My suggestion or question was could a suitable resistor be used to draw current on the negative half cycle, purely to "balance out" the flux in the transformer? This would obviously need a diode to make it work in one direction only - a "total loss" system if you like?I don't know what a diode resistor is.
You want full wave rectification. That requires 2 or 4 diodes depending on the transformer configuration.
Jack
Good link - some disagreement on the effect though. still not had my question answered - the answer to which my reading would suggest depends on how much reversed current the transformer needs to collapse the biased field? I confess I don't have a clue...There's an interesting relevant thread here:
Is a half-wave rectifier particularly hard on a transformer?
In the book Practical Electronics for Inventors, 3rd Ed., the authors recommend against using half-wave rectifiers because they're inefficient and cause "...the core to become polarized and to satu...electronics.stackexchange.com
I didn't read it in detail but it looks to be considering a small PSU. The effects are more when pulling 200 amp DC pulses through a transformer.Good link - some disagreement on the effect though.
still not had my question answered - the answer to which my reading would suggest depends on how much reversed current the transformer needs to collapse the biased field? I confess I don't have a clue...
If you place a resistor in series with a diode across the transformer, such that it conducts the opposite way to the required output, then you will get a "balancing" current on the negative half cycles no?I didn't read it in detail but it looks to be considering a small PSU. The effects are more when pulling 200 amp DC pulses through a transformer.
Some transformers are designed to have a DC current - "wet" transformers - these usually carry less than 100 mA, about 1000th of the current in a welding transformer.
If you add a diode (depending on the transformer configuration) you are back to a full wave rectifier. Adding a resistor to limit the output current would be pretty ineffective. Welding transformers (depending on CC or CV type) already have a significant output impedance to create the "slope" or "droop". Adding any significant additional resistance will kill the arc and, unless it has a large heat sink, would probably burst into flames.
So, I think the short answer is no.
Jack
It is an asymmetrical load so no, it will not balance.If you place a resistor in series with a diode across the transformer, such that it conducts the opposite way to the required output, then you will get a "balancing" current on the negative half cycles no?
If you use a 1 Ohm resistor in the circuit of a 200A welding machine, it will dissipate a peak of 40kW. That would require a no-nonsense resistor.The comments about "bursting into flames" is just nonsense you can apply to any power circuit, the resistor just needs to be suitable for use.
The current has to be balanced. If it is 200A in one direction, it needs to be 200A in the other direction.But, the question is, how much current needs to be drawn to reverse this flux saturation? Does it need to complement the current drawn by welding? I'm not trying to be awkward here, just picking someone's brain.
That's assuming 200V.If you use a 1 Ohm resistor in the circuit of a 200A welding machine, it will dissipate a peak of 40kW. That would require a no-nonsense resistor.
Ah but if this was a small tig welder you'd be down at say, 60A, and the resistor would only be working at half an ac cycle, so the wattage would be a fraction of 40kW. ..
But, if the current needs to be balanced to prevent saturation, then thats it, end of...
Why was this proposed over a bridge rectifier again?
Amps, like a welding machineThat's assuming 200V.
It was never proposed, it was a question out of curiosity. Perhaps a trade school "what if?".Why was this proposed over a bridge rectifier again?
Not really...Amps, like a welding machine
It was never proposed, it was a question out of curiosity. Perhaps a trade school "what if?".
What followed was the surprising part.
OK, back to basics. You said a 1 ohm resistor, passing 200A. That needs 200V. Voltage equals resistance times current.Amps, like a welding machine
You said I assumed 200 volts, I did not, I assumed 200 amps. I also said that the voltage drop would kill the arc.OK, back to basics. You said a 1 ohm resistor, passing 200A. That needs 200V. Voltage equals resistance times current.
Yes but that would mean we wouldn't learn anything. Yes it can be annoying to have some imbecile (me) asking the same question over and over until they get an answer that makes sense, but that's a positive.You said I assumed 200 volts, I did not, I assumed 200 amps. I also said that the voltage drop would kill the arc.
The original poster is long gone. That happens often and most say it is because they are ungrateful and only signed up to get one question answered.
That might be the case sometimes but look at this thread. It is as bad as the threads about "what generator do I need for my welder?", "what cable size do I need?" and "what size service and breaker do I need?"
Electrical circuits operate by the laws of physics. The original question should have been answered in less than 4 posts, not 54.
Jack
Fair enough, and I am not concerned with questions - I have lots myself.Yes but that would mean we wouldn't learn anything. Yes it can be annoying to have some imbecile (me) asking the same question over and over until they get an answer that makes sense, but that's a positive.
For this one I have learnt that core saturation is a real thing, but that there is some disagreement on it's consequences (present company accepted). If I ever need to try it, I feel confident I would know what to look for.
The point about the lawsof physics is a good one, but most of us here have forgotten most of what we've learnt...
It is an asymmetrical load so no, it will not balance.
If you use a 1 Ohm resistor in the circuit of a 200A welding machine, it will dissipate a peak of 40kW. That would require a no-nonsense resistor.
The current has to be balanced. If it is 200A in one direction, it needs to be 200A in the other direction.
Time to move on now I think.
Jack
Hmmm! Surely, if you say a 1 Ohm resistor in the circuit of a 200A welding machine will dissipate 40kW, that would imply 200V is being developed across that resistor? V = sqrt (P x R), and all that.You said I assumed 200 volts, I did not, I assumed 200 amps. I also said that the voltage drop would kill the arc.
The original poster is long gone. That happens often and most say it is because they are ungrateful and only signed up to get one question answered.
That might be the case sometimes but look at this thread. It is as bad as the threads about "what generator do I need for my welder?", "what cable size do I need?" and "what size service and breaker do I need?"
Electrical circuits operate by the laws of physics. The original question should have been answered in less than 4 posts, not 54.
Jack
Hmmm! Surely, if you say a 1 Ohm resistor in the circuit of a 200A welding machine will dissipate 40kW, that would imply 200V is being developed across that resistor? V = sqrt (P x R), and all that.
you two are getting things mixed up here. My resistor was a theoretical total loss resistor in series with a diode to make the current flow out of the transformer on a negative half cycle - not in series with the welding output. (to try to stop saturation of the transformer).As I said, and then said again, the voltage drop would kill the arc.
The point is that adding a resistor to reduce the welding current is futile. Constant current welding transformers are designed to have a "non ideal" (high) source impedance to create "droop". The significant mass and often fan cooling dissipates the resultant heat.
Of course, you can use Ohm's law to calculate a ridiculous voltage, but you can also use it to demonstrate the viability or otherwise of the original assertion.
Jack