Bi-polar step motor driver with MOSFETs

Discussion in 'The Projects Forum' started by nerdegutta, Feb 4, 2012.

  1. nerdegutta

    Thread Starter Moderator

    Dec 15, 2009
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    So I got inspired by the recent bi polar steppermotor drivers on the forum (this and this), using the TB6560 step motor driver. I search around, and found this on ebay.

    http://www.ebay.com/itm/Bipolar-Stepper-Motor-Driver-Kit-12A-60V-MOSFET-K158-/180660089974

    A bit more searching, and I found a schematic. I tried to make my own, and now I need some comments.

    The thought is that step and directions are controlled by signals applied to JP1.

    Will this work?

    I see now that I need decoupling caps on the IC's.

    Any comments are welcome, as long as they are constructive. :)
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    What version of Eagle are you using?

    It looks pretty much like K158, except you have Q7 and Q8 upside down (drain and gate swapped) - if you built it like that, they would seem ON all of the time due to the body diodes.
    http://www.kitsrus.com/pdf/k158.pdf
     
  3. nerdegutta

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    I got Eagle 5.7.0, Hobbyist Licence.

    That's the schematic I found.

    So if I turn Q7 and Q8 it will be ok?
     
  4. stirling

    Member

    Mar 11, 2010
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    Might do but personally I wouldn't have hard wired the opto cathodes to gnd. This means you have no choice but to source current from your indexer/PC/whatever and if it can't source enough you'll miss steps. If you brought the optocoupler cathodes out to jumpers as well, you could choose source/sink. I prefer sink with a PC parallel port - but that's just me.
    [/QUOTE]
     
  5. nerdegutta

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    OK, good thinking. Adding a jumper should be easy.
     
  6. t06afre

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    Will you use a PC for controlling this? In any case I would have made arrangement so that you can have the controlling device compleatly isolated from the motor controller. This can be done by not having the LEDs in the optocouplers sharing the ground with the rest of the board. But using the ground form the controlling device.
     
  7. SgtWookie

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    You need a ground below R5. Compare that circuit to the one below.
     
  8. PaulEE

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    The one below is not below :)
     
  9. nerdegutta

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    I will be using a PC, and this software.

    http://linuxcnc.org/

    Ah, I missed that one. Thanks.


    It's strange how you can see, check and update a schematic, and still overlook something...

    I've added jumpers to the octo's ground. With these I can wire ground to either PC, or the PCB. Will that work?
     
    Last edited: Feb 5, 2012
  10. SgtWookie

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    No, that is not good.

    Connect the output transistors' emitters back to ground and remove the jumpers.

    It is the IR emitters (LEDs) on the input side of the optocouplers that needs to be grounded at the controlling PC.
     
  11. nerdegutta

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    Like this?
     
  12. stirling

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    Mar 11, 2010
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    No - not like that. You've joined the cathodes - you need them separate if you want to follow my first suggestion AND the other's about grounding.
     
  13. nerdegutta

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    OK... I try again...
     
  14. stirling

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    Mar 11, 2010
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    Nope. Like this
     
  15. nerdegutta

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    Oh... Thanks.

    Better now?
     
  16. stirling

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    Mar 11, 2010
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    from the point of view of what I was getting at yes - but don't forget what other's have been saying though.
     
  17. t06afre

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    yes it looks OK. What you also could do. Is to add a schmitt trigger after optos and before the 4030N circuit. But this not strictly needed. Also check if your prallell port is 5 volt or 3.3 volt type output. What is your planned current draw to the LEDs in the optos
     
  18. nerdegutta

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    Do you think R1 and R2 are too big?
     
  19. Paul Kerry

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    Jan 9, 2012
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    I agree with t06afre, the LEDs in the optos may not have enough current through them to turn the transistor on.

    The datasheet quotes 10mA current flow and a forward voltage of 1.18V to light the LED.

    If your parallel port has a 5 volt output you will need a a current limiting resistor of :-

    (5V - 1.18V)
    ------------ =
    0.01A

    3.82V
    ------ = 382Ω The nearest preferred value is 390Ω
    0.01A


    If your parallel port has a 3.3 volt output you will need a a current limiting resistor of :-

    (3.3V - 1.18V)
    -------------- =
    0.01A

    2.12V
    ------ = 212Ω The nearest preferred value is 220Ω
    0.01A

    I could be wrong with the nearest preferred value, but I believe the maths is sound. Please shoot me down in flames if I'm wrong ;)

    Cheers

    Paul
     
    Last edited: Feb 5, 2012
  20. nerdegutta

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    I got the same values as you did, so now I'm going to check the motherboard port.

    Thanks. :)
     
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