71
Current controller optimization according to the absolute value optimum with a PI controller:
V
s
= K
t
* K
A
= 1.76 * 16.3 = 28.7
= 1.66 ms,
Tn = T
1
= 34.9 ms
Kp = T
1
/ (2V
s
) = 0.0349 / (2 * 28.7 * 0.00166) = 0.37
The rise time of the current controller is 4.7 = 4.7 * 1.66 = 8 ms
The settling time of the current controller is 8.4 = 14 ms
Equivalent time constant of the lower-level current controller:
= 4 (current control loop) = 4 * 1.66 = 6.64 ms (more suitable than the theoretical 2 for
the digital control 4 used in this case).
Speed controller optimization according to the symmetrical optimum with a PI controller:
Tn = 4 = 4 * Tn current controller = 4 * 34.9 ms = 0.14 s.
Let the integral-action time T
0
(mechanical ramp-up time of motor, e.g. measured in a ramp-
up experiment) be 4000 ms.
T
i
= T
0
/ V
s
= 4s / 28.7 = 4000 ms / 28.7 = 139 ms
Kp = T
i
/ (2 ) = 139 / (2*34.9 ms) = 1.9
The rise time of the speed controller is 3.1 = 3.1 * 34.9 = 0.109 s
Values measured for the 6RA80 were typically > 40 ms
The settling time is 16.5 = 16.5 * 34.9 = 0.9 s
Speed control with an oscillating mechanical system
The mechanical part of the drive is characterized by an I-element with the mechanical time
constant T
M
. T
M
contains the motor inertia and the inertias of the coupled load, including the
shaft, gearing, coupling, etc. In real cases, elastic connecting elements, such as couplines,
long flexible torsion shafts or elastic belts, often have an effect. In this case, it is a damped
multiple-mass oscillator. For this reason, the rise times calculated above can only rarely be
achieved in practice. The structure diagram of the control will also deviate if the mechanical
system is approximated to a two-mass oscillator.
M
M
= M
BM
+ M
F
where M
M
: motor torque, M
BM
: accelerating torque of motor, M
F
: torque of the springs
M
BM
= J
M
* dn / dt * 2 / 60; J
M
: inertia of motor in m
2
kg, n: speed in RPM
M
BM
= J
M
* 2 /60 * p * N
M
; n
M
= 1 / (pT
M
) * m
BM
, T
M
= J
M
* 2 * N
MN
/ (M
MN
* 60)
M
F
= C
F
* 2 * (N
M
– N
L
) / (p * 60); C
F
: spring rate in Nm / rad
m
F
= (n
M
–n
L
) / (pT
F
) where T
F
= 60 * M
MN
/ (2 * N
MN
* C
F
)
The torque transmitted by the spring is used to accelerate the load: m
F
= m
BL
load speed: n
L
= m
BL
/ (pT
L
) where T
L
= J
L
* 2 * N
MN
/ (60* M
NM
); J
L
: Load inertia
In the case of gearing, the load inertia converted to the motor speed is used.
The structure diagram shows a cascade connection of I-elements, which would result in
undamped oscillation once the system had been excited. The damping of the springs must
therefore also be considered.