Effectively designing and using portable, battery-based electrical circuits opens the door to creating autonomous hybrid systems. Understanding how to design a battery-powered, or direct current (DC), electrical circuit requires determining the following things:

- Voltage and amperage of the load
- Operational power requirements for each appliance (load)
- Power consumed over a duration of time
- Size of battery bank to meet the load requirements
- Power generation needed to replace energy used by the load

Operators should establish a power budget in order to select technologies that are appropriate for their given applications and available resources. Establishing a budget is the first step in ensuring the solution is effective and provides as much autonomy as possible for the operator.

When establishing a budget, the operator must know the power requirements of the application before the portable power system can be configured properly. Establishing a budget sufficient for the task is important, because the ironclad rule of power generation is power generated in the circuit must be greater than or equal to the power that will be consumed in the circuit.

In any electrical circuit, electricity flows dynamically. Electricity flowing through the circuit varies in accordance with both the appliance requirements (loads that use power) and the power generators (devices that produce power).

For example, communication devices, medical equipment, lighting, and refrigerators all consume energy in a dynamic fashion: refrigerators only consume power when the thermostat calls for colder temperatures; communication radios consume more power when transmitting than when simply listening. Generator power production can fluctuate widely based on factors such as temperature, altitude, and fuel.

The portable, fuel-driven generator is still the primary mechanism or tool used in portable power applications. Users often choose fuel-driven generators for convenience. Powering appliances with a fuel generator is simple: just turn it on, plug in appliances, and refuel as needed.

Many AC appliances have Energy Star ratings to demonstrate that they operate more efficiently, using less power.

When determining the peak load of an appliance, experience has taught us that a general rule of thumb is to take the rated power consumption and multiply it by three to get a good estimate of the appliance peak or surge power:

Peak or Surge Power = 3 ∙ watts

Therefore, if an appliance is rated for 300 W continuous usage, then it is reasonable that the start-up load may be 900 W.

There are a few, noteworthy types of appliances that require very high peak start-up power:

- Ordinary incandescent lamps and quartz halogen lamps need as much as 10 times the start-up wattage as their operating wattage
- Pumps or compressors that are under full load can require as much as eight times the start-up wattage or current as their operating wattage. This is sometimes labeled as locked rotor current.

Despite the highly refined mathematical methods and instrumentation we have today, it is still very difficult to calculate exact values for the power that is generated and consumed in a portable electrical circuit. Conversely, it is fairly simple to determine exactly how much power was generated and consumed in a circuit at the end of a period of time.

For example, many power generators today are rated, by their respective manufacturers, according to the maximum possible power output:

These ratings are often generated in a laboratory environment and do not take into account the derating (power degradation) often caused by operating environments (climate, altitude, location), type of fuel used, or a combination of factors. Similar effects lead to derating at the load, often for similar reasons. It is highly recommended the operator look at the total amount of power that flows through a closed electrical circuit over time as a game of averages.

The average load is the total energy delivered over a particular interval of time.

For a 12 VDC system, we can convert either watts (W) to amps (A) or amps to watts by using the following formulas:

A | = W ÷ V |

amps | = watts ÷ 12 |

W | = V ∙ A |

watts | = amps ∙ 12 |

STIKmann owns a recreational vehicle (RV) and wants to build a DC electrical system to power it. His camper has two appliances (the load) in the electrical system, a 12 VDC refrigerator and a 12 VDC lamp, with each appliance consuming different amounts of energy.

Most electrical appliances have a specification plate or tag that tells what type and how much power is necessary to operate the appliance. If no specification plate is present, consult the manufacturer’s product literature. Depending on the manufacturer, you may find the operating energy requirement rated in watts or amps. If you know the voltage, you can use either value (watts or amps) to determine the other.

STIKmann’s 12 VDC lamp consumes 25 W of power, and he needs to know how much current (amps) it draws. He applies the formula as follows:

W ÷ V | = A |

25 ÷ 12 | = A |

= 2.1 A |

Now consider STIKmann’s refrigerator. The power consumption rating given on the refrigerator’s specification plate is 60 W. He uses the system voltage in the formula to determine the refrigerator draws 5 A of current when operating:

W ÷ V | = A |

60 ÷ 12 | = A |

= 5 A |

Both watts and amps can be tallied as a sum by adding the values together. To determine the total load values for the RV’s DC system—or the system’s total load—he adds the values of the wattage or the amperage for all of the individual appliances. The total amp draw for Stkimann’s lamp and the refrigerator is 7.1 A.

2.1 A + 5 A | = A |

= 7.1 A total |

The total watt consumption for STIKmann’s lamp and refrigerator is 85 W.

25 W + 60 W | = W |

= 85 W total |

Lastly, he calculates the total values for the system. The same formula applies and should reflect the overall characteristics of the circuit.

V ∙ A | = W |

12 ∙ 7.1 | = 85.2 W |

It is common practice to round the numerical values for watts to the nearest whole number. In this case,

12 ∙ 7.1 | ≈ 85 W |

For STIKmann’s DC system, we determined the lamp and refrigerator draw a total current load of 7.1 A when both appliances are operating. However, this did not take into account that the lamp may not be in use all the time, or the refrigerator could be in operation for only 20 minutes every hour.

To get an accurate picture of his system’s total daily requirements, STIKmann must get the total hours per day of refrigeration and lamp operation time. Let’s say that he adds it all up and estimates that the total amp draw from the batteries is 7.1 A for 8 hours in a 24-hour period:

7.1 A ∙ 8 h | = 56.8 Ah |

This means STIKmann’s lamp and refrigerator consume about 57 Ah per day. In the same manner, the watt hours can also be calculated. STIKmann determines the refrigerator and lamp consume a total of 85 W and operate for a total of 8 hours daily:

85 W ∙ 8 h | = 680 Wh |

Then the total watt hours consumed by STIKmann’s appliances in a 24-hour period is 680 Wh. Even when labeled differently, the appliance energy requirements are exactly the same. With the appliances operating on 12 V multiplied by 56.8 Ah, a total load of 681 Wh is consumed:

12 V ∙ 56.8 Ah | = 681 Wh |

(= 0.68 kWh) |

The total daily energy requirement for the STIKmann’s light and refrigerator is about 0.68 kWh.

Now that he knows how much energy he needs for the light and refrigerator on a daily basis, his next steps are to determine the correct battery capacity and how to replace the 57 Ah or 0.68 kWh of power consumed from the battery during one day.