Cable loss and voltage drop
The current through an electrical circuit for a fixed load is different depending on the voltage of the circuit. The higher the voltage, the lower the current.
I = P/V
A wire has a certain resistance and is part of the electrical circuit that can be considered as a resistor.
When current passes through a resistor, it heats up. The same thing happens with wires, when current passes through them, they heat up. Power is lost as heat. These losses are called cable losses. The power lost can be calculated with the following formula:
Power = Resistance x Current²
P = R x I²
Voltage drop due to cable loss
OAnother consequence of cable losses is that a voltage drop will be generated along the cable. The voltage drop can be calculated with the following formula:
Voltage = Resistance x Current
V = R x I
In order to calculate the effect of the voltage drop of the cable, it is necessary to know two other electrical laws, Kirchhoff's first and second laws:
Kirchhoff's Law of currents (1st Law)
The current entering a node must be equal to the current leaving it.
An example of this is a parallel circuit. The voltage across each resistor is the same while the sum of the currents through each resistor is equal to the overall current.
Kirchhoff's law of voltages (2nd law)
The sum of all voltages of a closed loop of a circuit must be equal to zero.
Just the opposite is true here. In a series circuit, the current through each resistor is the same, while the sum of the voltages of each resistor is equal to the overall voltage.
Now let's use a practical example in which an inverter is connected to a 12 V battery to calculate the cable losses. The circuit diagram on the right shows a 2400 W inverter connected to a 12 V battery with two cables of 1.5 m length and 16 mm2 cross-section.
As we calculated earlier, each wire has a resistance of 1.6 mΩ.
With this data, the voltage drop of a cable can be calculated:
• A load of 2400 W at 12 V creates a current of 200 A.
• The voltage drop of a cable is: V = I x R = 200 x 0.0016 = 0.32 V.
• Since we have two wires, the total voltage drop of the system is 0.64 V.
Due to the voltage drop of 0.6 V, the inverter no longer receives 12 V, but 12 - 0.6 = 11.4 V.
Inverter power is a constant in this circuit. So when the voltage drops in the inverter, the current increases. Recall that I = P/V.
Now the battery will supply more current to compensate for the losses. In this example, this means that the current will go up to 210 A.
This makes the system inefficient because we have lost 5% (0.64 / 12) of the total energy. This lost energy has been transformed into heat.
It is important that this voltage drop is as low as possible. The obvious way to reduce it is to increase the thickness of the cable or shorten it as much as possible. But something else can be done, which is to increase the circuit voltage. The voltage drop of the cable varies with the battery (system) voltage. In general, the higher the circuit voltage the lower the voltage drop.
Consider the same load of 2400 W, but now the circuit voltage is 24 V:
• A load of 2400 W at 24 V creates a current of 2400/24 = 100 A..
• The total voltage drop will be 2 x 100 x 0.0016 = 0.32 V (= 1.3%)..
And at 48 V the current is 50 A. The voltage drop is 0.16 V (= 0.3%).
This brings us to the next question: what is the voltage drop that can be allowed? There are different opinions, but we recommend to aim for a voltage drop of no more than 2.5%. This is indicated in the table below for the different voltages:
|Battery voltage||Percentage||Voltage drop|
|2,5 %||0,3 V|
|24 V ||2,5 %||0,6 V |
|48 V ||2,5 %||1,2 V|
It is important to note that it is not only the wire that presents resistance. Any other element that the current has to pass through on its way will create additional resistance. Included in this list are items that can contribute to increasing the total resistance:
• Cable thickness and length.
• Cable terminal assembly
And pay special attention to:
• Loose connections.
• Dirty or corroded contacts.
• Incorrectly mounted cable terminals.
Resistance will be added to the electrical circuit with every connection that is made, or with every thing that is placed in the path between the battery and the inverter.
To get an idea of what these resistances may involve:
• Each cable connection: 0,06 mΩ.
• 500 A shunt: 0,10 mΩ.
• 150 A fuse: 0,35 mΩ.
• 2 m and 35mm cable²: 1,08 mΩ.
Source: Wiring Unlimited de Victron Energy