off a factor of two in my current draw guess for your light/adapter, you will be under 50% battery draw for a long night of use.
Now you need to of course charge this battery during the daytime. Your solar panel has a current output capability of 0.4 amps. For again, worse case during the short winter days of say only six hours of sunlight, this means that you will supply 0.4 * 6 = 2.4 amp hours back into the battery.
Here is where I’m estimating you may have a problem (in the winter months). You’re drawing from the battery at a rate of 4 amp hours, but only putting in charge at a rate of 2.4 amp hours. Your battery may go dead after a few days of use.
You may need a second solar panel in the winter time. Of course you might want to diode isolate the two solar panels, but I think I’ve gone far enough in this post as the first one seemed to cause you some (unintended) angst!
Anyway hopes this offers some help. The project does seem at least reasonably feasible with the parts you have purchased. You may have to tailor the run time of the light, amp hour rating of the battery, or as mentioned, charge capability into it.
You can see how important it is to actually know the actual current draw from the battery. Once you get this system in place you can easily measure it and fine tune the above calculations!
As always, if you need any help please ask. I also did the above calculations quickly so it would be nice if someone went over my math.
Regards,
BSR