Document Sample
DC-AC-Motors-Control Powered By Docstoc

            Overview and Control of DC and AC Motors
                                                                    Brian Bouma

                                                                              up of eight “T”-shaped parts) is the part of the motor that
    Abstract— an overview of the differences in form and function              rotates during operation, while the stator (the dark blue block
between DC and AC motors. An in-depth analysis of the general                  and light blue “fingers” around the rotor in Fig. 1) remains
differential equations and transfer functions of DC and AC                     stationary during operation, relative to the motor’s casing and
motors. Strategies for controlling DC and AC motors using gain
                                                                               mounting [3]. The stator is made up of either a winding or a
and PID control are discussed and analyzed in depth. Methods
for determining the time constants of DC and AC motors are                     magnet, which creates magnetic flux in the magnetic field
discussed and analyzed.                                                        formed between the stator and the rotor [3]. For the simple
                                                                               analytical purposes herein, it makes no difference whether a
 Index Terms – AC motors, Control systems, DC motors,                          winding or a magnet is used in the stator, so the use of a
Motor drives.                                                                  magnet will be assumed. The rotor has a winding on its
                                                                               surface, termed the armature, in which electromotive forces
                          I. NOMENCLATURE                                      are induced by the magnetic field formed between the stator
    PID        Proportional Integral Derivative                                and the rotor [3]. The armature winding is supplied current
                                                                               through the collector (the yellow cylinder attached to the rotor
                          II. INTRODUCTION                                     in Fig. 1), on which the brushes (the two brown tabs touching
                                                                               the collector) apply pressure [3]. The collector is mounted on
   Motors are an integral part of engineering in today’s society.
                                                                               the same shaft as the armature, and the fixed brushes are
They are used in a wide variety of applications, from running
                                                                               connected to the armature terminals [3]. Thus, the power to
fans to driving belts to turning wheels. Yet, despite their
                                                                               the motor runs through the brushes, into the collector, and
prevalence in the designs of undergraduate engineering
                                                                               through the armature winding to produce the electromotive
students, most such students have very little idea of how
                                                                               forces between the stator and the rotor (the power cables are
motors actually work, or of how to control them safely and
                                                                               the red lines attached to the brushes in Fig. 1). The brush-
dependably. This paper describes both DC and AC motors,
                                                                               collector assembly provides current to the armature windings
analyzes them from a control standpoint, and determines
                                                                               in such a way that the current flows in one direction when the
adequate strategies for controlling them in a manner which is
                                                                               windings are under a magnetic North pole (from the stator
both safe and reliable. Stepper and servo motors will not be
                                                                               magnet), and in the other direction when the windings are
discussed here, as their form, function, and application are
                                                                               under a magnetic South pole [3]. The rotor windings are made
considerably different from that of DC and AC motors, and the
                                                                               up of coils, called sections, all sections being of an equal
analysis of those four motor types would be too much
                                                                               number of turns (in Fig. 1, the eight colored sets of line
information to cover in this setting.
                                                                               segments, two segments per set, on the “front” of the T-shaped
                       III. MOTOR OVERVIEW                                     parts of the rotor are the ends of the coils; the coils run through
                                                                               the length of the rotor and wrap around the opposite end of the
   To the uninformed observer, DC and AC motors appear to                      T-shaped parts of the rotor) [3]. Each section has two sides,
be basically identical. Even though they seem to operate in                    which are inserted into two slots spaced apart a distance equal
essentially the same way, their physical structures, and thus                  to the distance between the two field poles [3]. This way,
their range of applications, vary significantly.                               when the conductors of one side of a section are under the
   The brush DC motor is arguably the simplest variable-speed                  North pole, the conductors of the other side of the same
DC motor design, in addition to being the most common. For                     section are under the South pole [3]. The sections of armature
these reasons, the brush design is the one being described and                 winding are all connected together, in series, with the end of
analyzed here. The brush DC motor (or all DC motors, for                       the last section being connected to the beginning of the first
that matter) is made up of a stator and a rotor (refer to Fig. 1               section, so that the winding as a whole is continuous, having
for all descriptions relating to the brush DC motor). As the                   no particular start or finish [3]. For this to work, each slot
names suggest, the rotor (the circular portion of Fig. 1 made                  must contain two sides of sections (half each of two different
                                                                               sections) [3]. As the rotor rotates, when a section changes
   This work was done for Engineering 315 at Calvin College in Grand
Rapids, Michigan, in the fall of 2004. All software used in this paper was     from being under the North pole to being under the South pole
supplied by Calvin College. This project was supported financially by Calvin   (and hence the current in that section reverses direction), that
College and Smiths Aerospace LLC.                                              section commutates. Commutation of a section of the winding
   Brian Bouma works for Smiths Aerospace LLC and attends Calvin
College in Grand Rapids, MI 49546 USA (e-mail:
                                                                               is the changing of the section from being under one pole to

being under the other pole [3]. Two sections (two sections
that are opposite each other on the rotor) commutate at a time,
one switching from North to South pole, the other switching
from South to North pole. Because the two poles “swap”
sections simultaneously, half of the windings are under each
pole at all times. The North and South poles are an effect of
the flux in the armature created by the current flowing through
the two sets of windings (each pole containing one set). When
a section commutates, the brushes, applying pressure on the
collector, short-circuit the two ends of that section together, to
release the energy stored in the coils of the section before the
direction of current flow in the section is reversed [3]. Despite
this ingenious design, sparks are still produced between the
brushes and the collector [3].

                                                                                 Fig. 2. AC Motor Stator and Rotor

                                                                        The inner face of the stator is made up of deep slots (or
                                                                     grooves, depending on the exact design, but that distinction
                                                                     makes no difference for this analysis), through which the
                                                                     windings run (the red arrow in Fig. 3 indicates one such slot)

                   Fig. 1. Brush DC Motor

   The primary advantage of a DC motor is that the magnitude
of the torque produced by the output shaft never changes [3].
This makes the DC motor perfect for applications that have
large startup loads, particularly automotive applications (like
the drive wheels in electric vehicles, for instance), where DC
power is readily available under most circumstances.
   Today, the asynchronous (or induction) motor is the most
commonly used electric motor in industry and in household                     Fig. 3. Winding Slot in AC Motor Stator
devices [3]. For this reason, the asynchronous motor is the AC
motor design being described and analyzed here.                      The motor’s number of poles is determined by the layout of
Electromagnetic induction is the creation of a current through       the windings within the stator (the number of poles will always
a conductor that that is within a magnetic field [3]. The            be a multiple of two, but, again, that has no impact here) [3].
magnetic field is capable of generating a large current in the          The rotor is also cylindrical, though this cylinder does have
conductor without requiring any physical contact whatsoever          a core. The rotor is made of steel disks (the long black
[3]. It is this principle that allows the induction motor to         segments in Fig. 4) slotted around the exterior of the cylinder
function without having any sliding electrical contacts (such as     [3]. A short-circuited winding is placed in the slots between
the brushes in the brush DC motor) [3]. The stator (the hollow       the steel disks, preventing the need for a supply to be
cylinder in Fig. 2) is essentially a hollow cylinder with no         connected to the rotor [3]. The currents in the rotor are
ends, and may be constructed out of either cast iron or              induced by the interaction between the magnetic fields of the
aluminum [3].                                                        stator and the rotor, leading to the name for this type of motor
                                                                     [3]. A copper or aluminum bar (in the case of Fig. 4,
                                                                     aluminum, judging by the color) is also placed in each slot
                                                                     between the steel disks [3]. A circular conducting ring is
                                                                     placed on the end of the cylinder (refer to Fig. 4), to connect
                                                                     the ends of the bars together (because the conducting ring and
                                                                     bars resemble a circular cage, this rotor design is called
                                                                     “squirrel cage”) [3].

                                                                   viscous damping, and θ(t) is the motor’s angular position [2].
                                                                   By assuming zero initial conditions and then Laplace
                                                                   transforming each of these equations, s-domain equations are
                                                                   reached. The developed torque is now
                                                                                           T(s ) K2 If(s )
                                                                                                             ,                (4)
                                                                   the field voltage is now
                                                                                       Vf(s )         
                                                                                                Lf s  Rf  If(s ),         (5)
                                                                   and the mechanical torque is now
                                                                                        T( s )        Is 2  Bs (s ),       (6)
                                                                   where all of the constants have the same meaning as in the
                                                                   time-domain differential equations, If is used in place of if and
    Fig. 4. Squirrel Cage and Steel Disks in AC Rotor              Θ is used in place of θ [2]. By substituting and solving, the
                                                                   transfer function of the motor is found to be
The interaction between the stator magnetic field and the rotor                    ( s )             Km
magnetic field forces the rotor to spin, relative to the stator,
resulting in a functioning AC motor [3].
                                                                                  Vf( s )                        
                                                                                                s  m s  1  e s  1         ,   (7)
   There are two varieties of asynchronous motors: three phase     where
and single phase. Three phase motors are used mainly in                                                       I
industry, while single phase motors are more common                                                           B                        (8)
household appliances [3].
                                                                   is the mechanical time constant of the motor,
   The primary advantage of an induction motor is that it
contains no sliding electrical contacts, resulting in a simple                                e
robust design that is easy to manufacture and maintain [3]. A                                       Rf
secondary advantage is that the available range of induction
                                                                   is the electrical time constant of the motor, and
motors is from only a few watts to several megawatts, making
the use of induction motors in a wide variety of applications                                 Km
physically feasible [3].                                                                              B Rf
                  IV. TRANSFER FUNCTIONS                           is another constant [2]. This is the transfer function that will
                                                                   be used for the control analysis of the DC motor in the next
   The differential equations and transfer functions of DC and     section.
AC motors are crucial to the analysis of the control of these         The differential equations and transfer functions for AC
machines. The transfer functions are derived from the              motors are considerably less complicated than those for DC
differential equations using Laplace transforms, a method all      motors, owing to the fact that AC motors only have a single
too familiar to most engineers.                                    time constant while DC motors have two. AC motors are
   The differential equations and transfer functions for DC        described by two differential equations: The torque (T(t)) is
motors are more complicated than those of AC motors, due to        described by
the fact that DC motors have time lags because of both the
armature inductance and the winding, while AC motors have                               T( t)         K v( t)  m  ( t)
only a single time constant. DC motors are described by three                                                      dt     (11)
differential equations: The developed torque (T(t)) is             where K is a constant, v(t) is the voltage provided to the
described by                                                       motor, θ(t) is the angular position of the motor, and m is
                         T(t) K2 if(t)                            described by
                                          ,                 (1)
                                                                                    "stall t orque (at rat ed volt age)"
where if(t) is the current through the field and K2 is constant               m
                                                                                   "no-lo ad speed (at rated volt age)" , (12)
[2]. The field voltage (vf(t)) is described by
                                          d                        where stall torque (at rated voltage) and no-load speed (at
                 vf( t)   Rf if( t)  Lf if( t)                  rated voltage) are characteristics of any specific AC motor [2].
                                          dt    ,           (2)    The torque is also described by
where Rf is the field resistance and Lf is the field inductance
[2]. Lastly, the mechanical torque (T(t)) is described by                                             d  d
                                                                                     T( t )    ( t)  B  ( t )
                                   2                                                       dt                     ,      (13)
                               d      d
                 T( t )     ( t)  B  ( t )
                                   2  dt                           which is identical to the third differential equation that
                        dt                     ,       (3)         describes DC motors, and has the same meaning [2]. By
where I is the motor’s moment of inertia, B is the motor’s         equating the two AC motor equations, assuming zero initial

conditions, and then taking the Laplace transform of the             PID controller only has two zeros, only two of the poles may
resultant equation, the transfer function of an AC motor is          be eliminated. This raises the obvious question: Which poles
found to be                                                          should be eliminated and which one should be left alone?
                      ( s )      Km                                 That depends on which pole, as the only pole in the system,
                               s    s  1 ,
                                                                     results in a system with the shortest rise time, the shortest
                      V( s )                                 (14)    settling time, and the least overshoot. This system was
where                                                                modeled and simulated in MATLAB/Simulink (see Fig. 5),
                                   K                                 using a unit step input as a standard input.
                                m B                         (15)
is a constant, and
                               m B                       (16)
                                                                     Fig. 5. Simulink Model of DC Motor Control System with
is the time constant of the motor [2]. This is the transfer
                                                                                          PID Control
function that will be used for the control analysis of the AC
motor in the next section.                                           The gain block in this system is used to dramatically speed up
                                                                     the response of the system, although it does not effect the
                          V. CONTROL                                 location of any of the poles or zeros, so it does not impact the
   Precise control of motors is vital to the use of motors in any    “control” aspect of this analysis (even though it decreases the
application. Without a system in place to prevent the motor          peak and settlings times for the system). The electrical time
from operating unchecked, a step increase to the inputs of a         constant of the motor (τe) was assumed to be 1 ms, the
motor would result in the motor accelerating until it literally      mechanical time constant of the motor (τm) was assumed to be
broke apart, costing untold amounts of money to repair and           100 ms, and the motor constant (Km) was assumed to be 0.050
replace damaged equipment, and undoubtedly ruining                   N*m/A, all of which are typical constants for a DC motor [1].
someone’s day.                                                       The first two poles to be eliminated were the poles that result
   The first step to safe control is the use of negative feedback,   from the electrical and mechanical time constants, leaving the
                                                                     third pole (at s = 0) alone. This was done by setting the
a concept so common that it warrants no explanation here,
                                                                     differentiator coefficient to the product of the electrical and
which has been employed in all of the simulations whose
                                                                     mechanical time constants, the proportional gain coefficient to
results are shown here.
                                                                     the sum of the electrical and mechanical time constants, and
   A control method which is popular because of its                  the integrator coefficient to one. The result of this simulation
robustness, its simplicity, and its reusability is PID control. A    is so worthless that it will not be shown here. By eliminating
PID controller contains a proportional gain, an integrator, and      those two particular poles (hand calculations would confirm
a differentiator (hence its name), all of which are summed           this, but are not necessary here), the overshoot of the system
together to produce the output of the controller. The transfer       became 100% (the output was “2” while the input step was
function of a PID controller has the form                            only “1”), the peak time became 0.44 seconds (not terrible, but
                                             2                       not great either), and the settling time became infinite. The
                     KI                KD s  KP s  KI
        PID KP     KD s                                           system never settled. The output of the system was sinusoidal
                 s                           s              ,(17)    and continued for all of eternity (until the simulation time of
where KP is the proportional gain coefficient, KI is the             ten seconds was reached). This is no good. In the next
                                                                     simulation run, the two poles eliminated were the pole at s = 0
integrator coefficient, and KD is the differentiator coefficient.
                                                                     and the pole created by the electrical time constant. This was
The proportional gain is used to amplify the input signal. The
                                                                     done by setting the differentiator coefficient to the value of the
integrator is used to improve the accuracy of the control
                                                                     electrical time constant, the proportional gain coefficient to
system, that is, to minimize the steady-state error (the             one, and the integrator coefficient to zero. The output of this
difference between the input value and the final output value)       simulation is shown in Fig. 6.
as much as possible. The differentiator is used to increase the
damping in the system, which will decrease both the peak time
and the settling time of the system.
   As can be recalled from above, the transfer function of the
DC motor is third order in the denominator, so it has three
poles (roots of the polynomial in the denominator). Likewise,
the transfer function of the PID controller is second order in
the numerator, so it has two zeros (roots of the polynomial in
the numerator). Thus, the PID values may be set so that the
zeros of the PID controller eliminate the poles of the DC
motor. However, since the DC motor has three poles and the

                                                                  while the system corresponding to Fig. 6 was simulated for 1
                                                                  second (with the step input occurring at time t = 0 seconds),
                                                                  the system corresponding to Fig. 7 was simulated for only 0.25
                                                                  seconds, because its time response was so fast that it could
                                                                  scarcely be seen on a 1-second time plot.
                                                                     From these findings, it appears as though the best way to
                                                                  use a PID controller in conjunction with a DC motor is to set
                                                                  the PID values so that the s = 0 pole and the pole created by
                                                                  the larger of the two time constants (typically the mechanical
                                                                  time constant) are eliminated.
                                                                     The transfer function of an AC motor is simple enough that,
                                                                  even though PID control may be implemented for an AC
                                                                  motor (as it will be later), acceptable time response values may
 Fig. 6. Time Response of System with Mechanical Time
                                                                  be attained by simply adjusting the gain in the system (see Fig.
                    Constant Pole
                                                                  8). For all of the AC motor control system simulations, a time
                                                                  constant (τ) of 0.1 seconds (100 ms) and a motor constant of
Fig. 6 shows the time response of the system when the
                                                                  0.050 N*m/A have been assumed.
mechanical time constant and s = 0 poles were eliminated by
the settings of the PID controller. The horizontal axis shows
time, in seconds, and the vertical axis is the output of the
system, normalized to the magnitude of the input step. Closer
analysis of this simulation output shows a settling time (using
2% criterion) of approximately 0.76 seconds, a peak time of          Fig. 8. Simulink Model of AC Motor Control System
approximately 0.14 seconds, and an overshoot of
approximately 49%. This simulation shows a better peak time       By running this simulation many times, and adjusting the gain
(0.14 seconds as opposed to 0.44 seconds), a better overshoot     value each time, it can be found that, between the gain values
(49% as opposed to 100%), and a much better settling time         of approximately 128.957 and 143.076, the overshoot is less
(0.76 seconds as opposed to not settling at all). However, the    than (or equal to) 10% and the peak time is less than (or equal
peak time could still improve (a little), the peak time could     to) 0.5 seconds. At a gain of 128.957, the overshoot is 8.21%
improve quite a bit, and the overshoot has the most room for      while the peak time is exactly 0.5 seconds. At a gain of
improvement. The simulation was run a third time, this time       143.076, the peak time is 0.461 seconds while the overshoot is
eliminating the pole at s = 0 and the pole created by the         exactly 10.0%. For gain values between the two just given,
mechanical time constant. This was done by setting the            the overshoot and peak time are between the two extremes
differentiator coefficient to the value of the mechanical time
                                                                  given above for overshoot and peak time. It just so happens
constant, the proportional gain coefficient to one, and the
                                                                  that the settling time, regardless of the gain value, is always
integrator coefficient to zero. This simulation output is shown
                                                                  0.8 seconds (two opposing values, both functions of the gain,
in Fig. 7.
                                                                  cancel out in the calculation of the settling time). These values
                                                                  are, of course, also dependent on the value of the motor’s time
                                                                  constant, so the overshoot, peak time, and settling time will
                                                                  change if the time constant is changed. For comparison
                                                                  purposes, the model of Fig. 8 was simulated (see Fig. 9), using
                                                                  a gain of 135, as that value is roughly midway between the two
                                                                  gain values discussed earlier which denote the two endpoints
                                                                  of the range in which optimum output values are produced.

  Fig. 7. Time Response of System with Electrical Time
                    Constant Pole

The simulation output in Fig. 7 shows that the elimination of
the s = 0 and mechanical time constant poles produces the best
system performance yet. Since the output does not oscillate,
the peak time cannot be characterized as having a unique
value, so the peak time will be considered to be the same as
the settling time. The settling time is approximately 0.073
seconds (73 ms), and there is no overshoot. Note also that        Fig. 9. AC Motor System Time Response with Gain of 135

                                                                    had to be used to attain a high enough resolution in the output
This simulation output shows a peak time of approximately           graph that the response could be seen without difficulty.
0.48 seconds, a settling time of approximately 0.73 seconds,           From these findings, it appears as though the best way to
and an overshoot of approximately 9.0%.                             control an AC motor is the use a PID controller and to set the
   Since the AC motor transfer function is only second order        PID values so that the poles of the motor are eliminated.
(as opposed to the third order DC motor transfer function),         When such equipment is not available, the gain can be
those values may be easily changed by adjusting the gain            adjusted to optimize the response, although that approach
value. However, better system response may be attained by           produces much less favorable results than those ensuing from
placing a PID controller in the control system with the motor       the use of a PID controller.
(see Fig. 10), and setting the PID values accordingly.

                                                                                         VI. TIME CONSTANTS
                                                                       A method to control both DC and AC motors with PID
                                                                    control has been determined. This method is heavily reliant on
                                                                    the time constants of the motor, and assumes that anyone
  Fig. 10. Simulink Model of AC Motor Control System                trying to implement PID control for a motor knows the values
                    with PID Control                                of the motor’s time constants. Since the nameplate on most
                                                                    motors only provides the name of the manufacturer, the
Because the denominator of the AC motor transfer function is        motor’s serial number, the motor’s input voltage, the
second order, both poles may be eliminated by setting the PID       maximum rotational speed of the motor’s output shaft (in
values strategically.       This was done by setting the            RPM), the power of the motor (in horsepower), and the weight
differentiator coefficient to the value of the time constant, the   of the motor, another questions arises: How can the time
proportional gain coefficient to one, and the integrator            constants be determined for a motor that one already has in
coefficient to zero. The output of that simulation is shown in      hand? This is a very good question.
Fig. 11.                                                               While the data sheets for some motors do contain the values
                                                                    of the time constants, this is not true in every case, so those
                                                                    values must be obtained experimentally.
                                                                       In the case of a DC motor, this is easier said than done.
                                                                    First, a system resembling that of Fig. 5 must be constructed,
                                                                    where the output being measured is the rotational speed of the
                                                                    output shaft of the motor. Keep in mind that, in this case, the
                                                                    transfer function of the motor is really a black box; all that is
                                                                    known about is for certain is that it has three poles, one of
                                                                    them at s = 0, and the other two created by the two time
                                                                    constants. Unfortunately, there is no cut and dried, step by
                                                                    step process by which to determine the time constants of the
                                                                    DC motor. As such, the system must be run many times,
 Fig. 11. Time Response of System with Both Motor Poles             adjusting the PID values each time. The integrator coefficient
                      Eliminated                                    should be set to zero, and the proportional gain coefficient
                                                                    should be set to one, and both values should left there for all of
   Closer inspection of Fig. 11 reveals a peak time of 0.14         the system runs.        Only the value of the differentiator
seconds, a settling time of 0.065 seconds, and an overshoot of      coefficient should be changed between runs. The electrical
0.4%. The settling time is less than the peak time because the      time constant will likely be around 1 ms, and the mechanical
peak value is less than 2% greater than the final value. In         time constant will be greater than the electrical. Recall from
comparison to the simulation (of Figs. 8 and 9) which was           earlier that the best system response will be attained when the
optimized by adjusting nothing other than the gain, this            differentiator coefficient is equal to the mechanical time
simulation exhibited considerable improvement, decreasing           constant. The mechanical time constant may range anywhere
the peak time by approximately 0.34 seconds (from 0.48 to           from only a few milliseconds to hundreds of milliseconds, so
0.14 seconds), decreasing the settling time by approximately        that full range should be tested. Thus, when a system response
0.665 seconds (from 0.73 to 0.065 seconds), and decreasing          closely resembling Fig. 7 (no oscillations, no overshoot, and a
the overshoot by approximately 8.6% (from 9.0 to 0.4%). A           very short settling time) is achieved, the value of the
simpler way to compare the two responses by inspection is to        differentiator coefficient for that run is the value of the
note the time scale in Figs. 9 and 11. The simulation               mechanical time constant.
represented in Fig. 9 was run for 1.5 seconds, while the               To determine the time constants of an AC motor, a system
response of the system whose simulation is represented in Fig.      resembling that of Fig. 8 must be constructed, where the
11 was so much faster that a time scale of only 0.5 seconds         output being measured is the rotational speed of the output

shaft of the motor. Choose an arbitrary gain value (it should      some reason, the value of the peak time is not available or
be somewhat high, to get a relatively fast time response, so a     cannot be gleaned from the time response graph, the damping
run will take no more than a couple seconds, but not so high       ratio and the settling time may be used in conjunction with
that separate oscillations are indistinguishable), and then do     (20) to find the value of the natural frequency, but since the
not change the gain value again. Keep in mind, again, that the     peak time may generally be observed with greater accuracy
transfer function of the motor is really a black box, and the      than the settling time, (20) should only be used as a last resort.
only thing that is known about it for certain is that it has two   Similarly, if there is some problem with the value of the peak
poles, one at s = 0 and the other created by the time constant.    time or percent overshoot, any two of (18), (19), and (20) may
The system must be run a single time (after running it several     be used to solve for the values of the damping ratio and natural
times to determine a good gain value), and the peak time,          frequency, but since the percent overshoot and peak time are
settling time, and percent overshoot all need to be observed.      the two that may be measured the most easily and with the
Fig. 12 shows the output of the model of Fig. 8, using a gain of   most accuracy, (18) and (19) should be used whenever
1000 instead of the gain of 135 shown in Fig. 8.                   possible. Finally, using the values of the damping ratio and
                                                                   natural frequency in conjunction with (21), the time constant
                                                                   may be solved for, in this case turning out to be approximately
                                                                   0.10 seconds. Considering that the value used for the time
                                                                   constant in the simulation was 0.1 seconds, it appears as
                                                                   though this method works with some degree of accuracy.
                                                                      When attempting to determine the time constants of a DC
                                                                   motor, the best method available is adjust the values of the
                                                                   coefficients of a PID controller until the result of a run bears
                                                                   resemblance to that of Fig. 7, and then to use those coefficient
                                                                   settings from then on. When attempting to determine the time
                                                                   constant of an AC motor, the system should be run, and the
                                                                   peak time, settling time, and overshoot observed and used to
                                                                   calculate the value of the time constant.
  Fig. 12. AC Motor System Response with Gain of 1000

Fig. 12 shows a peak time of approximately 0.14 seconds, a
                                                                                           VII. ACKNOWLEDGMENTS
settling time of approximately 0.76 seconds, and an overshoot
of approximately 49%. Next, several calculations must be              The author gratefully acknowledges the contribution of
made, using the following equations:                               Paulo Ribeiro, for his help in writing this paper. Without his
                                                                   support, this process would have been much more difficult.
                   PO 100 exp
                                                            The author also acknowledges the contributions of John
                                 1  2 
                                        ,                        Washburn, Frank Saggio, Paul Bakker, Matt Husson, and Nate
                                                                   Studer for their help in the research process.
where PO is the percent overshoot of the response and ζ is the
damping ratio of the system,
                                                                                                VIII. REFERENCES
                        Tp                                         [1]     Robert H. Bishop, Richard C. Dorf, Modern Control Systems, 9th ed.
                                             2                             Upper Saddle River, NJ: Prentice-Hall, 2001, p. 52-56, 227-231.
                               n 1                             [2]     D.K. Anand, Introduction to Control Systems. New York: Pergamon
                                               ,           (19)            Press Inc., 1974, p. 34-37.
where Tp is the peak time of the response and ωn is the natural    [3]     The ST Microcontroller Support Site Motor Control Tutorial,
frequency of system,                                             
                                    n
                                           ,               (20)
where Ts is the settling time of the response, and                                               IX. BIOGRAPHY
                           2  n
                                        ,                 (21)
                                                                                               Brian Bouma was born in Grand Rapids, Michigan
                                                                                               in the United States of America, on September 30,
where τ is the time constant of the motor. Using the percent                                   1982. He graduated from Grand Rapids Christian
overshoot (in this case, 49%) in conjunction with (18), solve                                  High School in 2001, and is currently pursuing his
                                                                                               undergraduate degree at Calvin College. He is
for the damping ratio, which in this case turns out to be                                      majoring in engineering, with a concentration in
approximately 0.22. Next, use the value of the damping ratio                                   electrical and computer engineering, and currently
and the peak time (in this case, 0.14 seconds) in conjunction                                  holds a minor in mathematics.
                                                                                                  His work experience includes Smiths Aerospace
with (19), and solve for the natural frequency, which in this
                                                                                               LLC where he works in Digital Design.
case turns out to be approximately 23.0 radians/second. If, for          He has been a student member of IEEE since 2001.

Shared By: