# Convert an AC to a DC Welder

• posted

I recently read that an AC welder can be converted to a DC welder, and all thats needed are either two or four very high amp diodes. The welder puts out up to 240amps. I never measured the voltage, but I think it's around 25 to 30. (I'll have to check). Two diodes will make a half wave rectifier, but if my memory is right, I'll only get half the voltage, whereas using 4 diodes as a bridge rect. I will get the full voltage output.

Now, my question is this. How do I figure the amperage of the diodes that I would need (for both methods), so I have at least 250amp capacity. I'd guess 300 would be better to cope with heating. Of all the years I puttered with electronics, I was never good at math. My guess is that each diode should be 150A to achieve 300A. But I might be wrong. Can someone please help.

Also, where could I get something like this?

Thanks

• posted

One diode will create a half wave rectifier. Two diodes will create a full wave rectifier, if the winding that you're feeding them from, has a centre tap, as the one in a welder almost certainly won't have. With a non-tapped winding, four diodes will be needed, connected into a bridge. You're kind of right in that two out of the four diodes will be conducting during any given half cycle, but the two diodes are effectively in series with either side of the load, and hence each other, so all four diodes have to be ratd for the full current that you wish to be able to draw.

A half wave circuit does not give you half the voltage. Rather, it gives you half of the total energy that would otherwise be available in the whole waveform, by chopping off one half.

As far as where to get such diodes, I'm sure that there will be others on here that know a lot more than I do about the specifics of electric welders, who will be able to better help you on that part of your question. FWIW, I had a feeling that long ago, the rectification and control elements of these units, changed to SCRs (thyristors) as they are readily available in sizes that can control electric trains (!) but I could be way off-beam there.

Arfa

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Note also, as Arfa stops at half-the-story (if not half the voltage):

One (1) Diode will give you 0.707 x the incoming AC voltage in DC voltage. a Full Wave BRIDGE (4 diodes and the best way to go, center-tap notwithstanding) will give you 1.414 x the incoming AC voltage in DC.

Welding-load rated diodes won't be cheap:

TE-ELECTRONICS-NTE6356

You will need four of them, an appropriate enclosure, connectors and so forth.

And be VERY, VERY careful with such a lash-up. You will have a lot of stuff designed to make actual real-live arcs quite close together. Make sure that you have appropriate insulators, adequate spacing, excellent connections and so forth before you apply power. And when you do so, make sure you are properly protected against a problem until the system is proven to both behave and operate properly.

Peter Wieck Wyncote, PA

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When selecting diodes, it's tempting to start with the 30V number. DON'T!!! You have a BIG transformer with LOTS of leakage inductance. You're shorting it then breaking the arc. Expect HUGE transients. You're gonna need diodes with WAY more breakdown voltage than you expect. And I'd also use snubbers. I can't give you any numbers from experience, but I'd expect to end up buying more than one set of diodes before you're thru. I'd go googling for DIY welders to see what others used successfully. mike

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• posted

Note also, as Arfa stops at half-the-story (if not half the voltage):

One (1) Diode will give you 0.707 x the incoming AC voltage in DC voltage. a Full Wave BRIDGE (4 diodes and the best way to go, center-tap notwithstanding) will give you 1.414 x the incoming AC voltage in DC.

Well Peter, I would contend that it depends on just what you are using to measure the voltage, and how exactly you define "DC". What you get from any kind of rectifier, be it full or half wave, is only "DC" in the loosest definition of the term. Unipolar Current would actually be a better term than Direct Current. With no load and some smoothing to get 'true' DC, either a half or a full wave system will return a voltage value of about 1.4 times the RMS value of the AC input voltage.

I don't follow what you are saying with your comment "- centre tap notwithstanding". The centre tap is irrelevant as his welder tranny will not have one. In the case of a power supply that does use a centre tapped tranny, then unless you require both positive and negative full wave rectified rails, there is no particular advantage I can see in using a 4 diode bridge over a two diode and tap arrangement.

Arfa

• posted

What kind of welder? Stick or wire? Stick welders can benefit from using AC or DC in different polarities. Wire welders are pulsed and at a high frequency. Just what are you trying to accomplish?

• posted

If there's a capacitive filter it will give the peak voltage . . . welders don't have capacitors (and finding a cap to take the abuse would be costly).

An AC welder might have an inductor in series with one welding lead as a reactive arc stabilizer but they don't raise the RMS to peak voltage.

He will have approximately the same voltage he started with as AC minus two diode drops with a FWB or ~1.5 volts less . . . .

Center taps won't be on welding transformers as a rule, and they waste some power since the secondary is only being utilized 1/2 the time. To compensate the transformer must be larger and cost more. CT transformers do make sense on very low voltage supplies where the diode drops are a significant percentage of the output voltage, and high frequency supplies where the diodes are costly.

• posted

FWIW Modern arc welders (not domestic types) use a high frequency square wave superimposed on the output voltage. This makes striking an arc easier at low amps and allows finer control over output current.

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Regards:
Baron.```
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In commercial ones that I have seen the IGBT's have been 600volt devices about the size of a coffee cup, with connections like rope and big ring lugs on them.

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Regards:
Baron.```
• posted

(SDI) R6110630 1000v, 800 amps 00-9 stud. Includes nut / washer. P/N: R6110630XXYA. NSN: 5961-01-276-9983. \$59.00

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(SDI) 1N3111 Semi-conductor device. 50v, 150 amps.

12.00 \$10.00
• posted

Now it's all coming back. It's been years since I did much with electronics. My eyes are not good enough to work on circuit boards. Most of the stuff I did work on was the old tube stuff with large parts and terminal strips. That stuff was easy to see and really built to last. Then I got into the old transistor stuff which was still ok to work on but harder to trace. When IC's came on the market, I pretty much quit working on the stuff. Just far too hard to test the stuff.

Anyhow, as soon as you said this, I recalled the old power transformers with their center taps. I can almost remember the whole power supply for the old tube guitar amps, which was one of my favorite things to play with. So, yes, without the center tap, I would need a bridge rect. Looking at the link to the part and price, I had no idea they would cost \$60 each. At \$240 I may as well buy a new welder with AC and DC. Oh well, it was a good idea while I thought about it.

By the way, I know that a welder is low voltage but high current. I wanted to measure the voltage with a standard digital volt meter. Then I pictured the whole meter getting it's circuits welded into a blob of junk. Is it safe to apply the leads of a common voltmeter to a welder? Logic tells me that it should be, but I am still not comfortable with it. For example, I have always been told to disconnect the NEG battery cable from the newer cars when welding anything on the car, or the car computer can burn up, as well as the alternator diodes. Knowing that, makes me question using a test meter.

By the way, what are SNUBBERS ?

Thanks

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There should not be a problem with hanging your digital meter across the output, set to an AC volts range, but never a current range, of course ...

Snubbers are networks, usually series R-C, connected to places where you want to knock the sharp tips off of voltage transisents, i.e. limit dv / dt excursions. So you might find one at the gate of, or wrapped around, a thyristor perhaps, to prevent false triggering. Or across the terminals of a DC motor to prevent commutation noise getting back into the drive circuitry, that sort of thing.

Arfa

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I made a converter using a bridge made up of diodes of 1/2 the required amperage, and a series coil made out of a 2KW variac core with a cut going all the way through the torroid. I used parallel winding for the output through the indictor.

greg

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Meters are safe to use on arc welders, as long as you don't try to measure current.

You might get by with the 150 amp diodes at \$10 each. I rarely set the switch on my buzz box welder higher than 120 amps but I don't know how that translates to actual amps at the diodes.

A better question might be how does one size snubbers You take a die hard, low inductance, low dissipation factor, AC rated cap, and put it in series with a good quality (flame proof) resistor, across the contacts of the AC line switch. Its purpose is to absorb some of the energy of the arc when the switch opens - and cancels some of inductive component. The purpose of the resistor is to lower the Q and prevent an unintentional series resonant circuit and spark transmitter. They can be bought ready made in one package.

Do they use snubbers across the arc in welder?

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that'll help protect the line switch. But you also need to protect the diodes when the arc breaks. To size them, probably need to make some inductance measurements and do the math.

Its purpose is to absorb some of the

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• posted

Before I did the mod, I just asked around like you did. The slotted inductor makes the current more constant. I don't imagine it would work well without it. It seems like you have the same basic figuring as I in making a bridge. If a bridge is outputting 150 amps, then each diode average current is 75 amps.

greg

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If what I learnt nearly 40 years ago is correct, then that isn't true. With a single untapped winding, at any given instant, one diode from the upper arm of the bridge is conducting, and one from the lower. When the phase at the winding ends reverses, during the following half cycle, the other two diodes will conuct instead. The two diodes are in series in the circuit, not parallel, so it's winding end - diode - load - diode - other winding end. Thus any current in the load will be the same current as in either of the diodes. If there is 150 amps in the load, then there will be 150 amps in both the diodes, not 75 amps in each.

Arfa

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Yep.

Peter Wieck Wyncote, PA

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If the input to the bridge is AC each set of two diodes conduct for 1/2 cycle. On the other half they are NOT conducting and cooling down. If in that 1/2 cycle two diodes carry 100 amps, then the average of 8.4 msec. zero amps, and 8.4 msecs 100 amps, I calculate 50.

How else do you do it ?

greg

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