Page 671 - Mechatronics with Experiments

ELECTRIC ACTUATORS: MOTOR AND DRIVE TECHNOLOGY 657
where the L ⋅ di∕dt term in the electrical model of the motor is neglected. Then, a drive
should be selected that can provide the required DC bus voltage and current requirements
with some safety margin. Notice that at a given torque capacity, the so-called head room of
supply voltage available limits the maximum speed capacity the drive can support,
V = V − R ⋅ i (8.203)
head-room DC,max max
= K ⋅ w max (8.204)
E
(V DC,max − R ⋅ i max )
w max = (8.205)
K E
Top speed is limited by the back EMF of the motor. At any given torque level T , the current
r
required to generate that torque is i = T ∕K . This means R ⋅ i portion of the available
r
r
r
T
bus voltage is used up to generate the current needed for torque. The remaining voltage
V DC,max − R ⋅ i is available to balance the back EMF voltage. Hence, the maximum speed
r
at a given torque output is limited by the available “head-room voltage.”
Example: Brush-Type DC Motor Consider a brush-type DC motor with stator coil
resistance of 0.25 Ω at nominal operating temperature. The DC power supply is 24 VDC.
The back EMF constant of the motor is 15 V∕krpm. The voltage to the motor is turned ON
and OFF by an electromechanical relay. Consider two cases: (i) motor shaft is locked and
not allowed to rotate, (ii) motor speed is nominally at 1200 rpm. Calculate the steady-state
current developed in the motor when the relay is turned on for both cases.
When the motor is locked and not allowed to rotate, there is not any back EMF voltage
as a result of the generator action of the motor. This is the stall condition. For steady-state
condition analysis, we neglect the transient effect of inductance, the electrical equations for
the motor give
di(t)
V(t) = R ⋅ i(t) + L ⋅ + K ⋅ w (8.206)
E
dt
≈ R ⋅ i(t) (8.207)
24 V
i = (8.208)
0.25 Ω
= 96 A (8.209)
When the motor speed is non-zero, the terminal voltage minus the back EMF is available
to develop current. Then,
di(t)
V(t) = R ⋅ i(t) + L ⋅ + K ⋅ w (8.210)
E
dt
≈ R ⋅ i(t) + K ⋅ w (8.211)
E
24 V = 0.25 ⋅ i + (15∕1000) ⋅ 1200 (8.212)
24 − 18
i = (8.213)
0.25
= 24 A (8.214)
Notice that under no load conditions (i = 0), the maximum speed is limited by the back
EMF and it is
di(t)
V(t) = R ⋅ i(t) + L ⋅ + K ⋅ w (8.215)
E
dt
24 ≈ R ⋅ 0 + (15∕1000) ⋅ w max (8.216)
24 ∗ 1000
w max = (8.217)
15
= 1600 rpm (8.218)