Section 6: Power Supplies
The choice of a power supply is determined by the voltage, current, and power supply type (i.e. switching versus linear, regulated versus unregulated, etc.). By far the most problematic and complicated factor is voltage, which will be discussed last.
The easiest factor in choosing a power supply is its current rating, which is based on your motor ratings. A motor control will always draw less than 2/3 of the motor’s rated current when it is parallel (or half-winding) connected and 1/3 of the motor’s rated current when it is series (or full-winding) connected. That is to say, a 6 amp per phase motor will require a 4 amp power supply when wired in parallel and a 2 amp power supply when wired in series. If multiple motors and drives are used, add the current requirements of each to arrive at the total power supply current rating.
When using multiple drives from a common power supply, use individual supply and ground wires to each drive and return them to a common point back at the power supply. This is called a “star” power supply distribution; never use a “daisy-chain” power distribution, where the supply and ground wires for the next drives are picked up from the previous one.
The voltage of your power supply is entirely dependent on the inductance rating of your motor, which we learned is translatable to the number of turns of wire in the stator. Every motor model will have a different inductance rating and will therefore have a different maximum voltage. To figure out what the maximum power supply voltage should be, use the following formula with the motor’s inductance in millihenries (mH) used for the L value.
32 * √L = VMAX
If you are using several different models of motors on the same power supply use the lowest inductance rating in the above formula. This will ensure that your motors will not overheat due to the voltage being too high.
Should a motor not list the inductance it will generally list the voltage rating of each winding, which will be very low. An acceptable way of determining your power supply voltage if this is the only information you have is to multiply this number by any number between 4 and 20. In Figure 14 you could use a power supply voltage anywhere from 8.8V to 44V if wired in parallel.
An unregulated power supply will be sufficient and is recommended for most applications because of its simplicity. If a motor with a large inertial load decelerates quickly it will act as an alternator and send voltage back to the drive which then sends it back to the power supply. Because many regulated power supplies feature protection circuitry this may cause the power supply to fault or reset; however, if the supply is unregulated it will simply get absorbed by the filter capacitor.
To make your own power supply you must have three key components: a transformer, bridge rectifier, and filter capacitor. The transformer’s current rating must be sufficient to adequately run all motors that will be run from it using the above current formula. The DC output voltage will be 1.4 times the transformer’s AC voltage rating of the secondary. For example, a 24VAC transformer secondary will provide about 34VDC at the output of the supply. The bridge rectifier’s voltage and current ratings must exceed what the supply will deliver. Finally the minimum filter capacitor size must be calculated. Use the following equation to do this:
(80,000 * I) / V = C
The result will be in microfarads for the capacitor if the value for “I” is amperes of current needed and “V” is the output voltage of the supply. When picking the capacitor, any value equal to or greater than the calculated value can be used. Be sure to use a capacitor with a voltage rating at least 20% higher than the output voltage of the power supply. A sample 68VDC 5A power supply is shown in Figure 15.
There is a special consideration if the power supply will be at or near the maximum voltage rating of the drive. If the motor will be rapidly decelerating a large inertial load from a high speed, care has to be taken to absorb the returned energy. The energy stored in the momentum of the load must be removed during deceleration and be safely dissipated. Because of its efficiency, the drive has no means of dissipating this energy so it returns it to the power supply. In effect, instead of drawing current from the power supply, the drive becomes a source of current itself. This current may charge the power supply capacitor to destructive voltage levels.
If more than one drive is being operated from the power supply this is not a problem since the other drive(s) will absorb this current for its needs, unless it is decelerating as well. For this case or for a single drive it may be necessary to place a voltage clamp across the power supply in the form of a Zener diode. The voltage of this diode must be greater than the maximum expected power supply voltage, yet low enough to protect the drive. A good choice would be either 82 volts or 91 volts as standard values.
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