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PSU Using copper rod for 24 pin ATX power

Discussion in 'Hardware' started by Goatee, 14 Aug 2018.

  1. The_Crapman

    The_Crapman Don't phone it's just for fun.

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    [​IMG]

    I once saw this video on how to straighten small bore copper pipe. Same should apply to the wire.
     
    Last edited: 18 Aug 2018
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  2. Goatee

    Goatee Well-Known Member

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    Its ALIVE!

    [​IMG]

    A new jig with really straight wire (thanks for the tips guys)
    [​IMG]
    Bend up to 90 degrees.....
    [​IMG]
    attach connectors and solder. Allowing a bit of extra time to make sure the solder breaks the enamel coating down. A little puff of smoke indicates a winner....
    [​IMG]
    Then a test with a multi meter to make sure a decent connection is present.
    [​IMG]

    23 x the above (with two resolders due to lack of enamel melt) helped create:
    [​IMG]
     
    Last edited: 18 Aug 2018
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  3. The_Crapman

    The_Crapman Don't phone it's just for fun.

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  4. Big Elf

    Big Elf Oh no! Not another f----ing elf!

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    I still think the one I saw used brass rods around 2mm in diameter but would brass be conductive enough?
     
  5. cobalt6700

    cobalt6700 Active Member

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    Nice job there mate :thumb:
     
  6. Dr. Coin

    Dr. Coin Well-Known Member

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    Spinning some 220 grit sand paper around the end of the wire is a better way to remove the enamel. That way you control how much is removed. Simple burning it off you don't know if the heat has damaged the enamel else where alone the piece of wire potential causing a short.

    It dose look very nice.

    Lets check. Typical PSU wiring is #18 AWG copper, so I'll use that as the base to compare to ⌀2mm brass.
    resistivity of copper ρ=17.2 x 10^-9 Ωm
    #18 AWG A= 0.823 mm² = 0.823x10^-6m²
    R=ρ*L/A R=17.2 X10^-9/0.823x10^-6*L=20 x10^-3 Ω/m

    resistivity of brass ρ = ~60 to 90 x 10^-9 Ωm depending on grade (according to random website, wish I'd noted the source)
    A=0.25*π*D² = 0.25*π*(0.002m)²=3.14x10^-6m²
    R=ρ*L/A R=60 X10^-9/3.14x10^-6*L=19 x10^-3 Ω/m less than the resistance of #18 AWG copper
    R=ρ*L/A R=90 X10^-9/3.14x10^-6*L=29 x10^-3 Ω/m

    From above assuming all my maths are correct then some ⌀2mm brass rods are an acceptable replacement for #18 AWG copper wire as the linear resistance is less than copper when the resistivity is ρ = 60 nΩm. Other grads of brass with higher resistivity are probably acceptable replacements despite having a higher linear resistance. The deciding factor is heat which is determined by current. Molex JR fittings are rated to Imax=9A. This is 56% of the rated ampacity of #18 AWG (16 A, for my jurisdiction). Now I get a little hand wavy as determining the rated capacity of a cable involve lots of variables that I don't know. When the grade of brass has an resistivity of ρ = 90 nΩm, the linear resistance is 145% of #18 copper wire. The max current draw will be 56% less than #18 AWG copper wire maximum rated value. As I said, hand wavy, it probably will work with out causing any thermal damage to the brass or the connectors but I make no grantees.
     
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