What is the maximum wattage (and maximum recommended wattage) for a single ribbon cable wire.

I believe the wire size is 28 AWG (maybe 26, but let's call it 28 to be safe), stranded copper. The length of the cable will be no more than 3', and it will either be 40 conductor or 50 conductor ribbon cable (don't worry, I will not be using any 80 conductor IDE cables or anything silly).

Anyway, while googling, I keep finding amp ratings and "Ampacity" charts, but they don't contain any references to the voltage those amps pertain to, and I think they might be AC specific, so it's not super helpful to me. Also, I keep finding completely conflicting information.

More Info:

I intend to run 12V DC on the cable and am trying to determine how many threads of cable I will need for particular amperages @ 12V DC.


It's my understanding that wattage (V * A) is what matters with wiring -- (i.e. because if the insulators melt at 5A @ 10V then it will also melt at 50A @ 1V, or 1A @ 50V, etc.)

If I know the wattage, I can determine the safe amperage at any DC voltage within the 300V rating.

Other Thoughts:

What is a reliable resource for finding out this kind of information? (Teach a man to fish...)

Is there a difference between "amps" and "Ampacity"? Wikipedia is kind of vague on the subject.


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    \$\begingroup\$ In the US and Canada, "Ampacity" is the maximum current permitted by regulation, and depends on the environment the wire is in and the type of insulation, as well as the wire size. In house wiring, the Ampacity of #14 wire in normal use is 15 Amps, but if you want to run several wires in a conduit, you may need #12 wire for the same ampacity. \$\endgroup\$ – Peter Bennett Mar 27 '15 at 19:29

The maximum voltage is defined by the insulation on the wire, not the gauge of the wire. It is the voltage (worst case) that the insulation can break down. The maximum voltage, as well as the maximum temperature, should be printed on the cable.

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The maximum current is defined by the cross-sectional area of the cable (AWG number).

The table I usually use is the one from PowerStream: http://www.powerstream.com/Wire_Size.htm

That table gives different current ratings for power cabling (long runs) and for "chassis" wiring (short wires between components or boards in a system). It also details the resistance of the wire, which is very very useful.

There are no exact precise values, only ballpark "rule of thumb" values. The actual value for your specific wire would only be within a range since it'll never be an exact area all the way along. The current rating is also dependent on the ambient temperature, and to some extent the type of insulation (some can handle the wires getting hotter than others). Often a more important factor is the resistance of the wire which, on longer runs and at higher currents, can cause unacceptable voltage drops.

The current, passing through the resistance of the wire, causes heating (\$P=I^2R\$) of the wire itself, and too much current will cause the wire to melt and "fuse". This may go hand in hand with the insulation melting, but not always.

There is no "wattage" for a cable, only limits of the voltage and current. I guess you could say the maximum wattage is the product of the maximum of the two, but that would be a misnomer, since you could be under the "maximum wattage" of the cable but have too much current, or too much voltage. The wattage of a cable only makes any sense when you are thinking about the amount of power dissipated over a specific length of the cable, which is a function of the current and the resistance. Remember - your "working" voltage isn't the voltage that is across the length of cable - that voltage is the voltage drop caused by the current flowing through the resistance (\$V=R \times I\$).

"Ampacity" is an Americanisationalism and could be construed as a portmanteau of Amp-Capacity. Most normal people use Amperage.

  • \$\begingroup\$ So the current -- in amps -- is the same maximum at any voltage? \$\endgroup\$ – BrainSlugs83 Mar 27 '15 at 17:58
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    \$\begingroup\$ Check out electronics.stackexchange.com/q/33522/49251 for some stuff on the current capacity of 28 AWG. \$\endgroup\$ – Greg d'Eon Mar 27 '15 at 18:10
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    \$\begingroup\$ There are no precise values. They are all dependent on cable length, ambient temperature, acceptable voltage drop, etc. \$\endgroup\$ – Majenko Mar 27 '15 at 18:43
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    \$\begingroup\$ 28AWG is roughly 1.4A for short runs like that. The resistance would be around 0.064Ω. The 300V is the voltage at which the insulation would break down (worst case). P=I²R, so 1.4A at 0.064Ω would be 0.12544W dissipated from a single conductor. The current causes fusing of the wire, not necessarily melting of the insulation, though that often does happen as well. \$\endgroup\$ – Majenko Mar 27 '15 at 19:00
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    \$\begingroup\$ The table is related to the cross-sectional area. The cross-sectional area is the same for stranded and solid core, but the outer diameter is smaller for solid core. The CSA for stranded is the sum of the CSAs for all the strands. \$\endgroup\$ – Majenko Mar 27 '15 at 19:12

I want to add something to Majenko's answer:

You are mixing up voltage with voltage drop!

If you are using a ribbon cable to supply your circuit with 12V DC, then the voltage drop across your cable is probably tiny - it's definitely a fraction of a volt, and it depends on the resistance of the cable.

In other words, you are trying to calculate

\$ P = IV \$

but you're using the wrong voltage. Instead, use

\$ P = I^2 R \$

where \$R\$ depends on the cable. This is why you're seeing a maximum number of amps, not watts, and no dependence on voltage!

EDIT: This is a response to your comments below.

You want to run a 60 W motor at 12 V. I'm going to make a silly assumption and claim that it's just a resistive load - that'll be good enough for my purposes. From \$ P = V^2 / R\$, your motor has a resistance of 2.4 ohms.

You're using 28 AWG wire. You didn't say how long it is, so I'm going to use this calculator and plug in 1 foot of wire, which gives me a resistance of 0.065 ohms.

Now, here is your (simplified) circuit:


simulate this circuit – Schematic created using CircuitLab

This is effectively a voltage divider, so you can find the voltage drop across each section of cable and the motor. If you do this, you find that

\$ V_{cable} = (12V) \frac{0.065 \Omega}{2.4\Omega + 2(0.065\Omega)} = 0.308 V \$

and the power heating either section of the cable is

\$ P_{cable} = (0.308V)^2 / (0.065 \Omega) = 1.46 W \$

This works for either section of cable - even though one of them is at 12V and the other one at 0.

Also note that the voltage drop across the cable is pretty small compared to the voltage across the motor, so a simpler way to find the power is to find the current in the motor while ignoring the voltage drop in the wires:

\$ I = \frac{12V}{2.4\Omega} = 5A \$

and just check to make sure that current won't make too much heat (per unit length, based on the resistance per unit length). No voltages involved. Does that make sense?

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    \$\begingroup\$ Weird. Is this a part of Ohm's law, or something else? And -- this is kind of a mind blown thing but -- am I reading this right -- you're saying 1A @ 1V will do the same amount of heat damage to the wire as 1A @ 10V? -- So I can just ignore the volts and just consider amps only? \$\endgroup\$ – BrainSlugs83 Mar 27 '15 at 18:37
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    \$\begingroup\$ This is Ohm's law. Imagine a resistor with 0V on one side and 1V on the other. Now, imagine a resistor with 9V on one side and 10V on the other. What's the difference? \$\endgroup\$ – Greg d'Eon Mar 27 '15 at 19:02
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    \$\begingroup\$ (Note that, as Majenko says, the voltage rating is just there so the insulation keeps your 10V from interacting with other stuff around it.) \$\endgroup\$ – Greg d'Eon Mar 27 '15 at 19:03
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    \$\begingroup\$ I added some more info with an example based on your comments. Am I making any sense? (PS: what do you mean by "dissipating 240 ma across the cable"? You dissipate heat, not current.) \$\endgroup\$ – Greg d'Eon Mar 27 '15 at 20:06
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    \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – Greg d'Eon Mar 27 '15 at 23:10

I want to add something to Greg's answer. Just a somewhat different approach to the calculations, which might be clearer. It's all just Ohm's law.

In the circuit shown in his answer, there are three resistors: two 65mΩ and one 2.4Ω. The total resistance in the circuit is the sum of those three: .065+2.4+.065, which gives 2.53Ω. The battery provides 12V, and 12V across 2.53Ω gives a current of 12/2.53, or 4.74A. That 4.74A flows through all three resistors: the motor and each of the two power leads. The power dissipated in each resistor is IIR (sorry, too lazy to figure out how to write superscript 2). So the motor consumes 4.74*4.74*2.4, or 53.92, Watts. Each power lead consumes 4.74*4.74*.065, or 1.46, Watts. The voltage drop along each power lead is IR, which is 4.74.065, or 0.31V; the voltage drop across the motor is I*R, which is 4.74*2.4, or 11.38V. If you add up the three voltages you get 11.38+0.31+0.31, which comes to 12V, as it should.


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