Electric Performance Dimensioning
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96 Performance Dimensioning
2,000 RPM
At 2,000 RPM, the motor requires a voltage of U = R × I + e.m.f. × n = 1 Ω × 30 A +
50 V / (1,000 RPM) × (2,000 RPM) = 130 V.
From this results a power of P = 130 V × 30 A = 3.9 kW.
At an intermediate circuit voltage of 300 V an input current results from the supply
voltage of I = P / 300 V = 13 A.
Thus, a considerable higher current flows in the power supply at 2,000 RPM than at
standstill.
5,400 RPM
At 5,400 RPM, the motor requires a voltage of U = R × I + e.m.f. × n = 1 Ω × 30 A +
50 V / (1,000 RPM) × (5,400 RPM) = 300 V.
From this results a power of P = 300 V × 30 A = 9 kW.
At an intermediate circuit voltage of 300 V an input current results from the supply
voltage of I = P / 300 V = 30 A.
Thus, an identical current flows in the power supply unit at 5,400 RPM as in the mo-
tor.
It must be considered that the currents flowing in the motor phases are lower by fac-
tor than the currents calculated above.
The examples clearly show that the expected motion profile must be considered
when dimensioning the power supply unit. An exact dimensioning can only be
achieved by integrating the motion profile.
The same applies for conceiving the output stage and the motor.
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