What power supply is required?
Support » FAQ » Stepper motor positioning controls
Through the application of the low-loss switching operation (chopper control or PWM) for the current control in all stepper motor drive of the SMC xx series, the specified phase current no longer needs to be multiplied by two. Instead, the power output of the power supply units must be calculated with the mechanical power of the motor P = Md* n * π/30 plus the efficiency of the motor, controller and power supply unit.
Determining the power supply size (in W)
The overall power of the power supply to be applied is thus made up of the kinetic energy (product of the torque required now in Nm at the desired speed in rpm and the factor π/30) and the power dissipation of the motor, final output stage and power supply:
Pges = Pmech + Pv (motor + end stage + power supply)
Examples of the respective torque characteristics
- a) Motor size 42 mm – ST4118M1206 (2000 rpm at 48 V 1wdg.) = 36 W
- b) Motor size 56 mm – ST5918M3008 (2000 rpm at 48 V parallel) = 140 W
- c) Motor size 56 mm – ST5918M3008 (1000 rpm at 48 V parallel) = 90 W
- d) Motor size 86 mm – ST8918L6708 (1000 rpm at 72 V parallel) = 315 W
The power losses are proportional to the phase current of the motor (I² * R) and must be taken into account, normally against the kinetic energy with max. 5-7%.
That is to say, in the example a) (motor size 42): kinetic energy * 1.07= 36 * 1.07 = 38 W.
As only certain standard sizes are offered on the market, a power supply of 50 W must be selected here (a further power consuming load can possibly be connected here).
Whereas in example b) a 150 W power supply may have to be sufficient, a power reserve of approx. 25% should already be planned for a stepper motor-drive solution. (which would mean a real power consumption of 0.67 Nm * 0.75 * 2000 * 3.14/30 = 105 W).
Thus, a power supply with 150 W would also be advisable here.
With a certain power reserve, not only the motor but also the power supply unit remains just warm to the touch. If you do not know or have not yet learned the final speed and torque values, but you also want to obtain the power supply on the same date of delivery as the motor order, in an initial approximation the following simple guiding value is sufficient: phase current * (0.7-0.8) * supply voltage, i.e. for a motor with 1A phase current *0.7 * 24 V = 16.8 W and, hence, a standard 20 W power supply.
Fuse
At least a factor of 1.5 times the motor phase current should be used as the value for the fuse (slow blow fuse)
Category and type of the power supply
Due to the greater efficiency as well as the construction size and weight compared to power supply units with transformers, rectifiers and filters via capacitors (these also often have the problem of an excessive open circuit voltage), switching power supplies are deployed today.
The 105-150% overload protection means that an overload peak is detected in this range and leads to a current limiter. The customer must, however, ensure that it does not come to a sustained overload (a sustained overload of approx. 120% may not be detected and the service life of the power supply would be greatly reduced.
Charging capacitor
The charging capacitor should designed in such a way that the voltage ripple remains less than 2.5 Vss and the minimum capacity lies very roughly at 2000 µF/A. (The minimum capacity also depends, however, on the difference of the admissible voltage of the drive minus the supply voltage as well as that of the delta brake application delay time and the size of the external flywheel mass). In any case, the capacity should be selected to be large enough so that the energy to be recovered during deceleration, braking or sudden stopping of the capacitor can be accommodated without any appreciable increase in voltage occurring compared to the admissible voltage of the power drive.
Appropriate Nanotec power supplies for our SMCI positioning controls and Plug & Drive motors can be found with good performance grades under Accessories on our website
