The AWG – American Wire Gauge – is used as a standard method of denoting wire diameter,
measuring the diameter of the conductor (the bare wire) with the insulation removed. The higher the number – the thinner the wire.
This AWG table is for a single, solid, round wire. Small gaps between the strands
in a stranded wire make it so it has to be slightly larger to keep the same current-carrying
capacity and electrical resistance as a solid wire.
Because a thick wire has less electrical resistance it will carry more current with less voltage drop than a thin wire.
| AWG | Diameter (mm) | Diameter (in) | Circular Mils | Square (mm2) | Resistance (ohm/1000m) |
|---|---|---|---|---|---|
| 14 | 1.63 | 0.064 | 4110 | 2.0 | 8.54 |
| 12 | 2.05 | 0.081 | 6530 | 3.3 | 5.4 |
| 10 | 2.59 | 0.10 | 10380 | 5.26 | 3.24 |
| 8 | 3.25 | 0.13 | 16510 | 8.30 | 2.2 |
| 6 | 4.115 | 0.17 | 26240 | 13.30 | 1.5 |
| 4 | 5.189 | 0.20 | 41740 | 21.15 | 0.8 |
| 2 | 6.543 | 0.26 | 66360 | 33.62 | 0.5 |
| 1 | 7.348 | 0.29 | 83690 | 42.41 | 0.4 |
| 0 | 8.252 | 0.33 | 105600 | 53.49 | 0.31 |
| 00 (2/0) | 9.266 | 0.37 | 133100 | 67.43 | 0.25 |
| 000 (3/0) | 10.40 | 0.41 | 168800 | 85.01 | 0.2 |
| 0000 (4/0) | 11.684 | 0.46 | 211600 | 107.22 | 0.16 |
hat good is all this data? I put it here so you can calculate the proper wire size for your
mobile installations. (yes I only included the wire sizes I thought were important)
The formula for voltage drop is:
VD = 2 x K x I x D/CM.
- K = Direct-Current Constant. K represents the dc resistance. (for a 1,000-circular mils conductor that is 1,000 ft long, at an operating temperature of 75ºC).
K is 12.9 ohms for copper and 21.2 ohms for aluminum. - I = Amperes: The load in amperes at 100%
- D = Distance: Where we specify distances here, we are referring to the conductor length.
- CM = Circular-Mils: The circular mils of the circuit conductor as listed in NEC Chapter 9, Table 8.
Let’s do an example. An amplifier rated 13.8VDC and 75A is wired
to the Batteries with 10 ft lengths of 4 AWG stranded wire.
Copper Wire
-
K = 12.9 ohms, copper
-
I = 75A
-
D = 10 ft
-
CM = 41740
-
VD = 2 x 12.9 x 75 x 10 / 41740 =
-
.46 vdc dropped
Aluminum Wire
-
K = 21.2 ohms, Aluminum
-
I = 75A
-
D = 10 ft
-
CM = 41740
-
VD = 2 x 12.9 x 75 x 10 / 41740 =
-
.76 vdc dropped
Just one more to those 16 pill-ers out there using 15 feet of 1/0 wire….this same formula for
350 amps would leave a voltage drop of 1.32 vdc and would leave 12.48 vdc at the amplifier.
1.32vdc x 350A x 65% = 300 watts lost
Reference : National Electrical Code (NEC)