I am designing an off-grid DIY solar power site where the distance between the solar panels and inverter is going to be several dozen meters, so I would prefer to put a thicker wire to minimize losses.

Though it seems pretty obvious that, in terms of losses, six isolated 6 mm²
copper cables should perform identically to one 36 mm² cable (to be precise, the closest size existing on the market is 35 mm² but let's assume 36 mm² for the purpose of this question), I am willing to double-check this with experts. Will DC flow evenly across all the six cables, or are there nuances/pitfalls I am not taking into account?

The reason why I would use 6 x 6 mm² instead of one 35 mm² is simply that the former is 1.5 times cheaper.


Just giving a bit more details as there are suggestions that the design might be flawed and I should put the inverter closer to the panels and run a few dozen meters of AC instead.

The set of panels will produce 92–112V. It is capable of generating up to 2900W (full sun at the right angle) so the current will be up to 32A. This calculator shows that for a 40m-long wire I would need 35mm² to keep losses within 2% (and I would actually hate them to be more than 1%). Yes I could possibly erect the power shed within 5m of the panels but that would not look very nice in terms of landscaping. Also, I would prefer to keep the battery closer to the house so that I could feed some DC appliances without double conversion.

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    The only improvement I'd suggest would be to consider 7 cables, for ease of twisting together (and a bit of heatshrink every foot or so to keep them tidy) – Brian Drummond May 12 at 11:57
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    @BrianDrummond just curious, why 7 would be better for twisting? I feel there is some interesting knowledge behind your sentence. – Vladimir Cravero May 12 at 16:58
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    Six strands fit neatly round one central one forming a 7 strand cable. (12 more would fit round them, and 18 more round them, for 19 and 37 strand cables respectively) – Brian Drummond May 12 at 17:01
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    @mkeith The battery is 48V. The inverter/charge controller will take 60–145V from the panels and output 48V for the battery. – Greendrake May 13 at 0:49
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up vote 16 down vote accepted

It's not just losses. The wire is being heat up because of the I^2R, and the heat flows out through the surface. Six wires will have much bigger surface to cool down, hence you can have more current.

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    Assuming, of course, that you don't cable tie them together in a bunch! – Finbarr May 12 at 11:18
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    Well... Good i didn't do any numbers :) – Gregory Kornblum May 12 at 11:19
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    Any wire material with NTC behaviour could give you bad surprises (current hogging, thermal runaway leading to cascade failure) here :) – rackandboneman May 12 at 22:27
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    Electrical wiring regulations have tables of #conductors vs current, and conduit size, so it is easy to know you are designing something compliant. – Henry Crun May 12 at 22:43
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    @rackandboneman, while what you say is true, the only real candidates for wiring are copper and aluminum which both have positive tempco with regard to resistivity. Basically all metals have positive tempco, including steel, stainless steel, tungsten, brass , etc. – mkeith May 13 at 1:37

One thing to consider is that some electrical codes prescribe minimum diameters for certain wiring patterns - eg, a minimum diameter for the PEN conductor in TNC / TNC-S system - for safety against mechanical trouble (in the example of TNC, a wrecked PEN could have disastrous results if a heavy load with a "grounded" metal enclosure is connected). How this applies or does not apply to off-grid DC circuits is a matter of code, too.

One other (slightly off topic but important) thing to consider with heavy DC circuits: Anything that connects/disconnects (switches, fuses, automats, contactors, connectors) and is rated for xx amps AC at 250V is NOT automatically rated (or suitable or safe!) for xx amps DC at even lower voltages. The reason is AC-only-rated designs relying on the fact that eventual arcs will be interrupted quickly by AC zero crossing.

Several dozen meters @ 36mm2 is crazy, except for 10's kWp setup. I guess it is not the case. It is heavy, expensive and generally shows something is far from optimal. Except for marginal cases, you can both save a lot of money and get more power by keeping the wires between 2.5 and 6mm2, using a higher voltage solar panel stack and a corresponding inverter.

  • What is a 'kWp'? – Transistor May 12 at 20:59
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    This isn't what the OP asked, but it is a fair point to raise. If possible, it would be better to put the inverter/charge controller and batteries near the panels, and make the long cable run at the higher AC voltge. – mkeith May 12 at 21:05
  • @Transistor kWp = kilowatt peak. The maximum power the panels can produce in direct sunlight. – Simon B May 12 at 22:13
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    You'd be better spelling it out for clarity. kW peak or \$ kW_{peak} \$. – Transistor May 12 at 22:16
  • It will be 92–112V with current up to 32A (full sun at the right angle) over up to 40 meters. This calculator shows that I need 35mm² to keep losses within 2% (and I would actually hate them to be more than 1%). As for the distance — I would rather put extra wires than erect the power shed where I don't want it to poke out. – Greendrake May 13 at 0:17

A consideration others have not addressed is that this is an outdoor installation subject to wear and tear from sun and rain. The six smaller wires will suffer a lot more damage that the single heavy conductor.

Solar Panels generate DC. There are some complex and fascinating effects of AC power transmission which are not applicable to your problem. The DC current carrying capacity of wires is dependent on their cross section, so use the cheaper method. Though Christoph's issue is technically correct, in real life it will never be a problem, partly because as a wire heats up its resistance increases. So there is negative feedback which balances the load. If you are concerned, use bare wires kept in contact with each other (twisted). It's more important to determine the max current (at max solar flux) and size appropriately.

For very high current DC application consider arc welding wire. I think 0000 size should work just fine. Notice the construction; hundreds or thousands of fine stranded wires.

That is too optimistic. If R of the six wires is slightly different (and it will be), the current will "prefer" to flow throught the wire with the lowest R and heat it up (more than the others). Eventually, this wire might burn up. I strongly suggest a single wire.

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    How different could it be? Current will flow on all of them in parallel, according to their R. – Gregory Kornblum May 12 at 16:16
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    And a wire with a lower R will be wider, hence it will have more current carrying capacity. – Wouter van Ooijen May 12 at 16:41
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    And the positive temperature coefficient of resistance will cause the hot wire's resistance to increase and the current will tend to balance itself out. – Transistor May 12 at 16:46
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    It may be true for parallel semiconductors like diodes or transistors, but it is not true for copper wires, their resistance increases with increasing temperature. – Uwe May 12 at 19:23
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    Copper wire conductivity is reasonably well controlled during manufacturing since the primary purpose of copper wire is to conduct current. And copper wire has a fairly high positive temperature coefficient with regards to resistivity. So there should be no problem as far as that goes. The only problem that could arise is if one or a few strands are somehow broken and become non-conductive. – mkeith May 12 at 20:58

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