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With the rapid development of power technology, AC motor frequency conversion speed control technology has made breakthrough progress and entered the stage of popularization and application. In China, the frequency converter is also being used more and more widely. At the same time, how to properly select and use it has become a very prominent problem for the majority of users.
1. About the capacity selection In the manual of the inverter, in order to help the user to select the capacity, there is a column of "with motor capacity". However, the meaning of this column is not accurate enough, which often leads to mis-selection of the inverter.
In various production machines, the capacity of the motor is mainly selected according to the principle of heat generation. That is to say, under the premise that the motor is driven, as long as its temperature rise is within the allowable range, short-term overload is allowed. The overload capacity of the motor is generally set at 1.8-2.2 times the rated torque. The temperature rise of the motor, the so-called "short time" is at least ten minutes or more. The overload capacity of the inverter is: 150%, 1 minute. This indicator, for the motor, only makes sense during the starting process, and in practice, it is actually not allowed.
Therefore, the exact meaning of the column "with motor capacity" is "the actual maximum capacity of the motor." When the inverter is actually selected, it can be selected according to the maximum current of the motor during the working process. For the blower and pump type load, because it belongs to the long-term constant load, it can be directly selected according to "with the motor capacity".
2. The transmission system is optimized to design the range of effective torque and effective power of the AC asynchronous motor after frequency conversion. When using the inverter, the transmission system must be optimized according to the mechanical characteristics of the production machinery and the requirements of the speed regulation range. The main contents and approximate methods of the optimization design are as follows:
2.1 Determine the maximum operating frequency of the motor (1) blower and pump load. The resistance torque TL of this type of load is proportional to the square of the speed n is TL=KTn2, and the output power PL is proportional to the speed of the power PL=KPn3. (KT and KP are constants). It can be seen that if the speed exceeds the rated speed, the torque and power of the load will increase by the square law and the cubic law, respectively. Therefore, in general, it is not allowed to operate above the rated frequency.
(2) Under normal circumstances, the strength, vibration and wear resistance of various machines are designed on the premise that the motor speed does not exceed 3000r/min. Therefore, the maximum operating frequency of the 2-stage motor should not exceed the rated frequency too much without redesigning the machine.
(3) When the asynchronous motor is running above the rated frequency, since the power supply voltage is constant, the electromagnetic torque Tx is almost inversely proportional to the square of the frequency adjustment ratio Kf when adjusted to fx, that is, T≈TN/Kf2 (and TN) It is the torque at the rated frequency fN). Therefore, the maximum operating frequency should not exceed the rated frequency.
(4) When the asynchronous motor is running at low frequency, torque compensation is often required in order to obtain sufficient torque. The torque compensation will make the magnetic circuit of the motor become saturated, which will increase the additional loss and reduce the efficiency. Therefore, as long as the situation permits, the upper limit of the operating frequency may be increased.
2.2 Determine the transmission ratio of the transmission system and check the capacity of the motor (1) The blower and pump load are generally direct drive, regardless of the transmission ratio.
(2) Constant torque load. First, determine the highest frequency and lowest frequency of motor operation based on the effective torque line and the required frequency adjustment range.
Assuming that the determined maximum operating frequency of the motor is fmax, the lowest operating frequency is fmin and the corresponding torque relative value is tTL, then the rated torque of the motor Tn = TL / qTL (TL load torque). If the original motor does not leave a margin, the capacity of the motor should be increased by 1/tTL times with the inverter. The transmission ratio of the transmission system is equal to the ratio of the maximum speed nLmax required by the nDmax load of the motor at the highest operating frequency.
(3) Constant power load: Similar to the constant torque load, firstly, according to the effective power line and the frequency adjustment range, the upper and lower limits of the motor running frequency are obtained.
Similarly, while determining the highest and lowest operating frequencies, the corresponding power relative value tPL is obtained, and the rated power of the motor PN ≥ PL / tPL (PL is the load required power).
When designing a constant power load, two points should be noted: (1) use as much as possible of the rated frequency; (2) When the adjustment range is large, try to use the two gear ratio. Because when the gear ratio is divided into two columns, the relationship between the frequency range αf and the αn ​​speed range is. It can be seen that in the case of the same range of rotational speeds, the frequency range will be greatly reduced, thereby reducing the capacity of the motor.
The mechanical characteristics of the load, because of the constant power load, the product of the abscissa and the ordinate of any point on the curve are equal and proportional to the load power, ie PL=KPTLnL=KPTLmaxLmin. The effective torque line when all the rotational speeds are adjusted below the rated frequency, in which case the required motor capacity PN = KPTNnLmax > KPTLmaxLmax = αnPL. This shows that the required motor capacity is larger than the On frequency of the load power, which is very uneconomical.
(1) When the maximum operating frequency is twice the rated frequency and the transmission ratio is only one gear. In this case, the capacity of the required motor is PN = KPTN1/2nLmax 1/2αnPL. It can be seen that the required capacity is only required to be greater than On/2 times the load power.
(2) When the maximum operating frequency is twice the rated frequency and the gear ratio is two. At this time, the required motor capacity is PN1/2 PL. It can be seen that for constant power load, this scheme is ideal when αn>4.
3. Self-matching external braking resistors Various frequency converters allow external braking resistors to speed up braking speed and external resistance. However, the matching braking resistor is expensive and difficult to buy. When it is automatically configured, its resistance and power can be determined as follows:
The voltage value of the DC circuit is UP=×380=53V; the braking current Is is generally based on the principle that the rated current IDN of the motor is not exceeded, that is, Is≤IDN, so the braking resistor Rs≥UD/Is.
Since the time for passing current in Rs is only a few seconds, the power PR can be selected according to (1/10-1/8) when working, that is, PR=(0.1-0.125) UD2/Rs.
When Rs is connected to the circuit, care should be taken to cut off the braking resistor inside the inverter. If it cannot be cut off, the value of Rs should be increased appropriately to avoid excessive braking current.
In the external braking circuit, in order to avoid burning the internal discharge of the inverter, the high-power transistor (GTR) can also be connected to the entire brake device (that is, including the braking resistor and the discharge transistor). In this case, the GTR should select VCEX ≥ 700. Volt; ICN ≥ (1.2-1.5) IDN security.