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Thread: DIY TIG Pedal

  1. #46
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    Hi Joe, Guys,

    From looking at the circuit you posted the two variable resistors are in parallel. I did add the caveat that the assumption was that the internal pot was at maximum. Because you are taking the supply voltage off the wiper of the internal pot, the resistance will be whatever the internal pot is set to in parallel with the pot in the foot pedal.

    For example:
    So if the internal pot is set at full value and the foot pedal is at half value, the total resistance will be 1K / 2 = 500 ohms in parallel with 1K. Using the sum of reciprocals, 500 ohms = 0.002 plus 1000 ohms = 0.001 = 0.003. 1/0.003 = 333.3 ohms.

    Note these figures only apply if the internal pot is at maximum value !

    You said the control voltage was 5 volts. so using Ohms law, 5v/1000ohms is 5 ma. This is the maximum available current through the internal pot when at maximum value. The full 5 volts will be applied to the foot pedal pot.

    The current at 5V/333Ohms now equals 15 ma, which gives us the available current through the foot pedal pot.
    So 0.015 ma across 333 ohms gives 4.995 volts across the pedal pot. The voltage at the pedal pot wiper will vary from zero to 4.995 volts depending upon where it is placed.

    These figures take no account of the impedance that is presented to the pedal wiper. So the control circuit could be either voltage or current operated.

    I have connected my effort in the following circuit - and would like some critique, because it has for some reason slightly increased the indicated minimum current of my welder....
    I hope that this explains why there is a difference in current values with and without the pedal connected.
    Best Regards:
    Baron J.

  2. #47
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    Pedal is complete bar some rubber on the top and wiring the cable. No worries about it scooting over the floor because it’s bloody heavy. I need to get some rubber feet for the base too.







    It’s not the prettiest thing but if it works I don’t care. Waiting on a new cable to arrive before I can test it out.

  3. #48
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    Default DIY TIG Pedal

    Pedal grip sorted. Forgot I had some skateboard grip tape I used for a different project.




  4. #49
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    Hi Neevo,

    Looks good, very professional !

    That skateboard tape is also used on steps to prevent slipping. I have some strips of it attached to short lengths of wood that I use like a file on the lathe for deburring and such.
    Best Regards:
    Baron J.

  5. #50
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    Quote Originally Posted by BaronJ View Post
    Hi Joe, Guys,
    ....
    I hope that this explains why there is a difference in current values with and without the pedal connected.
    John, that explains in principle why there will be a difference, but your example shows that the MAXIMUM current setting should be less - which it isn't.... maybe 0.05V isn't enough to change anything at that level...
    The MINIMUM current is set at 0V - no matter how many resistances are in parallel. I suspect the small resistance of the new cable is the culprit, perhaps, allowing a very small voltage to remain and set the welding current up a few Amps. What do you think?
    Cheers, Joe
    retired - less energy, more time to contemplate projects and more shed time....

  6. #51
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    Quote Originally Posted by jhovel View Post
    John, that explains in principle why there will be a difference, but your example shows that the MAXIMUM current setting should be less - which it isn't.... maybe 0.05V isn't enough to change anything at that level...
    The MINIMUM current is set at 0V - no matter how many resistances are in parallel. I suspect the small resistance of the new cable is the culprit, perhaps, allowing a very small voltage to remain and set the welding current up a few Amps. What do you think?
    If I'm understanding your diagram correctly, the 'X' on the 5v line on the left denotes that selecting foot control breaks the 5v line to the pot on the machine, which puts them in series rather than parallel. Which would make perfect sense given what you're experiencing.

  7. #52
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    Yes, the X denotes where the original connection is cut. It's the 0-5V output from the panel pot, now sent via the pedal pot for further voltage division. Vcc is the 5V line.
    Thanks for the confirmation.

    Sent from my WP5 Pro using Tapatalk
    Cheers, Joe
    retired - less energy, more time to contemplate projects and more shed time....

  8. #53
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    Quote Originally Posted by Jekyll and Hyde View Post
    If I'm understanding your diagram correctly, the 'X' on the 5v line on the left denotes that selecting foot control breaks the 5v line to the pot on the machine, which puts them in series rather than parallel. Which would make perfect sense given what you're experiencing.
    Hi J&H, Guys,

    No it doesn't, the 5V is taken from the wiper of the internal pot. The "X" indicates the break where the internal pot's wiper is broken and swapped for the foot pedal pot. If the internal pot's wiper is at any other position the voltage is going to be divided by the parallel resistance.

    John, that explains in principle why there will be a difference, but your example shows that the MAXIMUM current setting should be less - which it isn't.... maybe 0.05V isn't enough to change anything at that level...
    The MINIMUM current is set at 0V - no matter how many resistances are in parallel. I suspect the small resistance of the new cable is the culprit, perhaps, allowing a very small voltage to remain and set the welding current up a few Amps. What do you think?
    Hi Joe, The currents and voltages are so small in that control circuit I would suspect that the control circuit in the welder is a voltage sensing one, and I agree that there could be enough residual voltage to cause the effect that you are seeing. Disconnecting the internal pot from the 5 volt line and taking the external pot directly to the 5 volts supply would prove equivalence as would just opening the internal pot ground connection with its wiper set at the 5 volt end.
    Best Regards:
    Baron J.

  9. #54
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    Quote Originally Posted by BaronJ View Post
    No it doesn't, the 5V is taken from the wiper of the internal pot. The "X" indicates the break where the internal pot's wiper is broken and swapped for the foot pedal pot. If the internal pot's wiper is at any other position the voltage is going to be divided by the parallel resistance.
    I'm not seeing it (other than I did swap the 5v and the signal line before)? Far as I can tell from the diagram, 5v is supplied via VCC to the panel pot. Some portion of that is shunted straight to ground dependent on pot setting, the remainder feeds into the pedal pot, which again shunts some portion straight to ground and feeds the rest into the machine control circuit? To me it still looks like a series circuit....

    Did a crude diagram to hopefully illustrate how I'm looking at it, hopefully you can point out where I'm going wrong.

    schem.jpg

    Edit: Actually, I see where you're coming from - but I think it's both series and parallel. The first section of the panel pot is in series with a parallel section, one branch of which is the lower section of the panel pot, the other branch being probably the entire foot pedal pot.

  10. #55
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    Quote Originally Posted by Jekyll and Hyde View Post
    I'm not seeing it (other than I did swap the 5v and the signal line before)? Far as I can tell from the diagram, 5v is supplied via VCC to the panel pot. Some portion of that is shunted straight to ground dependent on pot setting, the remainder feeds into the pedal pot, which again shunts some portion straight to ground and feeds the rest into the machine control circuit? To me it still looks like a series circuit....

    Did a crude diagram to hopefully illustrate how I'm looking at it, hopefully you can point out where I'm going wrong.

    schem.jpg

    Edit: Actually, I see where you're coming from - but I think it's both series and parallel. The first section of the panel pot is in series with a parallel section, one branch of which is the lower section of the panel pot, the other branch being probably the entire foot pedal pot.
    Hi Guys,

    J&H, whilst your drawing is fundamentally correct, if the wiper of the internal pot is at the end where VCC is, then the resistance of the internal pot is effectively in parallel with the one in the foot pedal. But you are right insomuch that each pot is in series with VCC and ground.

    Somewhere I have a circuit that was used in a test paper that consisted of multiple series parallel resistors ! You had to work out what single value would replace the resistor network. If I can find it I'll sketch it out and post it.

    Neevo: Many apologies for taking your thread off topic.
    Best Regards:
    Baron J.

  11. #56
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    Quote Originally Posted by BaronJ View Post
    if the wiper of the internal pot is at the end where VCC is, then the resistance of the internal pot is effectively in parallel with the one in the foot pedal.
    Mostly true, except that from memory when I've bothered to measure a pot, they never go to zero. If we say the panel pot has a minimum value of 1 ohm, that is always going to be in series with the rest? Not much, but seems to me it's pretty likely to have enough effect to cause what Joe is seeing.

    And extrapolating out further - I assume Joe currently has the panel pot wide open, or mostly open, and I'm assuming 5v (ish) output on the signal line would be max amps. So if the foot pedal is set to minimum, and you then turn the panel pot down, the minimum current available should increase in value due to the increase in series resistance? Or have I located the wrong tree again?

  12. #57
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    I'll try it tomorrow and report what I find.
    Essentially I want to use the panel pot to limit the maximum I can reach with the pedal. That also means I should get better resolution with the pedal when I weld thin stuff. So instead of the pedal giving me 20 to 200A with a relatively slight foot movement, I'll get finer control.
    When I weld thicker stuff, say 5mm and up, I'm likely to set the panel pot to max and don't need to start at 20, so a small increase in the starting Amps won't matter....
    Cheers, Joe
    retired - less energy, more time to contemplate projects and more shed time....

  13. #58
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    Hi Joe, J&H, Guys,

    I've made the assumption that the internal pot causes the welder output to be maximum amps when the slider is at the VCC end.

    Based on your diagram (Joe) setting the internal pot at a lower value, say half way, would give the finer control of the output from the foot pedal that is wanted by Joe.

    This works because the resistances in parallel provides a lower value resistance with less voltage across it. Effectively a voltage divider supplied by a voltage divider. Which is essentially the circuit drawn by J&H.
    Best Regards:
    Baron J.

  14. #59
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    Quote Originally Posted by Jekyll and Hyde View Post
    Mostly true, except that from memory when I've bothered to measure a pot, they never go to zero. If we say the panel pot has a minimum value of 1 ohm, that is always going to be in series with the rest? Not much, but seems to me it's pretty likely to have enough effect to cause what Joe is seeing.

    And extrapolating out further - I assume Joe currently has the panel pot wide open, or mostly open, and I'm assuming 5v (ish) output on the signal line would be max amps. So if the foot pedal is set to minimum, and you then turn the panel pot down, the minimum current available should increase in value due to the increase in series resistance? Or have I located the wrong tree again?
    Hi J&H,

    There is often a residual resistance at the ends of a mechanical potentiometer ! This is caused by the moving contact surfaces within the pot itself. An ohm or so is a fraction of a percent of the pot's value and only causes a problem with very low pot values. Commonly these low value pots are used in high current circuits where an ohm would be a percent or more of the total resistance. Then you often will see that a wire has been welded to the wiper and taken to a terminal.

    Having the foot pedal connected and not in use should mean that the internal pot has very little ability to change the current output from the welder. It only starts to have an effect once the foot pedal is used to increase the output, because then the internal pot is supplying the external one and it is this that is now providing the control voltage.

    I may be wrong ! But I would bet that the welder output current is controlled by the voltage output from the foot pedal when it is connected. A higher voltage equals more current.
    Best Regards:
    Baron J.

  15. #60
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    Default DIY TIG Pedal

    That’s how mine works. The voltage flowing through the pedal is what controls the amperage on the machine.

    5v runs through the pedal and zero volts is pedal min and 5v is pedal max.

    My welder also appears to have a setting for min start volts. So arcs up at 20amps every time.

    My pedal is 10k to zero resistance for max amps. The double pot design reduces that min resistance with minimal impact on the starting resistance or the sweep. So effectively you can dial down peak resistance to get the adjustment.

    My adjustment pot is also wired across the wiper and max of the pedal pot.




    My Unimig TIG requires the connection of all 3 of the pot outputs. I assume because it uses these to sense the pedal. I tried to just connect the min and wiper but it wouldn’t work.

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