Subtleties of servoamps

Sept. 1, 2005
Properly sized power supplies eliminate erratic servomotor performance in motion systems.

Jim Woodward
Applications Mgr.
Copley Controls Corp.
Canton, Mass.

Many engineers are familiar with dc power supplies used in motion control systems. In fact, they're so familiar with these supplies that they tend to give them little consideration during an installation. Yet unexplained shutdowns, underspeed operation, and lengthy accelerations describe just a few of the problems encountered when there are mismatches between servoamplifiers and power supplies. It can take a long time to track and correct faults caused by the wrong power supply.

Batteries and dc power supplies behave about the same when connected to ordinary resistive loads. That similarity ends when it comes to servoamplifiers.

Servoamplifiers do not present a steady load to supplies. In fact, during regenerative braking, the servomotor itself becomes a generator delivering energy back to the supply. Supplies designed only to source power have difficulty coping with external power applied to their output leads.

Servomotors that draw 5 A at full speed typically require 15 A or more when accelerating from standstill. Power supplies may not have enough reserve capacity for this extra demand. Without sufficient reserves, motors take longer to reach full speed, may cut-out during acceleration, or may fail to reach speed at all.

Excessive regenerative braking energy may overheat reservoir capacitors, possibly to the point where the capacitor explodes. The good news is that exploded reservoir capacitors make diagnosis relatively straightforward. The bad news is that the same excess energy may only cause lingering capacitor death. Failure-diagnosis becomes far more difficult when the cause is hidden.

Most engineers know that the key components in a simple dc power supply are the rectifiers and a filter or reservoir capacitor to store energy. The rectifier converts sinusoidal ac power into dc pulses that power the load and charge the reservoir capacitor. When the rectifier output drops below capacitor voltage, energy stored in the capacitor supports the load's power demands. At 60 Hz, rectified-power line operation and capacitor charging occupies roughly 2 msec of each 8.33 msec half cycle. The capacitor becomes the sole source of drive power for the remaining 6.33 msec. In other words, the reservoir capacitor supplies power to the load about 75% of the time. As the energy in the capacitor releases, capacitor output voltage falls. The repeated charge and discharge cycles create an output ripple voltage. Amplitude of this ripple voltage rises with load current and more charge is removed from the reservoir capacitor. Excessive ripple reduces the supply's average output voltage. This voltage droop is a major factor in power supply and servoamplifier mismatch.

The reservoir capacitor resembles a storage battery only in its ability to store energy. Compared to a battery, however, its storage capacity is minuscule. Typical AA batteries the size of your little finger store orders-of-magnitude more energy than a capacitor as big as a can of tomato soup. More significantly, the battery maintains a relatively constant output voltage while delivering its energy. Not so for the capacitor that loses voltage as its stored energy is depleted.

The energy stored in a capacitor is expressed in Joules (J) or watt-seconds by the formula: J = 0.5CV2, where C is the capacitance in Farads and V is the voltage to which the capacitor is charged.

When fully charged, the output voltage of the filter capacitor equals the peak supply voltage, VPEAK. As the capacitor discharges, it's output voltage drops to a minimum value called VRIPPLE. The difference between these two values is called the ΔV.

If the size of the filter capacitor in Farads and the current demand of the load in amps is known, the ΔV voltage can be calculated with the formula: ΔV = (0.00633 x ILOAD) / C.

Average output voltage for the power supply is approximately halfway between VPEAK and VRIPPLE. Another method of calculating the average output voltage is to subtract half of the ΔV amount from VPEAK.

A power supply operating on 120 Vac supplies 5 A to the load. The power supply uses a 1,000 μFd filter capacitor. Calculate VPEAK by multiplying 120 Vac by the √2: 120 x √2 = 170 VPEAK. Then calculate the ΔV amount: (0.00633 x 5 A) / .001 Fd = 31 V. Subtract half the ΔV value from VPEAK to get the average output voltage: 170 - 15.5 = 154.5 V output.

To reduce the amount of ripple and boost average output voltage, technicians use a bigger filter capacitor. For example, increasing filter capacitance to 10,000 μFd reduces ΔV to 3.1 V and raises the average output voltage to 168.5 V. The influence of reservoir capacitance on power supply droop explains why so many "fixes" recommend more power supply capacitance.

However, raising filter capacitance involves its own set of caveats. Initial inrush currents to the larger capacitor may be higher than the rectifier diodes can handle. Some power supplies incorporate special limiting resistors minimizing initial inrush current before power is applied to the motor. When the capacitors reach full charge the limiting resistors are switched out of the line.

Demands on the power supply change depending upon motor acceleration, full-speed operation, and deceleration.

Motor acceleration demands significantly higher current than during steady use. One consequence is more output voltage droop. The sudden rise in load current also produces an immediate downward step in power supply voltage. The step is caused by a greater IR voltage drop in the diodes, isolation transformer, and other power supply components.

Although current remains constant for a constant acceleration rate, amplifier output voltage must rise with motor speed to overcome the motor's counter-electromotive force (cemf.) Consequently, faster speed is accompanied by steadily increasing power draw from the supply.

On reaching operating speed, the transition from accelerating to running current reduces the IR drop. There is a small upward step in output voltage. With less load current, supply voltage droop begins its recovery to normal running values.

Any supply must provide sufficient current for anticipated acceleration. There must also be sufficient output voltage headroom to encompass IR voltage drops, supply droop, and even a 10% reduction in the 120 or 240 Vac supply.

An interaction between voltage droop, IR drop, and amplifier protection circuits may significantly lengthen acceleration time. The combined losses may let supply voltage fall to a value that triggers the amplifier's under-voltage protection circuit, shutting down the amplifier. Amplifier shutdown leaves the motor free to coast until drive power is restored.

With the amplifier shut off, power supply voltage climbs back to normal values. With voltage values back to normal, the amplifier powers up and resumes motor acceleration.

However, the returning high-current demand also brings back the voltage droop and IR loss retriggering the amplifier protection circuit. Repeated accelerate-then-coast cycles may eventually bring the motor to full speed, but over a much longer period. An underrated power supply may also trigger a similar sequence of accelerate-then-coast cycles. This time, the power supply's own overload protection circuit trips when current draw exceeds supply specs.

During regenerative braking, the amplifier reverses motor drive current to produce a counter or slowing torque. In effect, the amplifier tries to drive the motor backwards. Kinetic energy stored in the load keeps the motor turning in the same direction, effectively converting the motor into a generator. Motor cemf now adds to the amplifier output voltage instead of opposing it. The motor-turned-generator feeds electrical energy back through the amplifier into the reservoir capacitors raising power supply output voltage in the process.

For small motors and those driving light mechanical loads the kinetic energy amounts to a few watt-seconds. Under most conditions the power supply reservoir capacitor safely banks that energy. Trouble arises when braking servomotors in the horsepower range or small motors that drive drums, centrifuges, and other high-inertia loads.

Most engineers are aware that it takes longer to accelerate high-inertia loads, absorbing more energy during start-up. Consequently, the motors return more energy to the power supply during braking. A reservoir capacitor that's too small will charge well beyond its voltage rating, possibly to the point of explosive destruction. Even if there's no bang, the recurring stress of overcharging weakens and eventually destroys the capacitor. To prevent this requires special protection against excessive reservoir capacitor voltage.

Several techniques help limit damaging voltage buildup in the reservoir capacitor. First, use capacitors and servoamplifiers that accept high voltage without harm. Second, use sufficient reservoir capacitance to limit voltage buildup to acceptable levels. Essentially, larger capacitors act like larger tanks to store the electric charge.

For applications involving substantial kinetic energy, a regenerative brakingenergy dissipater protects against excessive rise in power supply voltage. The dissipater connects to the amplifier's input terminals switching a high wattage resistor in parallel with the power supply when required.

Without these voltage-limiting techniques, servoamplifiers may create dangerous operating conditions if the over-voltage protection circuit activates. For example, the amplifier's over-voltage protection circuit triggers as voltage rises during regenerative braking. The amplifier shuts down, halting deceleration and leaving the motor to coast uncontrolled. With the amplifier shutdown, the high-voltage charge takes much longer to dissipate and restore amplifier control over the motor. Motor and load coast out of control until the load hits a stop, damaging the equipment and possibly harming bystanders.

Motion system designers have a choice between unregulated power supplies with transformer isolation, switching supplies, and servoamplifiers that come with built-in power supplies. For single-axis controls to about 500 W, a switching supply costs less and occupies the smallest space because it doesn't require isolation transformers. Beyond 500 W, and especially for two or more drive axes, isolated power supplies are the most economical.

Although more costly than simple amplifier/power supply combinations, servoamplifiers with built-in power supplies create an extremely compact controller.

A rule-of-thumb in sizing power supplies is to take cues from drive motor ratings rather than the servoamplifier. As an example, a motor needs 48 V at 3 A for full speed operation with short-term acceleration demands of 6 A at 50 V. It's paired with a 75 V servoamplifier with 3 A continuous and 6 A short term output. At first glance it might appear the power supply should be continuously rated at 75 V 3 A or 225 W.

A more appropriate power supply provides 48 V plus a 15% allowance for IR voltage drops and power supply droop. An output of 55 V rather than 75 V is adequate. A supply rating of 165 W (55 V 3 A) instead of 225 W would more than meet the need.

Copley Controls Corp.
, (781) 828-8090,

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