New to Forum with that old chesnut wire speed/amps question

[TobyOne]

Hi,

whenever I need an answer to a fabrication question I usually end up here, such a wealth of information!

So here's my first post, don't laugh!

So been using a Mig welder for a few years now, hobby use, done a bit of Tig and stick, so not really a newbie but not terribly
good at any of them! Treated myself to new Mig the other day and this particular model apparently calculates wire speed/volts
according to amps required, works quite well to although I think it actually calculates voltage from the set wire speed and amps
requested, however this got me to thinking

Why does wire speed affect the current?

As I understand it, in short circuit transfer, the wire
touches the metal heats up rapidly, melts, breaks the circuit then makes the circuit again because of the wire feed pushing the
wire back to the workpiece and this can happen several hundred times a second.

So why would pushing more wire in increase the amps? perhaps increasing the wire speed would increase the arc frequency ?

I think that this might possibly be where the answer lies, the definition of an amp is

"the ampere is a measure of the amount of electric charge passing a point in an electric
circuit per unit time with 6.241 × 1018 electrons, or one coulomb per second constituting one ampere."

so as an amp is defined as current flow/time then by increasing the frequency of the arc you are
increasing the current flow (amps)?

or

maybe it's too early in the morning and I'm overthinking this?

Please let this be the answer, it's been driving me nuts!!

 

[Mark 2]

Take a sear Sir, a boffin will be along shortly.

 

[TechnicAl]

Constant Voltage MIG set.......The set uses Wire Feed Speed and Voltage.....thats what you control......sometimes WFS is incorrectly marked Amps.........You set the WFS and the set gives you enough current to melt the wire (so its a constant burn off rate or deposition rate)....the stick out can influence the current.......eg 20mm ESO will have a higher Current than a 25mm ESO....because the Impedence has warmed up the wire.....So what you have is Constant deposition rate with variable penetration.

 

[TobyOne]

Constant Voltage MIG set.......The set uses Wire Feed Speed and Voltage.....thats what you control......sometimes WFS is incorrectly marked Amps.........You set the WFS and the set gives you enough current to melt the wire (so its a constant burn off rate or deposition rate)....the stick out can influence the current.......eg 20mm ESO will have a higher Current than a 25mm ESO....because the Impedence has warmed up the wire.....So what you have is Constant deposition rate with variable penetration.

yes, this seems to apply to my new mig, it displays amps but you adjust the wfs and it calculates the voltage to set the machine to the desired amps. it does have a fine tune for the voltage too, however that wasn't really the question more of a comment in passing what I'm trying to figure out is how wire feed speed affects current, not on any specific welder? in fact just ignore the first paragraph of my post

 

[TechnicAl]

yes, this seems to apply to my new mig, it displays amps but you adjust the wfs and it calculates the voltage to set the machine to the desired amps. it does have a fine tune for the voltage too, however that wasn't really the question more of a comment in passing what I'm trying to figure out is how wire feed speed affects current, not on any specific welder? in fact just ignore the first paragraph of my post

Sounds like you have a synergic set...there is a realtionship between WFS and Voltage.....synergic sets calculates the voltage for you and you do the fine settings

The faster the wire feeds it needs more current to melt it..........or do you mean the electronics side....

 

[Robert Weaver]

Found this which might help explain why...CV and CC

There has been discussion on this forum in the past, but I can't remember the context, so cant link to the thread.

Something along the lines of increasing wfs shortens the arc length, which reduces resistance and increasing current if voltage it constant.

If you're setting current, the set is either CC or perhaps duty cycle controlled waveform. The Uptime MIG160 seems to be the latter, and it seems works reasonably well.

 

[TobyOne]

allSounds like you have a synergic set...there is a realtionship between WFS and Voltage.....synergic sets calculates the voltage for you and you do the fine settings

The faster the wire feeds it needs more current to melt it..........or do you mean the electronics side....

ok, lets forget about my new welder, it was what got me thinking that's all.

On a mig where you have both variable wfs and voltage, we're constantly told that wfs controls the amperage, what I'm try to figure out is how?
if Voltage determines the max current that could flow, resistance determines the current that will flow, so on a short circuit all the current available set by the voltage will flow.
Now when the arc is struck the current flow will be maximum, when the wire burns back the resistance goes up and the current goes down, the wire feed is feeding the wire at a speed
less than the speed it burns back so eventually the circuit goes open (no current flow) and stays like that until the wire touches the work piece and another arc is struck.
This happens between 20 and 100 times a second depending on wfs.

I'm theorising that the reason the current flow increases when you increase the wfs is that it increases the frequency of the short, burn back, open circuit cycle.

This may seem a bit pedantic and possibly more involved than you need to know just to use a mig welder, but I'm one of those unfortunate people who can only
learn stuff if I know the whole process.

 

[TobyOne]

Found this which might help explain why...CV and CC

There has been discussion on this forum in the past, but I can't remember the context, so cant link to the thread.

Something along the lines of increasing wfs shortens the arc length, which reduces resistance and increasing current if voltage it constant.

If you're setting current, the set is either CC or perhaps duty cycle controlled waveform. The Uptime MIG160 seems to be the latter, and it seems works reasonably well.

yes, I think that this what, in my own ham-fisted way, I was trying say, but I think it's not just the arc length resistance. In mig short circuit mode
the arch breaks down and is struck again many times a second, if the frequency of that is increased then you use more amps?

 

[Justme]

Faster wire will be in short circuit for longer so more amps flow.

 

[eddie49]

In mig short circuit mode the arc breaks down and is struck again many times a second, if the frequency of that is increased then you use more amps?

Faster wire will be in short circuit for longer so more amps flow.

Yes, I think both these statements are correct. If you measured the DC output current with a Hall effect current probe and an oscilloscope, at 20 to 100 times a second you would see pulses/spikes that vary in width and frequency. If you use an Ammeter you will get an average of the current flow, and that will increase the faster you feed the wire into this short/arc/melt/repeat process.

 

[TobyOne]

Faster wire will be in short circuit for longer so more amps flow.

Good point, again it's based around time, time in short circuit and frequency.

 

[TobyOne]

Yes, I think both these statements are correct. If you measured the DC output current with a Hall effect current probe and an oscilloscope, at 20 to 100 times a second you would see pulses/spikes that vary in width and frequency. If you use an Ammeter you will get an average of the current flow, and that will increase the faster you feed the wire into this short/arc/melt/repeat process.

It's the realisation that amps is a measurement of current flow over time that makes it all make sense. So there are at least 3 factors which will change the current flow then, Contact time, frequency of cycle and arc length?