Questions re MIG Welder Circuit

[jackw19]

I'm hoping someone can help me understand a couple aspects of the attached circuit. It's a simple, low end MIG welder (Miller Cricket) rated at about 120 amps, 20vdc at 20% duty cycle. Even though it's a small machine, the basic design features are similar to arger transformer-rectifier MIG's, but without all the obfuscating bells and whistles.

BTW: VR1 is a 10 joule 68vdc varistor, R2 is a 20w 50 ohm wirewound and C1 is 46,000 uf 35vdc.

1) First of all, what's the purpose of (what I'm sure are) ceramic capacitors C2, C3 across the diodes of the center-tapped FWB? I understand these might have some role in supressing noise coming off the diodes--but what difference would that make in this simple machine? There are no IC's anywhere to be victims of noise. Anyway, some manufacturers use them, and some don't. I've seen them in 4-diode FWB's too. What would be a good size for these? .01uf, .1uf, 1uf, ... all of the above? I haven't been able to find a welder circuit diagram that calls out a value.

2) Some manufacturers also include ceramic capacitors from the output terminals (labeled work and torch in this diagram) to ground. What would be the purpose of such capacitors? What would be reasonable values?

3) What's the purpose of varistor VR1? Varistors are similar in some ways to diacs--at a threshold voltage they switch full on. So is this varistor to protect the big capacitor C1 against over-voltage spikes, or to protect the fellow with his hand on the work? NB: A varistor in this location doesn't seem to be a common feature in MIG circuits I've seen.

4) Finally, what's the purpose of the big resistor R2 across the output? Maybe to limit current when the electrode wire shorts to the work? Not every manufacturer uses them, but they are not uncommon on MIGs. A big 250 amp machine might have a 300w 50 ohm resistor in this location.

Thanks in advance; I really appreciate the help.

 

[wilker_wfd]

2) Some manufacturers also include ceramic capacitors from the output terminals (labeled work and torch in this diagram) to ground. What would be the purpose of such capacitors? What would be reasonable values?

the large use of this kind of capacitor is for work like a filter and his dimension , can be calculated use the work frequency and the required power

in the moment i dont remember but i will search 4 ya .. .

 

[user_88]

The purpose of the varistor might be to absorb energy in the secondary output circuit when the weld selector switch is moved from a higher to a lower increment. For a given welding range, there will be a certain amount of energy that is stored within the transformer core. As you go from a higher energy switch option to a lower one, the magnetic flux that is circulating in the core of the main transformer will have to dissipate ... until it reaches the level that is appropriate for the lower selector position. If you didn't have the varistor across the output circuit, you might get a spark across some other component that was not built to withstand such a high voltage.

The principle is basically the same as that of the ignition coil in an automobile.
A large primary current causes a magnetic flux to build up in the coil, and then the primary circuit opens .... points open or solid state version ....and the energy that has accumulated in the coil transformer generates a spark across the spark plug gap.... usually a very large voltage.

It seems that varistors switch on ..... or become conductive as a result of a voltage step or surge .... so this supports the argument. Varistors are also used for lightning protection of circuit boards, but the varistor value that you have given suggests that this is not the case here .... Lightning protection would require a much larger value.

 

[microtexan]

Me thinks R2 maybe a "bleeder" resistor to bleed off the charge of C1 when the machine is turned off. Although it seems kind of low value but it would discharge the capacitor.

 

[jackw19]

Thanks, gents.

BTW, I'm not so sure the resistor is there just to be a bleeder--although it will bleed the caps. I thought the bleeder rule of thumb was 100-500 uf per volt. This part just has too many watts and too few ohms to be an economical solution as a bleeder.

 

[user_88]

Thanks, gents.

BTW, I'm not so sure the resistor is there just to be a bleeder--although it will bleed the caps. I thought the bleeder rule of thumb was 100-500 uf per volt. This part just has too many watts and too few ohms to be an economical solution as a bleeder.

R2 ... as you suggest, might have a dual purpose.
It appears that there is a specific design effort in the secondary circuit so that there will be a ripple free voltage supply ... the large electrolytic capacitor .... and also a large choke or series inductor, designated with a 'Z', which would help create a significant current inertia to supply the weld tip. This current flow to the tip electrode would tend to build up a significant 'back voltage' at the instant when the weld trigger is sharply released, or if the weld spark were interrupted, as is frequently the case.
One analogy that stands out is the occurrence of water hammer, when you suddenly shut off a valve on a pipe that has a large volume of water going through it. The water pipe remedy is to install a surge damper ... usually a compressible pneumatic bulb of some sort .... so that the inertial energy of the flowing water stream can be absorbed ... essentially dissipated as heat .....
Without the presence of R2, the mean life of several circuit components would probably be shortened due to voltage spikes every time the weld spark was stopped or interrupted.

 

[Willbe]

1) First of all, what's the purpose of (what I'm sure are) ceramic capacitors C2, C3 across the diodes of the center-tapped FWB?

To protect the diodes from reverse voltage transients that must be all over the place when you're welding?
e=-L∆I/∆T, and I is huge.

50Ω and 46,000 uF gives a time constant of ~3 sec.
35 v is no shock danger so I guess R2 is there to actually turn the machine "off" when you switch it "off". For this cap, 1/2 CV² is quite a bit of energy lying in wait at the output terminals.

 

[jackw19]

Maybe R2 contributes to improved voltage regulation? MIG welding works best with "constant voltage" power supplies that have a very flat volt-amp curve. Charting volts on the vertical axis and amps on the horizontal, the slope of a typical volt amp curve would be on the order of -1/10.

Almost all MIGs use a constant speed wire feed, and this also contributes to good regulation. Assume the gun tip is held at a constant distance from the work. When wire stick-out is short, arc length and hence resistance is high; since voltage is roughly constant I=E/R forces current to be low. This reduces heat at the tip of the wire allowing it to extend further. Then arc length and resistance drop, current and heat increase burning back the wire and lengthening the arc.

I'm speculating now, but it seems the system would be most sensitive to current changes when the arc resistance was near 0, since small chagnes in R will cause large large changes in I=E/R for R small. Adding in R2 would set a lower bound on the denominator.

One question re user 88's damper theory -- would the damping action be the same if the resistor were on the _other_ side of the inductor?

 

[user_88]

One question re user 88's damper theory -- would the damping action be the same if the resistor were on the _other_ side of the inductor?

As far as using R2 for the trigger switch induced voltage spike remedy, it looks like you could physically place R2 anywhere between C1 and the weld tip electrode. .... Apparently, the advantage of placing it adjacent to C1 is to have it serve as a charge bleed when the selector switch is changed to a lower output voltage, or else the machine is turned off.

.... Strictly speaking, the inline inductor/choke, labeled 'z', will have a certain amount of energy stored in its magnetic field ... ½LI² .... that will have to be either used or dissipated ... depending on the trigger switch position. An excessively long transient period, related to this changing energy level could be inconvenient, or detrimental to the operation of the machine.

 

[chrisjpitt]

As mentioned by Willbe, capacitors C2 & C3, are there to protect the rectifier diodes against reverse voltage spikes generated by the transformer inductance, when the arc is initiated and ended (VR1 won’t clamp these spikes). The most important thing about these caps is their high frequency and pulse performance – hence the use of ceramic or plastic (Mylar) types. Value and voltage rating of the caps would depend on the size & energy of the voltage spikes you are trying to clamp, the diode voltage rating and the welder open circuit voltage. Generally, the larger the value, the more the caps will clamp spikes (for a given HF performance) – the upper limit being how much AC current you could accept being shunted by these caps.

The ceramic caps between the output terminals and ground are most likely there to stop RF (from the arc) being transferred back in to the welder, then in to the mains wiring. An arc welder is not that much different to a spark transmitter after all?

As also mentioned before, VR1 is there to absorb positive going voltage spikes and surges when the transformer and earth lead inductor are unloaded. C1 may be a very large capacitor, but being an electrolytic, it won’t have a very low impedance at HF and will not clamp fast rising spikes. Any high voltage spikes across C1 will also be seen by the rectifier diodes as extra reverse bias voltage - when the AC voltage on that diode goes negative.

The purpose of R1? It does not present enough of a load (0.5 amp?) to provide much in the way of regulation on a supply that may be loaded to 120A. It could be there to damp the reactive parts of the circuit (C1 & Z), but I think it’s too high a value to really have much effect, considering the welding currents involved. If it were there as a snubber (low impedance HF filter), it would normally be of a much lower value, with a bipolar capacitor in series with it. I suspect that R1 is there just to bleed the large amount of energy in C1 - not just for safety reasons (as the wire feed motor will bleed C1 away eventually), but to pull the output voltage down quickly, so the wire feed stops when the trigger is released?

 

[jackw19]

FYI gentlemen, I've done some additional internet research and found several MIG circuits w/ diode protection capacitors of .01uf and output lead to ground caps of .1uf. A few three phase machines also have caps from the mains to ground on the order of .003uf. Resistors in the R1 position when present are in the 5-50 ohm range.

chrisjpitt's point might be well taken. I've never quite understood the extent to which the drive motor was isolated from the welding current. Current is transferred to the wire at the exit wire guide, the wire is connected to the drive rolls and the drive rolls to the motor--I've just never gotten this. So if the big filter caps can dump current to motor post trigger release, that would cause a problem wire stickout. I do know that some machines have a solenoid operated brake on the wire roll ...