Now that EPU and EMA are more and more widely used, as a practitioner in the hydraulic field, it is necessary to have a basic understanding of motors.

Let’s talk briefly about the starting current of the servo motor today.

1 Is the starting current of the motor larger or smaller than the normal working current? Why?

2 Why is the motor stuck and easy to burn out?

The above two questions are actually one question. Regardless of the system load, deviation signal and other reasons, the starting current of the motor is too large,

Let’s briefly talk about the problem of starting current from the motor itself (not considering the problem of soft start).

The rotor of the motor (DC motor) is made of coils, and the wires of the motor will cut the magnetic induction lines during the working process to generate induced electromotive force.

At the moment when the motor is energized, because the induced electromotive force has not yet been generated, according to Ohm’s law, the starting current at this time is:

IQ = E0/R

Where E0 is the coil potential and R is the equivalent resistance.

During the working process of the motor, assuming that the induced electromotive force is E1 , this potential hinders the rotation of the motor, so it also becomes the counter electromotive force, according to Ohm’s law:

I = (E0-E1)/R

Since the equivalent potential across the coil is reduced, the current at work is reduced.

According to the actual measurement, the current of the general motor when starting is about 4-7 times that of normal operation , but the starting time is very short. Through the inverter or other soft start, the instantaneous current will drop.

Through the above analysis, it should be easy to understand why the motor is easy to burn out after being stuck?

After the motor stops rotating due to mechanical failure or too much load, the wire will no longer cut the magnetic induction line, and there will be no counter electromotive force. At this time, the potential at both ends of the coil will always be very large, and the current on the coil is approximately equal to If the starting current is too long, it will heat up severely and cause damage to the motor.

It is also easy to understand in terms of energy conservation.

The rotation of the coil is caused by the Ampere force on it. Ampere force is equal to:

F=BIL

The moment the motor starts, the current is very large, the ampere force is also very large at this time, and the starting torque of the coil is also very large. If the current is always so large, then the ampere force will always be so large, so does the motor rotate very fast, or even faster and faster. This is unreasonable. And at this time, the heat will be very strong, and all the energy will be used for heat, so why use it to push the load to do work?

When working normally, due to the existence of counter electromotive force, the current will be very small at this time, and the heat will be very small. The energy provided by the power supply can be used to do work.

Just like the servo valve, after the closed-loop operation, it is always near the zero position. At this time, the pilot current (or the current on the single-stage valve) is very, very small.

Through the above analysis, it is also easy to understand why the faster the motor speed, the smaller the torque? Because the faster the speed, the greater the counter electromotive force, the smaller the current in the wire at this time, and the smaller the ampere force F=BIL .

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Post time: Mar-16-2023