 Hello, myself Sunil Kalshiti, Assistant Professor, Department of Electronics Engineering, Vulturen Institute of Technology, SolarPore. Today, I am going to explain the three-phase dual converter, learning converter. At the end of the session, students can analyze three-phase dual converter. Up till now, we studied the three-phase semiconductor. Three-phase semiconductor exhibits one quadrant operation, means it provides the only one direction of voltage and one direction of current. That's why three-phase semiconductor called as a single quadrant converter. Three-phase full converter exhibits two quadrant operation. It provides the both polarity of voltage and only one polarity of current. Whenever we require the both polarity of voltage and both polarity of current, then we have to use the four-quadrant converter. For four-quadrant operation, which is generally required in many variable speedrides, the three-phase dual converters are used. It is used for the application up to 2 megawatt output power level, which is the circuit diagram of three-phase dual converter. T1 to T6 acts as a one converter and T1 dash to T6 dash acts as a another converter. LR by 2, LR by 2 is the circulating reactor. Dual converter is formed using two three-phase full converters connected back to back. Dual converters are widely used for speed reversal and regenerative braking of separately excited DC motor. When both converters are operated in rectifier mode, they produce average output voltage of opposite polarity. For getting output voltage of same polarity, one converter is operated in rectifier mode and other is operated in inverter mode. These are the waveforms for three-phase dual converter. These are the waveforms of converter 1, output voltage of converter 1 for alpha 1 is equal to 60 degree and these are the waveforms for output of converter 2 for alpha 2 is equal to 120 degree. In the three-phase dual converter, the value of alpha 1 plus alpha 2 is always 180 degree and these are the waveforms for circulating voltage, Vr is equal to V01 plus V02. In the interval pi by 6 plus alpha 1 to pi by 2 plus alpha 1, the line-to-line voltage VAB appears across output of converter 1 and VBC appears across output of converter 2. In this direction, pi by 6 plus alpha 1 to pi by 2 plus alpha 1, VAB appears across output of converter 1 and for the same duration, the VBC appears across the output of converter 2. The two converters are controlled in such a way that they produce equal average output voltage. If converter 1 is in rectifier mode and converter 2 is in inverter mode, therefore V01 is equal to minus V02, we define three line-to-neutral voltages as follows. Vrn is equal to VAN is equal to Vm sin omega t, Vyn is equal to Vbn is equal to Vm sin omega t minus 2 pi by 3. The phase difference between y and r is the 120 degree, Vbn is equal to Vcn is equal to Vm sin omega t plus 2 pi by 3. The corresponding line-to-line supply voltages are Vry is equal to Vab is equal to Van minus Vbn is equal to root 3 Vm sin omega t plus pi by 6, Vyb is equal to VBC is equal to Vbn minus Vcn, root 3 Vm sin omega t minus pi by 2, Vbr is equal to Vca is equal to Vcn minus Van is equal to root 3 Vm sin omega t plus pi by 2 expression for circulating current. If V01 and V02 are the output voltage of converter 1 and converter 2 respectively, the instantaneous voltage across the current limiting inductor during the interval pi by 6 plus alpha 1 to pi by 2 plus alpha 1 is given by Vr is equal to V01 plus V02 is equal to Vab minus VBC. Now, substitute the value of Vab and VBC in this equation. So, Vr is equal to root 3 Vm into bracket sin omega t plus pi by 6 minus sin omega t minus pi by 2. After solving this, we obtain Vr is equal to 3 Vm cos omega t minus pi by 6. The circulating current can be calculated by using above equation, Ir of t is equal to 1 upon omega lr and the limits of integration pi by 6 plus alpha 1 to omega t Vr d omega t. Ir of t is equal to 1 upon omega lr pi by 6 plus alpha 1 to omega t. Now, substitute the value of Vr in this equation. So, it becomes Ir of t is equal to 3 Vm upon omega lr into bracket sin omega t minus pi by 6 minus sin alpha 1 and for calculating the Ir max assume alpha 1 is equal to 0 and omega t is equal to 7 pi by 7 pi by 6 therefore, after solving this Ir max becomes 3 Vm upon omega lr. As we know, 3 phase dual converter is the 4 quadrant converter and this is the quadrant diagram for the 3 phase dual converter. In the first quadrant, the converter 1 operates in the rectifier mode. So, for this condition alpha 1 must be less than 90 degree or the second converter sorry for second quadrant the converter 2 operates as an inverter mode. So, for this alpha 2 must be greater than 90 degree. For third quadrant the converter 2 operates as a rectifier mode and for this the condition of alpha is alpha 2 is less than 90 degree and for the there are two different modes of operation non-circulating current mode of operation and circulating current mode of operation. Now, for non-circulating current mode of operation in this mode of operation only one converter is active at a time that is only one converter is switched on at a time. If we want load voltage positive that is P is positive with respect to Q with current entering at point P. So, for this convert for this condition the converter 1 operates in the rectification mode and when the converter 1 operates in the rectification mode the VDC becomes positive, IDC becomes positive, hence average load power VDC is positive and the power flows from AC source to load and converter operates in first quadrant. If we want load current in the same direction, but load voltage to be reversed that is Q is positive with respect to P. For this condition the converter 1 operates in the inversion mode and alpha 1 must be greater than 90 degree. So, for this condition the VDC becomes negative, IDC becomes positive and the power flows from load to AC source and the converter operates in fourth quadrant. If we want negative load voltage that is Q is positive with respect to P with current flow from Q to P converter 2 operates in rectification mode and for this condition alpha 2 must be greater than 90 degree. And for this condition the VDC becomes negative, IDC becomes negative and the power flows from AC source to load and for this condition the converter operates in third quadrant. If we want load current in the same direction, but load voltage to be reversed that is P is positive with respect to Q, the converter 2 operates in inversion mode, VDC is positive, IDC is negative and power flows from load to AC source and the converter operates in second quadrant. Circulating current mode of operation. In this mode both converters are switched on at a time to get the same polarity of average output voltage. One converter operates in rectification mode while the other operates in inversion mode. The delay angles are controlled such that the converter one operates in the rectification mode while the other operates in inversion mode. The trigger angles alpha 1 and alpha 2 are adjusted such that alpha 1 plus alpha 2 is equal to 180 degree. When alpha 1 is less than 90 degree converter one operates as a controlled rectifier and alpha 2 made greater than 90 degree and converter 2 operates as an inverter. So, VDC becomes positive, IDC becomes positive and average DC power is positive and the converter operates in first quadrant. When alpha 2 less than 90 degree converter 2 operates as a controlled rectifier, alpha 1 is greater than 90 degree and the converter one operates as an inverter. VDC is negative and IDC is negative and the average DC output power is positive and the converter operates in the third quadrant. Why? Circulating current mode is preferred. In this mode both converters are active at the same time. So, effect of this the instantaneous switch is possible. That is why circulating current mode is always preferred. These are references. Thank you.