Convert an AC to a DC Welder

[[email protected]]

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

 

[Arfa Daily]

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

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

 

[[email protected]]

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- Hide quoted text -

- Show quoted text -

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:

http://www.newark.com/29C8750/semiconductors-prototyping/product.us0?sku=NTE-ELECTRONICS-NTE6356

http://www.nteinc.com/specs/6300to6399/pdf/nte6354_65.pdf

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

 

[mike]

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

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

 

[Arfa Daily]

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- Hide quoted text -

- Show quoted text -

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

 

[Meat Plow]

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?

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?

 

[default]

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.

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.
--

 

[Baron]

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

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.

 

[Baron]

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

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.

 

[default]

Also, where could I get something like this?

http://www.surplussales.com/Semiconductors/DiodeSaleStud.html


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

-


(SDI) 1N3111 Semi-conductor device. 50v, 150 amps.
12.00

$10.00

--

 

[[email protected]]

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

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

 

[Arfa Daily]

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

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

 

[GregS]

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

 

[default]

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

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?
--

 

[mike]

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.

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

 

[Michael A. Terrell]

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

--
Service to my country? Been there, Done that, and I've got my DD214 to
prove it.
Member of DAV #85.

Michael A. Terrell
Central Florida

 

[GregS]

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

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

 

[Arfa Daily]

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

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

 

[[email protected]]

If there is 150 amps in the load, then there will be 150 amps in
both the diodes, not 75 amps in each.

Yep.

Peter Wieck
Wyncote, PA

 

[GregS]

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

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