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 |
3.4 |
|
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 |
What 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 75�C).
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
(So the voltage at the
amplifier will be 13.34 vdc) |
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
(So the voltage at the
amplifier will be 13.04 vdc) |
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)
|
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