The three-phase three-wire inverter circuit using the output filter is shown in Figure 1.

When the SPWM control method is adopted, the Fourier series expansion of 3 symmetric switching functions (interface signals) is

Using SPWM control, when n=3,5,7,11,13, A_{n}=0 can be set, then the inverter output line voltage is

In the formula, U_{d} is the DC power supply voltage.

The three-phase output line voltage of the inverter can be obtained from equation (2), and the input current of the inverter is

In the formula, I_{1}=SW_{1}·I_{a}; I_{2}=SW_{2}·I_{b}; I_{3}=SW_{3}·I_{c}.

therefore

In the formula, I_{a}, I_{b} and l_{c} are the output current of each phase of the inverter.

For three-phase three-wire inverters, there are

For a balanced linear load, the three-phase current i_{a}+i_{b}+i_{c} is approximately a sine wave with the same amplitude and a phase difference of 120°, so the input current of the inverter can be derived from equation (4). This current will mainly consist of a DC component and a harmonic component of the same switching frequency. These high-order harmonics will be effectively filtered out in the DC link.

**1. Unbalanced load**Unbalanced load is not an ideal working state. For the unbalanced load state that disconnects the inverter C phase (as shown in Figure 2), there is obviously

Assuming that the load current is approximately sinusoidal, that is

In the formula, φ is the load power factor angle.

In this case, the input current I_{in} of the inverter. It can be calculated. From formula (6), formula (7) and formula (4), when n=1, we get

Equation (8) shows that for unbalanced (such as phase C disconnection) load, the input current of the inverter is the DC component equal to (√3A_{1}1)/2·cos (φ-π/6) and twice the inverter The pulsating component of the working frequency of the device is composed of approximately (√3A_{1}1)/2·cos (2ω+φ+π/6). The pulsating component is caused by the unbalanced load, which generates a circulating current on the DC link capacitor and constitutes reactive power. If the capacitance value in the DC link is large enough, the distortion (ripple) of the DC voltage will not appear, and the overall performance of the inverter will be reduced. The reactive power in the DC link is directly proportional to the unbalance of the inverter output load. The greater the unbalance, the greater the pulsating component of the input current. The greater the value of the required DC capacitor, the greater the impact on the overall performance of the inverter. big.

The consequences caused by the non-linear load at the output of the inverter (such as the rectifier circuit) are the same as the unbalanced load.

**2. Filtering principle of DC side active filter**S7~S10 are controlled by the switch. The DC side active filter circuit composed of inductor L and L is shown in Figure 2. It and the passive filter composed of L

_{d}C

_{d}in the DC link form a comprehensive filter system to compensate for the effects of unbalanced loads and nonlinear loads, and active filters can be used to reduce the burden of L

_{d}C

_{d}passive filters. The following describes the working principle of the DC side active filter.

As shown in Figure 2, the total input current obtained by the DC active filter from the DC link is

Assuming that the aforementioned “C-phase disconnection” is used as an unbalanced load, the single-phase full-bridge filter is composed of the switch tubes S7~S10 in the filter, and the SPWM control method is adopted. The working frequency of the filter is the same as the working frequency of the following inverter, and the measured voltage between the P and Q points of the filter is

For SPWM modulation, when n=3,5,7,11,13,…, you can set Bn=0. The filter current I_{PQ} can be considered to be approximately sinusoidal, and the harmonics of the voltage u_{PQ} are fully attenuated to zero by the inductance L. Therefore, when n=1, from equation (10) we can get

The resistance of the inductance L is very small, and it can be considered that it is equal to zero, and then I_{PQ }lags u_{PQ} by about 90°. Next, calculate the input current of the filter by formula (4). This current is obtained by the single-phase full-bridge inverter that constitutes the filter, that is, obtained by formula (9). From this, the input current of the active filter can be expressed as

Therefore, when the active filter compensates for the unbalanced load (that is, the phase C is disconnected), the formula (12) must be used to offset the component at the input of the inverter, that is, the 1 of the formula (8). (√3A_{1}1)/2·cos(2ω+φ+π/6) component in. When fully compensated, the second harmonic component in equation (10) and equation (8) are equal to obtain

Solved by the above formula

Therefore, as long as the active filter works according to formula (13) , the purpose of eliminating the second harmonic wave component in formula (8) can be achieved. At this time, the input current of the inverter is

Equation (14) shows that the second harmonic component of la in Equation (8) caused by the unbalanced load such as “C phase disconnection” is effectively eliminated in the DC input current I_{i}, and the compensation is achieved. Purpose. Since other forms of unbalanced load or nonlinear load have the same influence on the DC input current as the “C-phase disconnected” unbalanced load, the DC side active filter method can also achieve the purpose of compensation.

The control circuit block diagram of the active filter on the DC side is shown in Figure 3, which adopts the two-state hysteresis current tracking control. The deviation between the reference current I_{ref}=0 and the DC input current I_{i} is filtered out by a high-pass filter. After the component, as the input of the two-state hysteresis comparator, its output controls the on and off of S7~S10, so that I_{ref} can track and eliminate the second harmonic component in I_{i}.

**3. Simulation results**The simulation results are shown in Figure 4, Figure 5 and Figure 6. Among them, Figure 4 shows the voltage and current waveforms of Figure 2, where Figure (a) is the waveform of the output voltage u

_{AB}; Figure (b) is the waveform of the line current I

_{a}, and Figure (c) is the waveform of the input current I

_{in}. Figure (d) is the input

The frequency spectrum of the current I_{in}. Figure 5 shows the waveform of the active filter in the three-phase three-wire inverter, where figure (a) is the waveform of the voltage uPQ, figure (b) is the waveform of the input current I_{inf}, and figure (c) is the input current la Spectrum. Figure 6 shows the waveform and frequency spectrum of the synthesized current I_{i}=I_{in}+I_{inf}, where Figure (a) is the waveform diagram, and Figure (b) is the frequency spectrum.