Another Welder question

I have a 240 amp welder and can add a full bridge rectifier to make it an AC/DC welder for about $50 worth of components. Why wouldn't I want to do this and why don't more people make this rather simple modification to their older AC buzz boxes?

Thanks for the observation. High frequency can really take out a silicon junction (like a diode.)

But the main reason this is not done more often is just exactly what is described below - your 240 amp AC welder becomes a 120 amp DC welder, and that's a little on the small side for a stick welder. Now, I have seen this done several times, very successfully, to make a lower-power scratch-start TIG welder. And no doubt there are other similar applications.

llater,

llamas

A single diode creates a half-wave rectifier and virtually cuts 1/2 of the current wave from the output. A full wave rectifier (bridge rectifier) routes both waves to the output and creates 1.44 times the input AC voltage.

Heat sinks are a must and a fan is strongly advised but I think I should get good results from this solution. Perhaps the only way to tell is to give it a whirl and see how it comes out.

$50 worth of diodes isn't too much for an experiment.

AC current is measured RMS, DC current is not. And you're talking about voltage, not current. Incidentally, I believe your factor is backwards - for a true sinusoidal AC, a full-wave rectifier can produce a DC voltage output of 1/1.44 = 0.7 times the peak AC voltage.

The rectifier on its own will not do that, of course, - all it outputs is full-wave rectification, or a series of wave segments all at the same polarity relative to ground. To get close enough to a "true" DC, additional smoothing is required - eg large capacitors.

I think you'll find that the DC current output of an AC welder rectified as you describe will come out roughly as I described - about 50-60% of the corresponding AC output current. That's the result for the various transformer-rectifier welders on the market - eg the Lincoln AC-DC 225/120, which has a full-wave rectifier for the DC side.

llater,

llamas

You would think a diode bridge would cause no significant loss. But for reasons I don't understand, that's what happpens. Welders that are built this way from the manufacturer have less output on DC than on AC. If anyone knows why that is, I'd be interested in knowing...Al English

If the current drops like that with a full wave bridge desige, I'd have to blame it on the .7v drop accross each diode. That would be 1.4v for a full wave bridge (only two work at a time). If working voltage goes really low during welding (think it does) then dropping 1.5 volts could explain the big drop in current.

Just a hunch.