The starting point for any calculation is knowing the load. Conservation is key! If you can conserve power by using a more efficient load you should. If you can cut your load in half you system cost also cuts in half. Conserve!

This can be the hardest step in the whole process so take your time and be accurate.

If this is a small system, for example a sign light, select loads of all the same operating voltage. Don’t try to have a 24Vdc load and a 12Vdc loads operate in the system system. While you can do this using a DC to DC voltage converter you are introducing 20% losses running power through the voltage converter, remember to conserve.

The ultimate goal is to know how much power you will use in a 24 hour day.

**DC Load Example**

Flood Light, 30 Watts @ 12Vdc

If this light will run dusk to dawn you need to know the maximum hours it will operate. In the USA, Google the sunrise and sunset times for December 21st. In my area it works out that I have 15 hours of darkness.

30W x 15H/day = 450 WH/day @ 12Vdc

For a DC system this is the load total to carry forward to step 3.

**AC Load Example, 120Vac**

Load #1, 100W x 4 Hours/day = 400 WH/day

Load #2, 50W x 6 Hours = 300 WH/day

Total = 700 WH/day @ 120Vac

To make the AC power for the DC batteries a DC to AC inverter is used. As power is converted you will have losses so the next step is to look for an inverter that can produce a minimum of 150 watts (100W plus 50W load being on at the same time) to find its efficiency. In this example let’s assume it will operate at 85% efficient (15% losses). On that same inverter specification look for the standby wattage rating. In this case the loads are off more than they are on so the inverter will be sitting with no load. Even without a load the inverter will consume power. Inexpensive small inverters can consume 5 to 10 watts or more without any loads at all connected to them. Large inverters for remote cabins (4000 watts and larger) typically will have a circuit to detect if there is a load connected and reduce the standby current to 1 watt or less. Small inverters don’t include this circuit as they are typically wired to energy sources like alternators that have unlimited supply of recharging power so there is no incentive to conserve power and reduce losses. In this example lets use 10 watts as the standby no load loss.

Total from above = 700 WH/day @ 120Vac

700 WH/day x 1.15 (inverter loss) = 805 WH/day

If we assume both loads operate at the same time this leaves (24H – 6H = 18H) 18 hours in the day.

10W standby loss x 18H = 180 WH/day

**New total AC load**

805 WH/day + 180 WH/day = 985 WH/day @ dc

Now that you have account for the AC losses think of this as a DC wattage hence the @ dc.

For an AC load the 985 WH/day @ dc is the load to carry forward to step 3a.

**DC to AC Inverter Selection**

A couple of notes about selecting an inverter to power your AC load. As you are determining your total load please keep in mind and total the wattage of all the loads that can be on at the same time. This total because the minimum rating of the inverters output. Also keep track of the surge wattage required to start your loads. For example if you have a well pump, the inverter must product enough power to overcome the locked rotor amps, LRA, of the motor to start it. Once the motor is running it takes less power to maintain its movement. This rating is also used to select your inverter. The total of any loads that could start at the same time should be the minimum surge rating of the inverter.

**Loads that Don’t Run Daily**

If you have a list of loads and notice that not all of them run daily there is one more step to your calculation. You need to calculate the daily average load. For example if you have a load that runs 3 days per week, multiply your WH/day for this load by a ratio of 3/7 to determine the average daily load.

[related-items]