Page 704 - Mechatronics with Experiments
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690   MECHATRONICS







                                                                                     T   e
                                                                                    a
                                                                      1
                                                                                     T   e
                                                                                      a
                                                                    1


                              FIGURE 8.66: Speed-torque characteristics of a DC motor in steady state with current
                              controlled amplifier.


                              motor with a constant terminal voltage (Figure 8.66). In other words, the steady-state
                              torque produced by a DC motor under constant current conditions (i.e., controlled by a
                              current feedback amplifier) decreases with increasing speed at a much slower rate that the
                              case when the motor is controlled by a voltage amplifier under constant terminal voltage
                              condition.

                              Example    Consider a DC motor with the following parameters, K Nm∕A = K V∕
                                                                                       T
                                                                                                  E
                                                                                    2
                                             −2
                              (rad∕s) = 6.7 × 10 , R = 0.5 Ω, L = 2mH, J = 4.8 × 10 −5  kg m , J = J . Determine
                                                 a        a         m                  l   m
                              the mechanical and electrical time constant of the motor (consider the case that the motor
                              and the load inertia are rigidly connected). If the current loop control gains are K = 10.0 and
                                                                                            1
                              K = 1.0, determine the speed-torque slopes under constant voltage and constant current
                               2
                              control conditions.
                                   The mechanical and electrical time constants of the motor with the load connected
                                                     J R    9.6 × 10 −5  ⋅ 0.5
                                                      T a
                                                   =      =                                     (8.322)
                                                 m               2   −4
                                                     K K      6.7 × 10
                                                      T E
                                                   = 0.010 s = 10 ms                            (8.323)
                                                     L a  2 × 10 −3  H
                                                   =    =          = 4 × 10 −3  s = 4 ms        (8.324)
                                                 e
                                                     R a    0.5 Ω
                              The slope of the torque-speed curve in steady-state under constant voltage and constant
                              current conditions is,
                                                           K K
                                                             T E
                                                   slope =                                      (8.325)
                                                        v
                                                             R a
                                                              2
                                                           6.7 × 10 −4
                                                         =                                      (8.326)
                                                               0.5
                                                         = 89.7 × 10 −4  N ⋅ m∕(rad∕s)          (8.327)
                                                         = 0.0857 N ⋅ m∕rpm                     (8.328)
                                                             K K
                                                               T E
                                                    slope =                                     (8.329)
                                                        i
                                                           R + K K
                                                            a
                                                                 1 2
                                                              2
                                                           6.7 × 10 −4
                                                         =                                      (8.330)
                                                             0.5 + 10
                                                         = 4.27 × 10 −4  N ⋅ m∕(rad∕s)          (8.331)
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