# Hybrid Harmonic Suppression Method at DC Link of Series-Connected 18-Pulse Rectifier
## 2 The Proposed Rectifier with Harmonic Injection Circuit  
### Phase-Shifting Transformer Design  
The proposed multipulse rectifier with multi-winding phase-shifting transformer and hybrid harmonic reduction method is shown in Figure 1. In Figure 1, the converter is composed of an 18-pulse isolation transformer, 3 diode rectifier bridges, and the hybrid harmonic injection circuits; each injection circuit consist of an injection transformer, 3 diodes, a switch, and its control circuit.  
Figure 1: The proposed multipulse rectifier using hybrid harmonic reduction method.  
According to the structure of the three-phase diode-bridge rectifier, the output voltage of each bridge has six pulse per supply cycle, and each pulse lasts 60\({}^{\circ}\). In Figure 1, three-diode bridge rectifiers are connected in series. To suppress the harmonics of input current and ripples of load voltage to the highest degree, the phase difference of the input three-phase voltages of the diode bridge rectifiers is 20\({}^{\circ}\), and Figure 2 describes the vector diagram.  
Take phase A as an example; from Figure 2, it can be obtained that  
\[\left\{\begin{array}{l}\cos\alpha=\dfrac{\left(N_{11}U_{\text{a2}}\right)^{2 }+U_{\text{a3}}{}^{2}-\left(N_{\text{32}}U_{\text{c2}}\right)^{2}}{2N_{11}U_{ \text{a2}}U_{\text{a3}}}\\ \cos 120^{\circ}=\dfrac{\left(N_{11}U_{\text{a2}}\right)^{2}+\left(N_{\text{32}} U_{\text{c2}}\right)^{2}-U_{\text{a3}}{}^{2}}{2N_{11}U_{\text{a2}}N_{\text{32}}U_{ \text{c2}}}\end{array}\right. \tag{1}\]  
where \(N_{11}\) and \(N_{32}\) are the turn ratio of the phase-shifting transformer; \(U_{\text{a2}}\), \(U_{\text{a3}}\), and \(U_{\text{c2}}\) are the amplitude of the input voltages of the three-phase bridge.  
From Equation (1), and Figures 1 and 2, when \(\alpha\) is 20 degrees, the turn ratio of the isolation transformer meets the following:  
\[N:N_{11}:N_{12}:N_{21}:N_{31}:N_{32}=\sqrt{3}:0.742:0.395:1:0.742:0.395 \tag{2}\]  
When the converter operates normally, the input voltage of the converter is described as  
\[\left\{\begin{array}{l}u_{\text{AN}}=UL/20^{\circ}\\ u_{\text{BN}}=UL/120^{\circ}\\ u_{\text{CN}}=UL/2-120^{\circ}\end{array}\right. \tag{3}\]  
where \(U\) is the amplitude of the input voltage.  
From Figure 2 and Expression (3), the three sets of output voltages of the isolation transformer can be calculated as  
\[\left\{\begin{array}{l}u_{\text{a1}}=\frac{U}{\sqrt{3}}\angle 10^{\circ}\\ u_{\text{b1}}=\frac{U}{\sqrt{3}}\angle 130^{\circ}\\ u_{\text{c1}}=\frac{U}{\sqrt{3}}\angle-110^{\circ}\end{array}\right.\left\{ \begin{array}{l}u_{\text{a2}}=\frac{U}{\sqrt{3}}\angle 30^{\circ}\\ u_{\text{b2}}=\frac{U}{\sqrt{3}}\angle 150^{\circ}\\ u_{\text{c2}}=\frac{U}{\sqrt{3}}\angle-90^{\circ}\end{array}\right.\left\{ \begin{array}{l}u_{\text{a3}}=\frac{U}{\sqrt{3}}\angle 50^{\circ}\\ u_{\text{b3}}=\frac{U}{\sqrt{3}}\angle 170^{\circ}\\ u_{\text{c3}}=\frac{U}{\sqrt{3}}\angle-70^{\circ}\end{array}\right. \tag{4}\]  
Figure 2: Vector diagram of the isolation transformer. a1, a2, a3, b1, b2, b3, c1, c2, c3 are the vectors of the output phases of the main transformer.