Difference between revisions of "Estimating power need of motor drive system"
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| − | Estimating the need of power supply wattage for motor drive can be a complex task. The needed power depends on the actual mechanical load and use of motor. The power demand is proportional to the product of motor speed and torque. The following table summarizes the relation between speed and torque vs power. | + | Estimating the need of power supply wattage for motor drive can be a complex task and may require experimentation. The needed electrical power depends on the actual mechanical load and use of motor. The power demand is proportional to the product of motor speed and torque. The following table summarizes the relation between speed and torque vs power. |
{| class="wikitable" | {| class="wikitable" | ||
| Line 5: | Line 5: | ||
! !! Moving slowly !! Moving fast | ! !! Moving slowly !! Moving fast | ||
|- | |- | ||
| − | | Producing low torque|| Very low power consumption || Low to medium power consumption | + | | '''Producing low torque'''|| Very low power consumption || Low to medium power consumption |
|- | |- | ||
| − | | Producing high torque|| Low to medium power consumption || High power consumption | + | | '''Producing high torque'''|| Low to medium power consumption || High power consumption |
|} | |} | ||
==Calculations== | ==Calculations== | ||
| − | Instantenous power | + | Instantenous electrical power demand of motor can be estimated from the equation: |
| − | <math>P_{ | + | <math>P_{electrical} [W] = \frac{Torque [Nm] * Speed [RPM]}{0.095488*efficiency [%]}</math> |
| − | + | Example: | |
| + | *we assume efficiency of motor being 80% and drive efficiency 97%, which makes total efficiency of 0.80*0.97 = 0.776 = 77.6%. | ||
| + | *motor rotates at 1000 rpm and during maximum instatenous load it outputs 2 Nm peak torque | ||
| + | Instantenous input power becomes: | ||
| − | <math>P_{electrical} [ | + | <math>P_{electrical} [W] = \frac{2 Nm * 1000 RPM}{0.095488*77.6}=259.9 W</math> |
| − | + | ==PSU sizing guide== | |
| + | In overall, PSU should be sized so that it does not overload or overheat during any condition in machine use. Since servo systems typically have greatly varying load, to find minimum sufficient PSU size might require measurements from the actual machine. To esimate minimum sufficient PSU size, measure RMS power consumption of the system during at least 10 second period of heavy use. | ||
| − | + | If measurements can't be done, then maximum RMS power load can be estimated roughly by summing rated power values of motors in the system. However, in typical machines the average power consumption is significantly less than summed motor power. | |
| − | + | ||
| − | + | {{tip|People tend generally to overestimate the power need. In most cases, required power supply wattage is significantly less than the combined power ratings of the motors in the system. This is the case if all motors are not running at their maximum speed and torque at the same moment.}} | |
[[Category:Argon_user_guide]] | [[Category:Argon_user_guide]] | ||
Latest revision as of 14:07, 4 December 2015
Estimating the need of power supply wattage for motor drive can be a complex task and may require experimentation. The needed electrical power depends on the actual mechanical load and use of motor. The power demand is proportional to the product of motor speed and torque. The following table summarizes the relation between speed and torque vs power.
| Moving slowly | Moving fast | |
|---|---|---|
| Producing low torque | Very low power consumption | Low to medium power consumption |
| Producing high torque | Low to medium power consumption | High power consumption |
Calculations[edit | edit source]
Instantenous electrical power demand of motor can be estimated from the equation:
$ P_{electrical} [W] = \frac{Torque [Nm] * Speed [RPM]}{0.095488*efficiency [%]} $
Example:
- we assume efficiency of motor being 80% and drive efficiency 97%, which makes total efficiency of 0.80*0.97 = 0.776 = 77.6%.
- motor rotates at 1000 rpm and during maximum instatenous load it outputs 2 Nm peak torque
Instantenous input power becomes:
$ P_{electrical} [W] = \frac{2 Nm * 1000 RPM}{0.095488*77.6}=259.9 W $
PSU sizing guide[edit | edit source]
In overall, PSU should be sized so that it does not overload or overheat during any condition in machine use. Since servo systems typically have greatly varying load, to find minimum sufficient PSU size might require measurements from the actual machine. To esimate minimum sufficient PSU size, measure RMS power consumption of the system during at least 10 second period of heavy use.
If measurements can't be done, then maximum RMS power load can be estimated roughly by summing rated power values of motors in the system. However, in typical machines the average power consumption is significantly less than summed motor power.
 | People tend generally to overestimate the power need. In most cases, required power supply wattage is significantly less than the combined power ratings of the motors in the system. This is the case if all motors are not running at their maximum speed and torque at the same moment. |
