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In a personal project, I need to power several 5V appliances which, together, use up to 12A.

I want to use a 60 inch cable between a case to which those appliances are connected, and the transformer with 230V AC in and 5V DC max. 30A out.

When I look at the specification for the cables and the connectors, they are all rated pretty low in terms of maximum amperage allowed. For instance, standard C13/C14 connectors used for personal computers are rated 10A.

I suppose that those ratings are likely to be for 250V AC.

Is it safe to simply scale the ratings down to 5V, so 10A for 250V would mean a power of 2500W, and so a maximum of 500A at 5V?

If not, what calculations should I use to determine the required AWG size of the cables and the type of connectors that I am allowed to use?

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  • \$\begingroup\$ This is a fairly uncommon sort of goal. You absolutely CANNOT trade voltage for current in the ratings of wiring or connectors - though you can and perhaps should redesign for higher voltage and lower current. What exactly are you trying to power here? You may have to source wiring and connectors intended for comparably high current 12v systems in cars, boats, etc. But before you can do that, you need to understand the nature of the loads as their current draw behavior may be more complex than a single number in ways that you must take into account. \$\endgroup\$ – Chris Stratton Jul 12 at 19:58
  • \$\begingroup\$ @ChrisStratton: I was afraid of that. To answer your question, I try to power a bunch of LED strips. Putting a transformer near the LED strips is not convenient, or should I say, impossible; therefore, my solution was to have the transformer in an enclosure, and have another enclosure which would receive the power from the transformer, and distribute it to the LED strips. \$\endgroup\$ – Arseni Mourzenko Jul 12 at 20:08
  • \$\begingroup\$ @ChrisStratton: I see. At the same time, the problem is not that uncommon, is it? I mean, laptop transformers use the exact same approach: the transformer is a separate box, with a rather narrow wire which could be 40 inch long which would transmit 12V. So how what I want to do is different from that (aside 60 inch vs. 40 inch, and a much larger amperage, I suppose)? \$\endgroup\$ – Arseni Mourzenko Jul 12 at 20:15
  • \$\begingroup\$ @ArseniMourzenko 10A*250V is how much your load takes. The wire experiences 10A and a very low voltage drop. The power lost in the wire (in the form of heat) is proportional to the current squared. So if a wire is rated for 10A*250V, it means that it has a certain wire thickness which allows 10A to flow through it, and that it is rated for 250V means that there is enough insulation and isolation so it's guaranteed that no spark will flow across them anywhere. If you want to pass through 500A you'll need to have 50 of those 10A cables in parallel. Or just get a copper bus system (busbar). \$\endgroup\$ – Harry Svensson Jul 12 at 20:38
  • \$\begingroup\$ Pick your wires first. Then pick your connectors. You probably need 10 or 8AWG wire to keep voltage drop acceptable with 30A. You probably didn't realize that yet. This will force you to use something like crimp-on ring terminals for connectors. If you can get by with 10AWG the types of available connectors will be easier to deal with. But 30A will give you a voltage loss of 0.4V after traveling through 6 feet of 10AWG cabling (down and back). \$\endgroup\$ – mkeith Jul 13 at 6:44
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Your best budget solution is to use 2 pairs of Molex 4 pin HDD contacts rated for 10A ea. for your 5V distributed loads.

These are used in PC towers to interconnect 5V/12V peripheral power with the centre pair joined at source and load, as ground return.

Use 1 connector pair for each peripheral and use wiki AWG tables for cable resistance per unit length of each cable. Voltage drop on each wire is I*R/m * m length.[ or ft]

LV contacts are always for current as temp rise increases with \$I^2Rs=Pd\$ for contact Rs. Hot insertion is not recommended to reduce pitting of contact surfaces and rise in Rs.

Use the largest AWG that you can to minimize the voltage drop or according to your load error budget, based on Ohms/ft or /m Vdrop=IR X2 for R length of cable.

If you care to use AWG 14 stranded or welding cable with jumpers to crimp or soldered pins that would be ideal, but if you can get away with xxx mV cable loss, lighter cable is O.K.

  • Use AWG tables Ohm/ft to compute drop and (x) for each wire length x Amps to achieve less drop than your voltage error tolerance. It’s not rocket science.

  • The contacts may be 5 to 30 mOhms max initially then may increase with aging abuse. So again from Ohm’s Law V=IR , each contact has a small,voltage drop.

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  • \$\begingroup\$ Any questions?...? \$\endgroup\$ – Sunnyskyguy EE75 Jul 13 at 4:07
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Crimped electrical terminals that you find at the hardware store are the simplest and most available high-current connectors that requires the cheapest and least amount of tools for most people. These include the ring, spade/forked connectors that get screwed down.

The hardware store also has similarly crimped Quick-Connect connectors that have a male and female end that you can plug into each other wire-to-wire. The female terminal is often just sitting inthe hardware store, however the male end is less common to find sitting in a hardware store, so you may have to order online.

The easiest thing is just to google an ampacity table and look up the AWG for current unless you want to do super-physics calculations and simulations for wire heating. Rule of thumb: if the wire fits in to the connector then the connector can probably support at least as much current as the wire can. There's little point to making a connector that can fit a 12AWG wire but can't carry the current of a 12AWG wire. Usually the limiting factor is the insulation or plastic on the connector or wire since that will melt/burn long before the metal conductor does.

Voltage and current ratings are independent. You can't trade one for the other.

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    \$\begingroup\$ Especially for low voltage DC applications, in addition to ampacity, you need to consider voltage drop also. Dropping 1V in a 240V circuit is no big deal. Dropping 1V in a 5V circuit is a big loss. \$\endgroup\$ – mkeith Jul 13 at 6:37
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Is it safe to simply scale the ratings down to 5V, so 10A for 250V would mean a power of 2500W, and so a maximum of 500A at 5V?

No, one limiting factor in connectors is the contact and conductor resistance. A ballpark number that I usually use is max 0.1Ω which works for most connectors. Many connector datasheets also have numbers for connector resistance.

P=I^2*R

If we have 30A, this means we will dissipate roughly 90W in a connector with 0.1Ω of resistance and 9W in a cable with 0.01Ω's of resistance. Most connectors will fall in that range. That is a lot of heat to dissipate in a small area. The connector can have a temperature rise and actually melt with enough current.

If not, what calculations should I use to determine the required AWG size of the cables and the type of connectors that I am allowed to use?

Always use the ratings on the connectors, they are tested and also tested in a variety of conditions including temperature. Dust\Dirt can also create resistance. So use the ratings found in the datasheet.

If you must go down to low voltages (there is a reason we use high voltages) then parallel connectors.

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