 Hello, Myself Sunil Kalshetti, Assistant Professor, Department of Electronics Engineering, Walchen Institute of Technology, SolarPort. Today, I am going to explain the Three-Phase Half-Wave Control Directifier with Inductive Load Learning Outcome. At the end of this session, students can analyze Three-Phase Half-Wave Control Directifier with Highly Inductive Load. This is the circuit diagram of Three-Phase Half-Wave Control Directifier with Inductive Load. In this circuit, three thyristors are used. The anode of T1, T2, T3 is connected to the source through the phase A, phase B, phase C. And the cathode of these three thyristors is connected combinely to the one terminal of the load and the another terminal of load is connected to the neutral terminal. If alpha is less than pi by 6, next SCR is fired before natural 0. Hence, VL and IL becomes continuous. If alpha is greater than pi by 6, next SCR is fired at the end of the half cycle. For large inductive load, the current continued to flow even after supply voltage has reduced to 0. Thus, the conducting SCR is continued to conduct with reversed phase voltage until next SCR is fired. These are the waveforms for alpha is equal to 0 degree. At omega t is equal to pi by 6 plus alpha. The gate pulse is applied to the thyristor T1. So, thyristor T1 conducts. Once the T1 conducts, the current flows through the source T1 load back to the neutral and phase VAN appears across the load here. From pi by 6 to 5 pi by 6, the phase VAN appears across the load. At omega t is equal to pi by 6 plus alpha, the gate pulse is applied to the T2 and once the T2 conducts, the phase VAN appears across the load and the T2 continuously conducts up to the 5 pi by 6. As long as T2 conducts, the VAN appears across the load. And before end of the half cycle, the next pulse is applied to thyristor T3 and once the T3 conducts, the previous thyristor turns off because of the acharal commutation. And once the T3 conducts, the phase VCN appears across the load. In this mode, the load current flows through the converter continuously. That is why the name is the continuous conduction mode. These are the waveforms for alpha is equal to 30 degree. Again, for alpha is equal to 30 degree, the load current flows through the converter continuously. That is why the mode is again continuous conduction mode. These are the waveforms for alpha is equal to 60 degree. At omega t is equal to pi by 6 plus alpha, the gate pulse is applied to the thyristor T1. So, once the T1 conducts, so phase VAN appears across the load. So, T1 is continuously conduct up to the end of the half cycle means in this duration pi by 6 plus alpha to pi, the T1 conducts and power is supplied by the source and the converter operates in the rectification mode. In this mode, the power flows from source to load. At the end of the half cycle, the negative half cycle starts. So, but the load is highly inductive load. This highly inductive load try to maintain the current in same direction. Therefore, it forces the current in the same direction through the conducting thyristor. Therefore, effect of this, the negative half set of this, the negative voltage is appears across the load. Therefore, effect of this, for the inductive load, the average output voltage is less than the gistive load. Here, in this negative half cycle means the duration from pi to 5 pi by 6 plus alpha, the power flows from load to source and the converter operates in the inversion mode. In this mode, the stored inductive energy is delivered to the source. So, because of this, it affects the power factor of the system. That is why it is essential try to reduce the circulation of reactive power. Now, at the next instant means that omega t is equal to 5 pi by 6 plus alpha, the next pulse is available for the T2 and once the T2 conducts, the phase VBN appears across the load and the process in the inductive load, the load current flows through the converter continuously. These are the waveforms for alpha is equal to 90 degree. Again, these are the waveforms for the different value of alpha. These are the current waveforms. Here, the load is highly inductive load, the current flows through the converter continuously and each thyristor conducts for 120 degree. Means the duration in this duration, pi by 6 plus alpha to 5 pi by 6 plus alpha T1 conducts, from 5 pi by 6 plus alpha to 9 pi by 6 plus alpha T2 conducts and from 9 pi by 6 plus alpha to 13 pi by 6 T3 conducts. In this way, the current flows through the converter continuously. Expression for average DC output voltage. If reference phase voltage is Vrn is equal to Van is equal to Vm sin omega t, the average or DC output voltage for continuous load current is calculated using the equation. Vdc is equal to 3 upon 2 pi and the limits of integration pi by 6 plus alpha to 5 pi by 6 plus alpha Vm sin omega t. In this duration, the phase Van appears across the load and this is the instantaneous value of the Van. After solving, we obtain Vdc is equal to 3 root 3 Vm upon 2 pi cos alpha. Vdc is equal to 3 Vlm upon 2 pi cos alpha, where Vlm is equal to root 3 Vm is equal to Piv is equal to maximum line to line supply voltage. Piv means peak inverse voltage. For the three-phase half-way control rectifier, the Piv is equal to root 3 Vm. The maximum average or DC output voltage is obtained at delay angle alpha is equal to 0 degree and is given by Vdc max is equal to Vdm is equal to 3 root 3 Vm upon 2 pi, where Vm is peak phase voltage and the normalized average output voltage is Vdcn is equal to Vn is equal to Vdc upon Vdm is equal to cos alpha. Generally, in the converter, when alpha is minimum, the average output voltage is maximum and when alpha is maximum, average DC voltage is minimum. The RMS value of output voltage is found by using the equation. V0 RMS is equal to 3 upon 2 pi, the limits of integration pi by 6 plus alpha to pi pi by 6 plus alpha, Vm square sin square omega t and we obtain V0 RMS is equal to root 3 Vm in the bracket 1 by 6 plus root 3 upon 8 pi cos three-phase half-way control rectifier with inductive load and freewheeling diode. Here, the diode is connected across the inductive load, that's why the name of the diode is the freewheeling diode, sometime it is also called as a bypass diode or commutating diode. For alpha is less than pi by 6, Vdc of converter is always positive and freewheeling diode does not come into operation. As alpha greater than pi by 6, IL starts to freewheel through the diode for certain period, thus cutting off input line current and preventing DC terminal load voltage from swinging into negative direction. The freewheeling diode serves two main functions, it prevents reversal of load voltage and it transfers the load current away from the rectifier. The circuit operates in two modes, mode one conduction mode, mode two freewheeling mode. These are the waveforms for alpha is equal to 60 degree, at omega t is equal to pi by 6 plus alpha the T1 conducts and in this mode the converter operates in the conduction mode. At the end of the half cycle, here the load is highly inductive load, so because of this the freewheeling diode conducts and once the freewheeling diode conducts, it protects the reversal of load voltage and it provides the circulating path for the stored inductive energy. As long as freewheeling diode is in conducting state, the load voltage remains zero and the freewheeling diode continuously conducts until the firing of next resistor. How average output voltage is improved using the freewheeling diode? When the freewheeling action takes place, it protects the reversal of load voltage, so indirectly it improves the average load voltage. How freewheeling diode helps in improvement of power factor of system? When the freewheeling action takes place, it protects the reversal of load voltage and at the same time it provides the circulating path for the stored inductive energy. So, effect of this the phi is reduces and as phi is reduces cos phi improved and indirectly the power factor of the system gets improved. Expression for DC voltage, VDC is equal to 3 upon 2 pi alpha plus 30 degree to 180 degree V0 d omega t. In this duration, the phase VAN appears across the load. Therefore, VAN is equal to Vm sin omega t and substitute this value in the main equation and after solving this, we obtain the VDC is equal to 3Vm upon 2 pi 1 plus cos alpha plus 30 degree. These are references. Thank you.