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  1. #1
    Join Date
    Jan 2010
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    2141
    Glad to hear that you got it working. :banana:

    The documentation for the 5056D is remarkably poor. According to figure 10 of the manual, it looks like step might activate on a negative-going pulse edge. But does not specify that the negative-going edge must be the leading edge of the pulse. Each step pulse has both a positive-going edge and a negative-going edge.

    The optoisolator can be hooked up in more than one way. The diagram in figure 9 shows the LED turning on when a steady positive voltage is applied to PUL+ and PUL- goes negative (in other words, if both PUL+ and PUL- are high, the internal LED is off, but pulling PUL- to ground turns it on). However it can also work the way that you have wired it, where the pulse going positive on PUL+ (and PUL- grounded) turns on the optoisolator.

    It was useful for me to go through the exercise, though...

  2. #2
    Join Date
    Jan 2005
    Posts
    136
    Quote Originally Posted by doorknob View Post
    The optoisolator can be hooked up in more than one way.
    That's what I was getting at. What is the benefit of doing it the other way?

  3. #3
    Join Date
    Jan 2010
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    2141
    Without more detailed information I can only guess.

    One thing that comes to mind is that some circuits are able to sink a large amount of current, but can only source a little bit of current, while others may be designed to do the opposite.

    An optoisolator input probably draws a fair amount of current, perhaps a few mA. or more (because it needs to turn on an LED). If you're driving that input with, for example, a TTL gate or any kind of open collector that is designed to sink current, it will do best with the hookup that puts a steady +5v on PUL+ with the signal on PUL- (so that when the signal goes low, it sinks current to turn on the opto).

    Judging by the C10 documentation, the output looks like it comes from a high-current buffer that can probably handle either sourcing or sinking sufficient current to drive an optoisolator input. So that probably explains why it will work either way for you.

  4. #4
    Join Date
    Aug 2008
    Posts
    1186
    I am having the same issue getting mine going and have wired it according to this diagram from Keling's site, can anyone shed some light on this for me, i am getting frustrated not finding a soilution and it seems like the more i search the more confusing it gets.......... HELP!!!

    http://www.kelinginc.net/DigitalDriverWithC10.pdf

    Chris

  5. #5
    Join Date
    Jan 2005
    Posts
    136
    While it's been a little while since I got my machine running, I think the setting I used here:

    http://www.cnczone.com/forums/930475-post11.html

    are the same I'm currently running. IIRC, the problem was the signals I was sending weren't going to the pins I thought they were. I used an oscilloscope to determine that by using Mach3 to move the motors, looking for series of pulses for steps as well as High/low for direction. Let me know if that helps, or if it doesn't...

  6. #6
    Join Date
    Jan 2005
    Posts
    136
    I just realized that post I linked may not be very clear. The part that says:

    Mach3 says / Output's on BOB

    2/9
    3/8
    4/7
    5/6
    6/5
    7/4


    Means that if you're putting out signals on pin 2 in Mach, they're showing up on pin 9 on the BOB. 3 on 8, 4 on 7 etc.

    I hope that clarified it a bit.

  7. #7
    Join Date
    Aug 2008
    Posts
    1186
    Tobybirch007,

    I don't have an O-scope available... wish i did. from what i gathered from you post below:

    Mach3 says / Output's on BOB

    2/9
    3/8
    4/7
    5/6
    6/5
    7/4

    are you saying that the way i have my board drivers wired up which is:

    X-axis driver to C10 board 2/3
    Y-axis driver to C10 board 4/5
    Z-axis driver to C10 board 6/7

    could be wrong because the board outputs are crossed up somehow such as in mach, the board chips etc? I think it is but im not clear on it totally.

    Thanks,

    Chris

  8. #8
    Join Date
    Aug 2008
    Posts
    1186
    http://www.kelinginc.net/DigitalDriverWithC10.pdf

    this is what i used from Keling's site to wire everything up, it is quite poor for someone of my level of wiring and newbishness. I know my wiring is pretty accurate according to the directions but it was limited in what it wanted me to do with the jumpers, I have tried it both ways and still no avail. I know mach works because i have a plasma table that uses a gecko G540 and it works flawlessly with the port address shown in the ports and pins screen. this is quite confusing... I wish Gecko made a higher amperage G540 style box, it would be a winner for guys like me trying to get their equipment up and running. I literally had 25 minutes of wiring after fabricating the enclosure with the power supply and G540 before i was up and going and tuning in the motors on my plasma table.... ugh...

    Chris

  9. #9
    Join Date
    Aug 2008
    Posts
    1186

    Seriously confused now...

    OK.... I found this wiring diagram on CNC4PC's site which is completely different then what i posted above from Keling's site... which is right and which should i be using? extremely confused now.....

    http://cnc4pc.com/Tech_Docs/1_C10R10.PDF

    it appears that on CNC4PC's diagram, they have the following listed for their wiring:

    from 5056 driver:

    DIR- to external 5v power Supply GRD
    DIR+ to C10 Board Pin 3
    PUL- to external 5v power Supply GRD
    PUL+ to C10 Board Pin 2


    Keling's diagram states:

    from 5056D driver:

    DIR- to C10 Board Pin 3
    DIR+ to comm between pin 2 and 3 on C10 board
    PUL- to C10 Board Pin 2
    PUL+ to comm between pin 2 and 3 on C10 board

    at first glance, that tells me that i would need to switch some settings in mach3 under ports and pins as well as set the jumper on the C10 board between pin 1-2 for GRD.

    or do i need to rewire it according to CNC4PC's diagram.... why can't this be made simpler very confusing...

    Chris

  10. #10
    Join Date
    Jan 2010
    Posts
    2141
    Quote Originally Posted by lcvette View Post
    OK.... I found this wiring diagram on CNC4PC's site which is completely different then what i posted above from Keling's site... which is right and which should i be using? extremely confused now.....

    http://cnc4pc.com/Tech_Docs/1_C10R10.PDF

    it appears that on CNC4PC's diagram, they have the following listed for their wiring:

    from 5056 driver:

    DIR- to external 5v power Supply GRD
    DIR+ to C10 Board Pin 3
    PUL- to external 5v power Supply GRD
    PUL+ to C10 Board Pin 2


    Keling's diagram states:

    from 5056D driver:

    DIR- to C10 Board Pin 3
    DIR+ to comm between pin 2 and 3 on C10 board
    PUL- to C10 Board Pin 2
    PUL+ to comm between pin 2 and 3 on C10 board

    at first glance, that tells me that i would need to switch some settings in mach3 under ports and pins as well as set the jumper on the C10 board between pin 1-2 for GRD.

    or do i need to rewire it according to CNC4PC's diagram.... why can't this be made simpler very confusing...

    Chris
    The confusion between the two sets of directions for hooking up the DIR +/- and the PUL +/- pins stems from the fact that there are actually two different ways to hook up the optoisolated inputs of the 5056D driver to the outputs from the C10 board, and the C10 is flexible enough to support both types of hookups depending on the jumper selections that you make.

    One way to hook it up (following the CNC4PC instructions) is to use the step and direction signals via the PUL+ and DIR+ inputs to directly drive the optoisolator inputs (namely, by supplying current to the LED inside each optoisolator from the step and direction signals). With that hookup, the minus terminal of the LED inside each optoisolator is connected to ground through the PUL- and DIR- terminals.

    The other way to hook it up (following the Keling instructions) is to supply current to the LED inside each optoisolator directly from the +5 volt power supply, by hooking the PUL+ and DIR+ terminals to the common pin (and setting the C10 jumper to apply +5 volts to the common pin). With the Keling hookup, the step and direction signals "sink" the LED current to ground to turn on the LED (by providing a return path to ground for the LED current that comes from the +5 volt power to the PUL+ and DIR+ terminals), or else they "float" the output up to +5 volts, which turns off the LED. So, the signals from the C10 must be connected to the PUL- and DIR- terminals of the driver.

    Some breakout boards are designed in a way where they can only work with one or the other of the hookup styles. The C10 is flexible enough to work either way.

    Hope that was somewhat clear.

  11. #11
    Join Date
    Jan 2005
    Posts
    136
    I think you understand what I'm saying. So, if I were you, I would change the pin assignments in mach.

    Example, if you have the XDir hooked up to pin2 of the bob, set XDir in mach to 9. Do the rest according to the numbers listed above and see if it works...

  12. #12
    Join Date
    Aug 2008
    Posts
    1186
    gonna try that right now! think i understand correctly now... will report back in a few minutes with results.

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