The simplest systems of solar energy can be easily dimensionados, as in the example below

Let us take our friend of the photo as example. We will calculate which components he will need in its system of solar energy to light a fluorescent lamp it compacts of 9W (12V) for 12 hours a day.

1st step - as the lamp works in 12V, its current will be 0,75 THE (9W / 12V)

2nd step - the necessary energy a day to work the lamp it will be of 9 ah (0,75 x 12 h) .

3rd step - including a factor of 10% of safety and battery recharge (extra energy to recharge the battery after very cloudy days) the total will be of 9,9 ah (9ah x 1,1). Para systems with several loads should add energy of todas.

4th step - the energy of the sun is measured in useful hours a day. In Brazil the heatstroke average is of 5 hours of sun (for calculations of systems the value of own heatstroke of the place should be used). the solar module will supply Like this its nominal current during 5 o'clock and, therefore, for we calculate the demanded current of the module we should divide the total energy for 5, finding 1,98 THE (9,9 Ah / 5:00) .

5th step - we sought a module then (dentre varies them options offered by Siemens) that gets to supply 1,98 THE. We can choose a module SP36, that possesses nominal current of 2,1 TO .

6th step - the load controller should support the 0,75 THE ONE that empty space for the load and the 2,4 THE corresponding to the current of short circuit of the module SP36. The solsum 6.6X is the ideal.

7th step - for we choose the battery we should observe the maximum discharge depth that she gets to have. We will assume a battery with 50%. we Defined how many days without sun that the system should support, for example 2 days. Then the chosen battery should supply 40 Ah approximately in two days (9,9 ah x 2 days / 0,5).