Base theory on reactor, inductor, choke is frying me
The question will be obvious to most of you, but not to me. I saw some people around the internet, a lot actually, say that you can limit AC current without dropping the voltage. I just can't understand that. Then I tried it to see if it works that way, but when I connect the inductor in series with the load, only ~60V reach the load, from a 127V wall socket. I'm using a 4.65mH inductor which at 60Hz, would be 1.753ohms.
The main question is if can I limit current without dropping the voltage that much? I need at least 120V, 60 is too low.
An inductor in the secondary would be expected in an SMPS to filter high frequency components from high frequency switching, but not in an electric network transformer. Apparently, do you intend to use the inductor as a current limiter? This direct approach without closed-loop control for such power is not appropriate.
Yes, I was trying to use as current limiter. It actually works, but with a lot of voltage drop, from 127 to 60VAC. Closed-loop control, you mean a feedback circuit not pure analog? It is not working anyway, so I'm not using it. I just wanted to have a transformer with a core where I could stick some magnetic stell in there and make a magnetic shunt, like welder transformers, but that core is already used to the limit. I can't fit a massive stack os plates in there, maybe one at the best.
Or rather, with a lot of heating.
Surelly not an efficient way to achieve the intent to limit the current, perhaps cheapest.
Yes, I meant using an inverter with double conversion, A/C --> DC --> A/C.
Perhaps there is a better way to do that, but it is the one that I would consider.
It has a lot of power loss, but since I'm using to spot weld some stainless steel. It only works for 1 second, sometimes 2 if the sheet is too thick. And usually I don't weld more than 3 spots on a work. I'm welding stainless "belts" that are used as clamps. I think I just don't know what to do haha. Will have to switch to another transformer I think. A less powerfull one or one in which I can insert a magnetic shunt.
What you might look into is a series saturable reactor. If you search " welder power supplies " on Wikipedia it states this as one method of control in welders.
An inductor is a valid way to limit ac current and it can do it theoretically without dissipating watts. Transformers in mains circuits often have consciously chosen (rather than purely parasitic) leakage inductance to limit potential short circuit currents.
However it can’t do it without reducing voltage. In-fact nothing, active or passive, can reduce current without reducing voltage (since loads ‘decide’ how much current to draw from a given voltage).
But a series inductor, like a resistor won’t have a voltage drop until current is flowing. So it won’t reduce open circuit voltage. I suspect that may be what some people are referring too or are confused by.
As a theoretical suggestion you could also try capacitive drop. It produces reactive impedance thus it does not generate heat. I call it theoretical because it is liable to be impractical. It's important to avoid certain LC combinations which might create resonant action. The danger is that high voltage amplitudes could develop across a component.
For the capacitor a reasonable value to try is 10 to 100 uF. It needs to carry several Amperes in both directions. Therefore it cannot be a polarized type.
I tried something similar to the saturable reactor with my homemade transformer. The transformer had 3 sets of windings. The primary had ac voltage applied. On one of the secondary windings I applied DC voltage. I was surprised at how little DC voltage it took to lower the ac output voltage on the secondary. I didn't really understand what was going on but I thought it was interesting nonetheless.
Based on the fast-acting requirements, it may be that what you're looking for is something like a PTC, that is, it changes its resistance in reaction to the "short-circuit" due to the arc of welding, but for such a peak current, I do not know if you will find it so easily, perhaps a parallel arrangement or something else.
What you have described is the basic principle of a magnetic amplifier (mag amp).
Although an American invention, mag amps were perfected by the Germans during WW2, and used as part of the control system in the V2 rockets.
For proper mag amp operation, one must choose the core material properly, such that it has a square hysteresis loop.
Common silicon-steel has a soft saturation characteristic.
Welders like this typically have high-ish open ckt voltages, and then the load R determines the on-load voltage for the current. For a transformer with series inductance the short ckt current is determined by the total L (leakage and series) referred to the pri, divided into the applied volts and then scaled to the sec side. For a short the voltage is defined at zero and the current is set by V/L (and the winding R) as the resistance tends to zero from some high value you cannot have a high voltage and a high current for a lowering value load - to keep a high voltage at the load the current would have to be very high, implying a lot more watts - power draw from the mains... which you simply don't have - the max current is defined by the total series L
I've done quite a few welding projects over the years but never really bothered with what makes them tick. According to Lincoln welders site a 1/16" diameter rod requires 25-45 amps. It doesn't give closed circuit voltage but it will probably range between 15-30 volts from info on other sites. The 120 vac wall receptacle is probably fed by a 15-20 amp breaker. The OP didn't say but I'm guessing his transformer is 1:1 ratio. In you guys opinion would it make more sense for him to use a transformer with a high turns ratio say 4 or 5:1 ratio for example. Even though the voltage on secondary is lower the higher current should give him roughly the same closed circuit voltage correct?
When I say closed circuit I mean with arc initiated and not dead short.