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Thread: need help with volt and amp

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    need help with volt and amp

    Hi all I just got 3 stepper motors the specs are:
    nema 34
    2 phase 1.8 degree
    8.8 rated current A/phase
    0.38 Resistance ohm/phase +-10%
    inductance 5.0
    4.8 rotor inertia (kg.cm2)
    My question is how many volt and amps should I run this motor with?
    thanks for the help


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    I was just about to ask the same question as well... Sorry everyone if this sounds like a very simple question, but my motor's datasheet states that it is rated at 1A/phase. I did some additional research online and read that this means the max RMS current each phase can take is 1A. Because this is a 2 phase motor, does this mean that my power supply current should be 2 A? Is there a problem with using a power supply with current output which is significantly higher e.g. 2 times higher than the rated current of the motor?

    I'm not too sure about the terminology here but is it better to run the motor with a supply current slightly below 2A (to avoid frying them), or to give them 1-2 A above the motor's rated current?


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    Registered Crevice Reamer's Avatar
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    Quote Originally Posted by 15mgtar View Post
    Hi all I just got 3 stepper motors the specs are:
    nema 34
    2 phase 1.8 degree
    8.8 rated current A/phase
    0.38 Resistance ohm/phase +-10%
    inductance 5.0
    4.8 rotor inertia (kg.cm2)
    My question is how many volt and amps should I run this motor with?
    thanks for the help
    Hi 15mgtar. Welcome to the Zone!

    I will assume that the motor is to be wired Bipolar Parallel, (which is best) and the 8.8A are for that wiring.

    Formula for Most efficient voltage is: 32 times the square root of inductance.

    32 x 2.24 = 72V,

    The best PSU for the MOTOR should be 72V. BUT, will the DRIVER accept that much voltage? will the driver accept 8.8A current?

    Formula for PSU Amperage: Total motor amps x .67.

    Assuming 3 motors: 3 x 8.8 = 26.4 x .65 = 18A.

    So you will need a 72V, at least 18A PSU.

    CR.
    Last edited by Crevice Reamer; 05-18-2009 at 09:50 AM.


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    Quote Originally Posted by davidYong View Post
    I was just about to ask the same question as well... Sorry everyone if this sounds like a very simple question, but my motor's datasheet states that it is rated at 1A/phase. I did some additional research online and read that this means the max RMS current each phase can take is 1A. Because this is a 2 phase motor, does this mean that my power supply current should be 2 A? Is there a problem with using a power supply with current output which is significantly higher e.g. 2 times higher than the rated current of the motor?

    I'm not too sure about the terminology here but is it better to run the motor with a supply current slightly below 2A (to avoid frying them), or to give them 1-2 A above the motor's rated current?
    Hi David. Welcome to the Zone!

    The motor only runs one phase at a time, so ONE Amp will be your maximum.

    Some of this info may be helpful to you:

    help with old parts

    CR.


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    Hi Crevice Reamer, thanks for the quick reply.

    I read the other forum thread you reccomended. I was wondering, how did you arrive at the 'magic number' 0.67 to obtain the optimum PSU current? Also with Voltage, how did you decide that PSU Voltage should be 20x more than the motor rating? It all seems rather daunting to accept without some mathematical explanation.

    Also, do these formulas for calculating PSU V and I work for both unipolar and bipolar steppers? I should have mentioned earlier, I am using a bipolar stepper. Voltage = 3.15 V/phase. Current = 1 A/phase. Here is my stepper's datasheet:
    http://www.active-robots.com/product...3h546-0440.pdf

    When you say the motor only runs one phase at a time, I intend to run my stepper in microstepping mode, will only one phase be used? I'm thinking of the concept of half stepping now, where two phases are turned on at the same time, so the rotor magnet will be halfway between the two phases.

    Thanks.


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    Oops sorry I meant to say, I am using a UNIPOLAR stepper, not bipolar as stated in the previous post.


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    Also, this is why I'm so confused. You're telling me that the max PSU current should be 1 A for a 1 A/phase unipolar motor. I want to drive the motor with the Linistepper which you may be familiar with at
    http://www.piclist.com/techref/io/st...step/index.htm

    Under the "Features" section of their website they say:
    Runs from your regulated +5v supply and controls motor supply which should be 2 to 3 times the rated motor voltage and 3 to 4 times the current of all motors.


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    It's not quite as simple as some would say. Power supply sizing depends on motor and driver. For example a linistepper is a linear driver, the power supply sizing will be the sum of all the motors and phases. i.e. if you have three normal 2 phase steppers rated at .8A per phase, then each motor needs 1.6A, so the power supply needs to supply 4.8V regardless of coil to voltage ratio. A chopped driver becomes more difficult to quantify, as the ratio of coil voltage to supply voltage alters the duty cycle and average current load. It doesn't alter the instantaneous current requirement, but if it's a simple smoothed linear supply the bulk capacitance handles the instantaneous requirements. For chopped drivers, I've found that computing total motor wattage and translating that to power supply wattage with a safety margin works well.
    Phil, Still too many interests, too many projects, and not enough time!!!!!!!!
    Vist my websites - http://pminmo.com & http://millpcbs.com


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    if you have three normal 2 phase steppers rated at .8A per phase, then each motor needs 1.6A, so the power supply needs to supply 4.8V regardless of coil to voltage ratio.
    Hang on, pminmo, didn't Crevice Reamer say earlier that only one phase will be running at a time, so why do you needx 0.8*2 = 1.6 A from the driver? Also, are you saying we need 4.8V based on the assumption that the Resistance is 3 ohms, therefore V = I*R = 4.8V ?

    So if we work based on your values i.e. I = 1.6A and V = 4.8V, my question is... is it safe for the PSU output to be slightly more or less than 1.6 A and 4.8V? Or do I have to place some resistors to get my V and I to be EXACTLY 1.6A and 4.8V? For example, still using your values you calculated, is it alright to supply those three motors with a power supply of 12 V and 5 A?


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    Quote Originally Posted by davidYong View Post
    Hang on, pminmo, didn't Crevice Reamer say earlier that only one phase will be running at a time, so why do you needx 0.8*2 = 1.6 A from the driver? Also, are you saying we need 4.8V based on the assumption that the Resistance is 3 ohms, therefore V = I*R = 4.8V ?

    So if we work based on your values i.e. I = 1.6A and V = 4.8V, my question is... is it safe for the PSU output to be slightly more or less than 1.6 A and 4.8V? Or do I have to place some resistors to get my V and I to be EXACTLY 1.6A and 4.8V? For example, still using your values you calculated, is it alright to supply those three motors with a power supply of 12 V and 5 A?
    Because the linistepper uses a linear driver and utilizes both phases on. At most angles it will use less than the full rated current per phase. I'm basing the info on the current waveforms posted in the linistepper website which shows in the full step mode, both phases are on. So if you pause at an angle where both phases are at 100% you will need the rated coil current per phase times 2.

    Your questions can't be answered based on the motor alone, they have to be answered based on the motor and the driver. BTW, the 4.8V i was referring to is the motor coil voltage, not a supply voltage. It was part of the explanation as to having 4.8V across the coil with respect to duty cycle impact.
    Phil, Still too many interests, too many projects, and not enough time!!!!!!!!
    Vist my websites - http://pminmo.com & http://millpcbs.com


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    Quote Originally Posted by Crevice Reamer View Post
    Hi 15mgtar. Welcome to the Zone!

    I will assume that the motor is to be wired Bipolar Parallel, (which is best) and the 8.8A are for that wiring.

    Formula for Most efficient voltage is: 32 times the square root of inductance.

    32 x 2.24 = 72V,

    The best PSU for the MOTOR should be 72V. BUT, will the DRIVER accept that much voltage? will the driver accept 8.8A current?

    Formula for PSU Amperage: Total motor amps x .67.

    Assuming 3 motors: 3 x 8.8 = 26.4 x .65 = 18A.

    So you will need a 72V, at least 18A PSU.

    CR.
    Thank you so much for the info I think there is a driver with a spec as : 36 - 80VDC 2.5 - 7.8A microstep upto 51,200 steps/rev.
    And as for the PSU can I just use PC psu connect 6 of them in series 12 * 6 = 72 VDC?


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    Quote Originally Posted by 15mgtar View Post
    Thank you so much for the info I think there is a driver with a spec as : 36 - 80VDC 2.5 - 7.8A microstep upto 51,200 steps/rev.
    And as for the PSU can I just use PC psu connect 6 of them in series 12 * 6 = 72 VDC?
    You CAN connect PC PSUs in series to get the Voltage, but I'm not sure you will get enough total Amperage that way.

    You can use less amps per motor, but that will detune the motor torque available to a lesser amount.

    For example, suppose you use a Gecko G203V driver rated for 80V and 7A. Just to make it simple i'll assume your motors are 1000 ounce inch. So, 7A/8.8A = .795.

    1000 x .795 = 795 ounce inches of torque that the motor is detuned to.

    CR.


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