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Trying to do a spreadsheet to figure out how much cable of different dimensions I will need for a low voltage DC electrical system, but got stuck on the formula for converting current draw, distance and acceptable voltage drop into the minimum cable area required. At the moment I have

((distance * amps * 0.04) / ((voltage * %drop) / 100 )) * 100) / 100

Which I copied from http://www.solar-wind.co.uk, but I have no idea where they got the 0.04 magical number from, and I suspect the results may be inaccurate. What is the correct way of calculating the (minimum) required (copper) cable area, when voltage, current, distance and acceptable voltage drop is known?

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    \$\begingroup\$ There's one parameter you're forgetting. Acceptable temperature increase. \$\endgroup\$ – Harry Svensson Jul 29 '17 at 17:20
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    \$\begingroup\$ 0.04 sounds like twice (so there and back) the resistance in ohms of 1m of 1mm2 copper cable at a modest rise above room temperature. As it's multiplied by amps, that seems reasonable. It seems that this formula is entirely based on voltage drop, not temperature rise. Not unreasonable for low voltage DC, as voltage drop tends to bite first, but check the temperature rise anyway. \$\endgroup\$ – Neil_UK Jul 29 '17 at 17:32
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    \$\begingroup\$ Also, in a low-voltage DC system, power loss is likely to be more restrictive than temperature rise. So choose your wire based on power loss, then just double-check the temperature rise after you do so to make sure it is within reason. \$\endgroup\$ – mkeith Jul 29 '17 at 18:35
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    \$\begingroup\$ @Neil_UK I have to look everything up every time and put it in a spreadsheet. I will try to remember your shortcut. It is much more intuitive. But I thought the OP might benefit from knowing the relatively simple derivation of resistance in wire. Also, here in the US, we use AWG for wire diameter, and I even have to look up the actual diameter. \$\endgroup\$ – mkeith Jul 30 '17 at 6:22
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    \$\begingroup\$ @mkeith Ah yes, AWG. I think the original reason for numbering wires was that's how many die passes it needed to get that size. Is it easier for the non-numerate? I think electricians can pick a reel of 4mm2 just as easily as a reel of xAWG. My Dad used to stock china teapots in his shop, the #16 size pot was called that because 16 fit inside a standard barrel, eegads! All we need is to get everybody away from inches to SI (I'm looking at you, US), and then it will all be slightly easier. I'm amused that AWG and SWG are slightly, like two numbers, different \$\endgroup\$ – Neil_UK Jul 30 '17 at 9:26
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I have a problem with equations like

((distance * amps * 0.04) / ((voltage * %drop) / 100 )) * 100) / 100

Too many terms, too many brackets. What if you transcribe it incorrectly? Where do all the terms come from? Do we need all those factors of 100?

I much prefer to build up understanding a step at a time.

The resistance of a length of pure annealed copper wire of cross section \$A mm^2\$ and length \$Lm\$ at room temperature is \$R=0.017\frac{L}{A}\Omega\$.

Note I'm not using strict SI here, I've put the area in square mm, the way the wire in the DIY store is sized. To stay in strict SI, the area would be square metres, and the resistivity factor would be 17n ohms, usually written as 1.7e-8.

Is 17m ohms the right resistivity to use? At a higher temperature, if the wire was work hardened, if it was slightly impure, if it was slightly undersize, it would be more. 20m ohms might be a better 'worst case' figure to use, and easier to do sums with.

The voltage drop when it's carrying a current of \$I\$ amps is \$V=IR\$.

We usually relate the voltage drop to the supply voltage, 1v lost in 12v is worse than 1v lost in 48v or 240v, so we need to divide by the supply voltage to get the fractional drop, and optionally multiply by 100 to get to percent.

To put it all in one expression, the percentage drop for a system is $$pcdrop = \frac{I\frac{0.02L}{Area}}{V_{supply}}\times100$$Obviously it can be simplified to \$\frac{20IL}{V_{supply}Area}\$, but the first form keeps the correspondence between the factors and the way we derived it. Personally, I prefer to work out resistance, then work out absolute drop, then work out percentage drop, but each to his own.

Generally with high voltage systems (240v), we work out cable size on temperature rise, and then do a voltage drop check as an afterthought. On low voltage systems (12v), voltage drop tends to bite first. It's worth checking that any calculated cable is OK for temperature rise, but it's usually a no-brainer. For instance, if your 10A circuit needs 16mm2 cable based on voltage drop, there will be absolutely no problem with temperature rise, as 16mm cable is rated 100A.

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  • \$\begingroup\$ Thank you! However, my goal is to work out the required minimum area, so the original form (length x amps x 0.017) / maxdrop is what's needed (or perhaps with 0.02 as the cable resistance constant). And no, none of the "factors of 100" are needed! \$\endgroup\$ – Ola Tuvesson Jul 31 '17 at 13:37
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There are two limitations on wire gauge- temperature rise and voltage drop.

Maximum allowable temperature rise depends on the power dissipation and the heat loss so it is not a simple formula. Typical factors taken into account are RMS current (of course) "bundling" (how many wires are in a bundle, and how many of those are carrying full current), maximum ambient temperature, insulation rating (or other limits on maximum wire temperature), and altitude. There may be other factors.

Voltage drop is simpler, it depends on current and resistance. Resistance, however, is a function of temperature, so you should use something like the wire resistance at the maximum temperature rating of the insulation to calculate the voltage drop. For example, if the wire is 150 degree C rated AWG 10 and 1mohm/ft at 20 degrees C, then it will be about 1.5mohm/ft at maximum rated temperature since copper increases in resistance by about 0.4%/degree C.

You can get the resistance from a wire table and adjust for temperature, or calculate the resistance from the voltage drop. Once you know the maximum resistance and length you can find the resistance per foot and use the table directly to find an available gauge (odd gauges are likely to be less available than even ones). Or calculate the area from the resistivity of copper and the equation R = rho * L/A.

It is possible to come up with a closed-form equation for voltage drop of a copper wire based on AWG and temperature since the AWG is related to copper area by a simple (but nonlinear) equation (which can be found in the above link).

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Adding to Spehro's answer:

Copper's resistivity has positive tempco. This means hot copper has more resistance than cool copper. If you stay below the melting point of PVC insulation, the effect will remain subtle.

Amp ratings of wires is also a subtle subject. It is easy to calculate how many watts a given piece of wire will dissipate into heat. It is very hard to know how much its temperature will rise, as this will depend very much on external conditions.

For example, a 1.5 mm2 copper wire can carry 50 amps or more if it is cooled by the exhaust of a fan. Or if it is in free air and not insulated (thus allowed to reach higher temperatures, which make convection more efficient).

If the same wire is insulated with PVC (low melting point) inside a conduit (enclosed space) trapped inside the fiberglass insulation inside a wall, then its rating will be much less. 10-16 amps max here.

If your wiring is inside conduits, stick to electrical code.

If you attempt to make a large low-voltage (like 12V) high-current installation, you will be limited by ohmic losses and will quickly find out that copper isn't that cheap...

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