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In DC welding, there are two configurations - direct current electrode positive (DCEP) and direct current electrode negative (DCEN). When the electrode is negative, the current flows from electrode to the workpiece. When the electrode is positive, current flows from workpiece to electrode.

It is said that in DCEN, 1/3 of the arc heat is produced in the electrode, and 2/3 at the workpiece.

Why is there such a difference? Why does the current direction matter here? I would intuitively assume that current is current and the direction it flows through the arc does not matter.

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    \$\begingroup\$ In a lot of circuit analysis interchange electron current and conventional (hole) current but when you're talking about arcs and things like vacuum tubes, there is nothing abstract flowing in the opposite direction since it's empty space. In these cases, electrons are being "fired" from one surface to another. I guess if you use a cannon analogy it would make more sense that the positive electrode in the arc heats up more. \$\endgroup\$
    – DKNguyen
    Sep 5, 2022 at 20:52
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    \$\begingroup\$ This might be a good, interesting topic to post in physics.stackexchange.com \$\endgroup\$ Sep 5, 2022 at 20:57
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    \$\begingroup\$ Arcs are plasma not just electrons. That gives you positive ions travelling in one direction, negative ions and possibly free electrons in the other. Often (MIG, TIG) an inert gas (the IG) is used : the choice of gas will affect the polarity and mass of the ions and the energy they carry. \$\endgroup\$
    – user16324
    Sep 5, 2022 at 21:47
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    \$\begingroup\$ The reason is two-fold: (1) the positive workpiece is heating due to the electron thermal energy in the plasma it is absorbing, while this same energy actually cools the negative electrode; and (2) the positive workpiece's thermionic work function also contributes heat to the positive workpiece, while again cooling the negative electrode. That said, both are heated by their respective fall-space voltages times the arc current. But note that the fall-space voltages are different for anode and cathode. Your question is good (+1) as these are still active topics in research literature. \$\endgroup\$
    – jonk
    Sep 6, 2022 at 0:24
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    \$\begingroup\$ If you want to study the topic, I can refer you to a number of key papers on the topic. Some of these go back to 1982 (so far as I'm aware right now.) But more recently, even in 2020 and probably still later, there are new papers covering still more about all this -- especially because of applying SMAW in underwater situations where everything matters. (I do SMAW, MIG and TIG welding, myself, by the way.) \$\endgroup\$
    – jonk
    Sep 6, 2022 at 0:27

1 Answer 1

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The motion of the molten metal droplets is different with polarity.

If the wire is negative, with the target is positive, then when the droplet separates from the wire, it will have an excess of mobile electrons. Immediately the electrons will move toward the positive target but still on the droplet. This forms a dipole. The negative end toward the positive target, and the positive end toward the negative wire, so there is an attractive force to both the target and the wire. The target wins because the droplet has a net negative charge.

If the wire is positive and the target is negative, the mobile electrons are almost completely removed from the wire. When the droplet separates from the wire it contains no (few?) mobile electrons so the dipole cannot form. The droplet is net positive for both the wire and the target, so it is pushed by the wire and pulled by the target. This polarity has more force propelling the droplet.

Which is better depends on the application. I suspect that for overhead welding the positive wire would better counteract gravity.

There are sevearal welding techniques. This answer may not apply to all. (Short circuit welding for example).

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  • \$\begingroup\$ The number of mobile electrons in the droplet s never "few" by any measure, let alone "no mobile electrons" -1 \$\endgroup\$ Nov 5, 2022 at 12:49

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