Simple Usage Method of Frequency Converter and Expanding Knowledge

Abstract

Since AC motors do not have brushes, the vulnerable parts of DC motors, and are easy to maintain, coupled with the rapid decline in the price of frequency converters, AC speed regulation has developed rapidly.

Since AC motors do not have brushes, the vulnerable parts of DC motors, and are easy to maintain, coupled with the rapid decline in the price of frequency converters, AC speed regulation has developed rapidly.

At present, the most widely used frequency converter is the voltage-type general-purpose frequency converter. This chapter takes this type of frequency converter as an example to describe the basic usage methods of the frequency converter. The usage methods of other types of frequency converters are slightly different, but the general situation is basically similar.

8.1 Basic Usage of Inverter

8.1.1 Selection of Inverter

• 1. First of all, the power of the inverter must match the rated current of the motor. Due to different power factors and different motor efficiencies, the rated current of the motor with the same power and the same voltage level has a large difference, and the inverter is affected by the current of the IGBT device. Due to class restrictions, the output current of the inverter cannot exceed the maximum allowable value, so the power of the inverter should be selected to match the rated current of the motor.

• 2. The nature of the load must match the type of inverter. Some manufacturers have two types of inverters: general-purpose and pump/fan. General-purpose inverters are suitable for constant torque loads, such as machine tools; pump/fan type frequency converter, only suitable for square torque load. The prices of these two kinds of frequency converters are different, and the price of general pump/fan frequency converter is lower.

8.1.2 Main Power and Control Wiring of Inverter

The main wiring of the inverter is shown in Fig. 8.1, where R, S, T are three-phase main power supply (or single-phase AC power supply), U, V, W are connected to three-phase AC motor, and the speed control input is analog 0–10 V or 4–20 mA signal, the start/stop control input (digital input) controls the start and stop of the motor, the forward/reverse control (digital input) controls the steering of the motor, and the alarm output (digital input) is used to notify the external control device of the frequency converter In the alarm state or running state. When the motor needs to be in the power generation state (such as emergency stop, heavy objects, etc.), it needs to be connected to the braking resistor. The analog signal output is mainly used to output the frequency, current or torque parameters of the current inverter. The communication port is used for data transmission and control with the controller. In most cases, the other wires of the inverter do not need to be used.

Fig. 8.1

Main wiring of the inverter

8.1.3 Basic Parameter Setting of Frequency Converter

There are displays and buttons on the inverter panel. The display can display the output frequency, output voltage, current, setting parameters, etc. The parameter input method will be different for different manufacturers. For specific methods, please refer to the manufacturer's product manual.

There are many parameters introduced in the instruction manual of the inverter, and some inverters can even have hundreds of parameters that can be set. But in the actual application of the inverter, there are only a dozen parameters that need to be input by the user.

• 1. Control mode parameters:

  Frequency control mode: (1) panel control; (2) external analog signal (0–10 V or 4–20 mA) control by terminal.

  Start-stop control mode: (1) The panel controls the start/stop of the motor; (2) Use the terminal to control the start/stop of the motor by an external switch signal.

• 2. Parameters of the driven motor: (1) rated power; (2) rated current; (3) rated voltage; (4) rated speed; (5) number of motor poles; (6) no-load current of the motor; (7) Motor impedance; (8) Motor inductive reactance;

  If there are no these parameters in the motor instruction manual, use the factory default value of the inverter (same power as the motor), or use the motor impedance and inductance online test function provided by inverters to test.

• 3. Main control parameters: (1) Power supply voltage (such as AC 480 V, AC380 V); (2) Output minimum frequency (such as 0 Hz); (3) Output maximum frequency (such as 60, 50 Hz); (4) Speed up time (Such as 0.1–3600 s); (5) deceleration time (such as 0.1–3600 s); (6) torque boost mode selection; (7) V/F mode selection; V/F, vector mode, direct torque control.

• 4. Other parameters: Under normal circumstances, the default value can be used. If there are special requirements, please refer to the manufacturer's inverter manual.

8.1.4 Outline of Frequency Converter

The appearance of the inverter is shown in Fig. 8.2.

Fig. 8.2

Appearances of several voltage-type general-purpose inverters

8.2 Basic Usage of ABB Inverter

8.2.1 Purpose

Quickly familiarize yourself with the use of ACS510 series inverters to realize stepless speed regulation of three-phase AC motors.

8.2.2 Essentials to Master

• 1. Correctly connect the power supply, motor and control wiring.

• 2. Correctly set the voltage, current, and power of the motor in the frequency converter to be consistent with the voltage, current, and power of the actual driven motor. When the motor power adapted to the inverter does not match the actual motor power, be careful to modify the inverter parameters, otherwise the motor protection function will not work well. Correctly set the basic frequency and maximum frequency. The basic frequency is the frequency when the rated output voltage is output (generally selected as 60 or 50 Hz), and the maximum frequency is determined by the allowable operating frequency of the motor. Some Dedicated variable frequency motors run at a higher frequency (possibly much greater than the base frequency).

• 3. Correctly set the acceleration time and deceleration time of the motor. If the acceleration time is too small, overcurrent will occur, and if the deceleration time is too small, overvoltage will occur.

• 4. Correctly set the torque boost of the motor so that the motor has enough starting torque, which is very important for constant torque loads (or non-square torque loads).

• 5. Select the setting parameters of the panel to control the start and stop or the external terminal to control the start and stop.

• 6. Select the setting parameter of panel control frequency output value or external terminal control frequency output value.

8.2.3 Inverter Appearance

The appearance of the ACS510 series inverter is shown in Fig. 8.3.

Fig. 8.3

Appearance of ACS510 inverter

8.2.4 Inverter Model

The inverter model number is arranged as follows.

8.2.5 Inverter Wiring and Floating Networks

• 1. According to the size of the power, the ACS510 inverter has several sizes of R1–R6, and the wiring method is basically the same; there are some changes in the wiring positions of the power cables of R5 and R6. Taking the inverter with R3 shape as an example, the basic layout of the wiring terminals is shown in Fig. 8.4.

  Fig. 8.4

  

  Layout of wiring terminals of frame R3 inverter

• 2. Floating networks (also known as IT, ungrounded, or impedance/resistance grounded networks) refers to the power supply system in which the neutral point of the power grid is not grounded or high impedance grounding. For example, the power supply network of a ship or a coal mine is a neutral point ungrounded system. The advantage of this kind of power supply system is: the impedance between the strong electric ground and the weak point ground is relatively large. It is not easily affected by the performance of the earth, and to a certain extent prevents electromagnetic interference caused by the common ground. The disadvantages are: Affected by the parasitic capacitance, the ground potential of the circuit changes, and it is also easy to cause charge accumulation and electrostatic breakdown, resulting in interference. In order to solve this problem, a high-impedance resistor is used to ground the neutral point to release the charge accumulated due to parasitic capacitance, which is the high-impedance grounded grid.

For the R1–R4 shape inverter, the filter capacitor in the internal RFI filter is grounded through the EM1 and EM3 screws in Fig. 8.4; for the R5–R6 shape inverter, the filter capacitor in the internal RFI filter is grounded through The F1 and F2 screws are grounded.

When the external power grid is a floating network, a high-impedance power grid, or an unsymmetrically grounded system, remove the EM1 and EM3 screws or the F1 and F2 screws.

• 3. U1, V1 and W1 are connected to the three-phase AC power supply, and U2, V2 and W2 are connected to the three-phase AC motor. The cross-sectional area of the power cable should be selected according to the maximum operating current of the motor. In order to reduce electromagnetic radiation, the motor side cable must be used Shielded cable, armored cable or metal conduit, the shielding of the cable is connected to the protective earth PE (Protective earth). Changing the phase sequence of U, V, W to the motor will change the direction of rotation of the motor.

• 4. When the motor is often in the state of braking and lowering heavy objects, for the frequency converter with the shape of R1 and R2, BRK+ and BRK− must be connected with a suitable braking resistor. Otherwise the frequency converter may cause an overvoltage alarm. For inverters with the shape of R3–R6, UDC+ and UDC− should be connected with appropriate braking units or braking choppers and braking resistors.

• 5. The functions of the control terminals are shown in Table 8.1.

  Table 8.1 Control terminal functions

Terminals 3, 6 and 9 are internally equipotential.

There are two connection methods of PNP and NPN for digital input signal, as shown in Fig. 8.5.

Fig. 8.5

Two connection methods of PNP and NPN

Cables connected to control terminals: (1) Do not share a single cable for analog signals and numerical signals; (2) Do not share a single cable for DC24V and AC115/230 V control lines; (3) It is best to use twisted pairs for relay control signals; (4) It is best to use twisted-pair braided shielded cables for analog signals, as shown in Fig. 8.6, the shield is grounded, and each analog signal uses a separate twisted pair; (5) The requirements for digital signal cables are looser, but it is best Also use braided shielded cable, with the shield grounded.

Fig. 8.6

Control cable

• 6. In order to quickly set parameters, ACS510 provides some conventional factory control modes. Different application types are available for users to choose in the form of application macros. Macros are a set of pre-defined parameters. Using application macros, the number of parameters that need to be defined in the process is reduced to the minimum. When different application macros are selected, the definitions of analog I/O, digital input and relay output are also different. The parameter item selected by the application macro is 9902, which has 8 application methods. Standard type, three-wire type, manual/automatic type and PFC control type (pump and fan), etc.

  Take the standard macro as an example, set the value of parameter 9902 to 1 (ABB standard), the definitions and connections of its analog I/O, digital input and relay output are shown in Fig. 8.7.

  Fig. 8.7

  

  Connection method of ACS510 standard application macro

  Other application macros have different definitions of analog I/O, digital input and relay output, please refer to the manual.

• 7. For a simple application, only 14 terminals is ok: U1, V1 and W1 terminals are connected to AC power supply, U2, V2, W2 and PE are connected to three-phase AC motor. 2 and 3 control terminals (0- 10 V) connected to the PLC analog output module AO1, to control the speed of the motor; the control terminals 11 and 12 are short-circuited, and the terminals 10 and 13 are connected to the PLC digital output (relay) module DO1, to control the start/stop of the inverter.

8.2.6 Parameter Setting

• 1. Selecting an application macro will set all parameters to the default value of the macro, and there are not many other parameters that need to be set.

• 2. The parameters inside the inverter can be set through the assistant control panel or the basic control panel through the keyboard panel, and the control panel can also be used to monitor the running status of the inverter. The appearance and functions of the assistant control panel are shown in Fig. 8.8.

  Fig. 8.8

  

  Appearance and functions of the panel

• 3. The parameters and steps to be set are as follows:

  • (1) Select language, code 9901 (LANGUAGE):0 English.

  • (2) Select different application macros, code 9902 (APPLIC MACRO): 1 = ABB standard, frequency or speed control; 2 = three-wire macro; 3 = alternating macro; 4 = motor potentiometer macro; 5 = manual/ Auto macro; 6 = PID control macro; 7 = PFC control macro; 15 = SPFC control macro.

  • (3) According to the data on the motor nameplate: set rated (nominal) voltage 9905 (MOTOR NOM VOLT); rated current 9906 (MOTOR NOM CURR); rated frequency 9907 (MOTOR NOM FREQ); rated speed 9908 (MOTOR NOM SPEED); rated power 9909 (MOTOR NOM POWER); power factor 9915 (MOTOR COSPHI): 0.01–0.97, 0 = automatic identification.

  • (4) Start/stop and direction: start/stop 1001 (EXT1 COMMANDS): 2 = use DI1 and DI2 to control start-stop and direction, 8 = control panel to control start-stop and direction, 10 = communication control; direction control 1003 (DIRECTION): 1 = forward, 2 = reverse.

  • (5) Frequency given 1103 (REF1 SELECT): 0 = control panel, 1 = AI1 given, 2 = AI2 given, 8 = communication given.

  • (6) the minimum frequency, code 2007 (MINIMUM FREQ): 0 Hz, the highest frequency, code 2008 (MAXIMUM FREQ): 60 Hz or 50 Hz.

  • (7) According to the load inertia, set the acceleration time, code 2202 (ACCELER TIME 1): the time required to reach the highest frequency from 0 Hz, and the shortest time is the time when the inverter does not stop due to overcurrent failure. The deceleration time, code 2203 (DECELER TIME 1): the time required to reach 0 Hz from the highest frequency, the shortest time is the time when the inverter does not stop due to overvoltage fault.

  • (8) Dangerous frequency control 2501 (CRIT SPEED SEL): 0-off, 1-on; the lower limit 2502 (CRIT SPEED 1 LO) of critical frequency 1, the upper limit 2503 (CRIT SPEED 1 HI) of critical frequency 1.

  • (9) U/f ratio control, code 2605 (U/F RATIO): 1-linear, 2-square torque;

  • (10) Operation enable 1601 (RUN ENABLE): 0- no need for external signal permission, 1-DI1 signal control;

  • (11) parameters save1607 (PARAM SAVE): 0 = DONE, value changes automatically when all parameters are saved; 1 = SAVE, saves altered parameters to permanent memory.

After the above 11 steps are completed, start the inverter again, and it can run according to the selected function.

• 4. Special tips

  • (1) If you do not know the setting value of this parameter, please choose the factory default value. If you accidentally mess up the parameters in the inverter, making the inverter unusable, the easiest way is to restore the parameters to the factory default values, and then set the necessary parameters above.

  • (2) If the load driven by the motor (such as a fan, long axis pump) has mechanical resonance at frequency f0 during operation, the lower limit of the dangerous speed is 2502, and the upper limit of the dangerous speed is 2503. The output frequency of the inverter skips the frequency f₀ (a certain frequency range) to avoid the resonant frequency of the machine.

8.2.7 Other Notes

• 1. Low-speed operation of the inverter:

  Ordinary three-phase asynchronous motors are designed according to the rated working conditions. Ordinary three-phase asynchronous motors rely on their own fans to dissipate heat. When ordinary three-phase asynchronous motors driven by frequency converters work at low speeds for a long time, the fans slow, the heat dissipation effect will be poor, and the motor may stop or burn out due to overheating. If you need to work at low speed for a long time, it is best to use a three-phase variable frequency motor. The heat dissipation of the variable frequency motor is solved by an independent power supply fan, so There is no low-speed heat dissipation problem, and the low-speed starting torque of the variable frequency motor is higher than that of ordinary motors.

• 2. Derating rating:

  The standard operating altitude does not exceed 1000 m. When the altitude of the place of use increases, the output current and power of the inverter need to be derated. When the altitude is 1000–2000 m, it needs to be derated by 1% for every 100 m increase.

  The ambient temperature does not exceed 40 ℃, when the temperature of the place of use increases, the output current and power of the inverter need to be derated for use. When the temperature is 40–50 ℃, it needs to be derated by 1% for every 1 ℃ increase.

  The increase of the switching frequency will also reduce the frequency converter. When the switching frequency is 8 kHz, the inverter needs to derate by 20%.

8.3 The Principle of Frequency Converter (Beginners Do not Need to Master)

8.3.1 Main Circuit Structure of General Frequency Converter

The main circuit of the general-purpose frequency converter is composed of three parts: rectification part, DC part and inverter part, the structure is shown in Fig. 8.9, adopting AC–DC–AC structure.

Fig. 8.9

Main circuit structure of a general-purpose inverter

The rectification part consists of 6 diodes to form a three-phase rectification bridge, which converts the three-phase AC power supply RST into direct current. The DC part consists of several large-capacity capacitors and equalizing resistors, the direct current passes through the filter capacitor to keep the DC voltage U_D stable. Since the current large-capacity electrolytic capacitors have low withstand voltage and the consistency of the capacitance value is poor. In order to avoid the difference of the voltage drop of each capacitor is too large, the voltage drop of one capacitor is higher than the withstand voltage value, causing the capacitor to break down. The voltage equalization effect of the resistor can be used to basically ensure that the voltage drop on each capacitor is consistent. In the figure, the capacitance values of capacitors C₁ and C₂ are equal to store electric energy and filter. Resistors R₃ and R₂ have the same resistance value and play the role of voltage equalization. The inverter part consists of 6 IGBT modules and 6 anti-parallel diodes, which convert the DC voltage U_D into a three-phase AC that can change the frequency and effective voltage.

In order to enhance the filtering effect, the capacity of the filter capacitors C₁ and C₂ is generally large. The voltage on the filter capacitor is zero when the inverter is first powered on, which will inevitably cause the charging current of the filter capacitor to be very large. Causing the instantaneous voltage of the power supply grid to generate the steep drop will cause other equipment on the same power grid to trip or malfunction and interfere with the normal operation of the power grid. In order to solve this problem, a current-limiting resistor R1 is added to limit the maximum charging current of the filter capacitor. After completion, in order to eliminate the voltage-drop and heat loss of R₁, use the KM contact or thyristor K₁ to bypass R₁ again.

In Fig. 8.9, the function of the DC reactor DCR is to use the restraining effect of the inductance on the current, smooth the input current of the power supply. Improve the power factor of the inverter, and at the same time reduce the charging current of the filter capacitor when the inverter is initially powered on. The DC reactor is a standard accessory in some frequency converters, and some frequency converters (or large-capacity frequency converters) are optional accessories.

In the occasions that need to stop quickly or lower heavy objects, the motor is in the state of generating electricity. The electric energy generated by the motor charges the filter capacitor through the anti-parallel diode, resulting in an increase in the DC voltage inside the inverter. If the energy is not properly processed, it will as a result, the DC voltage U_D exceeds the upper limit. Considering the withstand voltage of the IGBT and filter capacitor, the inverter will generate an overvoltage alarm and stop. If it cannot be stopped, it must be dealt with. There are two ways to deal with it. One is to feed back this part of energy to the grid, and the other is to use the braking resistor to consume the energy. In Fig. 8.9, the second method used in large quantities uses the braking resistor to consume energy. R₂ is the braking resistor, and V is the IGBT that acts as a switch in the braking unit. When the DC voltage U_D exceeds the upper limit, V When it is turned on, the brake resistor R₂ will consume the excess energy of the filter capacitor. The braking resistor and braking unit are built-in standard configurations in small-capacity inverters, and are optional accessories for large-capacity inverters.

In Fig. 8.9, the function of R₅ and HL is to indicate the presence or absence of voltage on the large-capacity filter capacitor. When the inverter is powered off, since the charge on the filter capacitor is not discharged immediately, the residual voltage is enough to cause personal injury. In order to avoid danger caused by people touching the external terminal of the filter capacitor before the discharge of the capacitor is completed, HL is used to indicate the presence or absence of the voltage of the filter capacitor. Only after the HL indicator is off can touch wiring or maintenance be performed.

8.3.2 Sine Wave Pulse Width Modulation (SPWM) Mode and Implementation

In Fig. 8.9, when V1 of the inverter part is turned on and V2 is turned off, the U-phase outputs V+, and when V1 is turned off and V2 is turned on, the U-phase outputs V-. The output voltage of the U-phase has two levels rectangular wave. V and W phase are the same, due to the influence of the inductance in the motor, the current rising speed is lagging behind the voltage, when the U phase output V+, and the U phase current is negative, D1 is turned on, and the current flows back to the DC side. D2 does the same.

The voltage waveform output by the inverter is a series of rectangular pulse waveforms with equal voltage amplitude and unequal width. The frequency and pulse width of the rectangular square wave are controlled by sinusoidal pulse width modulation (SPWM), which is equivalent to a sine wave. The principle of sine wave equivalence is that the area enclosed by the rectangular wave and the sine wave in each time period is equal. The area enclosed is equal to the area enclosed by the rectangular wave in the time period and the horizontal axis of time. Changing the frequency and pulse width of the pulse wave can realize the change of the frequency and amplitude of the equivalent sine wave, which is the variable frequency variable voltage output.

The more pulses that make up a rectangular wave, the closer its equivalent waveform is to a sine wave. The number of pulses is measured by the carrier frequency in the inverter. In Fig. 8.10, the carrier frequency is 5 times the frequency of the positive wave. The early generation method of SPWM is to use a triangular wave as the carrier, and use a sine wave that can change the frequency and amplitude to modulate the signal wave, and the frequency of the triangular carrier can be adjusted. Take the star connection method of the motor as an example, the voltage of the neutral point is OV, use a comparator to compare the signal wave of each phase with the triangular carrier wave. Use the switching signal output by the comparator to control the IGBT output of the corresponding phase. So as to U-phase as an example, when the sine wave of the U-phase signal is higher than the triangular wave, the U-phase IGBT is turned on, and the U-phase outputs U_D. When the U-phase signal sine wave is lower than the triangular wave, the U-phase IGBT is turned off and outputs 0 V. Assume that the triangular carrier wave is U_t, the U-phase signal wave is U_u, the U-phase output is U_U0, the V-phase signal wave is U_v, the V-phase output is U_V0. The line voltage of U-phase and V-phase is U_UV, the frequency of the carrier U_t is the signal wave U_u and the signal wave 3 times the U_v frequency, as shown in Fig. 8.11a).

Fig. 8.10

Sinusoidal pulse width modulation (SPWM) mode

Fig. 8.11

Relationship between signal wave and carrier wave

For the U-phase signal wave, when the amplitude of the signal wave U_U0 is higher than the triangular carrier U_t, the U-phase outputs a DC voltage U_D. When the amplitude of the signal wave U_U0 is lower than the triangular carrier Ut, the U-phase outputs 0 V, as shown in Fig. 8.11b) shown.

For the V-phase signal wave, when the signal wave U_V0 amplitude is higher than the triangular carrier U_t, the V-phase outputs a DC voltage U_D. When the signal wave U_V0 amplitude is lower than the triangular carrier U_t, the V-phase outputs 0 V, as shown in Fig. 8.11c) shown.

The line-to-line voltage U_UV = U_U0−U_V0 of U-phase and V-phase, the waveform of U_UV is shown in Fig. 8.11d), U_UV is the equivalent sine wave composed of rectangular waves. Because the carrier frequency is three times the frequency of the signal wave, so the wave head of each U_UV is composed of three rectangular waves with varying widths.

Changing the frequency of the signal wave changes the frequency of the output equivalent sine wave. Changing the amplitude of the sine wave changes the width of the rectangular wave in the equivalent sine wave, which also changes the effective voltage value, which will not be explained here. With the rapid development of digital technology, the current SPWM method no longer needs these complicated transformations, and can be easily realized directly with a dedicated chip or a calculation method.

Although the output voltage waveform of the frequency converter is a series of rectangular pulse waves, due to the suppression of the current change by the inductive load of the motor, the current waveform in the motor is a fluctuating and phase-lag approximate sine wave. The higher the carrier frequency, The smaller the current fluctuation, the smoother the current waveform. When the carrier frequency is high enough (such as 12 kHz), the current waveform output from the inverter to the motor is basically a smooth sine wave, as shown in Fig. 8.12. The current waveform lags behind the voltage, etc. When the carrier frequency is high, the low-frequency torque output is also stable, and the noise of the motor is small. It should be noted that the high carrier frequency will increase the loss of the inverter itself, increase the temperature of the inverter, and cause voltage glitches (du/dt) When the carrier frequency is low, the noise of the motor is large, the loss of the motor is large, the torque is reduced, and the temperature of the motor is high.

Fig. 8.12

Output voltage and current waveforms of inverter at high carrier frequency

8.3.3 V/F Control of Frequency Converter

When the output frequency f of the inverter decreases, the inductance XL of the motor also becomes smaller. If the output voltage U remains high, the current flowing into the stator winding of the motor will increase greatly, and the stator winding will be burnt out. In order to avoid this problem, occur, it is necessary to reduce the output voltage U while reducing the frequency f. Conversely, when the output frequency f of the inverter increases, the output voltage U of the inverter will increase accordingly, so that the inverter will output the maximum frequency, has a maximum output voltage, which can ensure that the motor can output rated power. In order to meet this requirement, it is necessary to keep U/f changing according to a certain law.

In order to keep the motor output constant torque at different operating frequencies (generally below the basic frequency), it is necessary to keep the air gap flux between the stator and the rotor constant. If the flux is too low, the output of the motor is insufficient, and the flux is too high, saturation occurs, and the winding will be burned out due to excessive excitation current. Therefore, keeping the air gap flux constant is one of the best operating modes for the inverter.

The effective value E of the electromotive force of each phase of the three-phase AC motor stator, such as Eq. (8.1).

$$E=K\times f\times N\times {\varphi }_{m}$$

(8.1)

where E is the effective value of the electromotive force induced by the air-gap flux in each phase of the stator winding, K is the coefficient. f is the frequency of the stator (Hz), and N is the turns of each phase of the stator winding in series Number, φ_m is the air gap magnetic flux of each pole.

Transform Eq. (8.1) to get φ_m as Eq. (8.2).

$$ \varphi_{m} = \frac{E}{K \times f \times N} = \left( {\frac{1}{K \times N}} \right) \times \frac{E}{f} $$

(8.2)

It can be seen from Eq. (8.2) that for a known motor, K and N are fixed values, in order to keep φ_m constant, Eq. (8.3) needs to be maintained.

$$ \frac{E}{f} = \left( {K \times N} \right) \times \varphi_{m} = constant $$

(8.3)

When the stator frequency f is high, the voltage drop on the stator winding impedance can be ignored, and it is approximately considered that the stator electromotive force E of each phase is equal to the stator phase voltage U of each phase. Then the Eq. (8.3) becomes:

$$\frac{E}{f}\approx \frac{U}{f}=constant$$

(8.4)

This is the constant control mode of the voltage-frequency ratio (U/f) of the frequency converter.

When the stator frequency f is low, the voltage drop on the stator winding impedance can no longer be ignored. In order to actually keep φ_m constant, the stator phase voltage U of each phase must be increased to compensate for the stator voltage drop and increase the starting current, as shown in the Fig. 8.13.

Fig. 8.13

Voltage-frequency ratio (U/f) curve

In Fig. 8.13, curve 1 is the U/f curve with low-frequency compensation when the motor is driving a constant torque load. Curve 2 is the U/f curve without low-frequency compensation when the motor is driving a constant torque load. Curve 3 is the U/f curve with low frequency compensation when the motor drives the water pump fan load, and curve 4 is the U/f curve without low frequency compensation when the motor drives the water pump fan load.

It can also be seen from Fig. 8.13 that since the motor cannot work at a state higher than the rated voltage, when the frequency converter is higher than the basic frequency f1, f increases, and the output voltage U remains unchanged at the rated voltage U1, so U/f curve becomes a horizontal straight line. According to the Eq. (7.9), this will cause the magnetic flux φ_m to decrease with the increase of the f, and the torque will also decrease. The motor becomes a constant power speed regulation, and a constant torque speed regulation (U/f is equal to a fixed value) as an example, draw the change curve of magnetic flux φ_m as shown in Fig. 8.14.

Fig. 8.14

Three types of U/f curves

In Fig. 8.14, curve 2 is the U/f curve without low-frequency compensation when the motor drives a constant torque load, and curve 1 is the flux change curve corresponding to curve 2. When f is lower than the basic frequency f1, it is a constant flux φ_me, it corresponds to the constant torque speed regulation mode; when f is higher than the basic frequency f1, the magnetic flux φ_m is the reciprocal change function of f, corresponding to constant power speed regulation; curve 4 is when the motor drags the square torque load, there is no The U/f curve of low-frequency compensation, curve 3 is the magnetic flux change curve corresponding to curve 4. When f is lower than the basic frequency f1, it corresponds to the square torque speed regulation mode. When f is higher than the basic frequency f1, the magnetic flux φ_m is f the reciprocal change function, corresponding to constant power speed regulation.

8.3.4 Vector Control of Inverter

The motor drives the load to move. If the no-load torque of the motor is ignored, according to Newton's second law, its motion equation is as Eq. (8.5).

$${T}_{M}-{T}_{L}=J\frac{d\omega }{dt}$$

(8.5)

where T_M is the electromagnetic torque of the motor (N · m), T_L is the resistance torque (N · m) of the load on the motor side. J is the total moment of inertia of the motor and the load on the motor side (kg · m²), ω is the rotational angular velocity of the motor (rad/s).

$$\omega =\frac{2\pi n}{60}$$

(8.6)

where n is the motor speed (r/min).

$$J=m{r}^{2}=\frac{G}{g}\times {\left(\frac{D}{2}\right)}^{2}$$

(8.7)

where m is the mass of the rotating body (kg), G is the weight of the rotating body (N), D is the diameter of the rotating body (m), and r is the radius of the rotating body (m).

Equation (8.5) becomes Eq. (8.8).

$${T}_{M}-{T}_{L}=\frac{G{D}^{2}}{375}\frac{dn}{dt}$$

(8.8)

It can be seen from the Eq. (8.8) that the acceleration, deceleration and constant speed operation of the load driven by the motor are directly controlled by the electromagnetic torque of the motor. The change of the electromagnetic torque of the motor directly controls the running speed of the load.

• T_M > T_L, the acceleration is greater than zero, and the load will run at an accelerated speed;

• T_M < T_L, the acceleration is less than zero, and the load will decelerate;

• T_M = T_L, the acceleration is equal to zero, and the load will run at a constant speed.

For a DC motor, if the positive and negative poles of the excitation power supply U_f and the armature power supply U are determined, the magnetic field direction of the main pole magnetic field (the excitation magnetic field) and the direction of the electromagnetic torque of the motor are also determined. The flow direction of the excitation current I_f and I_f form the direction of the magnetic field is shown in the figure. Due to the effect of the commutator and brushes, the direction of the armature winding current i_a under the main pole magnetic field remains unchanged. As shown in Fig. 8.15, the direction of the armature current i_a is under the N pole (The upper half) flows inward, represented by ⊙.The direction of the armature current i_a flows outward under the S pole (lower part), represented by ⊕. The direction of the armature current is perpendicular to the direction of the main pole magnetic field φ, and the direction of the electromagnetic force F on the conductor can be determined by the left-hand rule. The palm is facing the N pole (the magnetic force line is from N to S), the four fingers point to the direction of the current, and the thumb is the direction of the electromagnetic force (the force of the conductor), so that the torque direction of T_M is determined, as shown in the figure.

Fig. 8.15

Field current and armature winding current of a DC motor

The relationship between the electromagnetic torque T_M of the DC motor and the magnetic flux φ of the main pole magnetic field and the armature current i_a is shown in Eq. (8.9).

$${T}_{M}={C}_{M}\times \varphi \times {i}_{a}$$

(8.9)

where C_M is the torque constant, the excitation current if generates the main pole magnetic field φ, and the armature magnetic field generated by the armature current i_a is perpendicular to the main pole magnetic field. Since the excitation current if and the armature current i_a can be controlled separately, the torque control of the DC motor is relatively simple. The fixed excitation current if can control the armature current i_a to control the motor torque, and the fixed armature current i_a can control the excitation current if to control the motor torque.

In the AC motor, since the stator current contains both the excitation current and the armature current, a certain conversion is required to make the torque control of the AC motor the same as the torque control of the DC motor.

Taking a 2-pole AC motor as an example, the positions of the three-phase windings of the stator are fixed, and the spatial positions differ by 120°. The three-phase AC current with a phase difference of 120° is connected, and the rotational angular velocity of the three-phase AC is ω, thus a NS with an angular velocity of ω is produced. For the rotating magnetic field (magnetomotive force) of the poles, please refer to the previous chapters for the analysis process. The two-phase stator windings with fixed spatial positions have a difference of 90° in space, and the two-phase alternating current with a difference of 90° is connected. The rotational angular velocity of the alternating current is ω, and the same It can also generate a rotating magnetic field (magnetomotive force) of an NS pole with an angular velocity of ω. As long as the voltage amplitude is selected properly, this rotating magnetic field is exactly the same as that of a three-phase AC motor. For the analysis process, please refer to the principle of the single-phase AC motor. The spatial position of the excitation winding and the armature winding of the DC motor differ by 90°, see the principle of the DC motor. The direction of the magnetic field (magnetomotive force) of the DC motor is fixed, but if we let the coordinate axis of the DC motor rotate at the same angular velocity ω of the AC motor, and keeps the magnitude of the magnetomotive force of the rotating DC motor equal to the magnitude of the magnetomotive force of the above-mentioned two-phase AC motor, then the rotational magnetomotive force of the DC motor under the rotating coordinates is exactly the same as the magnetomotive force of the two-phase AC motor. The rotating magnetic field between the three-phase AC motor, the two-phase AC motor and the rotating DC motor is completely equivalent. According to the principle of the DC motor, we control the DC The excitation current of the motor excitation winding can control the strength of the magnetic field and keep the strength of the magnetic field constant. We can control the torque of the DC motor by controlling the armature current of the armature winding. By reversely pushing back through the coordinate transformation, we can have the current and voltage values of the three-phase AC motor that achieve the same torque control effect are obtained, which is the basic idea of vector control. For the simplicity of control, the magnetic flux of the rotor excitation is guaranteed to be constant, that is, to keep E₂/f constant, and E₂ is the induced electromotive force of the rotor.

Vector control is a torque control method that imitates a DC motor. It first decomposes the speed signal of a three-phase AC motor into the interaction between the excitation magnetic field and the armature magnetic field, calculates the DC excitation current i_f1 and the armature current i_a1. Then after a series of coordinate transformations, it is converted into control signals for controlling the three-phase currents i_a, i_b and i_c to control the inverter part of the frequency converter, as shown in Fig. 8.16. After the given speed signal n₀ changes, the excitation current i_f1 is maintained No change, only the armature current i_a1 is adjusted, so as to adjust the speed of the three-phase AC motor like a DC motor. The low-speed performance of the vector control mode is much better than that of the U/f mode, and it can provide sufficient torque even at 0 Hz.

Fig. 8.16

Schematic diagram of vector control

Vector control is the same as U/f control: vector control keeps the relationship E₂/f between rotor induced electromotive force and frequency constant, U/f control keeps the relationship between stator supply voltage and frequency U/f constant. Both of them can avoid under-excitation or over-excitation. The difference between vector control and U/f control is that vector control also adjusts the phase angle of the three-phase output current at the same time, which ensures that at low speed, only the armature current is increased to make the torque larger, while keeping the excitation current constant, avoiding overexcitation, so the vector control of the three-phase AC motor is better than the U/f control speed regulation performance.

8.3.5 Direct Torque Control of Frequency Converter

The torque of a three-phase AC motor is equal to the stator flux linkage F1 multiplied by the rotor flux linkage F2 and then multiplied by the sine of the angle between the two vectors. The flux linkage is in a derivative relationship with the voltage vector, and the two are perpendicular to each other. Direct torque control, the idea is to control the stator flux linkage in a basically constant range by controlling the voltage vector, so as to realize the direct control of the torque.

Let’s talk about the concept of six voltage vectors in the inverter part of the inverter. The inverter part of the inverter is composed of six transistor bridges, which supply power to the three windings a, b, and c of the three-phase AC motor, as shown in Fig. 8.9. If 3-bit binary numbers are used to represent the three bridge arms, 1 means that the upper transistor is turned on, 0 means that the lower transistor is turned on, such as 100 means that V1, V4 and V6 are turned on, and 011 means that V2, V3 and V5 are turned on There are 6 conduction modes of 010, 101, 001, and 100, plus 0 V short-circuit power supply mode 000 and 111, and there are 8 combinations in total. 000 and 111 are only applied when the torque is greater than the load torque. Considering that the three windings a, b, c differ by 120° in spatial position, for a 2-pole AC motor with neutral star connection, 100 conduction mode, the voltage vector of the three-phase winding is shown in Fig. 8.17.

Fig. 8.17

Synthetic vector U1 of three-phase winding voltage in 100 conduction mode

The amplitude of U_a is equal to U_D/2, the amplitude of U_b and U_c is also equal to U_D/2, the combined total voltage vector is U₁, the amplitude of U1 is equal to U_D, and the corresponding 101 synthesized voltage vector is U₂, U₂ is counterclockwise than U1 Rotate 60°, the voltage vector corresponding to 001 synthesis is U₃, U₃ rotates 60° counterclockwise than U₂, the voltage vector corresponding to 011 synthesis is U₄, U₄ rotates 60° counterclockwise than U₃, and the voltage vector corresponding to 010 synthesis is U₅, U₅ Rotate 60°counterclockwise than U₄, and the resulting voltage vector corresponding to 110 is U₆, and U₆ is rotated 60°counterclockwise than U₅. The six voltage vectors U₁, U₂, U₃, U₄, U₅, and U₆ are shown in Fig. 8.18.

Fig. 8.18

Six voltage vectors corresponding to six conduction modes

Direct torque control continuously measures the torque and flux linkage, applies different voltage vectors according to the direction and size of the current flux linkage, controls the amplitude and rotation direction of the stator flux linkage. The amplitude of the stator flux linkage changes within a small range and is basically constant, thereby achieving direct control of the torque. The advantage of this method is that the response speed is fast, and the disadvantage is that the switching frequency is not fixed, resulting in relatively large noise.

8.4 Expansion of Inverter Application (Beginners Do Not Need to Master)

8.4.1 Harmonics of Frequency Converter and Countermeasures

From Fig. 8.10, we can see that since the sine wave is approximately equivalent to the rectangular wave, the voltage waveform output by this type of inverter contains a large number of high-order harmonic components. When the motor is far away from the inverter, the distributed capacitance between the line and the ground becomes larger. The capacitive reactance is inversely proportional to frequency, capacitive reactance X_c = 1/(2πfC). The higher the carrier frequency, the smaller the capacitive reactance is, and the easier it is for high-order harmonic currents to flow into the earth. The line loss will become larger. The leakage current will cause the three-phase current imbalance, trip the leakage switch, and affect the nearby video signal through the ground. At the same time, the voltage glitch will also It will damage the insulation of the motor, so some measures need to be taken to solve these problems. Harmonic hazards can be dealt with by the following methods:

• 1. Reduce the carrier frequency and reduce the high-order harmonic current radiation. The disadvantage of this method is that the operating noise of the motor will become larger;

• 2. Add an output reactor between the output side of the frequency converter and the motor, as shown in Fig. 8.19. The output reactor is made of three wires wound several times in the same direction on an iron core, so that the synthesis of the fundamental wave current magnetic field is zero, which has a strong inhibitory effect on high-frequency harmonics, and can also extend the driving distance between the inverter and the motor, from tens of meters (such as 50 m) to hundreds of meters (such as 400 m). At the same time weaken Voltage spikes (du/dt) are also beneficial for motor insulation.

  Fig. 8.19

  

  Adding an output reactor to the output side of the inverter

• 3. Add output filter NF1 on the output side of the inverter, as shown in Fig. 8.20, the inductance L × 3 of the output filter is formed by winding 3-phase wires on the high-frequency magnetic core for 3–4 turns, and attention should be paid, the output filter is directional. The inductance side must be connected to the inverter, and the current limiting resistor R × 3 and capacitor C × 3 are connected to the motor. They cannot be connected in reverse, otherwise, there will be a current impact on the IGBT power module of the inverter.

  Fig. 8.20

  

  Adding an output filter to the output side of the inverter

  In order to reduce the interference of the frequency converter to the power supply grid and reduce the power interference to other frequency converters in the same network, the following methods can be adopted.

• 4. Add an input reactor on the input side of the inverter, as shown in Fig. 8.21. The input reactor is made of three wires wound several times on the same iron core in the same direction. The synthetic magnetic field of the fundamental wave current is zero. It can suppress high-frequency harmonics, and because the input reactor can suppress the current, the input reactor can also improve the power factor of the inverter.

  Fig. 8.21

  

  Adding an input reactor to the input side of the inverter

• 5. Add an input filter on the input side of the inverter, as shown in Fig. 8.22. The inductance L × 3 of the output filter is formed by winding 3-phase wires on the high-frequency magnetic core for 3–4 turns. It has a suppressive effect on high-frequency harmonics.

  Fig. 8.22

  

  Adding an input filter to the input side of the inverter

8.4.2 Estimation of Input and Output Reactors

In order not to significantly reduce the output capacity of the inverter, the voltage drop U_L on the input and output reactors of the inverter should meet the requirements of (2–5)% not greater than the rated voltage U_e, that is:

$$ U_{L} = (2 - 5\% )\;\;U_{e} = 2\pi fLI_{e} $$

(8.10)

where L is the inductance (H) of the AC reactor, I_e is the rated current (A) of the inverter, and f is the power frequency (Hz).

8.4.3 Heat Dissipation and Reactive Power Compensation of the Frequency Converter

• 1. The heat dissipation problem of the inverter

The ventilation conditions of the frequency conversion cabinet where the frequency converter is placed must meet the ventilation volume and ambient temperature requirements in the manual. However, the calculation of the ventilation volume of the frequency converter is generally complicated. In order to simplify this problem, the author can use the following simple method based on years of experience. Method to deal with: In the case that there is no fan above the frequency conversion cabinet, the ventilation area of the ventilation holes (including side ventilation holes) above the frequency conversion cabinet should be designed to be larger than the area of the inverter radiator, as shown in the Eq. (8.11).

$$ S1 + S2 + S3 > \alpha \times S $$

(8.11)

where the reserved coefficient α takes 1.2–1.5, S represents the total area of the air outlet of the inverter radiator, S1 represents the total area of the air outlet on the top cover of the inverter cabinet, S2 represents the total area of the upper cooling vents, S3 represents the total area of the side cooling vents, as shown in Fig. 8.23. If the total area of the cooling holes is not enough, you can also increase the height of the upper air outlet, or increase the fan to force suction.

Fig. 8.23

Layout of ventilation holes in the inverter cabinet

Due to the vague and general heat dissipation requirements of the frequency converter, the phenomenon that the frequency converter cannot work normally due to improper heat dissipation design has occurred in many projects, which should arouse sufficient attention.

• 2. The problem of reactive power compensation of the frequency converter

The power factor of the frequency converter is generally relatively high and no reactive power compensation is required.

Since the motor and the power supply are isolated through the frequency converter, no reactive power compensation is required on the motor side. On the power input side of the inverter, the power factor problem caused by high-order harmonics is difficult to be effective with capacitor compensation, and high-frequency fluctuations will also affect the life of the capacitor.

8.4.4 Calculation and Estimation of Braking Resistor

• 1. Calculation of braking resistor:

It can be seen from the Eq. (8.5) that when the motor driving the load needs to reduce the speed from the initial speed n₀ to n₁ within t₁ (s), if the resistance torque T_L of the load is not enough to meet the requirements, it needs the motor provides braking torque $T^{\prime}_{M}$, such as Eq. (8.12).

$$ T^{\prime}_{M} - T_{L} = \frac{{GD^{2} }}{375}\frac{{n_{1} - n_{0} }}{{t_{1} }} $$

(8.12)

Transform Eq. (8.12) to get the braking torque $T^{\prime}_{M}$ that the motor needs to provide, such as Eq. (8.13).

$$ T^{\prime}_{M} = T_{L} + \frac{{CD^{2} }}{375}\frac{{n_{1} - n_{0} }}{{t_{1} }} $$

(8.13)

The braking torque that can be generated by the internal loss of the motor itself is about 20% of the rated torque T_e of the motor. If this part of the braking torque is still not enough, the motor needs to provide an additional resistance torque T_R. It is provided through the braking resistor R on the inverter, as shown in Eq. (8.14).

$$ T_{R} = T^{\prime}_{M} - 0.2T_{e} $$

(8.14)

The power consumed by the braking resistor is P (W), as shown in Eq. (8.15).

$$P=\frac{{U}_{Z}^{2}}{R}$$

(8.15)

In the Eq. (4.15), U_Z is the voltage (V) on the DC bus of the frequency converter when the braking resistor is working. The power represented by the upper resistance torque T_R of the motor and the initial speed n₀ should be equal to the power P consumed by the braking resistor, such as Eq. (8.16).

$$\frac{{T}_{R}\times {n}_{0}}{9.55}=\frac{{{U}_{Z}}^{2}}{R}$$

(8.16)

solve R, such as Eq. (8.17).

$$R\left(\Omega \right)=\frac{9.55\times {{U}_{z}}^{2}}{{T}_{R}\times {n}_{0}}$$

(8.17)

Considering that there is a braking situation at the maximum speed, the initial speed n₀ in the Eq. (8.17) can be calculated by the rated speed ne. The calculation of the above Eq. (8.17) needs to know the moment of inertia converted to the motor side, sometimes this is difficult.

• 2. Estimation of braking resistor:

Generally, when the current flowing through the braking resistor is 50% of the rated current, the resistance torque T_R0 is equal to the rated torque T_e, and the current Eq. (8.18) and torque Eq. (8.19) are as follows:

$$ I_{R0} = 50\% I_{e} $$

(8.18)

$${T}_{R0}=T$$

(8.19)

The braking resistor value R₀ corresponding to this point is shown in Eq. (8.20):

$$ R_{0} = \frac{{U_{Z} }}{{I_{R} }} = \frac{{2U_{Z} }}{{I_{e} }} $$

(8.20)

Generally, the resistance torque T_R is (0.8–2) times of the rated torque T_e, and there is Eq. (8.21):

$$ T_{R} = (0.8 - 2)\;T_{e} $$

(8.21)

The torque is proportional to the current, and the current is inversely proportional to the resistance. Therefore, the value range of the braking resistor R is as follows (8.22):

$$ R = \frac{{R_{0} }}{0.8 - 2} $$

(8.22)

The calculation of the maximum power of the braking resistor is shown in Eq. (8.15). In actual selection, the actual power of the resistor should be determined by multiplying a proportional coefficient (0.11–0.4) according to the frequency and intensity of the motor in the power generation state.