Understanding your loads is essential for keeping your power system running smoothly. Many issues SunWize encounters stem not from the power system itself but from customers not fully understanding their loads.

For instance, an off-grid solar system may power a security camera, a radio, or both. The power consumed by these devices makes up the system’s loads. A security camera may use 10W, while a radio might use 12W, totaling 22W when both are running simultaneously. This is easy to understand.

But what if your load only operates 4 hours a day, uses amps for power consumption, and has efficiency losses to consider? What if you have multiple loads at different voltages with varying usage profiles? What if your loads change? If you’re unsure about any of these questions, read on to learn how understanding your loads will help you deploy off-grid solar and battery backup systems more effectively, ensuring optimal system performance and addressing limitations.

Calculating load consumption correctly is essential for sizing an off-grid power system. Since most off-grid systems rely on batteries to store energy, and battery capacities are measured in amp hours, you must calculate the load in amp hours to size the system accurately. You can learn more about this in our basic system sizing webinar.

In this Tech-Note, we’ll cover different types of loads and how to convert them into amp hours. All loads can be defined as follows:

• continuous or non-continuous
• at system voltage or at non-system voltage.

Continuous refers to loads that operate 24 hours a day, 7 days a week. Non-continuous, or intermittent, refers to loads that run for only part of the day or have varying consumption at different times. Typical non-continuous loads include lighting, radios on transmit, valves or actuators, pumps or motors, gate openers, and similar devices.

Off-grid systems typically use lead-acid batteries for energy storage, and the system voltage can vary from -10% to +20% of its nominal value. For example, a nominal 24VDC system can range from 21.6VDC to 28.8VDC, or even higher in colder climates with temperature compensation. Understand that this voltage range is for the health of the battery, not for the load.

• Raising the minimum system voltage prevents the system from using the full capacity of the battery.
• Lowering the minimum system voltage too much can over-discharge the battery, potentially causing damage.
• The maximum system voltage is set to properly recharge the battery. Increasing it too much can over-charge the battery, causing a VRLA to gas and lose capacity.
• Lowering the maximum system voltage results in the battery not being fully charged. If a lead-acid battery isn’t periodically fully charged, sulfation crystals will form on the plates, reducing capacity and causing premature failure.

Note: Adjust the maximum system voltage for temperature—raise it below 25°C and lower it above 25°C. This is temperature compensation, which ensures proper battery charging. For more details on temperature compensation, refer to the LinkedIn tech note here from our CEO.

Now that we understand the range and limitations of “nominal battery voltage,” let’s see how it relates to our loads.

If the load cannot operate within the system’s voltage tolerance, power conditioning equipment is necessary. Common equipment includes a DC to AC inverter for AC loads or a DC to DC converter for DC loads. These devices cause losses that must be factored into the load calculation. Typical efficiencies are 85-95% for inverters and 80-90% for converters.

Loads can be expressed in the following formats:

• Current (amps [A])
• Power (watts [W], kilowatts [kW])
• Energy (amp hours [Ah], watthours [Wh], or kilowatthours [kWh])

Most utility bills are based on kWh, making it a common unit for AC loads. Regardless of the load format, convert it to Ah for system sizing calculations. Use the power equation [Power(W) = Current(A) * Voltage(V)] to convert from watts to amps by dividing power by nominal voltage [Power(W) / Voltage(V) = Current(A)].

Some load calculations are simpler than others, depending on the load type and application. Ultimately, express all results in the total Amp-Hours consumed over a 24-hour period, for a given nominal voltage (Ah at 12V differs from Ah at 24V).

Let’s begin with the simplest example:
The simplest load is one that is:

• expressed in amps and
• is continuous and
• will operate within system voltage.

To calculate the load in this example you simply multiply the amps by 24 hours (the number of hours in a day) to get the daily load in Amp-Hours, often abbreviated “Ah”.

Example 1:
0.5A, 12VDC nominal continuous load
0.5A x 24h/day = 12Ah/day @ 12VDC

If that same load were expressed in watts instead of amps, we would just convert it to amps by dividing power(W) by nominal voltage (V) and follow the same procedure.

Example 2:
6W, 12VDC nominal continuous load
6W/12VDC = 0.5A x 24h/day = 12Ah/day @ 12VDC

If the load is non-continuous, we simply multiply it by the number of hours per day it is active rather than by 24.

Example 3:
8W, 12VDC LED light that operates only at night.

Let’s assume worse case the light will operate for 16 hours in the winter.

8W/12VDC = 0.67A x 16h/day = 10.67Ah/day @12VDC

Example 4:
A 12VDC radio that is 1A in standby and 10A in transmit. The radio transmits for 1 hour per day.
10A x 1h/day + 1A x 23h/day = 10Ah/day + 23Ah/day = 33Ah/day @12VDC

Example 5:
A 10A, 24VDC valve that operates for 15 minutes a day and a 20W, 24VDC continuous controller.
10A x 0.25h/day + 20W/24VDC x 24h/day = 2.5Ah/day + 20Ah/day = 22.5Ah/day @ 24VDC

If the load is at voltage other than the system voltage then the losses of the conversion equipment need to be included in the calculations.

Example 6:
8W, 120VAC LED light that operates only at night.

Let’s assume worse case the light will operate for 16 hours in the winter.

Let’s also assume a system voltage of 12VDC and a power conversion efficiency of 90%.

8W/90%/12VDC = 0.74A x 16h/day = 11.85Ah/day @ 12VDC

Example 7:
5W, 24VDC continuous sensors with a voltage tolerance of +/-10%.

As the load has a small voltage tolerance, a DC/DC converter will be required, as 24V +-2.4V is much narrower than the 21.6V to 28.8V we determined at the beginning may be present from a nominal 24V battery voltage.

Let’s assume the converter efficiency is 80% and the system voltage will be 24VDC.

5W/80%/24VDC = 0.26A x 24h/day = 6.25Ah/day @ 24VDC

If there are multiple loads, they can be calculated separately and then summed together to get the total load.

Example 8:
Telecommunications tower with a 500W, 48VDC continuous communication load and 50W 24VDC nighttime obstruction lighting load.

Again, let’s assume the worst case for the lighting load is 16 hours in the winter. We’ll also assume a DC:DC Converter will be required for the 24VDC load, with an efficiency of 85%.

Load1
500W/48VDC x 24h/day = 250Ah/day

Load2
50W/85%/48VDC x 16h/day = 19.6Ah/day

Total Load = 250Ah/day + 19.6Ah/day = 269.6Ah/day @ 48VDC

Some more complicated sites may have multiple combinations of the above examples. If necessary, calculate each load individually and then combine them to calculate the total load for the site. If necessary, a spreadsheet can help to keep track of the loads and the calculations.