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Arc welding transformer?

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

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  1. 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.

    Can anyone who knows more about the subject please help clear
    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.

  2. mike

    mike Guest

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

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  3. Yzordderex

    Yzordderex Guest

    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?

  4. Yzordderex

    Yzordderex Guest


    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.

    I hope this answers your question.

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