One formula you will use often is:

Power (Watts) = Voltage (Volts) x Current (Amps).

The more power, the more energy it produces. If only amps and volts are specified, you can calculate the power from that by multiplying them.

The rating of a solar panel in Watts will be the maximum for full sunshine. So when it is cloudy, the power produced will drop significantly, down to approximately 10-20%.

You can use any battery, as long as you have a circuit which controls the charging of the battery. However, this circuit needs to be compatible with your battery and the solar panel. The larger the battery, the more energy it can store, for example to store energy from sunny days so it's available on cloudy days or at night. Too small a battery and you might not even have enough power at night.

I will go through one example for you, so you can follow the process for whatever system you want to use.

First, decide how much electricity you will use. Say you want to run a TV from it, for 4 hours a day. And say the TV uses 100 watts, and runs off 230V.

The total energy your TV will use per day is 4 x 100 = 400 watt-hours.

And say you have 12 hours of sunshine every day. To get 400 watt-hours, you would need a panel that has a power of 400 / 12 = 33.3 Watts. This could, for example, be a solar panel that produces 1 amp at 33.3V, or 66.6V at 0.5A, or whatever, as long as amps x volts is greater than or equal to 33.3W.

If you want to use the TV at night, you will want the battery to store 400 watt-hours. So if you use a 6V battery, it must have a usable capacity of 400 / 6 = 66.6Ah (Amp-hours). Or if you use a 24V battery, it must have a usable capacity of 400 / 24 = 16.6Ah. If you use a car battery, be aware that most car batteries need to maintain a charge of between 50% and 75% if they are to have a long life. So only a quarter of its stated amp-hour capacity can be used. So say you buy a 100Ah battery, you will only be able to use 25 amp-hours! It would be more cost-effective if you buy a so-called "deep cycle" battery, so you can typically use more like 80% of its capacity without causing premature ageing.

And say your solar panel produces 48V. You will need a system which solar panel controller that can charge a 6v, 67Ah battery from a 48v, 34W solar panel. This will be hard to find: you may need to start with the controller, then buy a solar panel and battery accordingly. The "controller" is made up of two parts, and you may want/need to buy them separately: The first part is a voltage converter, which takes the voltage of the solar panel, and converts it into a voltage for charging the battery. Then the battery charger is essentially a circuit that uses the supply from the voltage converter to charge the battery without overcharging it.

Finally you will need an inverter which produces at least 100 Watts, from a 6v supply, to 230V AC.

## Best Answer

If you have the 'fixed 5v' version (LM2596-5.0) then you can connect the feedback pin to the output. But since it seems you actually have the 'adjustable' version (LM2596-ADJ) then you'll have to use the correct voltage divider between the output and feedback pins. See the instructions starting towards the bottom have of page 12 of the datasheet. You'll get 5v out as long as your solar panel gives you at least 7v in.

The 12v input for 5v output shown in the datasheet is just an example. This IC will quite happily run off 40v and give you a 5v output - check the 'LM2956-5.0 Electrical Characteristics' table on page 3, or the LM2956-ADJ table on page 4.