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  • Technical Question

    Hello,

    I would like an explanation as to why voltage and amperage are inversely proportional in arc welding. But in electricity, Ohm's law states that voltage and amperage are proportional. So I am confused as to why for example when voltage decreases, current increases and vice versa.

    Thanks!

  • #2
    So Ohms law states that voltage and ohms are directly proportional, where as with arc welding (assuming GMAW or FCAW) the relationship is inversely proportional and you're asking why this is?

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    • #3
      https://app.aws.org/forum/topic_show.pl?tid=5193

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      • #4
        You're not comparing the same things.

        In a typical circuit, voltage drives amperage through a resistance. The more voltage you have for the fixed resistance, the more amperage you can push through it.

        In a transformer, you are taking the available AC voltage and transforming it to another AC voltage. The total power (voltage x amperage) remains constant. So if you step down the voltage from 240 V to 48 V, you increase the available amperage that can flow on the secondary side by the same factor of 5, so your welder that's plugged into a 40 A circuit breaker can now weld with 200 A of current.

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        • #5
          Seems the spam filter is working overtime lately, let's try again...


          Are you asking about the voltage and amperage on the primary side or the secondary side of the welder?

          On the primary/input side, it's because the welder needs a fixed amount of power, and since power is voltage times current, when you raise the input voltage, you need proportionally less current for the same power.

          On the secondary/output side, it's because an arc isn't a resistor, and ohm's law only applies to resistors.

          An arc will pretty much let as much current as you give it flow. To control a typical stick welding arc, the welder acts as a constant current source, limiting the current through the arc. The voltage of the welder varies as needed to maintain a constant output current. For an ideal constant current source, the output current would stay exactly the same regardless of the voltage across the arc, but for a real-world welder, it changes a bit - the current will be allowed to drop a bit at higher arc voltages, and of course there's a voltage limit after which no current will flow at all. Some of the inverter units get close to a true constant current source, but for stability there is still some slope to the output curve even on regions where the ideal curve would be flat.

          For more details, you'll need to ask a more detailed question, so I know what you want answered...

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          • #6
            Originally posted by Bushytails View Post
            Seems the spam filter is working overtime lately, let's try again...
            Bushytails for what ever reason your posts are moderated, and i don't know why...Bob
            Bob Wright

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            • #7
              It happens whenever I edit a post... I fixed a typo and the post vanished. And now there's two of them. Oh well...

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              • #8
                Thanks for the feedback!

                I guess I need to read up on my electricity I'm fairly new to welding and I get confused with all of the electrical terms and how they are related.

                Essentially, I was wondering why in GMAW voltage and amperage are inversely proportional. I thought in any circuit ohm's law applies and I thought this is true for a welding circuit. I did not think about the primary and secondary sides of the welding circuit, but instead thought of just one continuous circuit. A simple example is a light bulb in a simple circuit. Voltage and amperage are proportional, but in GMAW welding circuits they are inversely proportional? Can you clarify what the primary and secondary welding circuits are?

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                • #9
                  Let me see if I can make this even more confusing. Voltage measurements are actually voltage drop measurements. That is the potential difference from point A to point B. Measured across the arc for a given stickout you have a certain resistance and current. If you lengthen the arc the voltage drop goes up because the resistance went up but less current flows because of the increased resistance, just like Ohm's law predicts.
                  Don't worry about the primary and secondary circuits. The primary is the input to the machine supplied by your utility. The secondary is the output of the transformer in the welding machine to lower the voltage output. Inverters are different more complex power supplies.

                  ---Meltedmetal
                  ---Meltedmetal

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                  • #10
                    Spend some time manipulating Ohm's Law algebraically:

                    E=IR
                    I=E/R
                    R=E/I
                    etc

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                    • #11
                      As someone who regularly uses ohm's law... I have absolutely no clue what that triangle is supposed to tell you to do or how it's supposed to help you in any way. lol

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                      • #12
                        That triangle represents Tesla’s victory over Edison.

                        Unless someone has a better explanation, because I just made that up.

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                        • #13
                          On the triangle, you put your finger over the value you want. The remaining two letters tell you how to get it.

                          So if you want to know resistance, you cover over the R. What's left is V / I. If you want V, what's left is I * R.

                          I've never used it. Knowing V=I*R is a lot easier that remembering the triangle. It's really simple algebra to get the other two.

                          Putting "V" in the triangle and then using "E" in the written expressions without explanation probably didn't help. Some schools use "E" instead of "V" now. Voltage = Electromotive Force. Of course, you all know "I" is current.

                          Current can't be "c" because that's the speed of light, which can't be "L" because that's inductance, which can't be "I" because that's current.
                          Last edited by MAC702; 07-06-2020, 08:29 AM.

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                          • #14
                            Trivia--I stands for INTENSITY of electron flow, in coulombs per second. Adding Mr. Ampere's name made it a lot easier to say.

                            Sorry, too old to remember how many electrons in a coulomb without looking it up. Probably lost that sometime in the 60s.
                            Last edited by Aeronca41; 07-06-2020, 08:02 PM.

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                            • #15
                              Originally posted by Aeronca41 View Post
                              . Probably lost that sometime in the 60s.


                              Richard
                              West coast of Florida

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