DTC Torque and Power

VFD Operation

When an induction motor is connected to a full voltage supply, it draws several times (up to about 6 times) its rated current. As the load accelerates, the available torque usually drops a little and then rises to a peak while the current remains very high until the motor approaches full speed.

By contrast, when a VFD starts a motor, it initially applies a low frequency and voltage to the motor. The starting frequency is typically 2 Hz or less. Thus starting at such a low frequency avoids the high inrush current that occurs when a motor is started by simply applying the utility (mains) voltage by turning on a switch. After the start of the VFD, the applied frequency and voltage are increased at a controlled rate or ramped up to accelerate the load without drawing excessive current. This starting method typically allows a motor to develop 150% of its rated torque while the VFD is drawing less than 50% of its rated current from the mains in the low speed range. A VFD can be adjusted to produce a steady 150% starting torque from standstill right up to full speed.^[16] Note, however, that cooling of the motor is usually not good in the low speed range. Thus running at low speeds even with rated torque for long periods is not possible due to overheating of the motor. If continuous operation with high torque is required in low speeds an external fan is needed. Please consult the manufacturer of the motor and/or the VFD.

In principle, the current on the motor side is in direct proportion of the torque that is generated and the voltage on the motor is in direct proportion of the actual speed, while on the network side, the voltage is constant, thus the current on line side is in direct proportion of the power drawn by the motor, that is U.I or C.N where C is torque and N the speed of the motor (we shall consider losses as well, neglected in this explanation).

(1) n stands for network (grid) and m for motor
(2) C stands for torque [Nm], U for voltage [V], I for current [A], and N for speed [rad/s]
We neglect losses for the moment :
Un.In = Um.Im (same power drawn from network and from motor)
Um.Im = Cm.Nm (motor mechanical power = motor electrical power)
Given Un is a constant (network voltage) we conclude : In = Cm.Nm/Un That is "line current (network) is in direct proportion of motor power".

With a VFD, the stopping sequence is just the opposite as the starting sequence. The frequency and voltage applied to the motor are ramped down at a controlled rate. When the frequency approaches zero, the motor is shut off. A small amount of braking torque is available to help decelerate the load a little faster than it would stop if the motor were simply switched off and allowed to coast. Additional braking torque can be obtained by adding a braking circuit (resistor controlled by a transistor) to dissipate the braking energy. With 4-quadrants recifiers (active-front-end), the VFD is able to brake the load by applying a reverse torque and reverting the energy back to the network.

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