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Multiply VDC by the rated current in terms of peak-of-sine to find the electrical power that the drive is capable of.
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2.A.) AC POWER SUPPLY (IN PHASE)
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The plot below compares the instantaneous and average power between the in-phase and out-of-phase condition. One can rather easily see that the phase-shifted average power is less.
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POWER BALANCE
We can now look at the efficiency of a system. Recall that the laws of nature require that power out cannot be greater than power in. Since they also dictate that a system cannot be perfectly efficient, there must be some term introduced that accounts for this efficiency loss.
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Therefore, the total available rotary output power is equal to the electrical input power times the total efficiency. Keep in mind that this available output power is a best-case scenario since other small losses have been neglected.
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WHERE TO FIND THE EFFICIENCY DATA
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Gearheads are commonly attached to motors for many reasons, namely to gain a mechanical advantage or reduce cogging effects. Thankfully, the same power equation that is applicable to motors also applies to gearheads; one just needs to account for the efficiency of the gearhead in addition to all the other components. The power balance equation is extended to the following equation where Pr refers to the gearhead output rather than the motor output:
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CALCULATING CURRENT DRAWN FROM POWER SUPPLY TO DRIVE
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And in the case of an AC power supply:
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SUMMARY
Power out of a system can never be more than power into a system.
Efficiency of each component must be accounted for to find the maximum power a system can output relative to the power brought in.
This can be a great step to take as a sanity check that you have a properly sized drive for the application since all drives have a power output limit. Typically, this limit is seen in the rated amperage limit.
Power out of a system can never be more than power into a system.