2.2.1. Conventional Space Vector Modulation

The current space vector diagram of conventional space vector modulation (C-SVM) for the AC–DC MC is illustrated in Figure 2. In order to synthesize the desired input current reference vector → *I re f* , C-SVM uses the two nearest active vectors and one zero vector among six active vectors → *I* 1 ∼ → *I* 6 and three zero vectors → *I* 7 ∼ → *I* 9, according to the sector location of the input current reference vector. Each sector is denoted as the area between two active vectors, for example, the area between two active vectors → *I* 1 and → *I* 2 is sector I, the area between two active vectors → *I* 2 and → *I* 3 is sector II, and so on. As it can be seen from Figure 2, the input current reference vector locates in sector I, thus two active vectors → *I* 1, → *I* 2 and one zero vector are used to synthesize the desired current reference vector. The selection of zero vector is based on the constraint to minimize the switching frequency of bidirectional switches so that the commutation between two switching states involves only two switches in two phase-legs of the converter, one switch is turned on, and one switch is turned off at the same time. Among three zero vectors, only zero vector → *I* 7 satisfies the requirements for switching devices. Therefore, zero vector → *I* 7 and two active vectors → *I* 1, → *I* 2 are selected.

**Figure 2.** Space vector diagram of conventional space vector modulation (C-SVM) for AC–DC matrix converter.

The duty cycles *d*1, *d*2, and *d*0 of the active and zero vectors, when the input current reference vector is in the sector I, are given by:

$$d\_1 = m\_l \sin\left(\frac{\pi}{3} - \delta\right) \tag{1}$$

$$d\varphi = m\_i \sin(\delta)\tag{2}$$

$$d\_0 = 1 - d\_1 - d\_2 \tag{3}$$

where *mi*: modulation index; *mi* = *ii*1/*idc mi* ∈ [0, 1]; *ii*1: the peak value of the fundamental-frequency component in *ii*; δ : input current reference vector angle, δ ∈ 0, π3 .

The durations *T*1, *T*2, and *T*0 of the active vectors → *I* 1, → *I* 2, and zero vector → *I* 0 are respectively expressed as:

$$T\_1 = d\_1 T\_s \tag{4}$$

$$T\_2 = d\_2 T\_s \tag{5}$$

$$T\_0 = T\_s - T\_1 - T\_2 \tag{6}$$

where *Ts*: switching period: *Ts* = 1*fs*; *fs*: switching frequency.

The calculation of the duty cycles and durations, when the input current reference vector passes through other sectors one by one, is obtained in a similar algorithm as sector I.
