Anyone who runs cables and connects electrical devices in their vehicle themselves should know what the right cable cross-section is. Too thick a cable is, at worst, unnecessary; too thin a one can, at worst, become dangerous. In this guide we’ll help you along.
To get this out of the way first: it does matter which cable an electrical load is connected with. Whether the full required power can be transmitted or not depends on the material, length and cable cross-section. We’ll limit ourselves here to the standard material in the automotive field: multi-strand copper cables, also called stranded wires. For one thing they’re mandatory in vehicles, for another they have advantages over a solid conductor wire. They are insensitive to vibration, easy to bend, and frequent bending doesn’t bother them much.
Cross-section is not the same as diameter
With the cable specifications you have to look closely. The cable cross-section can easily be confused with the cable diameter. The diameter, which plays no role here, is a measure of length. The cable cross-section, by contrast, is an area figure. And the area is responsible for the electrical properties of the conductor that matter here.
Voltage, current and power
Let’s briefly clear up the relationships. In a vehicle there is, in most cases, an on-board voltage of 12, sometimes also 24 volts. A load with a certain power is now connected to this voltage. Take the H4 dipped-beam headlights at the front, for example. Each bulb has a power of 55 watts, so 110 watts in total. For the load to be run at full power “P”, a certain current “I” has to flow at the given voltage “U”. From the voltage and the power we arrive at this required current.
The rule is P = U x I, so I = P / U. With two dipped-beam headlights we get 9.1 A (110 watts / 12 volts = 9.1 A).
The conductor needs a sufficient cross-sectional area to let the required amount of current flow. This compares well to a water pipe. The voltage is the force with which the water flows through. The current is the amount of water that flows through. The greater the force with which the water flows, or the larger the cross-section of the pipe, the more water arrives at the end of the pipe. That’s what the formula says too. The greater the voltage “U” or the current “I”, the greater the power “P” becomes.
The length
Now the length comes into play as well. The further the current has to be transported through a conductor, the more power it loses. The voltage drops over the length of the conductor. Too little voltage can also cause a device to switch off. The effect is caused by the material-dependent inherent resistance. This can again be compared to the water pipe. The longer the pipe, the more the water is slowed. Or, put the other way round, to get the same amount of water per second at the end, the water either has to be pumped through the pipe with more force or the pipe cross-section has to be increased. Applied to our example, the voltage or the conductor cross-section has to be increased. The voltage is fixed, so only the cross-section is left to reduce the loss caused by the cable length. So the higher the conductor cross-section, the less voltage we lose from the voltage source to the load, and the more power arrives at it.
What happens with too small a cross-section
While too large a conductor cross-section only makes the cable more expensive and more unwieldy, too small a cross-section has real negative consequences. In the simplest case the load simply doesn’t get enough voltage and runs at reduced power. If the load’s power draw is already clearly too great in relation to the cable cross-section, the conductor warms up noticeably. The first negative effect of this warming is that the conductor’s resistance rises and reduces the power still further. The second negative effect is that, in the worst case, the warming becomes so great that the insulation melts and starts a cable fire. The best example of this is the short circuit, in which the appetite for power is theoretically infinite.
Under certain circumstances this can mean the total loss of the vehicle. That’s why you have to size the cables correctly in terms of their cross-section.
The calculation
The top priority is to run an operationally safe cable that doesn’t heat up impermissibly. Other important points, which aren’t the topic here, are chafe-free routing and secure connections. There are a number of regulations that were drawn up for good reason. For example, 12- or 24-volt cables must not be routed alongside 230-volt cables either.
A power loss can’t be avoided, but it should amount to a maximum of 1 to 2 per cent. At 12 volts that would be a maximum of 0.24 volts. For the vehicle lighting that I run while driving, 2 per cent would be acceptable to me, for example. Loads that I run from my battery while stationary, however, should be supplied as efficiently as possible. There 1 per cent would be my upper limit.
More light output through higher voltage? Careful!
At this point a note about increasing the brightness of halogen lamps through improved cable sets. The more voltage is applied to a bulb, the more its lifespan shortens. Automotive halogen lamps are designed and tested for an operating voltage of 13.2 volts. Cable sets are offered on the market that provide less voltage loss and thus raise the voltage at the bulb. A higher voltage means brighter light.
The brighter light is produced by raising the temperature of the filament through the higher voltage. And that’s the catch. Raising the voltage by just 6 per cent, from 13.2 to 14 volts, increases the light output by roughly 25 to 30 per cent, but at the same time reduces the average lifespan by 55 per cent.
I’ve described this in a bit more detail here: Lighting.
Which values are needed for the calculation
First we need the voltage “U”. Then the power draw “P” of the device. From these we calculate the required current “I”:
I = P / U
The power is given in watts. If the power draw changes during operation, always take the maximum possible value. Both can be read off the device and in the manual. If several devices are to be run on one cable, all the power values are added together.
Next we need the length “L” of the conductor. It and the specific resistance of the conductor determine how far the voltage drops over the length. As the specific resistance we take that of the most common cable material, copper. It is 0.0175 ohms per mm2 per 1 metre. Finally we specify the accepted loss “DF”.
The complete formula for calculating the cable cross-section “A” is now made up as follows:
Cable cross-section A = ( I x 0.0175 x L x 2) / (DF x U)
Putting in the values for our example above, the dipped beam, for a 1.5-metre cable and a maximum of 2 per cent (0.02) loss:
x mm2 = ( 9.1 A x 0.0175 x 1.5 m x 2) / (0.02 x 12 V)
x = 1.99 mm2.
Now you have to round up to the next larger cable cross-section available on the market. In this case that would be 2.5 mm2. Here’s a cable calculator that works out the available cable cross-section for you straight away: cable configurator.
Common cross-section sizes for copper cables
Cables aren’t available in just any cross-section you like. The standard DIN EN 60228 defines the available cable cross-sections. How much current a cable of a given cross-section may carry at most depends on further factors (routing method, number of conductors, temperature, and so on). That’s why we find different figures on the internet, because only guide values can ever be given.
Anyone who wants precise information can find more in DIN VDE 0298-4. That is rather complicated, though. In the standard you have to work your way from table to table. From the routing method through the load capacity and temperature to the conversion factors for the number of conductors. So I’ve put together a few cable cross-sections in the following table to give a rough guide value. With it you can already orient yourself to the right order of magnitude. So if you’re dealing with 20 A, the cable lies somewhere between 1.5 and 2.5 mm2. In the end you should always follow the specification supplied with the particular cable.
Maximum current by cable cross-section (guide values)
| Cross-section in mm² | Current in amperes |
|---|---|
| 0.75 | 12 |
| 1 | 15 |
| 1.5 | 18 |
| 2.5 | 26 |
| 4 | 34 |
| 6 | 44 |
| 10 | 61 |
| 25 | 108 |
| 50 | 168 |
| 70 | 207 |
Fusing
An essential safety factor for people and equipment is the fusing of the cable. If a current level determined by the fuse is exceeded in the cable, the fuse blows and thus prevents damage. The fuse should sit as close as possible to the power source, so that the fused length of the cable is as long as possible. The piece of cable between the power source and the fuse is not fused. Here in particular, care must be taken to route it in a way that rules out damage to the insulation.
You have to match the sizing of the fuse to the cable cross-section. The rated value of the fuse may be at most the current level of the cable. It’s better if it blows even before that, so that the fuse’s maximum permissible current is below that of the cable. Then it’s ensured that the cable doesn’t reach its limit. If a 1.5 mm2 cable, for example, can carry a maximum of 15.5 A, the right automotive fuse would be one with a 15 A rating. One with 15.5 A would already be very close to the limit. Automotive fuses are defined in the standard DIN 72581.
If you don’t know much about electrics, it’s best to get help from a specialist workshop.