Using a solar cell to charge batteries used to be done like that in the old days, but nowadays we use charge regulators (not necessdarily expensive). There are a few problems with your approach :
- The voltage of a solar panel is not constant.
- The voltage of the battery will prevent optimal performance of the solar cell.
- The voltage of your battery will be different under load and when charging.
If you want to avoid using a regulator, you are going to need to get your multimeter out and check things are like they should be.
What is the charging voltage of the battery? The charging voltage and the operating voltage of the battery are not the same. If you are running your solar panel in parallel to your battery in parallel to your load, can your load handle the voltage difference? A charge controller will give you a (more or less) constant voltage reference.
What is the open-circuit voltage of the solar panel? You say you are measuring temperature, is it a particularly hot environment? If so, you'll need to check the voltage at the operating temperature of the solar panel. Even so the voltage of your solar cell will change as a function of light intensity and temperature.
What is the voltage drop across your diode? Your solar panel will operate at the charging voltage of the battery plus the diode voltage, about 14.6 V. You need to check that you are getting reasonable current out of your solar panel at this voltage, given temperature and light intensity. For reference check your manufacturer data sheet for a graph similar to this.
If you are confident that your solar panel is the right size for your battery you will need to connect the positive terminal of the solar panel to the positive terminal of the battery, with the diode preventing current from entering the positive terminal of the solar panel. The most reliable solution though is to buy a charge controller.
I assume your system "consumes" 0.5 Ampere of current and not 0.5 Amperehours of charge (that would be an unusual measure).
So you have calculated the power of your system to 5 V * 0.5 A = 2.5 W. It will run for 48 hours, so it needs 2.5 W * 48 h = 120 Wh.
So you'd need a battery with 120 Wh, but there is no ideal 5 V battery and no ideal 5 V solar panel.
You chose (I suppose) a 12 V battery and a solar panel with charger for that. And you want to charge the battery in one day so your system can run for two days without recharging.
In an ideal world you need a solar panel of 120 Wh / 6 h = 20 W (that is 12 V with a current rating of 20 W / 12 V = 1.67 A). And a battery with 120 Wh / 12 V = 10 Ah.
But the world is far from ideal:
To get from the 12 V of the battery to 5 V of your system, you'll use a switchmode power supply. Let's say you get a not very well build one (because it's cheap) and it has only an efficiency of 80%. The battery must have more energy to power the losses of the power supply. So with 80% you would need 120 Wh/80% = 150 Wh.
Next thing to consider is that for improved battery life (if that is an issue) you don't want to have it cycle from 0% to 100% often, but more like only from 25% to 90% (or even less). So you only use 65% of the rated capacity. For 150 Wh needed energy you'd use a 230 Wh battery.
The increased need alone now requires the solar panel to deliver at least 25 W. But sizing the solar panel based on peak power and sun hours is asking for trouble (except for some very sunny regions I guess). So to get a better estimate, you look up some statistics of the area where it is supposed to be used. At my place you have a measly 1.5 sun hours in December, and a day is roughly 8.3 hours long. Of course your solar panel will produce some power even when you don't have direct sunlight hitting it, but it's far from the peak power. So maybe it's 100% peak power and 30% power (I made that number up, no idea how much it is) for the rest of the day. So you'd get 25 W * 1.5 h + 30% * 25 W * (8.3-1.5)h = 85 Wh. We need roughly twice that amount. So better go for a 50 W panel.
I haven't even mentioned that you need a maximum power point tracking charger to get that, so the charger will have a certain efficiency which reduces the amount of energy available to charge your battery, think of another 90% efficiency and you need 55 W.
There should be better estimates around on how much solar energy is available. Like this graph:
(taken from Wikipedia by SechWatt)
It shows the total energy produces by a 1 kWp (kilowatt peak) solar panel per month somewhere in northern Germany. The average day (from sunrise to sunset) in December is 7.8 hours there, so you have 20 kWh in December, which averages to 20 kWh / 31 days = 645 Wh per day for a 1 kWp panel. With our needed 150 Wh we end up with a (150 Wh)/(645 Wh) * 1kWp = 232 Wp solar panel. So my estimation of 30% was probably way off.
Note: There should be calculators around for this kind of analysis.
If you plan to use the system for several years without replacing the components, you have to factor ageing in as well (battery capacity reduces, solar panel power reduces). So that makes things even larger.
Conclusion:
Use a battery rated for 230 Wh (12 V / 19.2 Ah), and a 232 Wp solar panel (12 V / 19.3 A), if you want your system to work in December in northern Germany.
If you plan to use it elsewhere, calculate again.
This should only be considered a rough guideline on what should be considered and I wouldn't consider it a complete analysis.
Best Answer
Use a buck converter as others have recommended. Car adapter to USB will work. This reduces the cameras current draw from the battery down to about 1 Amp. (5V*2A = 10W = 12V*0.8A) You must upgrade the battery though. AGM will give you longer life than regular sealed lead acid. 12aH is not enough as this only gives you 12 hours of run time when the sun goes down which means you will lose power on any rainy or cloudy day. You can get a 40Ah AGM battery for about $100. A 50W panel is probably passable with 6 hours of sun per day. 50W - 10W camera = 40W * 6h sun = 240Wh vs 24h day - 6h sun = 18h night * 10w camera = 180wH night usage. This means you gain about 60wH per day under good conditions. 100W panel would be preferred though if you have the space.