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- Check what is the available voltage to your application.
- Determine the needed motion profile:
a a) Repetitive operations, plot out motor speed throughout cycle, including acceleration/deceleration times.
b) For non-repetitive (like milling) operations calculate peak speed and acceleration
3. Define the amount of torque “muscle” needed (difficult to calculate accurately but most forgiving):
a a) Inertia J (influences to the acceleration)
b) Friction
c) External forces and torques.
4. Calculate inertial ratio, IR, (often overlooked but important for determining servo performance).
, where a is the gear reduction.
- Basic servo drives may require inertia ratios of 3:1 or smaller
- Advanced features (auto-tuning, vibration suppression, resonance filters, disturbance compensation functions) allow up to 30:1
- A ratio of 1:1 gives excellent performance (but usually oversized motor).
- Ratios less than 1:1 waste power with no performance advantage.
5. Find a motor and drive that:
- Matches supply voltage
- Has rated speed, continuous torque, peak torque, that exceed calculated values and
- Satisfies inertia ratio requirement in Step 4
6. If there is a motor which is a close match you are finished. Otherwise continue to the step 7 to add gearing.
7. Matching the motor to load with a gear. Gear matches the servo to the load by reducing speed, increasing torque and lowering inertia ratio:
- Gear box manufactures list the inertia of the gear box to be added to the load.
- Divide the motor speeds by required speed to get a starting gear ratio.
- Divide the required torque by the gear ratio to find the new required torque.
- This will narrow choices to a couple of motors. Find a motor with an acceptable inertia ratio. If two motors look equal choose the one with smaller inertia ratio.
- Repeat this step a couple of times.
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The different loads are reduced to the motor using the following equations, where the notations are obvious. The reduction is based to the equality of powers.
The inertial quantities are reduced to the motor with the equations. Here the reduction is based to the equality of the kinetic energies.
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