Regulated servo drives In operating cycles, all operating points must lie beneath the curve at a maximum voltage U max . Mathematically, this means that the following must apply for all operating points ( n L , M L ): Δ n Δ M k n · U max = n 0 > n L + M L When using servo amplifiers, a voltage drop occurs at the power stage, so that the effective voltage applied to the motor is lower. This must be taken into consideration when determining the maximum supply voltage U max . It is recommended that a regulating reserve of some 20% be in- cluded, so that regulation is even ensured with an unfavorable tolerance situation of motor, load, amplifier and supply voltage. Finally, the average current load and peak current are calculated ensuring that the servo amplifier used can deliver these currents. In some cases, a higher resis- tance winding must be selected, so that the currents are lower. However, the required voltage is then increased.
n
Speed-torque line high enough for all operating points
Speed-torque line too low for all operating points
accelerating
braking
M
Physical variables
and their units SI Catalog
Example for motor/gear selection The following speed curve is to be repeated cyclically.
Gear reduction* Motor current Stall current* No load current*
i
A A A A A
A, mA A, mA mA A, mA A, mA
I mot
I A I 0
n
RMS determined current Nominal current* Moment of inertia of the rotor* Moment of inertia of the load
I RMS
n = 100 rpm
I N
kgm 2 gcm 2 kgm 2 gcm 2 Nm/A mNm/A
J R J L k M k n M
Torque constant* Speed constant* (Motor) torque Load torque
rpm/V mNm mNm mNm mNm mNm mNm mNm
Nm Nm Nm Nm Nm Nm Nm Nm
0.5
2.5 3.0
3.7
M L M H
Time (s)
Stall torque* Motor torque Moment of friction RMS determined torque Nominal torque* Max. torque of gear* Operating speed of the load Limit speed of motor* Limit speed of gear* Speed
M mot
M R
M RMS
The accelerated load inertia J L is 300 000 gcm 2 = 0.03 kgm 2 . The friction torque is 400 mNm. The motor is driven with the 4-Q servo amplifier ESCON 36/2 DC for DC motors. The power supply has a maximum out- put of 3 A and 24 V. Calculation of load data The torque required for acceleration and braking are calculated as follows (motor and gearhead inertia omitted):
M N
Nm rpm rpm rpm rpm rpm rpm
M N,G
n
n L
n max
n max,G
Motor speed No load speed* Electrical power Joule power loss Mechanical power Terminal resistance Resistance at 25°C*
n mot
n 0 P el P J
W W W Ω Ω Ω
W W W Ω Ω Ω
100 · 0.5 = 0.628 Nm = 628 mNm
Δ n Δ t = 0.03 ·
π 30
π M α = J L · 30
P mech
R
Together with the friction torque, the following torques result for the different phases of motion. − Acceleration phase (duration 0.5 s) 1028 mNm − Constant speed (duration 2 s) 400 mNm − Braking (friction brakes with 400 mNm) (duration 0.5 s) -228 mNm − Standstill (duration 0.7 s) 0 mNm Peak torque occurs during acceleration. The RMS determined torque of the entire operating cycle is
R 25
Resistance at temperature T Heat resistance winding housing* Heat resistance housing/air* Max. winding temperature* Ambient temperature Winding temperature Motor voltage Induced voltage (EMF) Max. supplied voltage Nominal voltage* Resistance coefficient of Cu Max. angle acceleration Temperature difference winding/ambient Curve gradient* Time Temperature
R T
K/W K/W
R th1 R th2
s K K K K V V V V
s
t
°C °C °C °C
T
T max
T U T W
V V V V
U mot U ind U max
1 ·
1 +
2 ·
2 +
3 ·
3 +
4 ·
M 2
M 2
M 2
M 2
t
t
t
t
4
RMS =
U N
M
= 0.0039 rad/s 2 rpm/mNm
t tot
α Cu
α max
∆ n/ ∆ M
0.5 · 1028 2 + 2 · 400 2 + 0.5 · (–228) 2 + 0.7 · 0
∆ T W
=
≈ 486 mNm
3.7
K s
K
Run up time (Motor) efficiency (Gear) efficiency* Max. efficiency*
ms % % % ms
∆ t η η G τ m τ S τ W
The maximum speed (100 rpm) occurs at the end of the acceleration phase at maximum torque (1028 mNm). Thus, the peak mechanical power is: π P max = M max · 30 n max = 1.028 · π 30 · 100 ≈ 11 W
η max
Mechanical time constant* Therm. time constant of the motor* s s Therm. time constant of the winding* s
s s
(*Specified in the motor or gear data)
84 maxon
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