How to limit welder current (heat)?

Hi All, I admit to being electrically dumb, well one step above dumb, I'm a software engineer not electrical engineer, but I don't even know enough enough to be dangerous yet:) I'm going to be converting an ac welder to dc. It's a harbor freight 90 amp flux wire job, comes from the factory as AC for some unknown reason, but should be DCEN for FCAW welding. There are several walk throughs availabile online. I don't know the rules for linking to other forums, so I won't post a link, but you can google 90 amp welder dcen conversion if you want details. I attached an image of the circuit I got from a post by bluecatfish onweldingweb. I'll be taking the transformer output, running through a full bridge rectifier to convert to rippled DC, using capacitors to remove ripple, and lastly running through a torroid inductor. Another problem with this welder is it only comes with high/low power settings. The low is still too hot and burns through thinner metals.  I want a way limit current further, but maintain voltage for a stable arc. I'll be adding a bleeder resistor to drain the caps, which gave me an idea. At first I thought to use several more resistors in parallel with the bleeder to reduce current. I imagine this might work but be horribly inefficient. It would still use all available power from the supply, just converting some of it to heat, leaving less power for the arc. I've read that adding resistance in series will reduce current, but won't that drop voltage available for the arc? Then I was thinking, I have a motor speed control for a router. I believe this is a pwm. Could it be used to chop the mains input (120v 20a) to the transformer and reduce overall output without effecting voltage(I've read this is 38v 80a in the factory state before my mods)? since I'm adding capacitors downstream, will they just discharge too fast leaving me with ripple/pulsing? If so, how would I slow the discharge rate? (I'm looking at 3x or 4x caps 22,000uF @65v in parallel on a bus bar, so 66,000 or 88,000uF total). Sorry if in not making sense, as I said in the intro, I have a lot to learn, and right now, the more I read, the more confused I get.

Edit:  After more research, and to hopefully use correct terminology, I think I'm talking about using a current divider when referring to multiple resistors in parallel to the welding leads. I think I'm referring to a switching regulator when I suggested using a pwm to chop the transformer input.
Can either of these work like I'm hoping? Is there a better way, that is relatively simple? 
Thanks in advance for your patience and help. 

Picture of How to limit welder current (heat)?
sort by: active | newest | oldest
-max-1 year ago

Resistor dividers are really not useful for high power applications. In order to do what you want, you would probably need heating elements from a large stove or something, that would be very inefficient and impractical.

Ultimately your best bet is to change the secondary windings, or get a smaller (or right type of) welder for the job.

-max-1 year ago

Chopping up AC would reduce average power delivered to a load, and this is precisely what modern dimmers do. However, this does by definition reduce the RMS (special type of average) voltage available. Again, to reduce current or power without changing the load, you need to reduce voltage. However, inductive kickback or just the really large currents the welder draws will likely destroy the dimmer, so this approch will not work.

-max-1 year ago

To reduce current to a load, you either have to change the load, or reduce the voltage. That is one of the fundamental concepts of ohm's law. In this case, the load is the arc. The arc is quite complicated load (I think a arc exhibits some negative resistance properties), I do not know much about welding, but reducing the voltage is how you can reduce current.

However, most power supplies have some output impedance, which tells you how much the voltage will sag as more current is drawn. Something like a really large car battery or power outlet will have a pretty low output impedance, which means that the voltage will be pretty constant no matter how much current is drawn, while something with a high output impedance like a tiny watch (coin cell) battery, the voltage will sag quite a bit with a heavy load. That tiny battery looks electrically the same as a ideal voltage source with a resistor in series with it.

Shorin4 months ago

Question for folks out there - what about putting a variac on the primary side of the transformer to reduce overall voltage & current on the secondary? I have to do something about the wire feeder that comes off the secondary, yes, but do I have to worry about inductive kickback damaging the variac?

DonaldF9 (author) 1 year ago

I have a 100A single phase rectifier on its way from China. The the upper limits of the transformer is actually 120A when is switched to high. From what I understand, the rectifier's limits are a function of its ability to dissipate heat, so it will be mounted on a decent heat sink, and have a dedicated fan too. I didn't come up with this, it's been done quite a bit before with good results reported. My twist is to throttle it on low power even more to be able to weld thinner materials.

Yonatan241 year ago
Does that bridge rectifier need to have diodes capable of having more than 90A flowing through them?
iceng1 year ago

AC is used to weld Aluminum !

The ratio of 20:30 turns does 60A/38v to 40A/56v and works better then a resistor.

DonaldF9 (author) 1 year ago

ugh, I had a nice long reply typed up, but lost it redirecting to the login screen. This is the short version.

First, I really appreciate the responses, I'm learning a lot. Googling current limiting resistor helped me connect the dots. I wasn't grasping Ohm's law.

I think I get the resistor approach now, I'm sure I could make that work.

Max is right, I should just fork up for a TIG but then I wouldn't get to learn and create, but has given me another idea. I don't want to destroy this transformer or permanently alter it, I'm looking to get more range from it, so I was always planning to use either a manual jumper or a robust transistor to switch the mode for lower current.

But..the current transformer has I think 6awg on the secondary. What if I made a 2nd transformer with 20 turns 6awg in the primary and 30 turns of 8 awg on the secondary. If I understand correctly, that would step back up from 60A/38v to 40A/56v. My rectifier and caps should still work in that range. It would be lossy, but would it be any more inefficient than a resistor? It seems like it would be less heat to deal with too. Does this sound workable, or would it overload the main transformer?

iceng1 year ago

I use easily obtained used water heating elements in series or parallel to control that level of current.

Firstly you need to consider the right capacitors for the job as otherwise they can overheat, explode and cause big stinky mess.
Secondly and as said by Max you either need to reduce the load or add something to limit the curren - not that easy in the 50A region.
Your best and cheapest option to try might be 6V lantern batteries.
Remove the carbon rods, make holders for them and use the rods as current limiting resistors.
Even when glowin white they will not fail, just ensure the mounts and surrounds can take the heat.
Experiment with the amount of rods to get the current where you want it.