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The drive should have some means of adjusting the current.
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Hello all, I have an Anahiem Automation "Blue Box" that puts out 80V and 7A. It clips the current at 7A. My present Anahiem Automation NEMA 34 stepper motor is rated at 7A. I have an Anahiem Automation NEMA 42 stepper motor that is rated for 6.1A that I want to install in the place of the NEMA 34. How do I calculate the resistance for limiting the current, both for series and parallel? Both motors are unipolar. Thanks JP
The drive should have some means of adjusting the current.
Hello H500, It does, but it is not adjustable without changing discete componants. AA won't give me a schematic. Thanks JP
Are you sure? It implies that the drive was designed for a specific motor.
I don't think it is possible to current limit with resistors. The drive electronics would simply adapt and try harder to force 7 amps into your motor.
AA might not give you a schematic (which might reveal trade secrets), but might be willing to tell you which internal resistors to change.
Hello H500, It is designed for a specific motor. This is about a 1990 model. I think that I can run parallel resisters to bypass .9A but can't find a formula to do the math. Thanks JP
A motor is not a resistor, it is an inductor, so you can't just put a resistor in parallel to bypass the current. You would have to have an RL circuit that would track the motor impedance, and that would be expensive and nearly impossible to calculate without knowing the motor characteristics.
Here's the calculator that I have used.
http://llk.media.mit.edu/projects/cr...motorcalc.html
If it doesn't work BUY BIGGER MOTORS!!
That is for calculating series resistors. Series resistors would probably work, but no guarantee that it would. Since the drive current would never reach 7amps, the drive would most likely act just like a non-chopper drive. You would however lose the efficiency of the chopper drive. Don't know what voltage his motors are, but with 80v supply and 6.1 amps, you would probably need at least two 500watt resistors for each motor. Not cheap, and you might have to aircondition the shop even in winter.
JP, The equation for parallel resistors is R=V/I....in your case, r = 80/.9 = 90 ohms. The power rating might be tricky to determine since it depends on the PWM duty cycle at various rpm's. I'm guessing 10-20 watts, assumming a 10% duty cycle.
You will need 4. A simple, non-math, brute force method is to put in four 90ohm 20W's, run it for a while (with motor turning) and measure the resistor temperature to make sure they won't catch fire.
That would work with a modern drive where the voltage applied to the coil is always switched between zero and the full supply voltage. However, for many of the older Aneheim drives, they talk about a bi-level voltage, where they apply the full voltage initially and then switch to a lower voltage. I don't know if this is true, but if it is, when this lower voltage was switched in, the shunt resistors would not bypass enough current.
Hello all, Thanks for the help. I connected the motor to the driver without any resistors to see if it would run. I got it on Ebay. It ran as expected. I will start out with 90 ohm resistors and measure the motor temperature as well as the resistor temperature and make adjustments from there. Here are the specs from the AA website. Old motor- 34D314-Voltage per phase unipolar=2.2V-Current per phase unipolar=7A-Resistance per phase unipolar=.31Ohms-Inductance per phase unipolar=1.7mH. New motor-42D212-Voltage per phase unipolar=3.6V-Current per phase unipolar=6.1A-Resistance per phase unipolar=.59Ohms-Inductance per phase unipolar=5.94mH. Thanks JP
Jeff has a good point. If its an old bi-level drive, the resistors won't work. In that case, you won't need any resistors at all, since the drive will try to put 2.2 volt into the motor at standstill. The current will only be 2.2v/.59ohm = 3.7 amps.
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