Effect of two welders working on the same piece

Question

I am a beginner in arc welding and am fairly conversant in electrical concepts. I am told that two people working on the same piece do not affect each other. But my electrical background is telling me different, I just don't know why.

So I drew a simple diagram of two welders.

The diagram shows two welders using DCEP. There are some 12 different combinations of configurations, and therein lies my confusion. (Hopefully my combinatorial math is good?)

The welders could be using the same/ different voltage, the same/ different amperage, or the same/ different electrode configuration (DCEP, DCEN, or AC). Since this is electrical, not welding group, I won't assume knowledge of electrode configuration. DCEP is DC current, and the electrode (gray in diagram) is POSITIVE. DCEN is when the electrode is NEGATIVE. And AC is when the electrode is using AC current. The machines (orange) can have selectors for voltage, amps, and electrode configuration.

Assume both are configured for 10VDC, 10A. When they start welding, I see the power sources connected in parallel, 10VDC, so, total voltage seen by both is 10V. Amps, on the other hand, are added, so, 20A total (split across each welder, 10A? Does it work like that?)

Now assume Left adjusts to 20VDC, and 20A. The amps climbs to 30 across... each? What's the total voltage across both?

Now assume Left reverts back to 10VDC/10A, but switches to DCEN. Now I see the voltages connected serially, and so the voltages are added (10+10=20VDC), and amps are added I think 1/x = 1/10 + 1/10, or 10A apiece?

Assume Left changes again to 20VDC/20A, and DCEN. serial voltages added for 30VDC split between them. Amps are distributed similarly, 1/20 + 1/10 = 3/20, or about 6A apiece?

Finally, assume Left switches over to AC. How are these calculations made?

Perhaps there is circuitry within the welding machine which changes the nature of my electrical questions into one of the design of a welding machine, but bottom line, I can't but help seeing impact of two welders working on the same piece at the same time.

Am I missing something?

Answer 1

There's really no interaction between the two welders at all (assuming that the outputs of the welding machines are galvanically isolated). No current flows in the part of the workpiece between the two welders, so the workpiece behaves the same way that two completely separate pieces of metal would behave.

Assume both are configured for 10VDC, 10A. When they start welding, I see the power sources connected in parallel, 10VDC, so, total voltage seen by both is 10V. Amps, on the other hand, are added, so, 20A total (split across each welder, 10A? Does it work like that?)

That's preeeeeeetty much correct.

If both machines are set to 10 V, 10 A, then:

• The arc on the left will have a voltage of 10 V and will conduct 10 A.
• The part of the workpiece on the left will conduct 10 A.
• The arc on the right will have a voltage of 10 V and will conduct 10 A.
• The part of the workpiece on the right will conduct 10 A.

(Actually, either the voltage or the current will probably drop below the set value, but I don't know which or how much. Throughout the rest of this answer, I'm going to make the (false) simplifying assumption that each arc behaves like a 1 ohm resistor.)

You can add up these two currents and conclude that there's a total of 20 A going from the two electrodes through the workpiece into the two return connections. However, this number isn't particularly meaningful, since the two 10 A currents are flowing in completely separate parts of the workpiece.

Now assume Left adjusts to 20VDC, and 20A. The amps climbs to 30 across... each? What's the total voltage across both?

If the left machine is set to 20 V, 20 A, and the right machine is set to 10 V, 10 A, then:

• The arc on the left will have a voltage of 20 V and will conduct 20 A.
• The part of the workpiece on the left will conduct 20 A.
• The arc on the right will have a voltage of 10 V and will conduct 10 A.
• The part of the workpiece on the right will conduct 10 A.

It's as simple as that. (Again, I'm making the almost-certainly-false simplifying assumption that each arc acts like a 1 ohm resistor.)

The "total voltage across both" is a physically meaningless concept. It is possible to type "20 + 10" into a calculator and press the = key, and the calculator will give you a number, but that number will be meaningless.

(Voltage is a little bit like height. If there were 10 airplanes all flying 35,000 feet above the ground, would you say that they have a total height of 350,000 feet? Probably not, because "total height" isn't really a meaningful concept here.)

Now assume Left reverts back to 10VDC/10A, but switches to DCEN. Now I see the voltages connected serially, and so the voltages are added (10+10=20VDC), and amps are added I think 1/x = 1/10 + 1/10, or 10A apiece?

I'm not sure where you're getting that 1/x formula from. What will happen is:

• The arc on the left will have a voltage of 10 V (in the opposite direction) and will conduct 10 A (in the opposite direction).
• The part of the workpiece on the left will conduct 10 A (in the opposite direction as the right).
• The arc on the right will have a voltage of 10 V and will conduct 10 A.
• The part of the workpiece on the right will conduct 10 A.

At this point, you could take a voltmeter and connect the red lead to the left electrode and the black lead to the right electrode. There's no practical reason to do that, and the result doesn't tell you anything useful, but in any case, the voltmeter is going to read -20 V. The reason for this is that the left electrode is 10 V below the workpiece, and the workpiece is 10 V below the right electrode, so the left electrode is 20 V below the right electrode.

You could also add the -10 A on the left to the 10 A on the right to get 0 A, but this quantity is totally meaningless, since the reality is that the workpiece has 10 A going through it in one place and another 10 A going through it in another place.

Assume Left changes again to 20VDC/20A, and DCEN. serial voltages added for 30VDC split between them. Amps are distributed similarly, 1/20 + 1/10 = 3/20, or about 6A apiece?

Nope. The situation is analogous:

• The arc on the left will have a voltage of 20 V (in the opposite direction as the right) and will conduct 20 A (in the opposite direction as the right).
• The part of the workpiece on the left will conduct 20 A (in the opposite direction as the right).
• The arc on the right will have a voltage of 10 V and will conduct 10 A.
• The part of the workpiece on the right will conduct 10 A.

Finally, assume Left switches over to AC. How are these calculations made?

Again, it's analogous:

• The arc on the left will have a voltage of 20 V AC and will conduct 20 A AC.
• The part of the workpiece on the left will conduct 20 A AC.
• The arc on the right will have a voltage of 10 V DC and will conduct 10 A DC.
• The part of the workpiece on the right will conduct 10 A DC.

Answer 2

Assuming the output circuit is similar on your welders to my Thermadyne AC/DC TIG welder it should not be a problem, assuming the workpiece is grounded.

The output circuit is floating and current is sensed directly through the output terminal. It is recommended that the workpiece be separately grounded, meaning that the welder has no problem with some (small) random potential on the workpiece.

(I am not going to reproduce the output interconnection schematic here because it is difficult to link to a source to credit the document).

If the workpiece is not properly grounded then (especially) the HF start could conceivably cause problems.

Answer 3

Kartman has it right. Charge flows in a closed path. The power sources are not in parallel even when welding. They do have a common negative.

The voltages do not combine.