# Welding transformer question

Are there any EEs here who remember transformer design? :-)
I have a large old AC welder that far exceeds my current needs (pun
intended). Rated max output is 625 amps, half of that would be lots for my purposes. Rated input current at full load is 170A at 230V. I am concerned that the idle current will be rather high. Can anybody tell me what idle current I should expect?
It has two primary windings which can be connected in series for 460V and in parallel for 230V. Since I don't want the full output, can I reduce the idle current by using just one of the primary windings at 230V? I assume that I would not see a significant difference in primary current under load, for any given output.
What I remember of my electricity/electronics classes suggest this would work, but I'm no EE.
TIA, Rob
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Rob wrote:

Unless it has a separate inductor, welder transformers are designed to have fairly loose coupling so that they behave as constant (relatively) current sources. This makes it difficult (at least for me) to guess at the idle current.
Why are you concerned? I would expect the idle current to be almost entirely reactive and that won't turn your electric meter very fast. :-)

No. Paralleling the two primary wire reduces the resistive loss in the transformer as if it had one winding of thicker wire. It doesn't change the volts/turn.

Probably the transformer would just get hotter. Ted
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Thanks Ted.
Rob
On Sat, 01 Nov 2003 02:44:59 GMT, Ted Edwards

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Good answers Ted! Here are some thoughts.
Is this for welding ? - Is the output current only worry ?
Np/Ns = 170A/625A = .272 current transformer (pure numbers no loss)(there is).
Wp/Ws = 170*230 = 39,100 V-A / P(ower)secondary
Seems to me at this time of night the numbers above are ok - (I mean the ratios match like typed)
Now a yet another word - NYAW ? - The current in the primary is composed of two (mostly) elements. Secondary current (divided/multiplied through windings)and primary loss. There is magnetizing current needed to maintain a field - when the field is lossy more current.
Let us know what you are trying to do. That will help.
Martin
--
Martin Eastburn, Barbara Eastburn
@ home at Lion's Lair with our computer snipped-for-privacy@pacbell.net
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wrote:

Part of the problem is that like most tinkerers I want to understand anything I use. I would like to determine what primary current to expect at welding currents in the 150-200A range. I also want to avoid unplanned smoke and sparks.
When I bought the welder there was no indication of what input voltage it was strapped for. The data plate says it can take 230, 460, or 575V 1ph input. I have 230V 1ph power. A bit of wire tracing indicates that two primary coils were paralleled suggesting that it was connected for 230V.
There are six wires coming from the xfmr to the block where you make the primary power connections. I disconnected all the straps and bimetallic overload devices and took some continuity readings. From this I learned that there is one untapped winding, and another with two taps.
I put 12VAC onto the untapped winding and measured voltages. I determined that this was the configuration when I bought it... sorry for the ASCII art, you will need a fixed width font for it to make sense:
460V------+-||-+ 0 | 0 | 345V------+ | 0 | 0 | 230V--+---+ | L 0 0 | I 0 0 | N 0 0 | E 0 0 | --0V--+---+----+
There are four large capacitors rated at 460V, connected in parallel, which I assume are for power factor correction (indicated as a single cap in the diagrams). If they stay connected to the ends of the tapped winding, they will see a nominal 460V, for all valid input power configurations. The higher voltage inputs would look like this:
(690V)+-||-+ 0 | 0 | 575V------+ | 0 | 0 | 460V------+ | 0 | 0 | 0 | 0 | (230V)+----+ 0 0 0 0 0V------+
That way, even when connected for 575V input, the caps only see the 460V they are rated for.
Have I missed anything?
Rob
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Assuming reasonable transformer efficiency, 20 to 30 amps (real) primary current at 220 with a 150-200 amp output draw. With no PF correction, probably another 30 amps reactive, so you'd need a breaker sized for 60 amps.
But if the PF capacitors are sized correctly, then only a few amps reactive, and a 40 amp breaker would probably carry it. However, there might be a hefty turn on surge while the capacitors initially charge, so a motor rated breaker would be a good idea.
Gary
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The idle current is mostly reactive as Ted said. Some of it is real power due to the losses in the iron core, but most of it is reactive. If you want an idea of what the idle current is, you could apply less voltage and measure the current. And then multiply that by the ratio of the 230 volts to the voltage you applied. Say you applied 115 volts and measured 20 amps. Then the idle current would be about 40 amps at 230 volts. My guess is that it is somewhere around this. But that is a guess.
You might look inside and see if there are any large oil filled capacitors in there. I have an old Lincoln welder that has some caps in it to reduce the idle current. If it does not have power factor correction caps in it, you could add some. Have a look. And if it does not have any lange caps inside, let the group know and I will provied detailed instructions on how to add them. I did this on another welder and it reduced the idle current by a good bit. I forget exactly how much.
Dan

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On 1 Nov 2003 09:47:20 -0800, snipped-for-privacy@krl.org (Dan Caster) wrote:

With 11.7Vac across the untapped primary winding it draws 1.65A (7.1 ohms at 60hz).

There are four. I checked them with an ohmmeter. Only one gave the typical low resistance fading to high resistance as it charged, the others look like an open circuit. The ones that don't work are Sprague Clorinol capacitors (yeah, I know... they are filled with PCB laden oil). The one that works is a GE cap, presumably a replacement. They all appear to be 35 uf.
I tried connecting the caps across the coil. Only the GE cap had any affect, so I assume the internal fuses are blown in the others. This time, with 12.7Vac I got 1.56A (8.1 ohms at 60hz).
... It just dawned on me that the above were taken on the untapped coil, with the cap across the supply voltage. I just tried the tapped coil and got similar readings until I put the cap where it appears to belong, (across the 460 V taps) and the result was 16.2Vac at 1.24A, or 13.1 ohms at 60hz.

It appears that this welder is supposed to have about 140uf, does that sound reasonable?
If I replace the bad caps is the current likely to drop much more?
Can I calculate the ideal amount of PF correction capacitance from the information I have, or do I have to go by trial and error?
Rob
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The 140 ufd sounds reasonable to me. If you replace the bad caps the current will drop quite a good deal. The replacement caps need to be rated for 460VAC. If they are only rated as DC then the voltage rating needs to be at least 1.7 times 460. In both cases higher voltage rating is good but will mean bigger size. You could use lower rated caps if you connect them on the tapped winding where the voltage is equal or less than the caps are rated.
You can calculate the amount of capacitance to make the impedance entirely resistive ( no reactance ). But you really don't want to use that much capacitance. One reason is that the welder would be resonate at line frequency and you might get some really high voltages. Another reason is that it is expensive compared to the reduction in current to correct more than 90%.
The easy solution is graphic. From point A swing an arc equal to the current with no capacitance. From the same point A swing another arc equal to the current with some amount of capacitance. Now pick a point B on the first arc and swing an arc equal to the difference in the two currents and intersecting the second arc at point C. Extend this line and then connect a line from point A to the extended line so that it connects to the extended line at a right angle. The length of this line is the amount of current that the welder would draw with 100% power factor correction.
If you add another capacitor of equal size, it will reduce the reactive current the same amount as the original cap. That is it will establish a new point D that is down that BC line an amount equal to BC. Note this does not reduce the total current as much as the first cap did.
This is probably clear as mud without actually drawing the arcs and lines. So ask more questions if you are still confused.
Dan

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On 3 Nov 2003 09:35:28 -0800, snipped-for-privacy@krl.org (Dan Caster) wrote:

I have located some 440VAC caps. Is there any leeway on this rating, or do I have to keep on looking? I checked the supply voltage here and it is jst over 236VAC, so the caps would see about 472VAC.

Does the required PF correction change when the welder is in use?

I grabbed an old electricity text (Jackson) and reviewed the appropriate section. I think I figured it out.

Thanks,
Rob
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There is a lot of leeway. The voltage rating is based on the failure rate at some really large number of hours. It is not based on the voltage that will immediately punch thru the insulation. Years ago we took some caps that were rated for 2000 VDC and found out where they blew up. As I remember it was about 6000 VDC. So unless you are planning on four or five years of continuous use, I think you are okay using the 440VAC caps.
The power factor correction is to correct for the current thru the inductor in the equivalent circuit that is across the input line. This current is at right angles to the real power current when the welder is in use. So say the input inductance current is 40 amps and the real power at idle are 10 amps. The total current would be the square root of the two individual currents squared (pythagorean formular, that high school stuff is really useful ). Or close to 41 amps. In use you still have the 40 amps inductive current, but now say have sixty amps of real power current. So now the total is about 72 amp. Power factor correction make a big difference at idle, but not much in use. But low current during idle means the wire going to the welder and the breaker are cooler and the breaker does not pop while you are welding.
Dan
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