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Efficiency Curve 350 300 350 300 Torque, lb-ft 250 200 150 100 50 500 1000 1500 2000 Performance Curve 475 Torque, lb-ft Torque, Nm 407 339 272 204 136 68 2500 50 68 500 100 150 200 250 300 350 400 0 0 0 0 0 0 0 Speed, rev/min (b) 250 200 150 100 94% 136 80% 85% 407 339 272 204 Torque, Nm 90% 475
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Speed, rev/min (a)
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Figure 185 Performance and ef ciency characteristics of brushless DC motor (Courtesy Paci c Scienti c)
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drive electronics, since the switching sequence would be more complex Thus, brushless motors suffer from an inherent trade-off between torque ripple and drive complexity Among other applications, brushless DC motors nd use in the design of servo loops in control systems for example, in computer disk drives, and in propulsion systems for electric vehicles The comparisons between the conventional DC motor and the brushless DC motor are summarized in the following table:
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Conventional DC motors Advantages 1 Controllability over a wide range of speeds 2 Capability of rapid acceleration and deceleration 3 Convenient control of shaft speed and position by servo ampli ers Disadvantages 1 Commutation (through brushes) causes wear, electrical noise, and sparking
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Brushless DC motors Advantages 1 Controllability over a wide range of speeds 2 Capability of rapid acceleration and deceleration 3 Convenient control of shaft speed and position 4 No mechanical wear or sparking problem due to commutation 5 Better heat dissipation capabilities Disadvantages 1 Need for more complex power electronics than the brush-type DC motor for equivalent power rating and control range
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EXAMPLE 181 Sinusoidal Torque Generation in Brushless DC Motors
Problem
Show that the use of sinusoidal currents in a brushless DC motor can result in a ripple-free torque
Solution
Known Quantities: Coil (phase) currents Find: Total output torque, T Schematics, Diagrams, Circuits, and Given Data: Im1 = Im sin ; Im2 = Im cos Assumptions: The eld coil is wound in a two-phase circuit; each winding is sinusoidally spaced Sinusoidal currents can be generated by suitable power electronics circuits Analysis: Using equation 182, we determine that the torques generated by the currents
T(I2) T(I1)
in each of the two coils of the two-phase stator are: T1 = kT Im1 sin T2 = kT Im2 cos The sinusoidal form of the torques is due to the sinusoidal distribution of the stator windings in each phase, which are spaced 90 degrees out of phase with one another so as to produce sine-cosine components, and is shown in Figure 186 The net torque produced by the motor is the sum of the two phase torques: T = T1 + T2 = kT Im1 sin + kT Im2 cos = kT [(Im sin ) sin + (Im cos ) cos ] = kT Im [sin2 + cos2 ] = kT Im Thus, the torque generated by the motor is constant, or ripple-free
Comments: Note that this scheme requires two features: sinusoidally spaced two-phase
0 90 180 270 360 90 180 Electrical degrees
Figure 186 Sinusoidal torque-generation circuit and current waveforms for a brushless DC motor
windings and sinusoidal phase currents It is also very important that both the windings and the currents be exactly 90 out of phase
EXAMPLE 182 Selecting a Trapezoidal Speed Pro le to Match a Desired Motion Pro le
Problem
Determine the trapezoidal speed pro le required to move a load 05 m in 5 s Analyze the motion of the motor
Solution
Known Quantities: Desired load motion pro le Find: Required trapezoidal speed pro le
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Special-Purpose Electric Machines
Schematics, Diagrams, Circuits, and Given Data: The motor covers 05 m in 100 revolutions Trapezoidal pro le characteristics as shown in Figure 187
Speed (rev/min) Acceleration rate (aacc) Max speed (v) Deceleration rate (adec) Area = distance (d)
Time (s)
Figure 187 Trapezoidal pro le
Assumptions: Assume a trapezoidal speed pro le, and that the motor will accelerate for 1 s and decelerate for 1 s Analysis: De ne the following quantities:
d = motor travel (rev); v = motor speed (rev/s) T1 = acceleration time (s); T2 = time at maximum speed (s); T3 = deceleration time (s) a = acceleration or deceleration rate (rev/s2 ) From the above de nitions, we can calculate the maximum rotational velocity of the motor as follows For constant acceleration, the expressions for the motor displacement and velocity are: 1 2 and v = d = at at 2 From the above expressions, we can relate the maximum velocity to the acceleration and deceleration rates: v v adec = aacc = T1 T3 d= Now we can write an expression for the total motor travel (100 revolutions): d= = 1 1 v 2 1 v 2 1 aacc T12 + vT2 + adec T32 = T + vT2 + T 2 2 2 T1 1 2 T3 3 1 1 vT1 + vT2 + vT3 = v 2 2 1 1 T1 + T2 + T3 2 2
Note that the above expression is quite general, and could be used also for asymmetrical pro les Using the given numbers, we calculate the maximum velocity to be: v= d 1 1 T1 + T 2 + T3 2 2 = 100 rev = 25 rev/s (05 + 3 + 05)s
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