A standalone photovoltaïc system design


Consider an installation consisting of:

  • television → 40W
  • 2 lamps →20W each
  • 1 radio →10W

 Time of equipment’s’ functioning: television : 2.5h/day ; lamps : 3h/day; radio : 1h/day

 Calculate the total energy consumed by all devices in one day.

1) The system is powered by photovoltaic solar panels.The efficiency of MPPT + DC/DC converter is 75%. The efficiency of the battery pack with regulator is 70%. The efficiency     of DC/AC converter is 95%. 

2) Calculate the total energy consumed at the entrance of DC / AC converter (DC side).

3) Calculate the minimal capacity (in Ah) of the battery to install for a voltage of 48 V if we want an autonomy of 3 days without sun.

4) Calculate the new minimum capacity if we want that the battery does not discharge more than 50% during the 3 days of autonomy (to preserve its life).

5) Calculate the daily energy production required to charge the battery to its maximum and to supply the receiving facility.

6) We consider an equivalent exposure time of 0.6 h at 1000 W/m², Calculate the peak power of the PV generator needed to provide the total required energy.


  1. E(Wh)=Pabsorbed(W)×time of functioning(h),   Numeric application: ⇒television40x2.5=10h;lamps2 x 20 x 3Wh;radio10×10Wh
  2.  The efficiency of DC/AC converter is 95%. The total energy consumed at the entrance of the DC / AC converter (DC side) is then: Consumption DC side/Efficiency =240/0,95=242Wh
  3. We want an autonomy of 3 days. Therefore, we must be able to provide 3 times the energy daily consumed at the entry of DC/AC converter, i.e.: 3×242=726W. Considering the       battery efficiency, it is necessary to store: 726/0,7=1037Wh The capacity of the battery is therefore: 1037/48=22Ah
  4. If you do not want that the battery discharges more than 50% when you have 3 days without sun, you must store two times more energy: 2×22Ah =44Ah
  5. To charge the battery to its maximum SOC and to supply the receivers, photovoltaic panels must produce: Eto-prod=242+44×48/0.75=3138.7Wh/day
  6. In an equivalent exposure time of 0.6 h at 1000 W / m², to produce 3138.7Wh, we need a peak power of:    Pc=Energy to product/Ne at 1000Wh/m2 =3138.7/0.6≈5232Wc










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