Voltage Drop

Definitions

Voltage drop is defined as the amount of voltage loss that occurs through all or part of a circuit due to conductor resistance.

Conductor resistance is determined by conductor material, size, and ambient temperature.

Voltage drop highly depends on the total length of conductors that carry the electrical current. In DC systems, the voltage drop length is the total (round-trip) distance that current travels in a circuit. So the total length used in calculations is usually twice the length of the conductor run. In some AC systems, the distance equals the length of the conductor.

Reflection

Why is the conductor length different for AC and DC circuits?

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Voltage drop from PV array to inverter

NEC doesn’t require the calculation of voltage drop because it’s not a safety issue. However, it does recommend a maximum voltage drop of 3%. It is recommended to have up to 2% voltage drop at the DC side while only 1% is accepted at the AC side of the system for a total of 3% in voltage drop for the entire system.

Wires should be sized to reduce resistive (heating) loss to less than 3%. This loss is a function of the SQUARE of the current times the resistance, which is another manifestation of Ohm’s law:

V = I \times R

, or,

I = V / R

And the resistive loss is $I \times I \times R$ in Watts.

Note:

Use a wire-sizing table to choose the right wire size for the current and voltage you are working with. Visit Encorewire.com [1] for an example.

Example

Computing the voltage drop formula:

V_{d r o p} = I_{o p} \times R_{c} \times L

Where:
$I_{o p}$ is the circuit operating current, which for source circuits is usually taken as the maximum power current, Imp,
L is the total conductor length.
$V_{d r o p}$ is the voltage at which you want to find VD, and
$R_{c}$ is the wire’s resistivity in Ohms per 1000 feet and is found from NEC Chapter 9, Table 8 conductor Properties.

Example

If we have a PV array of 12 modules in string that is located 150’ away from the inverter (L=150 ft) and we are using wire # 14 AWG since it handles the current of 8.23A, and it has resistivity of 3.14 (Ω/kft).
$V_{d r o p} = 8.23 (A) \times 3.14 (Ω / k f t) \times 0.3 (k f t) = 7.75 V$ The operating Voltage is

V_{m m p} = 12 \times 37.2 = 357.6 V

The voltage drop then is calculated as:
$V_{d r o p %} = V_{d r o p} / V_{m m p} = 7.75 / 357.6 = 2.16 %$ , which is not within the limit of 2%, but this wire is running to a combiner box and to the inverter. In this case, the voltage drop should be less and the size of conductor must go up.

Upgrading to a larger conductor size for the same length and conductor type:
$L = 150 ft and # 12 A W G, R c = 1.98 (Ω / k f t)$
$V_{d r o p} = 8.23 (A) \times 1.98 (Ω / k f t) \times 0.3 (k f t) = 4.98 V$
The voltage drop then is calculated as:

V_{d r o p %} = V_{d r o p} / V_{m m p} = 4.98 / 357.6 = 1.37 %

, which is within the limit of 2%.
As can be seen, both conductors sizes # 12 and #14 work for ampacity, but the voltage drop calculation shows that both of them still not the best option for the long term. As a result, cable # 10 AWG is a more conservative design, but it will cost more.

Note:

There are some freely available tools that can be used for voltage drop calculation. This is an example of an Online Calculator. [2] If there is no DC option for the calculator, you can use single phase and choose the right length.