Cable/Voltage Line Loss-- Frequently Asked Questions

1. What is cable voltage line loss? What causes it, and Why is it important?


2. We power four 24v/250w lights on 100m 1.5mm2 cable, and we're experiencing voltage line loss. We prefer not to use high voltage. What are our options?


  1. What is cable voltage line loss? What causes it, and Why is it important?

    Cable line loss, or voltage line loss, is a loss of voltage over the length of an electrical line. It is caused by resistance in the line. It is important because voltage line loss of 10 or 20 volts, say, which are not uncommon, are significant when compared to the 12 or 24 volts often used in offshore systems.

    Conceptually, electricity in a cable can be thought of like water in a pipe. "Voltage" (also called 'potential') is analogous to the water pressure pushing the water along and "amperage" or electrical current is similar to water flow in the pipe (i.e. the amount of water which passes a certain point in the pipe during a certain time).

    Larger pipes, like larger electrical conductors, have lower resistance to flow. However, the output pressure will eventually become lower and lower over very long lengths of pipe, even when the pipe is large and has a very high pressure input. So it is with cable voltage line loss.

  2. We power four 24v/250w lights on 100m 1.5mm2 cable, and we're experiencing voltage line loss. We prefer not to use high voltage. What are our options?

    The formula used to calculate cable line loss is V = 2IRL, where 'V' is the voltage loss, 'I' is the current (derived by I = Watts/Volts), 'R' is the Resistance of the cable, and 'L' is the cable length. (Note: the '2' is because the line loss includes the voltage drop to and from the lamp.) We consider the resistivity of Cu to be 1.7 x 10-18 Ω-m.

    Using 24v/250w lamps, I = 10.4 A.

    • Typical 16 AWG cable has R = .0129 Ω/m
    • Typical 14 AWG cable has R = .0082 Ω/m
    • Typical 12 AWG cable has R = .0032 Ω/m

    Using 100m 1.5 mm2 cables, R = .0113 Ω/m, so V = 2(10.4A)(.0113 Ω/m)(100m) = 23.5 volts...so if you want to end up with 24 volts at the lights, you'd need to put in approximately 47.5 volts at the top of the cable.

    Options are:

    1. Reduce the lamp current. However, a low-current lamp at a low voltage means low wattage, i.e. low light output. We do have low-wattage lamps (e.g. 20w, 25w, 50w, etc), but the reduction in cable line loss gained by using a low-wattage lamp probably would not compensate for the overall light reduction caused by decreasing the lamp wattage.

    2. Increase the lamp voltage. High enough voltage (e.g. 120v or 240v) will solve the problem. BIRNS makes several good 120v and 240v lights.

    3. Decrease the cable resistance. Increasing the cable conductor size will reduce the cable resistances and, therefore, the cable voltage line loss. However, the extra weight of the cable should be considered.

    4. Install a voltage compensator. Generally the best solution is to install a device at the top of the cable to compensate for the voltage line loss. The BIRNS VTX, for example, automatically compensates for the voltage line loss created by any low-voltage BIRNS light used on any cable length and any cable conductor size.

      Alternatively, calculate the expected cable line loss for the cable size and length which you intend to use, then add that line loss number to the 24v for the light and use a transformer, say, to provide the total input voltage.

      One word of caution: a transformer or similar will provide the proper starting voltage for that particular cable. Make sure you adjust the input voltage if and when you decide to change the cable length or conductor size.