Electronics TIG Welder: Question about High Frequency Unit

[BlackMelon]

Hello,

According to this research paper: Optimizing the Transition Process from
Sparking for Non-contact TIG Welding
Inverters, a spark generator is used in order to start an arc between a tungsten electrode and a workpiece by generating a high voltage around 3.5KV.

Also, on this link: https://www.weldguru.com/tig-equipment.html, it says that the generator's output is at high frequency to let the current flow on the surface of a conductor. I know this as the skin effect.

However, I want to know why we need the current to flow only on the surface? Is it because we want the output current to heat the surface of our workpiece first? Can't we just pump voltage up alone, so we can start the arc?

Thank you

 

[FlapJack]

Skin effect is a property of materials, as a general rule we do not desire the effect.

The higher the frequency the less the skin depth is.

In the case of welding we are injecting a, high frequency / higher voltage low power signal over the, high current / low voltage, as a means to start the arc. If we increased the high current supply to 100 volts to start the arc from say 20 volts the total power put in to the weld would probably blast a hole right through the metal and the welder would have to supply all this power.

P=IxE
50A x 20V = 1000W
50A x 100V = 5000W

Using high frequency start saves us almost 4000 watts.

 

[BradtheRad]

My experience is with an AC welder. Whenever I tried to weld sheet metal, I vaporized it instead. I tried to figure out whether voltage was too high, or current too high. I tried to picture how I should reduce voltage, or Amperage, and make a combination that worked, but I could not figure it. (I believe the rods were thin enough, and new enough, which is very important.)

I want to know why we need the current to flow only on the surface?

Therefore I think this makes it easier to weld thin metal. An inexpensive welder sends current through the entire piece, and melts it, whereas we really want current just at the surface where it assists arc formation, and depositing of rod material. The more sophisticated TIG/MIG process achieves this.

 

[chuckey]

Brad, you need high voltage to start the arc and low current to stop the holes being burnt. So some series impedance some where will help a lot. A crude way is to have 5' of metal coat hanger wire in series with your LV side. It will get hot!!!
Frank

- - - Updated - - -

I am not sure that the writer has got this correct. The basic welder is a source of AC or DC with a very low source impedance so it can deliver 100s of amps at a low voltage, say 50V with out dropping a lot of voltage in side its self. So it might have an output impedance of .01 ohms. we need an HV source(> 1KV?) to break down the air and surface film so current can flow. So if we connected the HV source straight across the terminals of the welder, its .01 ohm impedance would kill our HV voltage source.
So we put a radio frequency choke in series with the welding current feed wound with very thick wire and couple our HV oscillator down stream of this. Because the choke has to carry a lot of LF current, it can not be too large in value, so the HV oscillator must be high frequency.
Frank

 

[BlackMelon]

1. it can deliver 100s of amps at a low voltage.
2. we put a radio frequency choke in series with the welding current feed wound with very thick wire and couple our HV oscillator down stream of this.
3.Because the choke has to carry a lot of LF current, it can not be too large in value, so the HV oscillator must be high frequency.
Frank

1.What is the word "100s of amps"? Is it 100Ampere or something?
2.In the newly attached picture, is your "radio frequency choke" an inductance across A and B? If so, I think it is also a secondary side of a high voltage transformer TX1 (like flyback transformer used in televisions).
3. I read some more papers, and found that the welding voltage generated by main inverter is about 125Hz. If the choke is too large, will it attenuate both inverter's voltage and that of high frequency unit? (I mean the cutoff frequency will go down... sth. like that)

Thank you

 

[chuckey]

1. 100s of amps is short for hundreds of amps.
2. Yes its the secondary of a high frequency transformer in this case, even so the main welding current has to flow through it.
3. Yes you have to keep the inductance within reason, but as the high frequency might be 1000 times higher then the low frequency it should not be to difficult. The other point is that when the very high current flows through the windings it could saturate the core of the windings at which point part of the the inductance disappears and it with its voltage drop .
Frank

 

[charliehill]

if the unit is electricity conductive, how is it possible to flow only on the surface?

 

[c_mitra]

if the unit is electricity conductive, how is it possible to flow only on the surface?

I assume you are seriously interested in knowing. The reasons are electromagnetic in nature and I will not be trying to draw diagrams- you need to imagine.

Consider a copper rod carrying DC current. The current will be uniformly distributed over the cross section of the rod. The current per unit area (perpendicular to the direction of flow of current) will be called the current density. In other words, the current density will be constant through out the cross section.

Now consider the same copper rod carrying the same current but this time it is AC. What difference does it make?

The DC current produced a magnetic field. This magnetic field has some effect on the current flow. The current (which is flow of charge) path will be affected. The current density will not be uniform any more. But how much will be the effect?

The magnetic lines of forces will be circular around the copper rod. Strongest magnetic field will be at the center and the magnetic field strength will decrease from the center. Therefore the current density will be somewhat less at the center and maximum near the edge (walls of the cylinder or rod).

What will be the effect when we have AC through the same rod?

Current flowing through the rod is not constant any more; it is changing with time and changing current gives rise to a changing magnetic field and a changing magnetic field gives rise to a changing electric field. The higher the frequency, the greater the rate of change and the electric field will be stronger. This is again going to push the current towards the surface of the conductor.

That is basically the skin effect; it causes the current at high frequencies to flow on the surface.

 

[Warpspeed]

The whole process of electric welding involves forming a plasma arc between the welder and the work.

Once that plasma arc is established, the current path through the arc has a negative resistance, and the current must be limited in some way or the arc will continue to grow to infinite proportions.

Once the arc is established, only a very few volts are needed to sustain the arc, typically around 15v ac or dc.

But getting the arc started in the first place is an entirely different thing.
A typical automotive spark plug with a gap of perhaps only 1mm might need 20Kv to break down the insulating air.

Very fast rise times, and sharp pointed or hot surface reduce the required breakdown voltage considerably. Some welding and plasma cutting torches use a fairly high power RF generator to ionise the air through corona discharge.
This is known as pilot arc.

Sometimes, as with stick welding, you just short the rod to the job, then break the connection, and the inductive back EMF begins the arc process.

While high frequencies, fast rise times, and skin effect may be important during forming the initial arc, although the current will be minimal. It has nothing to do with maintaining the arc, or the actual welding process.

In answer to your question, we don't need the current to flow on the surface.
But using a high frequency pilot arc will cause surface current flow just because it uses a high frequency.
Its just a by product of using high frequency. Its totally irrelevant really, its neither an advantage or disadvantage..