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Stepper Motor Power Supply Query

Discussion in 'Motors' started by Jaschreiber, Mar 11, 2021.

  1. Jaschreiber

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    I'm trying to determine the appropriate power supply to run some 12v 1.68amp/phase nema 17 motors on a camera rig. My project is a bit modular, so I'd like to have the ability to power up to 4 motors or as few as 1 depending on what is needed for the shot.

    I plan to use an arduino with a CNC shield and DRV8225 drivers. What I'm unclear about is how I need to calculate the amperage of the supply: since the amperage is listed as 1.68/phase for each motor, and they are 2 phase motors, do I need to have enough current for 2 times the listed amperage? I've seen threads online which indicate that I only need to supply 2/3x, 1x, or 2x the rated current and I'm not sure what is accurate. (I also gather that the driver will limit the current to whatever I set it to, which presumably also limits the amount the power supply needs to provide)

    So in short, for a single 12 volt 1.68 amp/phase motor, would I need a 12v 2 amp supply? What about for 3 or 4 motors?

    And is an off the shelf supply like this appropriate?

    Thanks!
     
  2. Peter Van Der Walt

    Peter Van Der Walt OpenBuilds Team
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    Say for example, our NEMA23 motors as an example (Sorry have the math on hand for a quick copy paste, and will demonstrate the concept)

    Datasheets says we have 3V coils, at max 3A rating per coil. V*I=W so 3x3=9w per coil x 2 coils = 18w per coil at max rating
    Lets factor in 80% driver efficiency = 18w * 1.2 = 21.6w per motor

    Now, if your input voltage was 12v, to get 21.6w you'd need 21.6/12 = 1.8A per motor total

    So its our motors (yay) which has these specs:

    upload_2021-3-11_20-19-56.png

    Doesn't list the coil voltage rating, but does list resistance and current rating. Ohms law says V=I*R

    So 3A * 1.65ohm = 4.95v coils. At a max rating of 1.68A = 4.95*1.68 = 8.316w per coil * 2 coils = 16.632w * 1.2 (efficiency of driver) = 19.9584w per motor
    19.9584 / 12v = 1.6632A per motor total * 4 motor = 6.65A

    So minimum of a 12v 6.65A PSU (or 3.33A at 24v - higher voltages = more efficient if you have the option)

    The chopper style stepper drivers are constant-current, so they lower the voltage output to match the resistance of the coils.
     
    Jaschreiber likes this.
  3. Rob Taylor

    Rob Taylor Master
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    1) Run NEMA 17s on 24V. 12V gives you pretty anaemic performance.

    2) 1.68A is per phase, and if you're microstepping you're using both phases simultaneously. Probably won't be using more than 1.2A per phase, but plan to allow 2A or so per motor. An 8A supply should be plenty for all motors, probably a 2.5A/60W supply for just one. The math can get a little complex here because of the way the drivers chop the voltage, but for peace of mind I like to just assume the motors are seeing the full 24V all the time and work from there, so I know there's a pretty good buffer built in. [edit: hahahah, there's the math above if you want it, we simul-posted]

    3) I would use a Mean Well supply. They're better built, more reliable, and just generally a far more known and used quantity all the way around. I'd probably use an LRS-350-24, but as we see above you can easily get away with an LRS-200-24 just fine. You might be able to find a smaller package if you look for a DIN rail mountable version like the NDR-240-24 but you pay for that compactness.

    4) CNC shield (well, maybe not the "real" Protoneer one) is pretty terrible, and DRV8225s are pretty well deprecated at this point. I'd look at Trinamic drivers if you really want to stick with the StepStick format.
     
  4. Jaschreiber

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    Thank you both for your speedy replies, will do some more research before I begin!
     
  5. Jay Webb

    Jay Webb New
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    Hi, Sorry I know this is an old thread however i came across it while researching for a CNC controller i'm doing for a uni project. I was wondering where you have the 3A * 1.65 ohm, have you doubled the current per phase to get the 3A if so, would you need to double the phase resistance for the voltage calculation?
     

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