Several people have already mentioned there is no direct relationship between wire feed and amperage other than the fact that they are proportional. That is, as the wire feed increases, so does the amperage for given filler metal alloy and diameter. Voltage also figures in, but because the arc is not constant state, it varies as the molten droplet transfers through the arc; it too is in a steady state of flux.
Ohm's law can be used to get a better understanding of direct current and the relationship between amperage, voltage, and resistance. Ohm's law is Amperage = voltage / resistance. Assuming for the discussion that voltage is constant for a given short time period, the relationship shows that amperage is reduced as resistance increases.
Resistance can be thought of the internal friction that inhibits the flow of electrons (current) through the electrical circuitry of the welding machine, but more importantly for this discussion, it is the distance between the end of the contact tip and the end of the electrode (starting point of the arc). The resistance through the arc is very low since the arc in essence is plasma which by nature is highly conductive, i.e., low resistance. As can be seen, as the length of the electrode beyond the contact tip is increased, the resistance increases and the amperage decreases assuming the voltage is unchanged (constant potential - same as - constant voltage). The resistance is proportional to the length of the electrode (electrode extension).
Melt-off rate can be assumed to be directly proportional to the amperage, thus if the amperage is doubled, the melt-off rate doubles in direct proportion. Thus if the resistance of the system is reduced by a factor of 2, the amperage doubles and the melt-off rate doubles. Unfortunately, we cannot neglect to consider the internal resistance of the machine, but assuming it is constant, we can ignore it for the discussion. After all, we are interested in the concept without all the details that complicate the theory.
Since most welders are AC, there are other considerations such as inductance and capacitance. Think of inductance as the equivalent of a shock absorber on the front-end of an automobile. Both inductance and the shock absorber dampen the rate of change. In the case of the automobile the shock absorber dampens the frequency and amplitude of the bounce of the car's ride. In a similar manner, inductance dampens the rate at which the welding machine is going to respond to changes in the circuit. Those changes include the resistance due to chances in the electrode extension as the welder moves the torch a little closer to the work piece or further away as the torch is moved along the joint and the welder breaths. Also changing is the resistance due to the transfer of metal droplets through the arc. With each change, there is a change in the resistance and a corresponding change in the welding current. If the rate of change is instantaneous, the welding machine becomes unstable and the arc becomes erratic. Hence the need for inductance to help stabilize the welding arc.
I said to assume the arc voltage is a constant, but that is an over simplification. Arc voltage will change. That change also influences the welding current as well as the melt-off rate. It all goes back to Ohm's law.
Melt-off rate is tied to current density. A certain current density is needed to heat the electrode to the melting point and to facilitate metal transfer. Since the electrode is round, the area is a square function, A = Pi x R x R where Pi is approximately 3.1416 and R is 1/2 the diameter of the electrode. The relationship between electrode diameter and the area is not a linear relationship, so doubling the diameter doesn't result in a doubling of the area. The following table shows the increase in cross sectional area as the electrode diameter increases.
0.035 inch dia.
0.045 inch dia. = 1.65 x increase in area
0.052 inch dia. = 2.29 x increase in area
0.0625 inch dia. = 3.19 x increase in area
0.125 inch dia.= 12.76 x increase in area
In order to maintain the same current density, thus the same melt-off rate, the increase in current would look something like the following:
If using 0.035 inch diameter electrode at 200 amps (spray mode w/ carbon steel), you would require:
0.045 inch dia. 330 amps
0.052 inch dia. 460 amps
0.062 inch dia. 640 amps
0.125 inch dia. 2250 amps
So, you can see that if the electrode diameter is held constant, voltage is held constant, and the electrode extension is held constant, the melt-off rate will double by doubling the welding current. However, if the electrode diameter is increased, the current has to increase rather dramatically to maintain the same current density, thus melt-off rate.
As most of us know, the welding system is very dynamic, so the electrode extension changes (especially if we are not aware of the importance of electrode extension which is similar to contact tip to work distance for our discussion), the resistance changes, the arc voltage changes, all of which provides us with some interesting happenings at the welding arc. Add in the influence of the inductance (some machines have variable inductance that can be set by the welder, some have dynamic inductance that changes as other variables change, and some machines have fixed inductance), and you really have some interesting affects. Last, but not least, we always have to consider the shielding gas and how it affects arc voltage, but as long as the gas mix is not changed, we can ignore it for the purpose of understanding the relationship of Ohm's Law when working with direct current.
I hope this helps.
Best regards - Al
Forum
