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Wire resistances. Copper wire figures. AWG Table. Wire Gauge Resistance per foot. Current ratings. Resistivities at room temp. Typical current ratings for mains wiring.
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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 inchThe 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 ~ 101.9If 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 3gauge 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.
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
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).
Element Electrical resistivity (microohmcm) 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.
W/cm C silver 4.08 copper 3.94 gold 2.96 platinum 0.69 diamond 0.24 bismuth 0.084 iodine 43.5E4This explains why diamonds are being used for high power substrates now. That's manmade 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.
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 43.4These Ohms / Distance figures are for a round trip circuit. Specifications are for copper wire at 77 degrees Fahrenheit or 25 degrees Celsius.
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
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.
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 highertemperature 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 0.5mm˛ 3A 0.75mm˛ 6A 1mm˛ 10A 1.25mm˛ 13A 1.5mm˛ 16A
mm^2 A 1.5 10 2.5 16
mm^2 A 0.5 3 0.75 6 1.0 10 1.5 16 2.5 25
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
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"
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 7.5Calbe ratings for 3A, 6A and 13A are based on BS6500 1995 specifications and are for stranded thick PVC insulated cables.
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
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 04 47 710 1013 1316 1619 1922 2228 020A 14 12 12 10 10 8 8 8 2035A 12 10 8 8 6 6 6 4 3550A 10 8 8 6 6 4 4 4 5065A 8 8 6 4 4 4 4 2 6585A 6 6 4 4 2 2 2 0 85105A 6 6 4 2 2 2 2 0 105125A 4 4 4 2 2 0 0 0 125150A 2 2 2 2 0 0 0 00
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 foilwound 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 multilayer windings. Fortunately, many of the new transformer shapes have a long and skinny window  good for low leakage L and low proximity effect losses.
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, CThe melting point of copper is 1083 C.
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 .0021For other materials the skin dpeth can be calculated using the formula:
s = 503.3sqrt(rho/(urf)) millimeters rho = resistivity in ohmmeters = 1.72x10e8 for copper or 2.83x10e8 for aluminum ur = mu r = relative magnetic permeability = 1 for both copper and aluminum f = frequency in magahertz
Last modified: May 28, 2002
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