Copper wire figures
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AWG Table
1 AWG is 289.3 thousandths of an inch
2 AWG is 257.6 thousandths of an inch
5 AWG is 181.9 thousandths of an inch
10 AWG is 101.9 thousandths of an inch
20 AWG is 32.0 thousandths of an inch
30 AWG is 10.0 thousandths of an inch
40 AWG is 3.1 thousandths of an inch
The table in ARRL handbook warns that the figures are approximate and may
vary dependent on the manufacturing tolerances. If you don't have a chart
handy, you don't really need a formula. There's several handy tricks:
Solid wire diameters increases/decreases by a factor of 2 every 6 gages,
" " " " " 3 every 10 gages,
" " " " " 4 every 12 gages,
" " " " " 5 every 14 gages,
" " " " " 10 every 20 gages,
" " " " " 100 every 40 gages,
With these, you can get around alot of different AWGs and they cross check
against one another. Start with solid 50 AWG having a 1 mil diameter.
So, 30 AWG should have a diameter of ~ 10 mils. Right on with my chart.
36 AWG should have a diameter of ~ 5 mils. Right on with my chart.
24 AWG should have a diameter of ~ 20 mils. Actually ~ 20.1
16 AWG should have a diameter of ~ 50 mils. Actually ~ 50.8
10 AWG should have a diameter of ~ 100 mils. Actually ~ 101.9
If you are more interested in current carrying ability than physical size,
then also remember that a change of 3 AWG numbers equals a doubling or halving
of the circular mills (the cross sectional area). Thus, if 10 AWG is safe
for 30 amps, then 13 AWG (yeah, hard to find) is ok for 15 amps and 16 AWG
is good for 7.5 amps.
The wire gauge is a logarithmic scale base on the cross sectional area
of the wire. Each 3-gauge step in size corresponds to a doubling or halving
of the cross sectional area. For example, going from 20 gauge to 17 gauge
doubles the cross sectional area (which, by the way, halves the DC resistance).
So, one simple result of this is that if you take two strands the same
gauge, it's the equivalent of a single wire that's 3 gauges lower. So
two 20 gauge strands is equivaent to 1 17 gauge.
Wire Gauge Resistance per foot
4 .000292
6 .000465
8 .000739
10 .00118
12 .00187
14 .00297
16 .00473
18 .00751
20 .0119
22 .0190
24 .0302
26 .0480
28 .0764
Current ratings
Most current ratings for wires (except magnet wires) are based on permissible
voltage drop, not temperature rise. For example, 0.5 mm^2 wire is rated
at 3A in some applications but will carry over 8 A in free air without
overheating. You will find tables of permitted maximum current in national
electrical codes, but these are based on voltage drop (not the heating
which is no problem in the current rating those codes give).
Here is a small current and AWG table taken from the Amateur Radio Relay
Handbook, 1985.
AWG dia circ open cable ft/lb ohms/
mils mils air A Amp bare 1000'
10 101.9 10380 55 33 31.82 1.018
12 80.8 6530 41 23 50.59 1.619
14 64.1 4107 32 17 80.44 2.575
Mils are .001". "open air A" is a continuous rating for
a single conductor with insulation in open air. "cable amp" is for in
multiple conductor cables. Disregard the amperage ratings for household
use.
To calculate voltage drop, plug in the values: V = DIR/1000
Where I is the amperage, R is from the ohms/1000' column above, and D
is the total distance the current travels (don't forget to add the length
of the neutral and hot together - ie: usually double cable length). Design
rules in the CEC call for a maximum voltage drop of 6% (7V on 120V circuit).
Resistivities at room temp:
Element Electrical resistivity (microohm-cm)
Aluminum 2.655
Copper 1.678
Gold 2.24
Silver 1.586
Platinum 10.5
This clearly puts silver as the number one conductor and
gold has higher resistance than silver or copper. It's desireable in connectors
because it does not combine well with other materials so remains relatively
pure at the surface. It also has the capability to adhere to itself (touch
pure gold to pure gold and it sticks together) which makes for very reliable
connections.
Thermal conductivity at room temp:
W/cm C
silver 4.08
copper 3.94
gold 2.96
platinum 0.69
diamond 0.24
bismuth 0.084
iodine 43.5E-4
This explains why diamonds are being used for high power substrates now.
That's man-made diamonds. Natural diamonds contain sufficient flaws in the
lattice that the phonons (heat conductors) get scattered and substantially
reduce the ability to transport the heat.
Copper wire resistance table
AWG Feet/Ohm Ohms/100ft Ampacity* mm^2 Meters/Ohm Ohms/100M
10 490.2 .204 30 2.588 149.5 .669
12 308.7 .324 20 2.053 94.1 1.06
14 193.8 .516 15 1.628 59.1 1.69
16 122.3 .818 10 1.291 37.3 2.68
18 76.8 1.30 5 1.024 23.4 4.27
20 48.1 2.08 3.3 0.812 14.7 6.82
22 30.3 3.30 2.1 0.644 9.24 10.8
24 19.1 5.24 1.3 0.511 5.82 17.2
26 12.0 8.32 0.8 0.405 3.66 27.3
28 7.55 13.2 0.5 0.321 2.30 43.4
These Ohms / Distance figures are for a round trip circuit. Specifications
are for copper wire at 77 degrees Fahrenheit or 25 degrees Celsius.
Wire current handling capacity values
A/mm2 R/mohm/m I/A
6 3.0 55
10 1.8 76
16 1.1 105
25 0.73 140
35 0.52 173
50 0.38 205
70 0.27 265
Information about 35 mm2 Cu wire
According Ströberg TTT 35mm2 copper wire can take continuous current
of 170A on free air and 200 A on ground. The wire can handle 5 kA short
circuit current for 1s. DC resistance of the wiure is 0.52mohm/m.
Mains wiring current ratings
In mains wiring there are two considerations, voltage drop and heat
buildup. The smaller the wire is, the higher the resistance is. When the
resistance is higher, the wire heats up more, and there is more voltage
drop in the wiring. The former is why you need higher-temperature insulation
and/or bigger wires for use in conduit; the latter is why you should use
larger wire for long runs.
Neither effect is very significant over very short distances. There
are some very specific exceptions, where use of smaller wire is allowed.
The obvious one is the line cord on most lamps. Don't try this unless
you're certain that your use fits one of those exceptions; you can never
go wrong by using larger wire.
This is a table apparently from BS6500 which is reproduced in the IEE
Wiring Regs which describes the maximum fuse sizes for different conductor
sizes:
Cross- Overload
sectional current
area rating
0.5mm² 3A
0.75mm² 6A
1mm² 10A
1.25mm² 13A
1.5mm² 16A
Typical current ratings for mains wiring
Inside wall
mm^2 A
1.5 10
2.5 16
Equipment wires
mm^2 A
0.5 3
0.75 6
1.0 10
1.5 16
2.5 25
We sizes used in USA inside wall
For a 20 amp circuit, use 12 gauge wire. For a 15 amp circuit, you can
use 14 gauge wire (in most locales). For a long run, though, you should
use the next larger size wire, to avoid voltage drops.
Here's a quick table for normal situations. Go up a size for more than
100 foot runs, when the cable is in conduit, or ganged with other wires
in a place where they can't dissipate heat easily:
Gauge Amps
14 15
12 20
10 30
8 40
6 65
PCB track widths
For a 10 degree C temp rise, minimum track widths are:
Current width in inches
0.5A .008"
0.75A .012"
1.25A .020"
2.5A .050"
4.0A .100"
7.0A .200"
10.0A .325"
Equipment wires in Europe
3 core equipment mains cable
Current 3A 6A 10A 13A 16A
Condictor size(mm) 16*0.2 24*0.2 32*0.2 40*0.2 48*0.2
Copper area (mm^2) 0.5 0.75 1.0 1.25 1.5
Overall diameter(mm) 5.6 6.9 7.5
Calbe ratings for 3A, 6A and 13A are based on BS6500 1995 specifications
and are for stranded thick PVC insulated cables.
Insulted hook-up wire in circuits (DEF61-12)
Max. current 1.4A 3A 6A
Max. working voltage (V) 1000 1000 1000
PVC sheat thickness (mm) 0.3 0.3 0.45
Conductor size (mm) 7*0.2 16*0.2 24*0.2
Conductor area (mm^2) 0.22 0.5 0.75
Overall diameter (mm) 1.2 1.6 2.05
Car audio cable recommendations
This info in from rec.audio.car
FAQ (orognally from IASCA handbook). To determine the correct wire
size for your application, you should first determine the maximum current
flow through the cable (looking at the amplifier's fuse is a relatively
simple and conservative way to do this). Then determine the length of
the cable that your will use, and consult the following chart:
Length of run (in feet)
Current 0-4 4-7 7-10 10-13 13-16 16-19 19-22 22-28
0-20A 14 12 12 10 10 8 8 8
20-35A 12 10 8 8 6 6 6 4
35-50A 10 8 8 6 6 4 4 4
50-65A 8 8 6 4 4 4 4 2
65-85A 6 6 4 4 2 2 2 0
85-105A 6 6 4 2 2 2 2 0
105-125A 4 4 4 2 2 0 0 0
125-150A 2 2 2 2 0 0 0 00
Skin effect
Skin effect is an effect that the electricity in high frequencies does
not use the whole condictor area. High frequencies tend to use only the
outer parts of the conductor. The higher the frequency, the less of the
wire diameter is used and higher the losses. Sin effect must be taken
care in high frequency coil designs.
The frequency dependency of the resistance of a cylindrical conductor
can be calculated by the following formula, which is surely valid for
high frequencies and radii of approx. 50 um:
R(f) = R(DC)* (1 + 1/3 * x^4) with x = Radius/2*sqrt(pi*frequency*permeability*conductivity)
The "formula" for skin effect is the same whether the conductor is rectangular
or cyclindrical. That is why the same value of "radius" used in wire size
in a switchmode transformer is used to determine half the thickness of
a flat foil conductor in the case of foil-wound secondaries.
An approximate equation for the resistance ratio for rectangular conductors
(from Terman) is:
rho = 1/(((8PI * f)/(Rdc * 10^9))^0.5)
Skin depth is not an absolute, but only the depth where current through
the wire or foil has fallen to a specific proportion of the current at
the surface. In fact, current falls off exponenially as you move inward
fromm the surface. The depth of the "skin" is also influenced by proximity
to nearby conductors (such as in a transformer) so is itself not absolute.
Also the formula has to be modified if you use wire that is ferromagnetic
(iron for example).
In addition to skin effect a lot of engineers doing their own magnetics
design don't consider the 'proximity effect' which 'crowds' the current
to one side of the conductor and increases losses. This condition is worst
in thick multi-layer windings. Fortunately, many of the new transformer
shapes have a long and skinny window - good for low leakage L and low
proximity effect losses.
Wire sizes used in fuses
The Standard Handbook for Electrical Engineers lists the following formula:
33 * (I/A)^2 * S = log( (Tm - Ta) / (234 + Ta) + 1 )
I = current in Amperes
A = area of wire in circ. mils
S = time the current flows in seconds
Tm = melting point, C
Ta = ambient temp, C
The melting point of copper is 1083 C.
See pp. 4-74 .. 4-79 of the 13th edition of the Handbook for more info.
Skin effect
At high frequencies there is one thing to consider on wire resistance
besides the DC resistence: skin effect.
The current intensity falls off exponentially with depth. The depth
of penetration (s=sigma) is the depth at which the current intensity has
fallen to 1/e of its value at the surface, where e equals 2.718.
Where the diameter of the conductor is large compared to the depth of
penetration, the total current is the same as if the surface current intensity
were maintained to a depth of penetration.
For example, for copper the depth of penetration is as follows:
MHz Depth of Penetration sigma (mm)
.1 .209
1 .066
10 .021
100 .0066
1000 .0021
For other materials the skin dpeth can be calculated using the formula:
s = 503.3sqrt(rho/(urf)) millimeters
rho = resistivity in ohm-meters
= 1.72x10e-8 for copper or 2.83x10e-8 for aluminum
ur = mu r = relative magnetic permeability
= 1 for both copper and aluminum
f = frequency in magahertz
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