Difference between revisions of "Estimating power need of motor drive system"
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Instantenous power output of motor can be estimated from the equation: | Instantenous power output of motor can be estimated from the equation: | ||
| − | <math>P_{mechanical} [ | + | <math>P_{mechanical} [W] = \frac{Torque [Nm] * Speed [RPM]}{9.5488} </math> |
As that is the ''output'' mechanical power of motor, the input electrical power should be estimated by: | As that is the ''output'' mechanical power of motor, the input electrical power should be estimated by: | ||
| − | <math>P_{electrical} [ | + | <math>P_{electrical} [W] = \frac{P_{mechanical}}{0.01*efficiency [%]} = \frac{Torque [Nm] * Speed [RPM]}{0.095488*efficiency [%]}</math> |
| − | Example: | + | 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 | ||
| + | |||
| + | <math>P_{electrical} [W] = \frac{2 Nm * 1000 RPM}{0.095488*77.6}=259.9 W</math> | ||
==PSU Sizing guide== | ==PSU Sizing guide== | ||
Revision as of 15:15, 3 December 2015
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.
| 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
Instantenous power output of motor can be estimated from the equation:
$ P_{mechanical} [W] = \frac{Torque [Nm] * Speed [RPM]}{9.5488} $
As that is the output mechanical power of motor, the input electrical power should be estimated by:
$ P_{electrical} [W] = \frac{P_{mechanical}}{0.01*efficiency [%]} = \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
$ P_{electrical} [W] = \frac{2 Nm * 1000 RPM}{0.095488*77.6}=259.9 W $
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, it might be necessary to find effective power consumption by measuring 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. Motor power consumption is proportional to product of actual torque and speed (Power=Speed*Torque). Following chart gives rough figure of power requirement in motion systems:
