Overall efficiency for a motor / drive system is not voltage dependent. But the voltage of the supply and the design of the motor and drive must match. Efficiency does go down as the load on a motor goes down. So if you have for example a 400VDC system and a motor designed for 500VDC, you will never be able get full output from that motor, so efficiency will be compromised because it will be running at partial load all of the time.

Also, are you referring to using an AC motor with a DC input to it? For that, you must understand that the DC input must match the PEAK voltage of the AC motor design, nit the RMS voltage. So for example if you have a 460V motor, the DC input must be higher by the sq. rt. of 2, so 650VDC if you want to get full output capability from that motor.

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One of the other things that (usually) happens when using a drive to power a motor is that the power electronics are optimized in terms of harmonic content and efficiency for a single load point - which typically corresponds to both a voltage and current rating at the drive output. As the voltage is reduced from this "optimum" point, additional harmonics creep in, causing extra thermal stress on the motor windings and therefore increased heating - which means more losses, and a negative impact on motor efficiency.

The system cost might be a factor as well: higher voltage results in lower currents for a given power rating. This means smaller conductor cross-section - but larger insulation thickness and/or spacing for the conductors. At the motor end, it may mean a physically larger machine (to accommodate the increased clearance and creepage requirements for the higher voltage). It also typically means more turns in a given winding, since the magnetic force required to do the actual mechanical work is directly proportion to the product of (amps x turns).

Running a high-volt low-amp drive can sometimes mean a lower drive cost as well, depending on whether the limiting factor in the design is the voltage at discrete points in the circuit or the current carried through those points.

The bottom line is to try and optimize the SYSTEM (drive plus motor plus conductors between) in terms of cost, physical space, efficiency, and other performance-related criteria.

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All else equal, and it never is, higher voltage will be more efficient because there will be lower I^2R losses.

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