Arc welder and power factor caps - Page 1
Forum section: electronics / projects

author: Circlotron
author_rank: Super Contributor
timestamp: January 26, 2017, 12:06:43 pm
content: I have a small transformer type arc welder, and these things can draw quite a bit of current from the mains at times. Does anyone know what the approximate input power factor might be? My plan is to add some power factor caps across the input to try and reduce the mains current draw and so lessen the chance of blowing a fuse. Also it should help things if I am running the welder off a small portable generator. I think I read somewhere the the power factor can go as low as 0.4 so that would mean it would be drawing 2-1/2 times the amps actually needed for the wattage consumed.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 26, 2017, 12:38:20 pm
content: Somewhere between 0 and 1...Think I'd be surprised if it was more than ~0.7 under normal load. Should be less when fully shorted (< 0.5?), and even less than that when unloaded (< 0.1?). The equivalent circuit (as seen by the mains) is a parallel inductor (magnetizing current) and an inductor (leakage) in series with the load resistance (which is winding DCR, cables, and whatever equivalent resistance the arc has).Not sure if that's useful enough to spec a cap. Mind that the cap needs to draw on the order of the same load current as the welder (i.e., 5-10A?), so it's quite large, and will draw stupendous inrush (maybe enough to trip the breaker right away, anyway?) unless you have some means of starting it.Tim

author: Zero999
author_rank: Super Contributor
timestamp: January 26, 2017, 06:06:57 pm
content: Is this an mains frequency AC, high frequency AC or DC arc welder?If it's high frequency AC or DC, then it will be very non-linear and slightly capacitive, so a parallel capacitor will only make it worse.If it's mains frequency AC, then it will still be non-linear but not as bad as the above and a capacitor may help. You need to know the power factor and apparent or real power to calculate the correct value.

author: Circlotron
author_rank: Super Contributor
timestamp: January 26, 2017, 09:12:11 pm
content: It's a 240v 50Hz mains frequency AC thing.I expect I am going to need something like 100-150uF.Maybe I should just use a moving coil ammeter and switch in a number of different values of cap and look for the lowest current while doing a typical weld. Or the lowest short circuit current while the rod is stuck. I'm good at doing that.

author: Benta
author_rank: Super Contributor
timestamp: January 26, 2017, 10:56:54 pm
content: A welding transformer is just that: a transformer.The only inductive part of the current draw is the primary magnetizing current. Everything else goes to the load (parasitic effects ignored here).So, you can neutralize the magnetizing current with a cap. But the magnetizing current is negligible compared to the load.Frankly, I wouldn't bother.

author: Circlotron
author_rank: Super Contributor
timestamp: January 27, 2017, 03:20:46 am
content: The primary and secondary are widely separated on the core so there is a high leakage reactance. Wouldn't that add to the lagging power factor? What I might do is power it up with say 30 or 40vac and measure the primary current with the secondary shorted, and how many degrees it lags, and also with several values of lowish resistance across the secondary. That might give me some kind of starting point.

author: calexanian
author_rank: Super Contributor
timestamp: January 27, 2017, 04:10:00 am
content: too many factors to make a suggestion. Especially if its a variable transformer time. The power factor will be mostly determined by the saturation characteristics of the transformer and that will be effected by voltage under load, and the size of the load (The stick and weld pool)

author: james_s
author_rank: Super Contributor
timestamp: January 27, 2017, 04:54:39 am
content: Typically an arc welder has a movable core, with a screw attached to the current adjust knob that slides portion of the core in and out to adjust the coupling.I haven't measured an arc welder but I have worked with an electromechanical 6.6A series streetlighting regulator. What we found with that is that the primary current is constant, which makes sense since the secondary current is constant as that's the point of a constant current regulator. What does change with the load factor, ie the number of series lamps on the output is the power factor of the input. Near full load it's close to unity, running a single lamp it's quite low.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 27, 2017, 07:58:26 am
content: Adjustment doesn't matter, to the basic problem at hand.Adjustment does matter to optimization, such as reducing average operating mains current.Consider:Mains current is highest at full output, shorted load. The power factor under that condition is largely inductive (due to the leakage inductance). Therefore, compensating that to unity power factor will have the greatest impact on welder capacity.Note that I'm defining capacity as the magnitude of amps drawn by the welder. You can't draw any more amps than the breaker will allow you to. So if you draw, say, -20A of reactive power through the capacitor, you can draw +40A of reactive (i.e., inductive) power through the welder, which comes at a much higher power level (and more fully shorted load) than without the capacitor. The worst case is still just as bad (now you're drawing full rated amps at idle!), but now you can actually sustain full load.Tim

author: Circlotron
author_rank: Super Contributor
timestamp: January 27, 2017, 08:42:06 am
content: It's not an adjustable core type transformer, it has two primary tappings for 2.5mm and 3.5mm rods. Very basic welder but it does what I need it to do. I'll work on the tapping for 3.5mm rods. Having said that, I just realised that if I put the capacitors from neutral to the 2.5mm tapping then when the switch is set to 3.5mm because of autotransformer action the caps will see a higher than mains voltage so I could more fully utilise them. Provided I don't take them over voltage.

author: Siwastaja
author_rank: Super Contributor
timestamp: January 27, 2017, 11:41:11 am
content: Quote from: Benta on January 26, 2017, 10:56:54 pmNot true at all. Just a transformer wouldn't work very well for welding; it would be hard to initiate the arc and keep it going.These simple mains transformer welding machines either need a big inductor in series with the transformer, or, as often done, they implement it as a leakage inductance in the transformer; this makes the power factor poor.

author: Benta
author_rank: Super Contributor
timestamp: January 27, 2017, 04:34:32 pm
content: QuoteOK, let me qualify my statement: it's a transformer with piss-poor primary-to-secondary coupling, which gives it a "soft" voltage/current curve.That still has nothing to do with power factor.

author: oldway
author_rank: Super Contributor
timestamp: January 27, 2017, 07:12:41 pm
content: Before asserting anything about the power factor of a welding transformer, we must first distinguish the different possible operations:1) no load: the current is the magnetizating current.It is in theory purely inductive (cos phi = 0) but there are losses by hysteresis and eddy currents ... the cos phi is therefore not 0 but generally close to 0.2.2) short circuit: the value of the short circuit current depends on the short circuit impedance of the transformer. In an ordinary transformer, this short circuit impedance is about 7%, ie the short circuit current is 14 x the nominal current.In a welding transformer, this short circuit impedance is increased by leaving a distance between the primary and secondary windings (short circuit impedance of approximately 20% and short circuit current = 5 times the nominal secondary current) and even by adding a magnetic shunt between primary and secondary. (short circuit impedance up to 50%)In short circuit, the impedance is mostly inductive with a low resistance coming from the resistances of primaire and secondaire wires.Cos phi is then also more or less 0.2.3) with load: This is a more difficult situation because an arc is a non linear load.Therefore, current is also not sinus and it became more difficult to determine the power factor.Something like 0.4 or 0.5 would be a correct value for power factor during welding.

author: oldway
author_rank: Super Contributor
timestamp: January 27, 2017, 07:31:10 pm
content: To answer the question of Circlotron, first whe have to know what's the related power of his welder in VA or KVA.Whe can then calculate what's the max. current of primary : Iprim = Pva / Vmains (240V in Australia)To compensate partially the low inductive power factor, whe can use a capacitive current of 40% of the primary current.NB: as power factor vary, it is not possible to compensate the power factor for every situation...Icap = 40% Iprim = Vmains/ZcZc = 1/2 x pi x frequency x COr C= 0.4 x Iprim / (2 x Vmains x pi x frequency)

author: Benta
author_rank: Super Contributor
timestamp: January 27, 2017, 07:35:51 pm
content: Quote from: oldway on January 27, 2017, 07:12:41 pmSpot on, oldway!When talking about PF, it consists of two components:- Reactive power. That's the part you can improve with a capacitor/inductor (transformer magnetizing current), but it's mostly futile, the ROI is too small.- Current draw distortion. No capacitor in the world is going to help here, but it's the main reason for poor PF from welders.

author: oldway
author_rank: Super Contributor
timestamp: January 27, 2017, 08:22:01 pm
content: Quote from: Benta on January 27, 2017, 07:35:51 pmCos phi is for sinus current and PF is for not sinus current.In this case, Fase interval is between fondamental of current and voltage.In case of non sinus current, the capacitor used for power factor improvement act as a filter and help a lot to not inject harmonics in the net feeding....What you wrote is non sense....you don't really know what you are speaking about.....

author: Benta
author_rank: Super Contributor
timestamp: January 27, 2017, 08:47:11 pm
content: QuoteThank you very much, I'll return the compliment."the power factor of an AC electrical power system is defined as the ratio of the real power flowing to the load to the apparent power in the circuit,"Why you suddenly bring cos phi into this is a mystery to me. Obfuscation? I specifically mentioned current draw distortion, a feature of non-linear loads, which we are talking about here.Get off your high horse.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 28, 2017, 03:08:04 am
content: Like I said, all you need to do is provide enough capacitive reactance so that the worst-case current draw is reduced. This will probably be around half the total current draw.Nothing more complicated than this.Tim

author: oldway
author_rank: Super Contributor
timestamp: January 28, 2017, 04:44:54 am
content: Quote from: Benta on January 27, 2017, 08:47:11 pmhttps://support.industry.siemens.com/cs/document/51402748/7km-difference-between-cosine-phi-and-the-power-factor?dti=0&lc=en-WWThe formula S * cos phi = Pis only valid with sinusoidal valuesIn the case of non-sinusoidal currents, the calculation of the power factor correction capacitor is based on the value and phase shift of the fundamental (50 or 60Hz) of the current, not the harmonics.If the aim is to prevent a very distorted current with a high harmonic content from being injected into the power grid, other techniques must be used.For example, the current of an SMPS is impulsive and highly distorted.The current content at the fundamental frequency of 50 or 60 Hz is practically in phase with the voltage (cos phi = 0.93 approximately).It is therefore not necessary to correct this with a capacitor.But the content in harmonics is very high.To correct it, either an active PFC or a passive PFC is used.This passive PFC is not a capacitor but an inductor in series with the AC supply voltage.This inductance reduces the amount of harmonics injected into the power grid and makes the current closer to a sinusoid.In high-power installations, it is also possible to use LC filters tuned to harmonic frequencies H3, H5, H7 ....to reduce the harmonic content of the current.NB: We do not claim to be competent, we prove it

author: SeanB
author_rank: Super Contributor
timestamp: January 28, 2017, 05:20:26 pm
content: Just to add, the capacitor with non linear current flow ill see a lot more heating as it will be passing a very high harmonic current. This is often a lot more than the capacitor designers have considered in design of things like the foils, the leadout tabs and the allowed temperature rise under full load. You will find that they will puke the guts with monotonous regularity, and this can be rather spectacular for the dry impregnated units, the oil filled ones ( most common ones) do rather strongly note that you need a good size gap at the terminal end for the cap to safely blow out to disconnect the internal element on failure. Very messy, and the oil goes everywhere in doing so as well. You might consider a few in series parallel to implement this, as you need to reduce the heating and dissipate it as well. Bonus for most motor caps is they are easy to daisy chain, as there is almost always 2 faston connectors per connector. Just use a cable capable of handling the transformer primary current, the capacitor can have 25A or more flowing through it, so 1.5mm cable at the minimum for short runs and low duty cycle.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 28, 2017, 06:29:59 pm
content: Not really. Two reasons:1. This isn't a strongly nonlinear load, like a rectifier. The harmonics will be a small fraction of the total.2. Capacitors draw current based on the applied voltage (namely, its change). If the mains impedance is low, then the voltage harmonics will also be low -- even if the load distorts quite strongly.To get good harmonic filtering, you need to construct an LC filter. (If you wished to build one -- use a typical LC filter design, setting the characteristic impedance equal to the amount of regulation you need, i.e., for 240V supply and 5% regulation (12V drop at full load), use Zo = 12V/(FLA). It should also be terminated with a R+C at one end or the other, else the frequency response will be all kinds of wrong. Ideally: because the harmonics are fairly tight, and don't move around, a few stages of notch filters should be used, getting better performance than an all-pole filter can.)Tim

author: Benta
author_rank: Super Contributor
timestamp: January 28, 2017, 06:48:30 pm
content: QuoteAt least you're able to misread my posts. Whether it's intentional or due to "competence", I don't know.'Nuff said.

author: oldway
author_rank: Super Contributor
timestamp: January 28, 2017, 06:59:19 pm
content: I really do not believe that for a simple welding transformer. Circlotron is ready to invest to mount sophisticated filters. L / C to avoid injecting harmonics on the power grid.This would be completely absurd because the correct strategy is to act on the source of the harmonics by choosing a welding inverter with active PFC instead of a welding transformer.Moreover, all those who have had the opportunity to compare the MMA welding with a transformer (ac welding) and that with a DC inverter know that there is no comparison possible: welding with an inverter is 100 x better and more convenient.

author: GettinBetter
author_rank: Newbie
timestamp: February 10, 2018, 12:31:44 pm
content: What a brilliant thread. Really enjoyable reading.... thanks guys

author: najrao
author_rank: Regular Contributor
timestamp: February 10, 2018, 03:14:17 pm
content: Circlotron,Please understand that the replies you get are a mixed lot of valuable suggestions by some and utter crap by others. Just ignore the latter.You asked a simple straightforward question. It is a 60Hz copper and iron power transformer, with a primary winding 'tapped'* for a lower current output from the secondary. Current, even into a short circuit, is limited by deliberately enhanced leakage between the two windings. It is the most common way of making cheap and reliable welding power source.* the 'tapping' gives the *higher* current; the primary is overwound to reduce the output.The only significant cause of wasting reactive current draw is the leakage reactance. The arc voltage is essentially in phase with its voltage, all be it distorted from true sine. The open circuit voltage would be at least 50V , to be dropped to 20 or 24 at the arc. This needs a reactive drop of --- not 30 to 26, but 45 to 40, because of the quadrature between them. So, it can be said that the current is largely at low power factor, not much over 0.4. At the lower 'tap', it is even lower.There is NO reason to discredit the time honored pf corrector, i.e., capacitor. The current draw is bound to reduce if this is fitted.The exact value is not critical, but it is common practice to use rather less microfarads than are needed to raise the overall pf to unity. An overall pf of 0.9 to 0.95 would be optimum. PF correction caps are rated directly by their VAr, not microfarads. Self healing MPP caps are needed, with built in discharge resistors. They can be left on, connected to the mains through a fuse, and not switched repeatedly. The leading current is generally not harmful to the mains. Very large caps will need in rush limiting switching, but your case of under 5kVAr should get through with just a switch and fuse/s. Please check if special wiring codes apply.You can expect to reduce the mains current draw by about 30% while welding at full power. The caps will continue to draw almost the same current even when the welder is not feeding an arc, but this is a leading reactive current. Your purpose of not blowing mains fuse/s should be well met.Good luck!

author: ahbushnell
author_rank: Frequent Contributor
timestamp: February 10, 2018, 05:22:38 pm
content: Quote from: Circlotron on January 27, 2017, 03:20:46 amGood idea