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I have 40 loads connected to the same power supply. If every load draws 0.5 A in current and all of them 20 A.

According to this I need cable dimension of 2.5 mm² for the main cable and then 1.5 mm² cable for every single load.

Is this correct or do I need 2.5 mm² for everything everywhere?

it is for 12 volt DC from a power supply like this The cable wont be more then maybe three decimeters.

UPDATE: so according to this table regarding 12 volt and I need a 10 awg cable for the main from the power supply (20 A, 7ft long), distributed to 40 pcs of 16 awg cables (0.5 A, 1ft long) with a 1A fuse on each 0.5 A path right? Not sure if I need the fuse though since I think driver a4988 will be dead long before the fuse is opened?

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    \$\begingroup\$ The wire used must be able to withstand the maximum fault current the power supply can generate without getting too hot. Otherwise you need to add protection like a fuse. You also need to consider voltage drop. \$\endgroup\$
    – Kartman
    Commented Feb 28, 2022 at 12:30
  • \$\begingroup\$ 'Cable dimension' is two words, not all one. Have corrected the question. \$\endgroup\$
    – TonyM
    Commented Feb 28, 2022 at 12:33
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    \$\begingroup\$ @acroscene - What's your application? And I'm not sure I would use that reference you referred to. \$\endgroup\$
    – SteveSh
    Commented Feb 28, 2022 at 12:54
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    \$\begingroup\$ cable gauges only make sense under a condition – for example "my 5m long cables must not drop more than 0.75 V" or "my 20cm cabling must not produce more than 10 W in heat". The more you're relaxed about temperature, voltage drop, the thinner your cables could be. Realistically, though, you don't got that many cheap cables to choose from, so this might limit your options. Also, note that things can become more complicated with AC vs DC, and with highly dynamic loads with their own decoupling caps. I think you'd be best off describing what this all is from a higher-up perspective! \$\endgroup\$ Commented Feb 28, 2022 at 14:39
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    \$\begingroup\$ First, when talking about cable sizes, be careful about the unit i.e. if it's millimetres squared or millimetres. For example, according to the reference you linked, for 20A the cable should be 2.5mm², not 2.5mm. Second, those tables are based on some assumptions such as temperature rise, length, voltage drop, load type etc. For example, for applications with 220V, a voltage drop of up to 3V may not be that important. But for a 12V application, 3V voltage drop is huge. Give more detail about your application e.g. voltage, load type (resistive or reactive), load switching states, temperature... \$\endgroup\$ Commented Feb 28, 2022 at 15:55

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It is very common to reduce the size of the current carrying conductor, whether it be cables or printed circuit board traces, as you progress further from the power source to the loads.

Residential power distribution is a good example of this. Coming into your home you may have a 200 amp service that goes into a breaker panel. This panel is protected by a circuit breaker with a rating of 200 amps or less. That panel then branches off to multiple #14 , #12, or other size cables, protected by 15 A or 20 A breakers (to protect the wiring). At the individual outlets, a device that's plugged in (like a TV set) may have it's own fuse to protect that appliance. That fuse's size (1 A, 2 A, etc) may be much lower than that of the supplied branch circuit.

Here's another example. We have a system with a 200 amp power supply that feeds multiple (several hundred) 1 A loads. The individual loads are designed in such a way that they disconnect themselves from the power in the case that a short develops within them with a fuse-like element. This is to protect the system wiring and allow the system to continue to operate with one or more of these loads removed. There are no discrete fuses as such in this power path, though the power supplies are design with overload protection. One of the trickier parts of this design was ensuring that you had enough current at the faulted load to open the protection mechanism, but not a high enough current that the interconnections would be damaged.

Edit 1 - Added Sketch Below

enter image description here

Without knowing more about your system, it's hard to say definitively what you need. But as a first cut, I would start off by sizing your conductors for a 10 deg C temperature rise or, alternately, the max voltage drop you can tolerate. This means much smaller conductors for the 0.5 A paths than for the 20 A path. Then I would fuse each 0.5 A path with a fuse or breaker with the appropriate time*current trip characteristic. This may turn out to be a 1 A, 2 A, or 5 A fuse or breaker.

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  • \$\begingroup\$ thanks. ok. so according to this table regarding 12 volt and I need a 10 awg cable for the main from the power supply (20 A, 7ft long), distributed to 40 pcs of 16 awg cables (0.5 A, 1ft long) with a 1A fuse on each 0.5 A path right? Not sure if I need the fuse though since I think driver a4988 will be dead long before the fuse is opened? \$\endgroup\$
    – acroscene
    Commented Mar 4, 2022 at 13:52
  • \$\begingroup\$ 16 gauge for a 1 ft run carrying 0.5 A seems like overkill. What's your thinking there? \$\endgroup\$
    – SteveSh
    Commented Mar 4, 2022 at 14:03
  • \$\begingroup\$ Yes it is. I just went according to the table. The highest gauge happened to be 16. but maybe around 28 awg would be enough \$\endgroup\$
    – acroscene
    Commented Mar 4, 2022 at 18:44

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