Nice to see someone else in this field! I'll echo your question.
We haven't paid much attention to the topic of transformer sizing. The
equipment we sell (10-100 kA welding current) we size on experience and
estimates rather than computer analysis. (We don't make the machines,
we only sell them and put together parts from different suppliers to
form a working system.)
Obviously, we have to verify that the transformers can supply enough
welding current with the selected inverter.
The maximum current can be found in the transformer's datasheet. In our
experience, the limit is often the diodes on the transformer in manual
applications. Long times between each weld when the user moves the gun
or the work piece allows the transformer too cool off.
With automated welding, the limit is more likely to be the kVA of the
transformer and the cooling of the electrodes. The robot moves rapidly
from weld to weld and never lets the transformer cool off. The
necessary kVA can be calculated if you know the current the machine
will normally need and how many spots per minutes will be made. In
reality, we find that the kVA is choosen either by the buyer or from
experience from similar projects.
Let's say I have a inverter that outputs 100 kVA @ 500V and 20% duty
cycle. I then see little point in choosing a transformer larger than
100 kVA. A larger transformer gets you a a higher secondary voltage and
correspondingly lower current. This is rarely desired.
In most cases, a smaller transformer is quite sufficient and will give
a higher secondary current due to the different turns ratio. The turns
ratio is determined by the transformer manufacturer and is as much a
manufacturing and marketing choice as a techical one. I think the
common ratios are something like 40:1, 50:1, 60:1, 70:1 but don't quote
me on that.
When I know how large a transformer is needed, I pick a cable for the
primary to match this: If I use a 100 kVA transformer per the above
data, I get a nominal current of 200A at 20% duty cycle which gives
the same thermal load as a constant current of "square root 0.20" x 300
= 89 A. Thus, a 25 sq.mm. cable (or 3 AWG) should be enough.
The conductor on the secondary side is sized by the manufacturer of the
Is this reasoning right or wrong? I don't know, but it seems to work.