The issue to be tackled in this primer is high powered wiring needs in a boat. Often boat manufacturers run a couple of extra circuits to the helm for DC powered devices that allows a marina or boatowner to add elecronics at the helm. This may be adequate for many devices, but high-power devices, such as radar systems or high-powered stereo systems may exceed the capacity of these circuits. This concept is also true for other areas of the boat, such as running wiring for a windlass, macerator, or other high current demand item.
The scenario presented here is that a boat has a helm power curcuit, but with the intended installation of radar and other electronics devices, that the capacity of those circuits will be exceeded.
I will present a real-world case study to bring home this point. My boat has a nice 10AWG auxiliary circuit to the helm. One might at first believe that this should be sufficient to run any radar system, but I will show that is a marginal setup for these purposes.
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 disadvantage is the primary reason that your homes are AC powered - because there is no similar voltage drop when using AC power.
This phenomenon was discovered in the late 1800s by Nikola Tesla. In conjunction of Westinghouse, Tesla developed the concept of the AC power grid, which was much more effecient than Edison's DC systems. Few people today know of the "current war" that developed between Tesla and Edison for control of the power grid. Due to the superior nature of AC systems in long distance power delivery, Tesla won the battle, and we have the AC power system today as a result.
For less sensitive items, such as a windlass, no more than a 10% loss should be allowed. USCG regulation 46CFR183.340 requires that no circuit shall have more than a 10% loss along a wire.
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.
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.
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
E = (K x I x L) / CM
10.75 x 10 x 36 = 3,870
3,870 / 9343 = 0.41V
Unfortunately, 0.41V is more than a 3% voltage drop (3% = 0.36V). While this could be considered marginal, I also need to power a VHF radio and other electronics I plan on installing, such as a depth finder, NMEA instruments, fuel flow instruments, and so on. When considering these additions, the voltage drop will be excessive, and will lead to problems.
But wait you say - this is confusing, there are charts that say that 10AWG wire can carry 50 Amps!
Well this may indeed be true, but not with the limit of a 3% voltage drop. Lets use the formula again to see just how much of a voltage drop 50 Amps will produce on that 10AWG wire:
10.75 x 50 x 36 = 19,350
19,350 / 9,343 = 2.07V
Therefore, a 12V powered circuit will have 9.93V at the instrument, which likely will prevent proper operation.
You should realize that the 50Amp rating is a safety rating, and specifies the maximum amount of current that can safely flow within the wiring in perfect conditions. It has nothing to do with the amount of voltage that is lost across the wire at this high current flow rate.
It would be interesting to note however, should that same 10AWG wire be used with a 120VAC circuit, it could safely carry 30Amps without any significant voltage loss. Such are the problems with high-current DC powered devices.
Therefore, the reason for the large wire size in this application is its series resistance (which causes the voltage drop), not its current carrying capacity.
To meet my needs, I will use the next larger wire for the helm wiring; 8AWG.
I have determined that 8AWG has 14,810 Circular Mils. If you recall, 10AWG wire had 9,343 Circular Mils. Therefore, 8AWG is 60% larger than 10AWG.
This time, I want to know what capacity 8AWG is going to give me, in Amps, yet stay within the 3% voltage drop range.
I = (E x CM) / (K x L)
(0.36 x 14,810) / (10.75 x 36)
The answer: 13.8Amps
This will be enough to allow me to power the GPS/Radar/Chartplotter and other miscellaneous items. I can also power the VHF radio either from this circuit, or the 10AWG circuit that is already installed, which will help "split the load".
It cost $100 just for 20ft of 8AWG marine grade duplex cable, so this is not an inexpensive venture.
One lesson in all of this is that distance is a wire killer. Since the boat has a flybridge, a much longer run is required than if the boat were simply an express cruiser where the helm was only a few feet away.
By the way, many car-stereo folks know about the voltage drop problem, although they may not know all of the details. They often wire up a large capacitor next to their power amps. The capacitor serves as a temporary storage cell, not unlike a battery, and can help stabilize the voltage at the amp. Since music only requires high amplifier power during short term transients, this tends to work.
However, I would caution against its use in a boat since the capacitor - even though it might not be physically located in the engine room, could generate a spark when connecting/disconnecting the battery in the engine room. A spark in the engine room is never a good idea in a boat.
I created a voltage drop chart showing the various distances, AWG, and such for a 3% voltage drop.
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):
Note that the USCG allows either AWG or SAE rated wire to be used under 50V, but only AWG wire may be used for circuits over 50 volts.
|18 AWG = 1,600||18 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 = 83,690||1 AWG = 77,790|
|0 AWG = 105,600||0 AWG = 98,980|
|00 AWG = 133,100||00 AWG = 125,100|
|000 AWG = 167,800||000 AWG = 158,600|
|0000 AWG = 211,600||0000 AWG = 205,500|
Now that you have the basic fundamentals of determining the proper cabling, you can use the Wire Sizing Calculator by clickin on the icon below: