Motors connected to VFDs receive power that includes a changeable fundamental frequency, a carrier frequency, and very rapid voltage buildup. These factors can have negative impacts, especially when existing motors are used.

There are a number of potential problems that can become real when a variable frequency drive (VFD) is used to power an existing induction motor. As such, you should carry out a careful study to determine if these problems could be sufficiently bad to cause reconsideration of such an installation. With a VFD, an existing motor normally having a number of useful years left in it could abruptly fail.

Existing motors are designed for 60 Hz only, 50 Hz only, or 60/50 Hz service. As such, you have to question whether or not a new VFD can be matched to your existing motor and still have the motor perform reasonably well. In other words, will the motor be able to handle additional factors that may cause greater vibration, heat rise, etc., and a possible increase in audible noise?

High Frequencies Can Cause Problems

You should be aware of possible side effects caused by high pulsing frequency when applying a VFD to an existing motor. These negative effects include additional heat, audible noise, and vibration. Also, pulse width modulation (PWM) circuitry (see sidebar "Basics of Adjustable Speed Drives," on page 38), which causes a high rate of voltage rise of the carrier frequency, can cause insulation breakdown of the end turns of motor windings as well as feeder cable insulation.

The carrier frequency, a by-product of obtaining current at a variable fundamental frequency, is the cause for having additional watts in the motor; this power is essentially wasted energy that adds heat to the motor. The amount of such loss varies, depending upon the motor's stator and rotor designs and frequency of the carrier wave.

With frequencies other than the fundamental, a motor runs at very high slippage and, therefore, is running somewhat inefficiently. (Slip is the difference between the rotational speed of the stator's magnetic field [the synchronous speed of the induction motor] and the speed of the rotor.) Also, numerous lines of magnetic flux are being cut by the rotor; this phenomenon produces additional watts and additional heat. (Note that the high-frequency ripples in the current are at low magnitude, and the additional heat is in the order of 5% to 10% above that produced by a pure sine wave).

The synchronous speed of a four-pole motor served by 60-Hz power is 1800 rpm. This same motor, when considering "overtones" or ripples in the current's fundamental frequency caused by a voltage carrier frequency of 4 kHz, will have current flowing through it based on that high frequency. Thus, the rotor of a four-pole, 60-Hz designed motor (with a rated full-load speed of 1750 rpm) being supplied power by a VFD adjusted to 10-Hz output will be turning at 1/6 rated speed. If the load's torque requirements are constant at low- through full-rated speeds, slip rpm remains constant. (For more information on slip, see "Terms to Know," on page 38, and "Some Motor Basics," on page 46.) For the motor above, which is operating at 10 Hz, the shaft will be turning at 250 rpm.

The rotor, while turning at 250 rpm and crossing lines of flux (the magnetic field) based on the 10-Hz fundamental frequency and the synchronous speed of 300 rpm (1/6 of 1800 rpm), is also crossing lines of flux due to the carrier frequency voltage of 4 kHz. The synchronous speed at 4 kHz is 120,000 rpm ([120 x 4000] [divided by] 4).

Based on a synchronous speed of 120,000 rpm and a shaft speed of 250 rpm, you can see that the magnetic lines of flux being cut due to the carrier frequency (4 kHz) are substantial compared with a synchronous speed of 300 rpm caused by the 10-Hz frequency. This additional current, which is transmitted to the rotor bars by the cutting of additional magnetic flux caused by carrier frequency, produces very little useful power. Most of this current is dissipated as heat, adding to the temperature rise of the motor. This additional heat represents about another 5% to 10% thermal buildup in the motor and can place an additional thermal strain on the motor's rotor bars and stator windings, if it is running at full load. This high frequency power is an inefficient producer of torque.

Because of these and other conditions mentioned, you may wish to derate an existing motor when it's connected to a VFD. The amount of energy from the carrier frequency that's dissipated by the motor depends upon the amplitude and the frequency of the voltage, and the reactance and resistance of the motor at the resultant frequency. The amplitude of the current is determined by the ratio of the voltage over the impedance, while the watts lost is a product of the current squared times the resistance.

Other Undesired Side Effects

You also should be aware of other potential side effects caused by high frequency. These include undesirable audible noise, harmful vibration, and bearing problems.

Vibration and noise problems. To avoid noise and vibration problems, it's recommended that the motor being used not have components that can resonate at the frequencies the motor (and its load) will generate. This is possible on systems where the frequency of the power is known, such as with 60 Hz. However, today's VFDs have no standard carrier frequency, and the fundamental frequency can range from less than 10% of 60 Hz to 100% of 60 Hz, and beyond. Depending on which brand and model number of VFD is mated to the existing motor, and other factors such as the characteristics of the electrical system at the site, resonances in certain components may or may not be excited.

You must also consider that when a 60-Hz designed motor is operating at a different electrical frequency, various components of the motor might go into mechanical resonance, such as the fan or shaft. Each component has its own natural mechanical frequency, and an electrical frequency going through the coils and rotor bars can cause mechanical vibrations that are different from the initial design parameters. When an electrical frequency matches the natural frequency of a mechanical component, serious problems may occur. This may include the disintegration of a component.

Bearing problems. Another possible problem, which still isn't fully understood, is the slow disintegration of the roller/ball (antifriction) bearings that support the shaft. It appears this is caused by bearing current and static discharge. What happens is that pitting occurs on the roller/ball surface and, when accumulated, causes the bearing to make noise. If not addressed, vibration will begin to develop.

Air flow problems. An additional factor you should consider when operating a standard 60-Hz motor at very low speed is that the fan, which is fixed and attached to the rotor, may not create enough air flow to effectively cool the motor. This is true because air flow is proportional to shaft speed. Thus, at half shaft speed, the air flow is half normal flow. To compensate for low-volume air flow at low motor speeds, if installation is possible, the attachment of a constant velocity air blower package to the back of the motor will usually provide adequate cooling.

Conductor Insulation Breakdown

As mentioned, PWM circuitry, which causes the high rate of voltage rise at the carrier frequency, can cause insulation breakdown of the end turns of the motor windings, as well as possible breakdown of the feeder cable insulation. This relates to the very high rate of rise of the voltage (rate of voltage change with respect to time) in combination with the very rapidly repeating voltage pulse caused by the VFD. [ILLUSTRATION FOR FIGURE 3 OMITTED]. Conductor insulation failures in motors have occurred because of this phenomenon. This subject is not completely understood and is presently being researched. The known facts about the matter are summarized as follows.

  • Switches in the inverter section of VFDs used today cause instantaneous turn-to-turn voltage inside a motor's windings to be significantly higher than what an equivalent normal sine wave supply produces.
  • Each cycle of the fundamental voltage consists of numerous pulses of voltage.
  • Long distance between a motor and its VFD causes the turn-to-turn voltage to get even higher.

There are different approaches in explaining why there's an increase of voltage at the motor terminals. Some explain it in terms of resonant capacitance/inductance (LC) circuits; others explain it in terms of standing wave theory. Both approaches end up with a similar result. When the distance between a motor and its VFD exceeds a critical distance (which may be as low as 30 ft), there is a voltage overshoot that may exceed twice the amplitude of the voltage pulse originally delivered at the VFD output terminals.

This higher voltage comes at the motor at such a high rate of change for each of the PWM pulses, from zero volts to its peak value, that it's unevenly distributed across the winding, causing high turn-to-turn voltages in the turns connected closest to the power leads. The result places very high stress on the conductor insulation, which can cause early breakdown of the insulation.

Special inverter duty motors are available that are designed to meet or exceed the voltage amplitudes and rise times defined in NEMA Standard MG1, Motors and Generators, Section .31.40.4.2., Voltage Spikes. When connecting existing motors to VFDs with long cable lengths, you should consider using a filter to reduce the effects caused by the long cable.

Skin effect contributes to losses

In addition to the problems described above, there's yet another loss component you should be aware of: skin effect. Skin effect induces the current in an AC system to crowd to the outside surface of a conductor. This phenomenon causes resistance to be directly related to the square root of the frequency of the current. In other words, the greater the frequency, the greater the resistance due to skin effect. The carrier frequencies are usually between 800 Hz to 15 kHz, and currents at these high frequencies will cause [I.sup.2]R losses. While the high frequency currents are relatively nominal, the loss relates to the current's square power. And the carrier frequency, even at its square root, can be somewhat effective because of its basic high value. The geometry of the rotor bars also determines the degree to which the skin effect impacts rotor losses.

Motor application is very important

You should remember that a motor is a constant torque machine. In other words, at rated speed and rated torque, it will produce a certain horsepower. When speed is reduced through frequency and voltage reduction, the motor, by consuming more current, will try maintaining constant horsepower, if called for by the load. This can be done to a limited extent. As more current flows, more heat is produced, and it will not take long for the motor to overheat.

For situations where, throughout the speed range being used, there is a constant horsepower requirement, it's critical that the motor be sized to match the horsepower requirement at the lowest shaft speed anticipated. For example, if the required speed range is from 50% to 100% of rated speed and the load's horsepower requirement is 100 hp, then the motor must still be able to produce 100 hp at 50% speed. This also means that at 100% speed, the motor's horsepower output, as called for by its load, also will be 100 hp; however, the load's torque requirement will be reduced by 50%. At full-rated speed, the motor will be capable of producing 200 hp, meaning the motor will be larger than normal.

Using a VFD, with fundamental frequency being reduced to achieve lower speed, the voltage also is reduced in direct proportion to the speed reduction. As mentioned earlier, a 460V motor at half rotor speed will have 230V across its lines. Thus, if the motor's rating is 100 hp at full speed, its output would only be 50 hp at half speed.

Certain loads, like lathes and grinders, require constant horsepower throughout their operating speed range. Let's assume a VFD is serving a 20-hp lathe motor that's operating at a 25% reduction in speed (3/4 rated speed). The lathe's rotating chuck, which holds some material being worked by a cutting tool, will need constant horsepower over the entire speed range being used. If speed is reduced by 25%, voltage will be reduced by 25%. For the motor to maintain constant horsepower output, it will draw 33% more current (4/3 of normal amperage). Because current produces heat (primarily [I.sup.2]R losses), the motor will have to have sufficient thermal capacity to handle the extra amperage.

Some motors can withstand a certain amount of excess thermal load based on the motor's service factor (SF). Usually an SF ranges from 1.0 to 1.15; beyond that point, motor damage will occur. Because voltage is reduced using a VFD, a motor's horsepower rating must be increased to match the load requirement at the lowest speed used, should constant horsepower be required. Of course, this means the motor is overbuilt when used at higher speeds and will have higher losses and lower power factor (PF) at the higher speeds when operating at less than full load. However, the lower PF is compensated for by the VFD. This is a condition that must be accepted. Otherwise, you're asking for trouble.

When working with motors, you'll find it helpful to remember the following relationships:

1 hp = 0.746kW = [3 ft-lb x 1750 rpm] [divided by] 5250

Any of these numbers can be changed. When doing so, however, the equality of both sides of the equation must be maintained. Torque is ft-lb. If horsepower remains constant and speed (rpm) is reduced, obviously torque must be increased. Thus, in the above motor application (where there is a 25% reduction in speed), the motor's torque output must be increased by 33%. And if kW remains constant and the voltage is reduced (which will happen using a VFD to reduce speed), current must be increased. This could lead to overheating. Incorrect application of motors is one of the main reasons why they fail.

If someone recommends getting a VFD for your existing motor, with the idea of making adjustments that would cause the output voltage to be set to any value (with a limit up to the VFD's incoming voltage) for any particular fundamental frequency, use caution.

Such an adjustment can be made; for example, you can adjust a VFD to produce 460V at 30 Hz. If 460V is the line voltage (thus the maximum voltage), then as the fundamental frequency increases beyond the set point, the voltage going to the motor remains constant.

Let's look at one of the above examples again. Say 100 hp is required at half speed and the VFD is adjusted for delivering 460V at 30 Hz. If you use an existing motor rated for 100 hp, what will happen? Well, the motor will try to deliver 100 hp at half speed and will continue to try should the fundamental frequency be increased while the voltage remains constant at 460V. (Note that when the fundamental frequency gets below the set value [say 15 Hz], the voltage will be reduced proportionally, in this case to 230V.) At 30 Hz and 460V, the iron in the stator of this existing motor is magnetically saturated, which causes more current to flow and the motor to become excessively hot. This condition may destroy conductor insulation as well as negatively affect other motor components. Motors usually have enough iron in their stators to handle a certain ratio of volts to frequency (V/Hz). But when the ratio increases extensively, more iron is needed; otherwise, there will be overheating.

Still, using 30 Hz at 460V is an effective way of getting adjustable speed at constant horsepower, providing the iron in the motor's stator is designed to take a higher V/Hz ratio. This means that more iron has to be placed in the motor's stator. There are certain motors built today that have extra iron in their stator for operation at high V/Hz ratios. You'll have to pay a premium for them. But for certain type applications, such as above, such motors can be cost effective when compared with using an existing motor of twice the capacity. This is because the premium motor can operate at 30 Hz, 460V, and normal current, whereas the high-capacity existing motor, operating at 30 Hz, 230V, will have to use double the current, and will experience the losses associated with high-current operation.

Summary

In applying motors to loads requiring constant horsepower over wide speed ranges, you'll often find it beneficial to work with a person knowledgeable in motors. When. using an existing motor for such use, a compromise is often made between the motor's capability and the actual horsepower output, in other words, derating of the motor. In these situations, it would probably be better to purchase a new motor having the requirements you need.

When you use a motor for an application where torque requirements remain constant or reduced over the total speed range being applied, a VFD will be a good means to achieve speed control, providing the motor is able to handle the distorted electrical power being delivered to it by the VFD. Applications where torque requirements remain constant or reduced over a motor's total speed range include fans, pumps, and conveyer belts.

There are certain loads, such as centrifugal pumps and fans, where as speed decreases, torque will usually decrease as the square of the speed, and horsepower will decrease with the cube of the speed. Thus, if the horsepower is established at the low end of the speed requirement (say 50% rated speed at 10 hp), the horsepower requirement at full speed will be eight times as much, or 80 hp. As you can see in this type of situation, the deciding factor for horsepower requirements must be based at full-rated load.

TERMS TO KNOW

Inverter. A machine, device, or system that changes DC power to AC power. In regard to VFDs, inverter operation is carried out by devices such as insulated-gate bipolar transistors (IGBTs) and gate turnoff (GTO) thyristors.

Rectifier. A machine, device, or system that changes AC power to DC power. Rectification in PWM-type VFDs is performed by diodes in so-called "bridge circuits."

Synchronous speed. In regard to induction motors, the rotational speed of the stator's magnetic field is called the synchronous speed, which is equal to (in rpm): [120] x [f (the line frequency in Hz)] [divided by] P (the number of poles).

Slip. This term reflects the difference between an induction motor's synchronous speed and its rotor speed. Slip ratio in percent is equal to [(synchronous speed - rotor speed) [divided by] synchronous speed] x 100.

RELATED ARTICLE: BASICS OF ADJUSTABLE SPEED DRIVE

There are different types of VFDs, but all of them use the principle of changing the fundamental frequency (60 Hz or 50 Hz) to vary the speed. The basic components of a VFD are rectifier/inverter-equipment (the latter includes the electrical switching apparatus) and electronic controls. A VFD changes (rectifies) the incoming 60-Hz supply to DC and then changes (inverts) the DC back to AC, but at an adjustable fundamental frequency.

For a motor with a constant number of poles, there's a direct relationship between the fundamental frequency and the motor's shaft rpm. Thus, a 60-Hz-rated motor operating at half-rated speed will be powered by a VFD producing 30-Hz power.

The most popular type of VFD being manufactured today produces AC via pulse width modulation (PWM), which chops up the sine wave into DC segments of constant amplitude every half cycle. A full cycle consists of one-half positive and one-half negative voltage segments. This modulation processing of the electric power produces very high frequency (in the range between 800 Hz to 15 kHz) DC voltage pulses for each half cycle. These pulses are rectangular shaped when seen on an oscilloscope and are wider (longer duration) at the center of the one-half cycle and narrower at the ends of the one-half cycle. [ILLUSTRATION FOR FIGURE 1 OMITTED], on page 41.)

The high frequency of the voltage pulses is of ten referred to as the carrier frequency. These numerous "shots" of DC volts every half cycle have a zero-to-full-DC-bus-voltage rise time amounting to tenths of microseconds. This rate of rise of the voltage pulses (voltage change from zero volts to peak volts) is much higher than the voltage rate of rise for a normal sine wave. Because of the high rate of voltage rise, voltage spikes are produced.

As the PWM-type VFD is producing high frequency pulsating voltage, the resulting current, which is inversely dependent upon impedance, is actually in the form of a sine wave, but with numerous small irregularities, such as the frequency of a tone produced by a musical instrument with the overtones associated with the instrument. Typically, the total harmonic distortion (THD) at these high frequencies (800 to 15 kHz) is about 5% to 10%.

The output resulting from the PWM process is a current having a sine wave that can vary from 1 to 60 Hz (and sometimes the high number can exceed 60 Hz for obtaining rotor speeds in excess of rated speed). But, the current waveform is actually the summation of the fundamental frequency plus all the ultra high frequencies produced during the modulation processing, which is used to create the pulsating voltage. Due to motor winding inductance, the current waveform appears sinusoidal with a high frequency "noise" superimposed. This causes the current sine wave to have numerous small ripples or "overtones," as in Fig. 2, on page 41.

The output of a VFD has two components: the adjustable fundamental frequency and the carrier frequency. Thus, a motor connected to a VFD is being supplied power with these types of frequencies.

The impact of the high carrier frequency on a motor can be harmful. To reduce this impact, a number of VFDs are now using asynchronous switching, which causes the carrier frequency to constantly change frequency at very high speeds. In so doing, the VFD limits at least one negative impact (noise) caused by the carrier frequency.

You should note that as the fundamental frequency (which determines a motor's speed) is reduced to lower the rated shaft rpm, the voltage also is reduced, and in the same ratio. This means that, for a VFD serving a 460V motor operating at 50% rated speed, the VFD is supplying 30-Hz power to the motor, and is doing so at 230V. Thus, the motor is producing the same torque but only at half speed and at half its rated horsepower.

RELATED ARTICLE: TYPICAL MOTOR/DRIVE INSTALLATION

An understanding of motor characteristics and load characteristics is very important when using a motor for a certain load. And, such understanding becomes more important when using VFDs.

Induction motors, which typically are rated for a specific speed, are the most popular drive mechanism used today. For a given steady-state load, these motors will maintain a constant shaft speed. In other words, except for slight changes in the rated speed due to load variations, the motor's shaft speed does not change. Instead, it's determined by the frequency of the incoming power supply to the motor and the number of motor poles.

Today, when a drive system requires speed adjustment, the usual approach is to purchase a VFD and use a motor that's been designed to handle the additional parameters beyond those required for operating with a normal 60-or 50-Hz sine wave. Thus, VFDs and motors are often ordered as a package from one source. Not only do you get a motor and VFD of matched design this way, but you also get the benefit of having just one manufacturer to deal with should problems arise. (See photos on page 38.)

However, cost considerations may promote the matching of an existing motor to a new VFD. In this situation, a careful analysis of this matching is required.

RELATED ARTICLE: SOME MOTOR BASICS

The synchronous speed of an induction motor is represented by the following equation:

[N.sub.s] = (120xf) [divided by] P

where [N.sub.s] = synchronous speed in rpm

f = line frequency in Hz

P = number of poles

For example, a four-pole, 60-Hz induction motor has a synchronous speed of 1800 rpm ([120 x 60] [divided by] 4).

The rotor speed of a motor is always less than the synchronous speed because the former takes electrical/magnetic power by rotating slower than the stator's magnetic field. (With a generator, the opposite is true; the rotor, rotating faster than synchronous speed, delivers electrical power to the terminals.) This difference in speed is called slip. Slip ratio is represented by the following equation:

% Slip ratio = [([N.sub.s] - [N.sub.r]) [divided by] [N.sub.s]] x 100

where [N.sub.r] = rotor speed in rpm

Thus, an unloaded, high-efficiency, four-pole motor operating at a shaft speed of 1790 rpm would have a slip ratio of 56%. This is derived as follows: [(1800 - 1790) [divided by] 1800] x 100. The same motor, fully loaded, might have a shaft speed of 1750 rpm. Under this condition, the slip ratio will be 2.78% And the derivation is: [(1800 - 1750) [divided by] 1800] x 100. Full-load shaft speed for an 1800 rpm synchronous speed induction motor usually varies from about 1780 to 1730 rpm. This speed depends upon motor efficiency/design characteristics.

A slower speed (amount of slip) allows the rotor to cross the magnetic lines of flux established by the stator, thus creating the electric power for its magnetic needs. As a motor's toad increases, the rotor's speed decreases, resulting in the magnetic flux lines being cut at a higher rate. This induces more voltage in the rotor bars and more current, thus producing more electrical power to counteract the increase in torque requirements. When lines of flux are crossed at a higher rate, more current flows, and there is an increase in heating because of [I.sup.2]R losses.

Lester B. Manz is Leader, Adjustable Speed Engineering for GE Motors, Fort Wayne, Ind. Robert B. Morgan is Senior Editor of EC&M.