# Arc welding transformer?

Discussion in 'Electronic Design' started by Eric Y. Chang, Sep 3, 2003.

1. ### Eric Y. ChangGuest

Hi. I was wondering how an arc welding transformer works. According to
many posts on newsgroups, the output has a constant current behavior due
to weak coupling between the primary and secondary.

A typical small hobby buzz box welder is rated at 120 V @ 20 A input and
70 A output at an unspecified voltage. This voltage is probably
approximately 20 V for the typical metal arc. At first glance, the
power in and power out seem reasonable, since this looks like 1400W
out, if the load is approximated as purely resistive. But, a real
transformer does not multiply current (past that done by the ideal
transformer embedded within it). In other words, it should not be
possible to increase output current by drawing down on the output without
also increasing primary current. Note that this behavior is fundamentally
different from a buck converter. By decreasing the duty cycle, the
buck converter can actually multiply current when the output is drawn
down. The magnetic transformer cannot, aside from the amount produced
by the turns ratio.

These small buzz boxes typically have an open circuit voltage (OCV) of
about 80 volts, to aid in starting the arc. Obviously, the rated 70 A
is not delivered at 80 V. Instead, the voltage sags according to the
non-ideality (regulation) of the transformer, down to the arc voltage.
But, if one considers the ideal transformer model, the internal ideal
transformer is still producing 80V (and the voltage drop is across the
leakage inductance). There is still 70 A flowing through the output
winding of the internal ideal transformer. Therefore, there is 70*
80/120 = 47 A flowing through the primary. Since there is no current
multiplication in the primary either, this means that 47 A is being
drawn through the mains circuit. This will surely trip the breaker?

Conventional magnetic transformers, no matter how they are constructed
(leakage vs. magnetizing inductance) cannot multiply current above the
amount given by the OCV (ideal transformer) rating.

A simple example can be given by the 1:1 ideal transformer. The
schematic will be omitted, since the transformer model is already
familiar to the newsgroup. There is a leakage inductance in
series with the parallel connected magnetizing inductance and load.
The current is Im + Ir = IL (magnetizing, load and leakage, resp.).
By the triangle inequality, which applies to complex addition:
|Ir| <= |IL|, so the current flowing in the mains portion is always
greater than that flowing through the load, regardless of the non-
ideality of the transformer. Thus, no current multiplication
occurs due to drawdown.

things up? Does the welder draw these large currents from the mains
without tripping breakers, or do they just produce less than their
nameplate rating? Note that this does not apply to the newer
inverter powered welders, due to the true current multiplying effect
alluded to earlier.

Thanks,
Eric

2. ### mikeGuest

I don't have the answer, but am pretty sure you need to account
for phase angle in the AC analysis.
mike

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3. ### YzordderexGuest

The sears buzzbox I use has a moveable core. When you turn up the
amps the core goes inwards to increase the reluctance. If I've got
that right.

Does this help at all?

regards,
Bob

4. ### YzordderexGuest

Eric,

Sorry for the two posts. I had a work related interruption that
prevented me from a full answer.

It's really very simple. Assume you have a 120v:80v transformer -
ideal if you wish. Now if arc voltage is 20v and xfmr secondary is
80v, you have to drop 60v. If you need say 60amps then you need to
put an inductor worth an ohm at 60hz in series with this ideal
transformer. With me so far?

Ok, so if you move the iron slider into or out of the core, you can
change the leakage inductance. Your 'ideal' transformer becomes less
ideal as you move the core out. The leakage reactance exists because
some of the flux lines which are generated by the primary turns do not
intercept the secondary turns. As you move slider out some of the
flux lines go outside of the core, and leakage inductance increases.
Current goes down on BOTH primary and secondary.

So what you've in effect done (when moving the slider outwards) is
place an inductor in series with the transformer - added impedence to
tame the current.