Foot pedal mods for PA160-STH

[Rambozo]

Yes, I'm sure what you want could be done. You would have to check with Everlast about warranty issues. I would guess that you would be on your own there, as you might be able to mess things up. If the warranty is important to you, I'd stop as you mentioned that you don't have a lot of experience.
You can change the way the pot works, but it's a bit beyond the scope of forum posts. You would have to reverse engineer what is going on and you might have to open up the welder to get that info. I'm not even completely sure that just opening the pedal sense pins takes everything out of circuit. It might take out the panel amperage but leave the pedal parts in circuit, that could make your adjustments strange or non-functional. You have to think about the original design. The purpose of those pins is to bypass the panel control and route it to the pedal. A little testing would be required to determine that the reverse is true. So you might need a few more contacts on your switch to get the job done. I believe someone has done a breakout box like you mentioned, but I'm not sure for what model welder. You can search the forum for more info. It would be handy if switching to stick mode bypassed the pedal. However, I have stick welded once using a foot pedal for fine amp control. I was filling in a varying gap from a bunch of previously torch cut pieces way up under a roof where I really couldn't do much to improve the fit. The only machine available was a TIG and it turned out having the foot amp was kinda cool. HF start was nice, too. Just another tool in your belt for the odd job. I know Miller has a belt mounted wireless remote amp for pipeline stick welders so you don't have to climb out of a trench to switch your machine.

[joshuab]

That's a good point about cutting the panel amp control out, but maybe not cutting the pedal entirely out of circuit. Depending on the number of pins involved, it could still be doable to buy an multi-pole switch and just cut the lead entirely.

Regarding the pot, I'd love it if you had any search terms or links to more information.

[Rambozo]

That's a good point about cutting the panel amp control out, but maybe not cutting the pedal entirely out of circuit. Depending on the number of pins involved, it could still be doable to buy an multi-pole switch and just cut the lead entirely.

Regarding the pot, I'd love it if you had any search terms or links to more information.

There are both log and linear pots, and also solid state pots that can produce almost any curve you want. Post a schematic of the two pot pedal and I'll have a look at it.

[joshuab]

Post a schematic of the two pot pedal and I'll have a look at it.

I'll get back to you on this. Have to open the pedal up first. And have dinner with my family ;-)

[joshuab]

Well, who doesn't love data? Just for kicks, I measured the resistance on pins 4 and 5 of my pedal. Here's the results.



It occurs to me that one other useful thing would be to add the welder's power output onto the same graph, to see how it is mapping resistance to weld output. There's no reason to assume a linear relationship there.

[joshuab]

Okay. One more data point. This adds the front-panel output value, mapping it across the resistance values. The pedal was full-down for the front-panel series. The welder was plugged into 110v power for the exercise, so I could have it in the same room as my computer.



I think this really clearly demonstrates the issue. More than half of the dial's travel range is spent between 27 and 19 amps, which is only 10% of the usable range of the pedal (96 to 19 amps).

[joshuab]

Ok. I opened up the pedal and poked at it with a multimeter. Here's my wiring diagram.



To sum up: The gray wire goes to pin 5. The brown wire goes to pin 4. The green/yellow wire goes to pin 4, but only when the pedal is pushed. Hopefully, this is enough information for a person familiar with electronics to suss it out. I'm a bit surprised to find that the max-output pot doesn't have three wires going to it, but maybe it's just being used as a rheostat. Or maybe there's some nuance of electronics there that I'm missing--quite likely.

I have been thinking about the graph of amp output relative to potentiometer resistance. The pot seems to have basically a linear response. The amp output line seems basically logarithmic. It seems that if the max output pot could be caused to be inverse-log, then the amps output would become linear. Any intelligent people have thoughts about this?

[Rambozo]

Ok. I opened up the pedal and poked at it with a multimeter. Here's my wiring diagram.



To sum up: The gray wire goes to pin 5. The brown wire goes to pin 4. The green/yellow wire goes to pin 4, but only when the pedal is pushed. Hopefully, this is enough information for a person familiar with electronics to suss it out. I'm a bit surprised to find that the max-output pot doesn't have three wires going to it, but maybe it's just being used as a rheostat. Or maybe there's some nuance of electronics there that I'm missing--quite likely.

I have been thinking about the graph of amp output relative to potentiometer resistance. The pot seems to have basically a linear response. The amp output line seems basically logarithmic. It seems that if the max output pot could be caused to be inverse-log, then the amps output would become linear. Any intelligent people have thoughts about this?

Kinda hard to tell things from your diagram, maybe pictures would be better. You have two leads going to pin 4 and that isn't correct. Where the wires go on the pot are important. A pot has a fixed resistance between two leads, with a wiper that can sweep between them, connected to the third. That one photo of the max pot is an odd one, too. It has three leads but they are placed in a way that I would check everything with an ohm meter to verify just how it is constructed, it may be something a little out of the ordinary.

[Mr120]

Kinda hard to tell things from your diagram, maybe pictures would be better. You have two leads going to pin 4 and that isn't correct. Where the wires go on the pot are important. A pot has a fixed resistance between two leads, with a wiper that can sweep between them, connected to the third. That one photo of the max pot is an odd one, too. It has three leads but they are placed in a way that I would check everything with an ohm meter to verify just how it is constructed, it may be something a little out of the ordinary.

A picture of where the leads connect to the potentiometer would be good. That odd lead off to one side could be a ground like the one below:



I'm thinking there is another lead hidden in the picture?

[joshuab]

Kinda hard to tell things from your diagram, maybe pictures would be better. You have two leads going to pin 4 and that isn't correct.

There's no question that that's how it's hooked up. With the pedal in the neutral position (full up), there is a 0-ohm connection between the G/Y lead and pin 4, and an open condition between the Red lead and pin 4. As the pedal is pressed, the resistance on the G/Y increases and resistance on Red decreases. Brown is always 0-ohms to pin 4, regardless of pedal position.

Where the wires go on the pot are important. A pot has a fixed resistance between two leads, with a wiper that can sweep between them, connected to the third. That one photo of the max pot is an odd one, too. It has three leads but they are placed in a way that I would check everything with an ohm meter to verify just how it is constructed, it may be something a little out of the ordinary.

The max pot has two wires hooked up to it. The third lead is disconnected. My understanding is that this causes it to act as a rheostat instead of a potentiometer.

I did a lot more poking around inside the pedal with a multimeter tonight. I read that you can convert a linear pot to semi-log by attaching a resistor in parallel. I have some resistors up to 10 kOHM, but given that the max-output pot is something like 500 kOHM, they didn't even budge it. After quite a lot of thrashing about, I hit upon the trick of rubbing some pencil lead on a piece of paper and using paper clips to create a variable resistor (based on the distance of the clips). Actually, I knew that trick, but I was pretty sure we didn't have any pencils in the house. Anyway, this allowed me to at least characterize the effect of putting a resistor in parallel, even if the exact resistance wasn't particularly precise. Basically, what I found is that you can approximate a semi-log curve from a linear pot by putting a resistor in parallel, BUT you do this by decreasing the max output of the pot. The less linear you want the curve to be, the more you have to reduce the max output of the pot. This had exactly the opposite effect that I was looking for! By the time the dial got sufficiently non-linear, the max output was only around 50 amps (on 110v). I had a TON of resolution within the range of 20-50 amps, but that's hardly what I'm looking for. I have no trouble dialing in 50 amps with it the way it was shipped! It's stuff in the 70-96 amp (on 110v) range that's the issue!

Also, I discovered that the pot used in the pedal appears to have a "runoff zone" at either end. What I mean is that when the pot is hard-over one way, it is full open, and when it is hard-over the other way, it is 0-Ohm. What this means is that, no matter what kind of tomfoolery I might pull with resistors, the pot is always going to jump abruptly to max in the last 1/8" or so of travel. I think this "runoff zone" is part of what's causing the problem--the pot goes very quickly from close to its max value RIGHT on up to wide open, and the welder interprets that literally. I wonder if the situation could be addressed by putting a resistor in series, thereby basically pulling the whole curve down to a lower point on the pot's travel. By the time the pot got to the threshold of its "runoff zone," the welder would already be at max output and you wouldn't notice the pot jump. If this worked, it would have the effect of losing amps off the bottom of the dial. Instead of being able to turn the dial down to 20 amps, you would only be able to turn it down to 30 amps, or something like that.

The long and short of it, IMO, is that the only way to even attempt to address the issue would be to swap in a new pot with the same resistance, but a different curve. I still have some hope that if an inverse-log pot could be used, additional sensitivity in the upper output range could be gained without compromising lower-end range.

[joshuab]

A picture of where the leads connect to the potentiometer would be good. That odd lead off to one side could be a ground like the one below:
I'm thinking there is another lead hidden in the picture?

No. The max-output pot only has two leads. My (amateur) understanding is that this turns it into a rheostat (burns off excess electrical energy as heat) instead of a pot (acts like a voltage divider, cutting current between two leads). I agree that the fourth lead is ground, as it is a part of the back cover of the rheostat. It appears not to be used, and my best guess is that's because the circuit is deriving reference from the welder itself, via the pins in the connector.

[joshuab]

... and here's the marking on the back of the pot, which I'm sure will be very informative to a knowledgeable person.