[bullethead67]

I have a mill i bought cheap that does not work. I would like to use the existing steppers

POWERPAC
Model N34HRHM-LNK-NS-00
Bipolar Paralel 1.8°
Holding Torque 2725(19.23) (2 phases on) oz-in (Nm) ±10%
Rated Current/ Phase 11.3 (amps DC)
Phase Resistance 0.2 (ohms) ±10%
Phase Inductance 2 (mH) Typical
Detent Torque 65 oz-in (Nm)
Thermal Resistance 1.3 (°C/watt)
Rotor Inertia oz-in-S² 0.075
Weight lbs 15.1


with the G203V but the G203V spec sheet says
"The motor’s rated phase current must not be more than 7 Amps"
as you can see in the specs these are 11.3.can i still use these and just run them at a lower power or derate them somehow.

I also need a power supply and i find the "formula" confusing can someone tell me whitch numbers i am supposed to be doing what with. I would like to understand selecting the power supply better using 3 of these steppers.

would derating these change the power supply i needed.

[79TigerPilot]

Stepper motors are current driven and the torque is only a function of the current you can get to flow through the winding. With the rotor stalled the current is just going to be defined by the winding and wiring resistance based on ohms law. The 203V will limit the current to 7A and this is limitation imposed by the components used in the design of the 203V. Since the motor resistance is low it doesn't take much voltage to draw 7A, only a volt or so.

Things change when the rotor turns, the torque is still defined by the current but now you are pulsing the coils. Two things happen, first the inductance of the windings, tends to impede the pulsing current and second the rotor generates a back EMF (like a generator) that bucks the input voltage. Now to get you torque based on 7A you need to supply a much larger input voltage because you need to overcome the back EMF and the winding inductance.

The Gecko site sets it at 32 times the sq root of the inductance in mH but I have also seen it specified at 20 to 30 times the motor voltage specification.

Generally the faster you run the motor the more voltage you need to get 7 amps into the motor. The actual voltage is not all that important as long as you have enough. The Gecko's are limited to 80V input. If you use an unregulated PS, you would want to be 10 to 15% below 80V so you have some margin so as to not over voltage the Gecko from high line voltage or from back emf. I think a lot of folks shoot for around 65 volts. Using Gecko's formula on your motor I calculate 45.25 V but I would opt for something a little higher say between 60 and 70V. The transformer is going to set your voltage in any case. For transformers see www.parts-express.com

Craig

[SCzEngrgGroup]

Stepper motors are current driven and the torque is only a function of the current you can get to flow through the winding. With the rotor stalled the current is just going to be defined by the winding and wiring resistance based on ohms law. The 203V will limit the current to 7A and this is limitation imposed by the components used in the design of the 203V. Since the motor resistance is low it doesn't take much voltage to draw 7A, only a volt or so.

Things change when the rotor turns, the torque is still defined by the current but now you are pulsing the coils. Two things happen, first the inductance of the windings, tends to impede the pulsing current and second the rotor generates a back EMF (like a generator) that bucks the input voltage. Now to get you torque based on 7A you need to supply a much larger input voltage because you need to overcome the back EMF and the winding inductance.

The Gecko site sets it at 32 times the sq root of the inductance in mH but I have also seen it specified at 20 to 30 times the motor voltage specification.

Generally the faster you run the motor the more voltage you need to get 7 amps into the motor. The actual voltage is not all that important as long as you have enough. The Gecko's are limited to 80V input. If you use an unregulated PS, you would want to be 10 to 15% below 80V so you have some margin so as to not over voltage the Gecko from high line voltage or from back emf. I think a lot of folks shoot for around 65 volts. Using Gecko's formula on your motor I calculate 45.25 V but I would opt for something a little higher say between 60 and 70V. The transformer is going to set your voltage in any case. For transformers see www.parts-express.com

Craig

Craig,

Higher is not better in this case. Once you get above the voltage suggested by the Gecko equation, the additional energy goes into generating heat in the motor, with no significant increase in performance. I'd go with a 45V supply.

Regards,
Ray L.

[Crevice Reamer]

Craig,

Higher is not better in this case. Once you get above the voltage suggested by the Gecko equation, the additional energy goes into generating heat in the motor, with no significant increase in performance. I'd go with a 45V supply.

Regards,
Ray L.

I concur, and will add that at 7A, the motors will run with only 62% of rated torque.

CR.

[79TigerPilot]

Not sure I fully understand this. I would assume the Gecko will keep the average current to 7 amps. no matter what the voltage is. The heating should be mostly from the IR losses in the windings which hasen't changed. As the pulse voltage is raised the dt to max current decreases. I guess there could be some hystersis or eddy current heating in the laminations because of the shorter pulse widths but that should be small compared with the IR losses.

Craig

Craig,

Higher is not better in this case. Once you get above the voltage suggested by the Gecko equation, the additional energy goes into generating heat in the motor, with no significant increase in performance. I'd go with a 45V supply.

Regards,
Ray L.

[SCzEngrgGroup]

Not sure I fully understand this. I would assume the Gecko will keep the average current to 7 amps. no matter what the voltage is. The heating should be mostly from the IR losses in the windings which hasen't changed. As the pulse voltage is raised the dt to max current decreases. I guess there could be some hystersis or eddy current heating in the laminations because of the shorter pulse widths but that should be small compared with the IR losses.

Craig

Craig,

From Mariss Freimaniss' excellent "Stepper Tutorial":

MOTOR HEATING AND POWER SUPPLY VOLTAGE

There are two major causes of motor heating; copper losses and iron losses. Copper losses are the easiest to understand; this is the heat generated by current passing through a resistance, as in the current passing through the motor’s winding resistance. Often this referred to as “ I squared R” dissipation. This cause of motor heating is at a maximum when the motor is stopped and rapidly diminishes as the motor speeds up since the inductive current is inversely proportional to speed.

Eddy current and hysteresis heating are collectively called iron losses. The former induces currents in the iron of the motor while the latter is caused by the re-alignment of the magnetic domains in the iron. You can think of this as a “friction heating” as the magnetic dipoles in the iron switch back and forth. Either way, both cause bulk heating of the motor. Iron losses are a function
of AC current and therefore the power supply voltage.

As shown earlier, motor output power is proportional to power supply voltage, doubling the voltage doubles the output power. However, iron losses outpace motor power by increasing nonlinearly with increasing power supply voltage. Eventually the point is reached where the iron losses are so great that the motor cannot dissipate the heat generated. In a way this is natures’
way of keeping someone from getting 500 hp from a size 23 motor by using a 10,000 volt power supply.

At this point it is important to introduce the concept of overdrive ratio. This is the ratio between the power supply voltage and the motor’s rated voltage. An empirically derived maximum is 25:1. That is to say, the power supply voltage should never exceed 25 times the motor’s rated voltage. Below is a graph of measured iron losses for a 4 Amp, 3 Volt motor. Notice how the iron losses range from insignificant to being the major cause of heating in the motor compared to a constant 12 Watt copper loss (4 Amps times 3 Volts).

Regards,
Ray L.