Page 623 - Mechatronics with Experiments

ELECTRIC ACTUATORS: MOTOR AND DRIVE TECHNOLOGY 609
this so-called regenerative energy is large, external resistors are added for the purpose of
dumping it.
In load driven applications, such as tension controlled web handling or gravity driven
loads, the motor continuously operates in regenerative power mode (generator mode). The
tension, hence the torque, generated by the motor is always in the opposite direction of
the speed. If all of the regenerative power is to be dissipated as heat using external resistors,
the resistors should be sized based on the following continuous power dissipation cabability,
P cont = RMS(F tension (t) ⋅ ̇ x(t)) (8.10)
In a given application, the amount of regenerative energy is a function of inertia,
deceleration rate, and time period. In this example, the regenerative energy is is the time
integral of the regenerative power as follows,

t dec
E (t) = P (t) ⋅ dt (8.11)
reg
m
∫
0
t dec
= F(t) ⋅ ̇ x(t) ⋅ dt (8.12)
∫
0
x dec
= F(x) ⋅ dx (8.13)
∫
0
x dec
= m ⋅ ̈ x(t) ⋅ dx (8.14)
∫
0
x dec d ̇ x
= m ⋅ ̇ x(t) ⋅ dx (8.15)
∫
0 dx
x dec
= m ⋅ ̇ x(t) ⋅ d ̇ x (8.16)
∫
0
1 ( 2 2 )
= ⋅ m ⋅ ̇ x − ̇ x (8.17)
2 1 2
If we consider the tension control application, then regenerative energy is always increasing
which must be either stored or used or dissipated. Let us consider a tension control case
where the web tension is constant F(t) = F and the web speed is constant, ̇ x(t) = ̇ x . Then
0 0
for any period of time Δt, the regenerative energy is
E reg (Δt) = F ⋅ ̇ x ⋅ Δt (8.18)
o
o
This energy (E ) must be dissipated at the “regen” resistors and the motor winding
reg
due to its resistance (E ) and partially stored in the DC bus capacitors (E cap ).
ri
E = E + E (8.19)
reg cap ri
Let us assume that the regenerative resistors will be activated by an appropriate logic circuit
whenever DC bus voltage reaches a voltage level V , where V nom < V reg < V max . V max
reg
is the maximum voltage level above which the amplifier control circuit would disable the
transistors and go into “fault” mode. Then, the amount of energy that can be stored in the
capacitor is
1 ( 2 2 )
E cap = ⋅ C cap ⋅ V − V (8.20)
2 reg nom
E cap
C (8.21)
cap = 2 ⋅ ( )
V 2 − V 2
reg nom
where C cap is the capacitance of the capacitors. Clearly, the capacitor can store a finite
amount of energy and its size grows as the required energy storage capacity increases.