High Power DC Wiring

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DC wiring, especially high-power DC wiring has one huge problem... excessive voltage drop. This voltage drop is a function of the wire length, and RV owners may encounter such issues as they add electrical or electroinc systems to their RV.

Fortunately, the automotive style wiring (chassis ground) makes this issue a bit less critical than say in a boat, but the RV project enthusiast must still be mindful of this phenomenon.

The issue at hand is high powered 12VDC wiring issues in a RV (although this discussion also applies to boat and other 12V systems). When the dealer or RV owner wants to add high-power devices to their rig, whether it be a high-powered stereo system, a leveling jack system, or even a solar power system, unexpected problems can occur if the DC wiring is not up to task.

I will present a real-world case study to bring home this point. The scenario is adding a 12VDC powered air compressor to one of the storage bays in the RV's basement area.

The problem.

When powering devices via DC, there is a significant problem concerning voltage drop along the wire. The DC voltage drop can be so severe that it must be accounted for anytime you power DC devices over long distances. This is a major reason your homes are wired using AC power. True, AC wiring has it's issues too, however - a 120VAC power feed only needs to deliver 1/10th of the ampacity as a 12VDC power feed does to obtain the same wattage (notwithstanding other losses). Since less current flows through an equivalent AC power line, there is less voltage drop along that line.

I suspect that some of you may have experienced voltage related issues after you wired in a device that takes significant current. Many times, the typical RV owner is just not aware of the needs for wiring high current DC loads.

The universally accepted rule-of-thumb, as well as recommendation by various industry groups, is that for proper operation of electronic devices, you should not have more than a 3% loss of voltage due to voltage drop along the wire.

For less sensitive items, such as lighting circuits, no more than a 10% loss should be allowed.

Say for ease of discussion - your battery is producing 12VDC - then a 3% voltage drop means that the voltage found at the electronic device should never be less than 11.64V. Just where does the 3% loss in voltage go? It is dissipated along the wiring itself, and is a function of wire size, current flow, and distance.

Note that in the typical RV wiring scheme, the chassis may or may not be part of the negative side of the circuit. Regardless of whether the chassis or a dedicated wire carries the negative side of the battery, the concept of voltage loss is the same.

We could draw an equivalent circuit showing the wiring and the load as resistances. Normally, the values of R1 and R3 are insignificant, and can be disregarded. That is, until we deal with high-power requirements. Only then do these resistances become significant enough to lower the voltage at the load to an unacceptable level.

It should also be realized that if a chassis is used for the negative side, the "equivalent" resistance of the chassis will be much lower than if a dedicated wire were used. You will see in subsequent calculations that there will be a difference whether a chassis or dedicated wire is used.

However, when high currents are introduced, the values of the resistances of R1 and R3 become significant, and can be large enough to interrupt the proper operation of the device. It should also be noted that this is not an un-safe condition - as the wiring can be large enough that it is not overheating (overload condition), but not large enough to prevent excessive voltage drop.

For example, should the wiring size not be adeaquate, a 20% or more voltage drop may occur at R1 and R3. This voltage drop results in a lower voltage being applied at the load (R2).

The solutions to compensate for this voltage drop are:

Of the above options, often the only solution is the last solution increase the wiring size. It may, however, be possible to decrease the load requirements by running multiple circuits, and placing each device on its own circuit but this is often a more expensive option.

Determining the extent of the problem

To determine the extent of the problem, and whether in your application, you will have a problem, we can use a formula to determine the voltage drop:

CM = (K x I x L) / E


CM = circular mils of the wire (the copper cross-section).
K = 10.75 (Constant representing the mil-foot resistance of copper)
I = Current (amps)
L = Length (feet)
E = Voltage drop (in volts)

Note: for negative chassis ground systems (using the chassis as the negative path), the length is the distance of the positive wire. For two wire systems (using both positive and negative wires without the chassis), then the length is the round trip distance. Hence, using the chassis as the negative side can reduce the size of the wire needed for the positive wire.

The scenario.

I want to power my air compressor. I see from the manufacturer's specification that it requires up to 15Amps @ 12VDC.

I want to therefore select a wire that will adequately power the air compressor. My options are to wire both positive and negative wires from the battery to the air compressor, or use the chassis for the negative side.

I next need to find out how large of a wire I need to carry 15 Amps and still provide less than a 10% voltage drop. I have also determined that I will need 10ft of wire to connect the air compressor to the battery... so I now start plugging values into the formula.

So, to determine the circular mils I need, the formula is re-arranged as follows:

CM = (K x I x L) / E

10.75 x 15 x 10 = 1,612.5

1,612.5 / 1.2(v) = 1,343.75

So then, I need a wire having 1,343.75 circular mils. Since I used 10ft in the formula, then this would suffice only if I used the chassis as the negative side. If, however I wanted to wire both positive and negative wires to the air compressor, I would have needed to use the round trip distance... in this case 20ft - which would have resulted in needing a wire having 2,687.5 circular mils.

The last step is to look at the circual mils conversion chart below. Obvously, you need to select the wire size larger than 1,343.75 circular mils, which would be 18AWG if using the chassis for the negative side.

And finally, if not using the chassis for the negative side, then we would need to increase the wire size to 14AWG.


The following chart shows the voltage drops at various distances for a 3% voltage drop. This chart was created for "round trip" distances - that is, not using a chassis for the negative lead. If using the chassis, you can double the distances.

Here are a few references:

Iterations of the formula:

CM = (K x I x L) / E

I = (E x CM) / (K x L)

L = (E x CM) / (K x I)

E = (K x I x L) / CM


CM = Circular Area of Conductors
K = 10.75 (Constant representing the mil-foot resistance of copper)
I = Current (amps)
L = Length (feet)
E = Voltage drop (in volts)

Circular Mils (1 mil = 0.001" dia):

18 AWG = 1,60018 AWG = 1,537
16 AWG = 2,600 16 AWG = 2,336
14 AWG = 4,100 14 AWG = 3,702
12 AWG = 6,500 12 AWG = 5,833
10 AWG = 10,500 10 AWG = 9,343
8 AWG = 16,800 8 AWG = 14,810
6 AWG = 26,600 6 AWG = 24,538
4 AWG = 42,000 4 AWG = 37,360
2 AWG = 66,500 2 AWG = 62,450
1 AWG = 83,6901 AWG = 77,790
0 AWG = 105,6000 AWG = 98,980
00 AWG = 133,10000 AWG = 125,100
000 AWG = 167,800000 AWG = 158,600
0000 AWG = 211,6000000 AWG = 205,500

Final Thoughts.

Even though you may be able to use smaller wire in these calculations, you should never use wiring less than 18AWG stranded for RV projects. For this reason, the circular mil chart does not show wire smaller than 18AWG. This advice is because of the smaller the wire, the more sensitive it is to the varous vibrations and flexing the wire will be subjected to in the RV as it travels down the road. Smaller wire or non-stranded wire may have more of a tendancy of breakage due to these stresses.


Last reviewed and/or updated May 10, 2017