maxon

maxon motorRegulated servo drivesIn work cycles, all operating points must lie beneath the curve at a maximumvoltage U max . Mathematically, this means that the following must applyfor all operating points (n L , M L ):k· U max 0> L+ LWhen using servo amplifiers, a voltage drop occurs at the power stage, sothat the effective voltage applied to the motor is lower. This must be takeninto consideration when determining the maximum supply voltage U max . Itis recommended that a regulating reserve of some 20% be included, sothat regulation is even ensured with an unfavorable tolerance situation ofmotor, load, amplifier and supply voltage. Finally, the average current loadand peak current are calculated ensuring that the servo amplifier used candeliver these currents. In some cases, a higher resistance winding must beselected, so that the currents are lower. However, the required voltage isthen increased.nMExample for motor/gear selectionA drive should move cyclically according to the following speed diagram.n0.5 2.5 3.0 3.7The inertia of the load to be accelerated J L is 140 000 gcm 2 . The constantcoefficient is approximately 300 mNm. The motor should be driven with a4-Q servo amplifier from maxon ESCON 36/2. The power supply unit deliversmax. 3 A and 24 V.Calculation of load dataThe torque required for acceleration and braking are calculated as follows(motor and gearhead inertia omitted): M = J L = 0.014 30 30600.5= 0.176 Nm = 176 mNmTogether with the friction torque, the following torques result for thedifferent phases of motion.– Acceleration phase (duration 0.5 s) 476 mNm– Constant speed (duration 2 s) 300 mNm– Braking (friction brakeswith 300 mNm) (duration 0.5 s) 124 mNm– Standstill (duration 0.7 s) 0 mNmPeak torque occurs during acceleration.The RMS determined torque of the entire work cycle isM RMS==t 1· M 2 1 + t 2 · M 2 2 + t 3 · M 2 3 + t 4 · M 2 4t tot0.5 ·476 2 + 2 ·300 2 + 0.5 · 124 2 + 0.7 · 0 285 mNm3.7The maximum speed (60 rpm) occurs at the end of the acceleration phase atmaximum torque (463 mNm). Thus, the peak mechanical power is:P max= M max n 30 max= 0.476 60 W30Physical variablesand their unitsSI Catalogi Gear reduction*I Motor current A A, mAI A Starting current* A A, mAI 0 No-load current* A mAI RMS RMS determined current A A, mAI N Nominal current* A A, mAJ R Moment of inertia of the rotor* kgm 2 gcm 2J L Moment of inertia of the load kgm 2 gcm 2k M Torque constant* Nm/A mNm/Ak n Speed constant* rpm/VM (Motor) torque Nm mNmM L Load torque Nm mNmM H Stall torque* Nm mNmM mot Motor torque Nm mNmM R Moment of friction Nm mNmM RMS RMS determined torque Nm mNmM N Nominal torque Nm mNmM N,G Max. torque of gear* Nm Nmn Speed rpmn L Operating speed of the load rpmn max Limit speed of motor* rpmn max,G Limit speed of gear* rpmn mot Motor speed rpmn 0 No-load speed* rpmP el Electrical power W WP J Joule power loss W WP mech Mechanical power W WR Terminal resistance R 25 Resistance at 25°C* R T Resistance at temperature T R th1 Heat resistance winding housing* K/WR th2 Heat resistance housing/air* K/Wt Time s sT Temperature K °CT max Max. winding temperature* K °CT U Ambient temperature K °CT W Winding temperature K °CU Motor voltage V VU ind Induced voltage (EMF) V VU max Max. supplied voltage V VU N Nominal voltage* V V Cu Resistance coefficient of Cu = 0.0039 max Maximum angle acceleration rad/s 2n/M Curve gradient* rpm/mNmT W Temperature difference winding/ambient K Kt Run up time s ms (Motor) efficiency % G (Gear) efficiency* % max Maximum efficiency* % m Mechanical time constant* s ms S Therm. time constant of the stator* s s W Therm. time constant of the winding* s s(*Specified in the motor or gear data)42Key informationMay 2012 edition / subject to change