How consistent is current shared in parallel conductors?

Discussion in 'General Electronics Chat' started by strantor, Apr 21, 2015.

  1. strantor

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    I've made a resistive spot welder from a pile of microwave transformers and it's a real beast. My clamp meter only measures up to 1000A, so I can't measure how many amps this thing is puking out. According to the efficiency that I can measure at lower input/output amperages (~80% efficiency), I calculate it's putting out over 3000A at direct short.

    I want to put an output current meter on it (fast-deflecting analog meter movement preferably). All my attempts at winding a torroidal current transformer for it have failed. Looking at 3000:5 and 4000:5 commercial CTs, I can see why I failed. The CT would have to be the size of a Bible in order to work. I don't want one of those ginormous iron bricks weighing down my machine, nor do I want to pay for one.

    So I'm considering placing a smaller CT around one of the parallel conductors of my welding leads. My welding leads/transformer(s) secondary are made of all the scrap wire I had in my garage, in parallel. They are comprised of:
    ( 1X) 3AWG stranded, bare, building ground/bonding wire
    (11X) 12AWG stranded, insulated, machine panel wire
    (16X) 12AWG solid, insulated (recovered from ROMEX)
    ( 8X) 12AWG solid, bare (recovered from ROMEX)
    ( 4X) 14AWG stranded, insulated, building wire

    So, 40 parallel conductors of varying gauges and geometries, all in parallel sharing the same load. If I eeney-meeney-miney-moe and pick one conductor to put a CT around it, can I reasonably be assured that the current going through that specific conductor will be a fixed ratio of the overall current, in all instances?

    I'm not looking for a high degree of accuracy, just a reasonable degree of consistency. I will use the amperage readout to adjust system parameters like input power, clamping force, weld time, etc.

    It is my understanding that the conductors will tend to automatically share the load as they have a positive temperature coefficient; if one carries more current than the others, it heats up, increasing its resistance, pushing the other conductors to carry more of its load. But I have no idea the time scale involved. Do you have to be pumping thousands of amps for 10+ minutes in order for this miracle of nature to automatically balance the load? Or does it work out nicely beginning from the first AC half-cycle? My weld times are measured in mS, so I need the correct current indication right away.
     
  2. mcgyvr

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    Oct 15, 2009
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    "balancing" will NOT happen fast enough at all and the different gauges isn't helping either.
    Time for the trial/error phase.. pick one.. try it.. pick another..compare.

    What about a hall effect mounted on a bus bar?
    http://www.gmw.com/magnetic_sensors/ametes/BBM.html
     
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  3. gerty

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    If I'm reading you right these mw xformers are in parallel? If so can you not measure just one of them , then multiply by total number of xformers. That might give you a ballpark figure.
     
  4. gerty

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    Since you know the ratio, measure the primary current and multiply, another ballpark figure..
     
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  5. MikeML

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    At 3000A, the voltage drop across any conductor in the circuit will be enough to deflect a sensitive AC voltmeter...

    (i.e. use the conductor as the shunt, and just measure the voltage drop between two points)
     
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  6. AnalogKid

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    PHOTO OF THE BEAST!!!!!
     
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  7. #12

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    NEC says equal wires of equal lengths are acceptable in house wiring. The amp meter in my 1973 Ford used a volt meter on a length of wire (to which I immediately added 10 feet). I'm saying your instincts are good and MikeML has a good idea. At 1000 amps, I would be calculating the voltage drop across a foot of of 1/2 inch copper pipe or 5/8 ground rod. Scrounge, rattle, crash..."Where's the worm off that old still, ma'?"
     
  8. strantor

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    I thought to myself "a hall effect sensor mounted anywhere within 10ft of the thing should be able to pick up on the current" but when I went to search for hall current sensors, all I found were torroidal rings with hall sensors embedded, and ICs where the current is to be passed through the IC. I did not see anything like this bus bar sensor you linked. thanks. I will study it and see if it's something I think I can DIY.

    The transformer primaries are in series and the secondaries are in series as well.
    How accurate is that? If my transformer(s) are 80% efficient at 800A output, can I assume they will be 80% efficient at 3000A output? That is after all where I get the 3000A estimate from, by making that assumption; but I do not know if the assumption is correct.
    That's another idea I considered, but I am unsure of the effects of the conductor heating up. Despite being one helluva cable, after 20+ welds the cable does start to heat up, I assume changing its resistance, which I think will skew my readings.
     
  9. strantor

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    FEAST THINE EYES!!!!!
    20150415_172827_resized.jpg
     
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  10. #12

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    So? Make a rack of copper pipes for a shunt and blow a fan on them.
     
  11. WBahn

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    If your cycle time is measured in milliseconds, how reasonable is it to expect an analog meter to respond reasonably to such a short burst?

    Can you run your current through a large solid conductor (a piece of pipe or bar stock) placed between the machine and the cable? If so, then you can use the pipe as a current sense resistor.
     
  12. WBahn

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    So how do you control the weld time. It's been 35 years since I did any spot welding, but I'm pretty sure that the unit I used automatically stopped the current after a fraction of a second. Do you have a similar feature or do you have to manually stop the current by opening the clamp jaws?

    Nice project, by the way!
     
  13. WBahn

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    Probably not by enough to matter that much -- but use a rigid pipe or rod that is big enough to not heat up appreciably.
     
  14. strantor

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    I tried using copper tubing early on in the build (with transformers in parallel). I found the resistance of 1/4" copper tubing to be much higher than I expected. you can see in the picture below I have 4 hammered copper pipes in parallel. I found that what I had assumed to be negligible resistance "bus bar" actually had higher resistance than 3 parallel 12AWG wires. It was limiting my current output. So I ditched the hammered copper pipe idea.

    20150412_085511_resized.jpg

    I would like to avoid anything that limits my current output. I am actually going to shorten the leads so that I can get more current out of it. Shortening the leads and then addin a shunt seems like one step forward and two steps back.
     
  15. strantor

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    EDIT: also, I went to home depot to pick up some ground rod and found that it's actuall copper plated STEEL! WTF? I have been making a bad assumption my whole life!
     
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  16. WBahn

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    I'm thinking more like a piece of 2" copper pipe. What you want to start with is figuring out how much voltage you are willing to drop across your shunt. Then pick a piece of pipe accordingly. I'd recommend pipe over a rod since you are dealing with high frequencies (pulses) and want to minimize issues associated with skin-effect.
     
  17. strantor

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    Good point. I should be looking for a digital meter with peak capture.
    Right now I'm tapping the foot pedal control as fast as my foot can move. But I have a .01-10 sec decade-selectable timer relay I am going to install.
    Thanks!
     
  18. MaxHeadRoom

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    Opening the electrodes will cause a splash usually, it is best to keep pressure on, most welder use either a timer/contactor on the primary or a couple of SCR's back to back.
    I would have tended to use multi-strand welding cable for the secondary.
    Max.
     
  19. strantor

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    I'm not really willing to drop any voltage. Voltage dropped across a shunt is voltage not dropped across my workpiece, which is less welding power. As I said above, I'm trying to get even more out of it by shortening the leads.
     
  20. #12

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    1/4 inch copper? You're thinking to small.
    I'm thinking, 35 millivolts for the last analog meter movement I worked with. @1000 amps, that's 35 watts lost in the shunt.
    1 gauge copper @ 8072 ft/ohm
    35 micro-ohms needed
    math...math...math
    3.39 inches
     
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