Foot pedal mods for PA160-STH

[joshuab]

I've got a couple questions about mods to the foot pedal for the PA160-STH. Another thread got me thinking that it'd be nice to be able to cut the pedal out of circuit without having to disconnect it entirely when I switch over to stick mode. Ideally, this would happen automatically when the welder was in stick mode (hint hint: future feature request!) but that would probably require mucking about inside the welder, and might not be compatible with my warranty. I don't know, though--would it be as simple as jumping a pin of the pedal jack over to a pin of the TIG/switch button, causing a certain pin to be pulled to ground when the welder was in one mode or another? Hmm... maybe so. An alternative that would work without opening up the welder would be to install a breakout box on the cable or something like that. The breakout box could have an on/off switch and a pot on it for adjusting the max output, just like the pot that's currently on the pedal, but I would be able to adjust max output without having to bend over and pick up the pedal and push it with my hand.

The second mod that I'd like to make would be to decrease the sensitivity of the pedal's max-output pot towards the high end of its range. As I've mentioned previously, the pedal's max-output pot goes basically from 120 amps right to 160 amps and pegs there for the last 1/8" or so of the dial's travel. Mark has explained to me that the pedal's pot needs to be logarithmic because... resolution. Or something. Sorry--here's my ignorance showing. I don't really have a strong electrical background, but I can solder components together if given instructions! So I'm wondering if there was some mod I could make to the max-output pot to change its response curve to have less sensitivity at the top end. For example, I once read about how you could build a semi-logarithmic pot by putting some resistors in circuit with a linear pot in the right way. So maybe there's a way to change the response curve of this one.

Thanks!

[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]

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]

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]

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?

[joshuab]

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

[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.

[Rambozo]

The wiper of the pot is what is hooked to pin 4. You are reading the resistance of the pot when you move the pedal. All pots have a section near each end of travel where the wiper goes to zero ohms to that pin, that is normal. Some photos or a better drawing would really help. What you are dealing with is a voltage divider the max amp pot is wired in series with the pedal pot, and goes to zero ohms for max amps. What you are looking for is a combination that will give you max amps at full travel, and about half amps at half travel. You do not want a log output, as that is what you have right now. You need to make the circuit more linear, in it's behavior. If you parallel a resistor across your max amp pot, you will lower the resistance. The taper will not change very much. Most pots are labeled with the taper A = log and B = linear. It looks like what you have there is a 470 kilo ohm log taper pot. A photo of how the leads are connected would help determine if the taper is wired in a way that makes it more touchy at high settings.
An no a pot never goes open, you are just not using the correct range to measure it. Yours should go from aprox 470k to 0 but I have to see exactly how the legs are wired as often when a pot it only using two leads, the wiper is shorted to one end. This makes it behave slightly different.

[joshuab]

Thanks for sticking with me here, Rambozo. I'm attaching some pics that will hopefully help make things clearer for you. It's not quite visible in the pic, but the pedal pot is labeled 22k A. Thanks for telling me what the A was, by the way. The only way I could think to read it was Kilo-amps, but that made no sense at all.

On all pots that I have used, the middle lug is the wiper and the outer two lugs are wiped-between. That seems consistent with what I observe when I put a multimeter on these two. On the pedal pot, resistance between Y/G and Brown goes from 2.4 Ohms to 22 kOhms when I push the pedal down. Resistance between Red and Brown goes from 22 kOhms to 2.4 when I push the pedal down. So my reading of it is that Brown (middle lug) is the wiper and Red and G/Y are being wiped-between. If that's true, then a more formal wiring diagram would be as shown below:



You're right that both Brown and G/Y are not attached to pin 4. I was confusing myself. When the pedal is pushed, G/Y shows connectivity to pin 4 because it is being connected to Brown through the pot. G/Y is actually connected to pin 3. Brown to pin 4. Gray to pin 5. You're also right that the pot doesn't go to zero Ohms. I just got sloppy in my speech. It's actually about 2.4 Ohms.

[joshuab]

I got to thinking... changing the output of the pedal's potentiometers isn't going to address the problem, I don't think. Let's say that we were able to change the pedal's pots to a curve such that the output of the welder became linear relative to dial position. That would look like this:



Problem is, you've still lost a ton of sensitivity when the pedal pots' resistance is so close to the top of its logarithmic curve. So it seems like this is kind of an intractable problem--either the pedal's resistance is more or less linear and the welder's output is logarithmic (this is the default), or you make the welder's output linear by converting the pedal's resistance to a log curve. Either way, you've got the "tail" of that logarithmic curve where you lose sensitivity.

[joshuab]

Can you tell I love data? To get a better understanding of the pots' behavior, I mapped resistance vs. percent of dial travel. Here's the result:



Two things I don't understand. First, since the pots are logarithmic taper (A type), why does their Ohms-vs-Travel look linear? If this was a logarithmic unit, like dB, then it would make sense that a logarithmic output mapped on a linear scale would look linear, but I don't think Ohms are logarithmic, so if the pot's output is log, and we map it on a linear scale, it should look logarithmic. No?

Second, notice that the two pots' response is basically identical, as you would expect. So why is the pedal pot soooo smooth and easy to control, but the max-output pot is all jumpy at the end of its travel?

This is the problem with loving data. I'm going to bury myself in data and be no closer to an understanding...

[Rambozo]

Two things I don't understand. First, since the pots are logarithmic taper (A type), why does their Ohms-vs-Travel look linear?

Because it is.
With Chinese pots, it's like that box of chocolates, you never know what you're gonna get. The problem is that the pot codes are different in Europe. Japan usually uses US coding, but China uses whatever they are copying, or maybe they flip a coin. I don't know, I just know they use different codes all the time.

In the US and Japan:
A - Audio or Log taper
B - Linear taper
C - Reverse Log taper
G - Graphic taper

In the EU (mostly, there are exceptions):
A - Linear taper
B - Audio or Log taper
C - Reverse Log taper
G - Graphic taper

So when in doubt you just have to measure. Gotta love when they use the same code letters with different meanings.

[joshuab]

Because it is.
With Chinese pots, it's like that box of chocolates, you never know what you're gonna get. The problem is that the pot codes are different in Europe. Japan usually uses US coding, but China uses whatever they are copying, or maybe they flip a coin. I don't know, I just know they use different codes all the time.

Well, at least I'm not crazy!

[joshuab]

Can anybody tell me the exact connector that is on the pedal? I have found some 7-pin microphone connectors that look almost the same, but the radial pins are not evenly spaced around the center pin like they are on the Everlast pedal.

Here's an example of what I've found, that I don't think is right. http://www.vetco.net/catalog/product...FY5r7AodpwwAzg