I get this one asked so many different ways, because the subject of "KVA" is so grossly misleading.
People ask this because KVA is used to describe the size of a welder.
After all, everything else is sized by KVA;
  Generators
  Power transformers
  Even I.C. engines !
Soon some marketing novelist will describe software that way!

To determine the amount of current a welder can produce,
you must consider the secondary voltage and the secondary impedance of the welder.
This requires:
    Physical measurements and a lot of math,
    A well trained observation,
    Or an actual measured test, fired tip-to-tip.
Yes you can fire this way, if the tips are clean and come together tightly.
But, the KVA of the transformer won't help you determine maximum welding current.

Here is what KVA determines in a weld transformer:
The measure of its ability to process heat.      Wow that sure helped...NOT!

Whole lotta good that does us... But, we must pay attention to heat,
so let's say we have an assembly operation with 10,000 amp welds being made at the rate of 20 every minute.
---This is normal for an automotive assembly line.---

We use KVA to determine if the transformer will overheat.
KVA means "Kilo-Volt Amps" and is closely related to "real power", or "Watts".

But Watts is actual power, it is Volts times Amps in a purely resistive circuit.

Welders don't have a purely resistive circuit.
They have a lot of inductive reactance in their secondary, so we use "KVA"
This is just a way to avoid doing all the math required to express "Watts" accurately.
(To use "Watts" would require some complex phase-angle calculations, and this could hurt
our already over-taxed frontal lobe...)

We need to know how much heat is being generated in the transformer.
So we can take the current times the voltage, and get pretty close.
      What voltage?
The voltage of the secondary, go look at the name tag on the transformer.
Oh, I forgot, they mount that down against the mounting plate, so it won't get damaged...
Pretty clever eh?
--- Tell your favorite build shop to stop doing that. ---
Well then, let's go to the welder verification sheet, where it lists all the parts to the welding secondary, you DO have this right?
No?, well then you need to get modern and do "WGV" on your NEXT start-up, too late for this one...

It is difficult to measure the secondary voltage because the current's huge magnetic field induces a voltage into the meter probes.
If you don't believe this, make a reading with both probes on the same point, you will get at least a few volts.

So call the manufacturer of the transformer and ask him. But he will need a part number, and will have to know what tap you are on.
Do you see why we needed the "WGV" ("Weld Gun Verification") sheet?

Let's say our transformer has a 12 volt secondary, this will give about 16 kA in a secondary about 20" long by 8" high.
If you have a tiny little trans-gun, it may be only need 3 to 8 volts.
If you have a big portable gun with a 10 foot kickless cable, it will need 24 volts.
Note at this point that the physical size of the secondary is what determines the voltage needed to do the job.
This is an important concept to remember...

But we are using 12 volts here, so 12 Volts times 10,000 Amps is 120,000 or 120 KVA.
    So we're done right?
No, this is only the Peak KVA, during a weld. There is OFF time between welds.
So now we must figure the duty cycle. If the welder is on for 10% of the time, it produces less heat than for 100% of the time.
So we must figure out how long it carries current over an "averaging time" of one minute.

Our weld time, "10 cycles", means "10 cycles of a 60 cycles-per-second AC power source".
So one weld is 10 cycles, or 10/60th of a second. and twenty of these welds are performed each minute,
so 10 times 20 equals 200 cycles "ON" time every minute, and a minute is 3600 cycles.
So our Duty Cycle is 200 divided by 3600
= 0.055 or 5.5%

Now we can figure the equivalent heat, which is the square root of the duty cycle times the peak heat.
The square root of .055 is 0.236
0.236 times the 120 KVA is only 28 KVA
So we only needed a 28 KVA transformer to do this job!
Well... No, it's more complicated, welding transformers are rated at 50% duty cycle... No, I don't know why, maybe they sound bigger that way...
So lets scale this up to see what NAMEPLATE rating we need...

If we had a 100 KVA load at 50% duty cycle, that would be 70.7 KVA equivalent load.
So a 70 KVA transformer would get a "Nameplate Rating" of 100 KVA...
Why do they do this? I don't know, ask them, it makes my life difficult though...
So we need to take our requirement of 28 KVA and divide it by .707 to find the "Nameplate" size of the transformer we need.
28 divided by .707 is almost 40 KVA, so that is the transformer we need.

So will an 80 KVA transformer weld a part twice as big?
No, but it can do these welds at a much faster rate without overheating...
The 80 KVA "Nameplate Rated" transformer can handle 56 KVA at 100% duty cycle
and 56 KVA divided by the 120 KVA of each weld is 0.46, which is the square root of the duty cycle we can run at.
If we square 0.46 we get 0.21 or 21% duty cycle. This means we can do more welds in a minute.
How many in a minute?
Well, 3600 cycles in a minute, we can be "ON" for 21% of them or 756 cycles.
This is 75 welds, almost 4 times as much...What's up with that?
We are dealing with squares and square roots, so it won't be a linear relationship...


    Well all this is fine, but it doesn't really answer my question...
The answer is that you can't tell a welder by its KVA. If actual current capability is what you are after
then you must judge the secondary voltage capability against the secondary impedance. That is not easy...
So if you are buying a machine, ask what the max secondary current is tip-to-tip.
Then on a medium sized secondary (20" x 8") that will be only a little higher than your actual weld current.
    How much higher?
That depends on:
   If it is a large secondary, then the inductance of the part in the gun will load down the welder more.
     The resistance of the part will be a small faqctor.
   If it is a DC machine, then there is no reactance, but the resistance of the part is
      a bigger consideration. Aluminum may pass a lot of current, steel may not (it has higher resistance).
   If it has a kickless cable, then the resistance of the part will mean almost nothing
     compared to the huge resistance of the kickless cable.
    If it is a little trans-gun, then the resistance of the part is a major factor.

So you just can't go by KVA. If you just have to know, contact me, and I will ask you a bunch of questions,
but we can estimate the machines performance pretty well.

Hello Bill, Nice to hear from you. Is your lab still in the ARO plant?

I'll have to drop by sometime...

Anyway, yes the fused area stops growing when the current stops. There is a video that GM had made (that Dave Kelly should be able to get for you) that shows the growth of a nugget during welding. The video is developed at a high rate, I think 420 frames per second, and the view is of a pair of tips and two coupons of about 1mm each, all sliced in half at the center of the weld area. This video shows both the growth of the nugget, and the conduction of current at the same time. It is easy to see that the growth of the nugget corresponds directly to the application of current. In fact, the growth rate speeds up with the peak of the AC current, and slows down to a stop as the current approaches zero, at the end of each half cycle. I was asked by Dr. Karagoulis of the GM Weld Council (Tech Center) to show this video to all the GM students that I held welding classes for. I am not allowed to copy or distribute this video, but if you are working on GM projects, I can show it to you. (as Dave Kelly can also)

The hold time would not affect the fused area, but it could affect the weld quality, in such a way as to appear to affect the "button" size revealed with a peel test ("Coach-peel"). It can do this in a number of ways:
1. If the part is springy, and the hold time is short (shorter than 1/2 the heat time) the weld may be disrupted when the part moves while still hot enough to be in the plastic state.
2. If the water cooled tips provide too much quenching, and you have quench-sensitive HSLA, the nugget will fracture as the steel returns to room temperature size (shrinks) while it cools through the plastic state too quickly. This is why there are no 1983 Corvettes, it took extra time to engineer the weld heat profile to avoid this on their new (at the time) heavily galvanized (G-90) HSLA frame material.
3. If another weld is done nearby and disturbs the part while this weld is just entering hold time. This is pure conjecture on my part, but I did study this relationship while working on what Medar calls "Thermal Force Feedback" with Mr. Ariel Stiebel at the Pertron lab, before he went to Medar because Square D purchased and closed up Pertron.
4. Another conjecture on my part would be changes that might occur in the rich chemistry of AHSS during different cooling rates after the weld. There will be much discovered about this over the next 10 years as the auto plants start getting AHSS to work with.

Anyway, if the coach-peel is used to reveal the nugget size, anything that affects nugget strength will show up here. A better test if you really wanted to dig into this relationship would be done ultrasonically, such as with GE Krautkramer equipment, which can identify nugget size before any peel test is made. The new ultrasonics can identify the difference between metal that has been heated and cooled from that that has not, by measuring the losses in the sound reflection as it passes through the grain structure of the metal.

Presently they identify the nugget size by controlling the diameter of the sound beam, and identifying if anything within that beam was not welded. So to actually read nugget size you would have to try increasingly larger probes until lack of fusion is indicated. The equipment is designed to simply tell if the fused diameter was larger than the probe, which it does a good job of. They make no claim to find cracking, as the cracks are usually "seen" on edge by this equipment, but we actually did get indications of cracking when we tested welds with this condition at a demo GE Krautkramer did at AZ Automotive. The GE people did not know we "set them up" with known bad welds, and were pleasantly surprised when we "fessed-up" to our little trick, but they still will not tell you it will find cracks unless they are perpendicular, like "de-lamination" cracks.

I hope this long-winded answer helps, please respond on your thoughts.

Digital Error
This is a well known problem with all digital displays, if you get a display of 12.3KA, displayed as just three digits, that means it could have been 12.3000 to 12.3999...  The reading has to be cut off somewhere.  This is over and above all other accuracy limitations any instrument may have.
A good way to explain how the last digits get eliminated, even if they are .9999 is to take two pieces of paper, on one, draw square waves from right to left across the width of the paper, put about 12 of them there, so it looks like:

Cut out a window in the other page big enough to see only ten of these pulses.
Lay this "window" over the pulses drawn on the first sheet, what do you see?  You may see ten full pulses, or you may see only nine full pulses, because of the way the window is placed.  This is how meters register digital values, they count all the full pulses and ignore any incomplete ones.  This is just a very simple example of digital error.

As this is trying to illustrate, three windows displaying 10.1 can only be trusted + or - "0.1" so we cannot expect to hold the measurement to "0.1"
Adding extra windows does not help, unless the instrument has the greater accuracy to deal with those windows.  Since a current checker is customarily + or - 3% and those windows could be used to display 99.9, you could be off + or - 3.0 right there!
A popular gauge has an extra window for the last digit which always displays "0", and people believe it to be accurate to the last digit!  The eyes play tricks on people, especially if they are unaware of the facts.  Engineers rule!

Question 20 does requires a lot of explanation, but you asked for it so I assume you will sit through the answer.
1. When we weld metal, the resistance of the workpiece changes during the process. The average resistance is 100 micro ohms, and it is often presented as just that, but it's not that simple. First, when the tips first contact the part, and the force stabilizes, we have 200 micro ohms, due to the rough surfaces meeting each other in three places; upper tip to work, faying surface, and lower tip to work.
2. During the first 3 cycles the surfaces soften, after all they get most of the heat, because that is where the bulk of the resistance is at this time. When they soften, the surfaces flatten and contact each other better, lowering the contact resistance. At the end of the 3rd cycle the resistance will be 50 micro ohms.
3. As we continue adding heat, the resistance rises, as steel (and I think all the metals) has a positive coefficient of resistance with temperature. So as we add heat, the resistance rises up to around 100 micro ohms. Many resistance feedback systems look for the fall, then this rise in resistance, to signal that the weld was successful, but it is only a fair approximation, we can't tell for sure that the weld is good, so resistance feedback never amounted to anything but a lab experiment. Believe me, at Pertron we pushed the technology of resistance feedback to the very limit, finally building a control for GM Indy Stamping that could tell if it was bare or galvanized, then what category of weld it was turning out to be in galvanized. There were three categories (for galvanized) , some dropped resistance right away, some never dropped resistance, some had multiple fluctuations. We didn't understand why, but we did recognize that all three could be made into a good weld, if the heat was controlled properly with feedback and we did that. But the system required even more care, because any unusual variables would throw it out of control.
4. Then the heat is stopped, and the part cools. In roughly the same time as the heat time was, the resistance drops to 10 micro ohms.

So, knowing this, with a fitup problem of unknown amount, we can try a weld heat. It will either weld, or if it was a bad fitup, soften and bend together. We can tell which by reading the resistance after a bit of cool time. I like to use the same time as the heat time just to be sure. If the resistance is high, it didn't weld. If the resistance is low, it did.

But we aren't reading resistance, so what can we do? Well, if it didn't weld the resistance will be 100-200 micro ohms, and we need to try again, and if it did weld the resistance will be 10 micro ohms, and we don't want a lot of heat. So we give it another weld heat. If the resistance is 100-200 micro ohms, this will generate heat and possibly weld together. If the resistance is 10 micro ohms, it doesn't get much heat because heat = I squared R, so it gets roughly one tenth of the heat because it has one tenth of the resistance..

Classically we give it three heats, the 3rd one will solve even the toughest of fitup problems reliably.

The saying: "There ain't no free lunch" applies here. Tip life will be 1/3rd of normal, and the cycle time is 5 times longer (for the weld-cool-weld-cool-weld sequence that now exists between squeeze and hold). But it does solve your fitup problem, by using the dynamic resistance change of the weld to regulate the heat automatically.

This would be based on the life expectancy of each component. Your maintenance records will be the best guide, but I would start with the following, based on number of welds made. Note that a weld gun on an automotive assembly line may make 160,000 welds in a week on a robot, or only 7000 on a dedicated "hard tool", such as a framing line or press welder.

125,000 welds:
    Measure resistance of all water cooled cables, replace at 20% increase from new.
    Perform cooling check on tips (They must return to original temperature within 5 seconds)
    Check tip force, being careful to avoid false reading from "impact" of gun closing.
        (Do not use "latching" feature of weld force gauge).
    Check schedules in use, investigate all changes from original setup standard.
        (Keep a copy of all standard settings and readings at each weld station)

250,000 welds:
    Measure air cooled cables, replace if 50% increase.
    Measure all bolted connections, repair (silver plate) if greater than 10 micro ohms.
    Measure water flow to each gun arm, must be =1.0 GPM (3.8 Liters per minute)

1,000,000 welds:
    Rebuild straight-acting gun cylinders

4,000,000 welds:
    Rebuild rocker gun weld cylinders
    Measure water flow to transformer, cables, and jumpers.
    Water flow to transformer =1.0GPM
    Cables =2.0GPM,
    Jumpers =2.0GPM,
    SCR=1.2GPM (4.5LPM) but this I think is very excessive, SCRs generate 10-100 watts of heat.

The "Stepper" is used to keep current density the same, as the tips get larger. The variables differ with each application, but the main ones to consider are:
Q9-       1. How much larger do you want to allow before changing tips?
      2. What current is required at that tip contact area?
      3. What current is required when the tips are new?
      4. Can your process tolerate the contamination that the tip has at that point?
      5. What is the weld count at that point?

I   have experience in automotive welding of zinc coated sheet metal, in the 0.8mm to 2.5mm range. It is generally accepted that the tips will have too much contamination (Zinc + Copper = Brass) at the count of about 5000 welds. Testing a welder with tips in that condition, we find that we need about 5000 more amps. Now, this will have to be done with tips that were allowed to wear to this point, making reasonably good welds. You cannot move these tips into another welder, as they will not act the same due to critical duplication of alignment. Also, we must use the same welding schedule (Squeeze, Heat time, Hold time) as was used in the beginning, all we can change is current.

Here is an opportunity to find out if your "starting" weld schedule is good enough to be used as your "ending" weld schedule.
To explain: The schedule you setup with new tips will probably be optimized for that condition, BUT, when the tips get larger, you will have to make a larger weld, this takes a longer heat time. Now the problem... The weld controller manufacturers offer the "stepper" option for this, but it is NOT GOOD ENOUGH! You need more heat TIME. So, what you must do, is find a heat schedule that will weld with old tips, and see if a lower-current version will work for new tips. It usually does, but it upsets people who want short cycle time for high speed assembly lines.

So the answers to the questions above FOR THIS APPLICATION would be:
1. Ending tip size?=about 50% larger (contact area).
2. Ending heat?=about 16,000 amps.
3. Starting heat?=about 11,000 amps.
4. Is contamination OK?=Yes, but not if it goes any further.
5. Weld count=about 5000 welds.

Now that we have our information organized, it is easy to see we need to change tips in 5000 welds, we need 5000 amps above our starting current for extra transformer capacity.
Now we have one last question to answer: How much extra capacity do we need for regulation of weld bus voltage? Few people have actually measured the voltage dips on their weld bus, it takes a "line disturbance analyzer", such as the one made by Dranetz. Weld busses vary, they usually have drops of only 10%, but I have seen the bus drop so low all the relays would drop out when welding. The customer's cure for this was to power the welder controllers separate from the weld bus. These are drops of 40-50%!

You should make measurements if you are having welds missing every now and then. Modern weld controllers often will give you this information in "last weld data". You can trust this, and use it as a substitute for the Dranetz test. If you have a reasonable weld bus, allow for a 10% drop, so the transformer (in the above case) would be able to put out 16,000 amps with 90% of the line voltage. Roughly this would mean we want a capacity of about 18,000 amps. Now, be careful, too much capacity means you will be welding at a low "phase angle" on the SCR firing point, when the tips are new. This means the pulses of power at your AC line rate will have high peaks, and long inter-cycle cooling time. If your starting heat is below 60%, you will have difficulty avoiding expulsion. In this case, we are at a starting point of 61%, which is acceptable. In cases where the transformers will produce 25,000 amps, you may have more trouble with weld flash and sticking tips.

As you can see, the numbers you need must be gained from experience in your type of welding, so go back and learn from what you already have working.

When expulsion occurs, the metal that was conducting the current between the sheets is suddenly gone. The effect is to cause current flow to be lower than usual for the rest of that half-cycle. The current feedback reports that current was below the "target" level, and the control responds with more current.
This can be observed in the single-cycle data, watching the measured current and the gain factor that the control applys for each new cycle of heat. These values are easily obtainable in the ADC control, because I specified that to Toshiba, who built the control.
Some controls limit the amount of heat gain allowed cycle-by-cycle, and this feature makes up for a lot of this. Medar does this, I suspect that WTC does, as I met their engineering group, and they really impressed me with the way they control the SCR to regulate current. I cannot divulge the slick things they do, just trust me....
ATek does something to limit this too, because that control will fault if the current is only off by one percent! I have personally measured this at a Harrison plant, and it really works! Another thing that works for them is that the president is an Ohio State Welding Engineer and he won't let you set-up the control wrong... Those controls are on the real tough applications, like gas tanks that weld through coated metal, and all the tough jobs at Delco Kettering.
Too much limiting of current adjustment will reduce the control's ability to make up for a really bad weld bus. But if it is that bad, you should fix it.

The tip temperature must never cause permanent softening of the copper. Zirconium tips (usually the preferred alloy) soften at 930 degrees F, Cupal (D.S.C.) is 1475 F.
To measure your tip cooling, measure the temperature before welding a part, then again 5 seconds after the last weld, the temp should be the same. If it takes longer, you have a water flow problem. Note that one hot tip can cause the nugget to be "off center" (closer to the hot tip) which can cause a peel test to give a small button! Tip cooling is more important than most people realize.

Expulsion occurs when the expanding metal escapes the containment force. The metal expands about 10% as it becomes molten, and the forces rise as high as 30,000 psi. If the tip force cannot contain this, it gets away, and it does so violently. It causes cosmetic problems, and is very irritating to the workers. It causes a lot of maintenance problems with alignment pins and prox switches getting coated with slag.

There are a number of reasons, first make sure your tip force is at the proper setting, as per your company standards. Make sure that the force is fully applied at the time of welding. Slow guns are a popular cause of expulsion. Check that the tips are aligned on axis, and on center. A 2-mm misalignment is too much, you must fix it. "Normalize" the gun to the part (make it fit square, not at an angle).

Is the current being applied too rapidly? You can heat the part two ways, a lot of current for a short time, and a little current for a long time. The latter method is less likely to expel metal. Again, check your company standard settings.

Well, the LAST thing you should do is dream up your own schedule, the liability is too high… Refer to your Welding Engineer, or contact the Standards Group. Without your feedback, they cannot correct for application problems. Don't be too proud to call for help, it is the smart thing to do. Give me a call, maybe I can figure a way to charge you for my time.
Hey, did I mention the part has to be clean? You can't weld dirty parts, sorry, but the voltage on a resistance welder is very low and it just doesn't fire through CRUD!

You must have good measuring equipment available. I stress available because if someone has it locked up so they can find it at calibration time (remember the ISO9000 guidelines…) then you might as well not have it.