AC to DC Buzbox conversion plans?

I've got a Miller Thunderbolt 225 (your basic AC transformer welder), it's
been a very reliable unit for general welding but I sure would like to be
able to get DC out of it.
Does anyone sell a diode bridge and choke assembly that goes on the output
Any plans for such an add on accessory?
Will a bridge of 4 300amp diodes (2+ and 2 -base on a heatsink ground) work
without a choke?
Since a welder is regularly shorted ( striking arc, stuck rod etc.) on the
output, is a special type of diode or bridge needed to handle the overload
short conditions?
Spud: ve7ifd and that would be at telus dot net

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I have an old Miller Thunderbolt 225 and the voltage/current curves show a maximum current of about 280A when the voltage is 0 volts. This would be the voltage when the rod is stuck. In normal use say running at 150A instead of full power the stuck down current would be less because of the action of the constant current shunts in the transformer. So, a bridge of 4 300A diodes is quite adequate. Also, each diode only conducts half the time so you have even more of a safety factor.
The choke will add stability to your arc. One way of looking at it is that the field collapses and provides some current when the voltage goes through zero to keep the arc alive. I did some internet searching on this and recommendations range from "there is enough inductance in a set of long leads to achieve this" to "try and find an old one somewhere". Another poster recommended wrapping a few turns around a large bolt.
I built a rectifier for mine and I opted for another solution: Get an old microwave oven transformer (biggest you can find). Remove the windings by grinding the weld holding the E laminations and the I laminations together. Microwave transformers are assembled cheaply so the Es and Is are not interleaved thankfully. I wrapped as many turns of welding cable as I could around the center of the Es; only about 5 turns I think. You don't want to put the I laminations back on as they were. You must have an air gap or the high welding current will saturate the core. I put a piece of panelling between the Es and the Is to ensure a space.
I actually used 2 old transformers in series giving me about 9-10T total or thereabouts. No scientific measurements done on it but it works fine. The choke can be in either lead. My diodes are 12 35A 600V bridge rectifiers in parallel with equal lengths of #18 supply wire on each leg of the bridge to give some load balancing resistance. Your 4 diode method is better.
Your diodes should be rated at 150V or better. Put a 0.01-0.1ufd capacitor with a voltage rating of 200V or more across each diode. This will protect the diode against voltage spikes caused when the arc is broken and the fields collapse. You can also put a 0.1ufd across the + and - output leads as well.
Make sure you mount your diodes on a heat sink and try and get it in the fan's air flow. I built mine in an old desktop computer box and put a couple of muffin fans in to blow air on the bridges. I don't think they get too hot given the intermittent nature of welding but heat is the enemy of semiconductors so good cooling is cheap insurance.
Hope this helps, Billh
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Hello Bill and TMT, A chap on another group gave me a few clues on how to set the gap.
Connect the primary of the stick welder to a variac.
mains supply ----variac---stick welder primary
Connect say a 100 amp meter shunt and the choke under test across the secondary of the stick welder like this.
welder terminal------meter shunt--- choke-----welder terminal
Across the shunt, connect a digital voltmeter and an oscilloscope.
Turn up the variac from zero volts and look at the DMM and interpret the amps from the DMM (eg 50mV may equal 100 A)
Look at the oscilloscope waveform as you turn up the current flowing in the shorted out secondary of the stick welder. If you see the waveform changing from a nice sine wave to a slightly distorted sinewave at say 40 amps you will know that the choke core gap isn't big enough.
Reset the gap with thicker packing and test again.
Like you two guys, I found very little information on the web on how to build and test chokes. Well, information that wasn't written in mathematical gobbledegook for hobbyists like me.
Regards, John Crighton Sydney
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John Crighton
John, Thanks for sharing the test method. I liked electricity but magnetism and its relation to electricity sure took the fun out of it! bill
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This old article on adding a welder to a vehicle contains a description of a pretty simple means of arc stabilization. The homebrew inductor should be easy to make.
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billh wrote:
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Gerald Cooper
Wow sounds simple & complex. CAUTION on the variac - make sure it can handle the current the stick welder needs. I have half a dozen around the house and only the one I will get from Dad in a few months can handle that!
What is the Mains current - Mains voltage - simple VI = Power lets say a 220 at 25 amp - 5.5KW That is a big Variac. Maybe your mains current isn't that high.
billh wrote:
Reply to
Martin H. Eastburn
Hello Martin, by your comments and CAUTION I can tell that you have not read what I said.
Turn up the output of variac slightly, just above zero volts, is maybe what I should have said and then you would not have issued cautions. I apologise for not making myself crystal clear.
We are only using the stick welder as a piece of high current test equipment NOT as a welding machine.
We are testing chokes for the saturation point NOT WELDING.
The variac output voltage is adjusted by hand from zero to low volts while watching the digital mutimeter. The primary voltage to the testing transformer (which just happens do be a stick welder) is low. OK?
The secondary resistive load is the shunt.
Power is volts times amps as you pointed out. Voltage across the shunt is 50 milli Volts ( 0.05 V ) 0.05Volts times 100 amps equals NOT MUCH.
We are NOT welding! Just using the welder as a high current source at near zero volts output and with low voltage input to the transformer (which just happens to be a stick welder because it is convenient to use).
A small variac used as above will do nicely for this choke gap setting excercise.
Regards John Crighton Sydney
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John Crighton
John -
The transformer has copper loss and core loss so the unloaded transformer would be drawing some current and has power loss. When you shunt out the output except through a choke (aka coil with an iron core...) the secondary delivers current. This makes the primary deliver power through the core and draws current through the variac. The current is proportional to the turns ratio. Without the transformer in the normal operational magnetic curve position, the reactance will be different.
I would fuse the variac lead to protect it. Inductive inrush currents and reactive currents can be hard to deal with.
I have designed and made transformers and have a background in Tube designs.
Caution means to be 'cautious' not to 'stop doing this stupid'. Science lost hundreds of highly educated professionals to the unknown inrush of current from this or that flowing in their body.
John Cright> >
Reply to
Martin H. Eastburn
Excellent Martin! Now, how about giving the original poster and hobbyists like me a few clues on how he should build the choke assemby for his AC to DC conversion.
In case you missed the original request, here it is,
"I've got a Miller Thunderbolt 225 (your basic AC transformer welder), it's been a very reliable unit for general welding but I sure would like to be able to get DC out of it."
Reply to
John Crighton
Hi. How much is it worth to you? :-)
Seriously, I am planning to write up a document on how to make a micro-welder out of a defunct microwave oven.
The design of the reactor is straightforward if it is taken step by step. The key reference is by Eric Lowdon.
Estimate the energy storage required by drawing the plot of sine squared on graph paper.
For professionals: 1. Use Hanna curves for the required core size as a function of energy per cycle.
2. Design around this core size according to the book. This is an iterative process with the wire size and the gap thickness as variables. Start with a fixed wire size, calculate filling factor and turns. Select a gap to prevent saturation. Compute inductance and heat dissipation. Adjust wire size to obtain desired result.
For amateurs: 1. Use a junk 60 Hz transformer from an old battery charger or TV or flea market. I saw somebody selling them for 10 cents per pound.
2. Disassemble the core and measure the path length and cross section.
3. Bifilar wind the thick wire and quadfilar wind the thin wire, or more, and fit as many turns as possible.
4. Compute a gap size to avoid saturation.
5. Compute the resulting inductance.
6. If insufficient, repeat process and place inductors in series.
7. If multifilar winding is too much of a drag, single wind the primary, and bi or trifilar wind the secondary and place it in series. Then, parallel the inductors to obtain sufficient current capacity.
Warning: this procedure takes a long time. Make sure it is what you want to do.
Also, the above (very brief) discussion above makes it quite clear that the inductor design given in the aircraft alternator welder article will not work, since its inductance will be too low. The inductance is lower because these alternators typically produce much higher frequency. Also, if you draw it on graph paper, the output is 3-phase, so you are adding sine squared + sine squared alpha + sine squared beta, which has much less area (graph paper squares) in the dips. Any interest in a booklet?
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Hello Eric, Yes! Count me in. I am definitely interested. Regards, John Crighton Sydney
Reply to
John Crighton
In reference to the pdf file from hobartwelder, I believe there is a gross error. On page 21 (the plans) figure 7-1 is not right! The two upper Diodes are connected together at their bottom end, which shorts out the transformer, as well as the fuse/circuit breaker. You get the point.
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What I would do first is to measure the output open circuit voltage. Closed is close to zero. - close enough anyway.
I'd find out how it was measured - likely RMS and then generate Peak and if I had a Drantz power line sensor/logger I'd hook it up to the output and do some welding. What is the maximum peak hits. Baring that, some meters have peak holding or maximum holding. Ratio up from that for protection.
Find the Current the Thunderbolt can generate on a short - one considers 250 amps and down rated to 225 for specs. So the current would be 400 amp rectifiers.
The 400 amp would handle the peak inrush current from 0 to max stick current. (e.g. when the stick sticks and the transformer hums...)
The voltage is rated by the peak inverse voltage that would be seen - this is really tough to guess on my half - what is the Thunderbolt open voltage ? Does the Thunderbolt surge or oscillate when the process is started ? or does it have a controlled voltage signature that drops from open to short in a linear or near linear method and returns when the arc is cut. Transformer regulation and core design determines how 'wild' it is.
A good amp probe attached to a scope could show so much. Mine is packed now for a move.
Now for waveform shape. What a single or ganged (parallel) rectifier (power diodes) would provide is a half sine wave. The Positive or Negative pulse would be seen while the other alternation is almost flat. Duty cycle is about 50%. During the other 16ms of 'off' time the rectifier, mounted on a large copper plate or large Al plate - best with fins and a blowing fan to cool - perhaps a demand type based on temp.
This would be a DC without a choke - used to smooth the waveform. Not much good with one diode.
When four diodes (Rectifiers) - this is expensive as 400 amp rectifiers are expensive - one would build a Full Wave Bridge and this would take the typical sine voltage and convert it into a double positive or double negative waveform and the duty cycle is about 100%.
If one wanted a constant current output (or something like it if we keep a blind eye closed :-) ) we would add a choke in series or between the bridge and the output lead. A +/- switch exactly prior to the bridge would be best - then both - the line out of the switch then proceeds to the choke. What this is trying to do is simply keep the current the same value by building an electric field on the higher currents and clasping the field and inducing current to make up for the loss. Since the currents we are talking about are so huge compared to electronic power supplies, the regulation has to be tremendous. It is used to aid and not smooth the waveform. In other words, it is a smoothing device. But it will not produce anything like a battery or a near pure constant current.
My guess is this : The coils used are to generate a high voltage oscillation due to being out of regulation. I believe these are just to improve stick arc by making a higher voltage alternation when the stick current drops (as the hand moves...).
A regulation coil or choke to control a stick welder is a dream. Consider running a stick at 90 amps and another at 150. What coil will be useful on both ? Now kick it up a bit to 200.
You can consider the core power of a transformer or choke to equal 100 watts in a core the size of a large mans fist. Not much power considering a welder. What is the size of the transformer core ?
Reply to
Martin H. Eastburn
The Miller OCV on low current range is about 80V which is a safety maximum for welders. High range it is about 55V.
Max stuck down current on the high current range is about 280A according to the Miller VI curves.
Assuming a 300A max current limited by the constant current action of the shunted transformer a 150A diode in a bridge (50% conduction) will just make it.
Any transient is likely to be produced only when the arc is broken and the transformer and any reactor in the DC circuits fields collapse. I think commercial stick welders only use 200 PRV diodes.
be attached to a scope could show so much. Mine is packed now for a move.
The shunted transformer design provides for constant current for stick and tig welding. The regulation for stick welding is overly important and the transformer provides sufficient CC regulation. The inductor is not really there for smoothing in the "nice DC" sense; it is there so the current does not go to zero when the voltage goes to zero thus keeping the arc lit. You can look at this in the sense that the collapsing field generates current to keep the arc going when the voltage is zero or you can consider the inductor provides a few degrees of phase shift for the V and I. Note that once the arc is struck the voltage needed to keep going is only a few volts, nothing like the OCV.
The ideal reactor would be variable to match the current.
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Meant to say "regulation for stick welding is NOT overly important" Sorry, Billh
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Remember it isn't simple math - Imax(secondary) = Pi*Izero / 2 = 1.57 I0 = 1.57 I output.
Reply to
Martin H. Eastburn
Well if you want to quote the formula relating the average DC current for a full-wave rectifier as a function of the peak current per cycle in a bare rectifier circuit. The formula for the peak current in a half-wave rectifier is the same but without the amazing factor of 2, ie, Imax= 3.14 I output. So the full-wave diode only has to work half as hard.

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Bill - That was from a Motorola power supply design book. Martin
Reply to
Martin H. Eastburn
It shows the maximum (peak of the sinusoid) current in the secondary as a function of the rectified average DC current. The principal current rating for a rectifier diode (the one typically listed) is average current, not peak. My original point is that the average current through each diode in a bridge is 1/2 the average output current. billh
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Most folks are worried about calculations for these diodes because, the higher the amperage, the more expensive the diode. I have a simpler solution. If you live near a large city, go and visit the local scrap metal yards. Usually they will let you look around for stuff on your own. Find the biggest diodes that you can and make your own bridge rectifier. The reason that I am mentioning this is that most metal scrap yards only have prices for metal. i.e. aluminum might be a buck a pound, etc. When you walk up to a metal recycling yard, they don't have a price for electronic components. They see you with this tiny piece of metal in your hand and usually, you walk away with a small fortune in semi-conductor components for almost free. You can test them with a multimeter right on the premises, so you won't be taking any scrap home with you. I have built many things like this. My plasma cutter came from the metal scrap yard. It cost me $30 for the unit and only $6 for the components that needed replacing. It pays to dig around a bit. Big business is very wasteful and most items in the scrap yard really don't need to be there. Motors usually fail because of bad bearings or brushes. Bearings and brushes are cheap. Therefore, if you need that 5 hp motor for your shop air compressor, you can have it for usually less than 20 bucks.
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