 Hello! Myself Sunil Kalshatti, Asset Professor, Department of Electronics Engineering, Valchand Institute of Technology, Solapur. Today, I am going to explain the three-phase semiconductor learning outcome. At the end of this session, students can analyze three-phase semiconductor. It is also called as a three-phase half-controlled bridge converter. Three-phase semiconductors are used in industrial DC drive applications up to 120 kW power output. Single coordinate operation is possible. Power factor decreases as the delay angle increases. Power factor is better than that of three-phase half-view converter. This is the circuit diagram of three-phase semiconductor. Load is highly inductive load and DM is the freewheeling diode. It is connected across the load. It consists of three SCRs and three diodes. The load is connected in between the cathode of SCR and anode of diode. Only one SCR from upper group and one diode from lower group conduct at a time, providing line voltage across load. The SCR, which is connected to phase, having higher phase voltage than any other phase is fired. The phase voltage Va becomes higher than Vb and Vc after omega t is equal to pi by 6. T1 is connected to phase A, so T1 is fired at omega t is equal to pi by 6 plus alpha. At omega t is equal to pi pi by 6, Vb becomes equal to Va and after Vb becomes higher than Va. So, T2 is forward bias and T2 can be fired at omega t is equal to pi pi by 6 plus alpha. The phase voltage Vc becomes higher than Va and Vb. After omega t is equal to 9 pi by 6, T3 is connected to phase C, so T3 is fired at omega t is equal to 9 pi by 6 plus alpha. The conduction period of D1 is from pi by 2 to 7 pi by 6. Conduction period of D2 is from 7 pi by 6 to ln pi by 6. The conduction period of D3 is from ln pi by 6 to 15 pi by 6. The conduction period of each diode is 120 degree. We define 3 line neutral voltages. As follows, Vrn is equal to Vn is equal to Vm sin omega t, Vyn is equal to Vbn is equal to Vm sin omega t minus 2 pi by 3 and Vbn is equal to Vm sin omega t plus 2 pi by 3. Here the phase difference is 120 degree. Vm is peak phase voltage of y connected source. Vrb is equal to Vac is equal to Van minus Vcn root 3 Vm sin omega t minus pi by 6. Vyr is equal to Vba is equal to Vbn minus Van is equal to root 3 Vm sin omega t minus pi by 6 and Vvy is equal to Vcb is equal to Vcn minus Vbn is equal to root 3 Vm sin omega t plus pi by 2 and Vry is equal to Vab is equal to Van minus Vbn is equal to root 3 Vm sin omega t plus pi by 6. These are the waveforms for alpha is equal to 30 degree. That omega t is equal to pi by 6 plus alpha t1 d3 forward bias. So, pulse is applied to the t1 and the current flows through the source t1 load d3. So, if you have to have this Vab appears across the load and the t1 d3 continuously conduct up to pi by 2. After pi by 2 the phase c becomes more negative. So d3 becomes reverse bias and d1 becomes forward bias and the current flows through the t1 d1 and when the t1 d1 conducts Vac appears across the load and the t1 d1 are continuously conduct up to the end of the half cycle. At omega t is equal to pi by 6 plus alpha t2 d1 becomes forward bias and the phase Vbc appears across the load. In this way the operation repeats. These are the current waveforms for I t1, Ia, I t3 and phase A. Average or DC output voltage of 3 phase semi-converter for alpha is less than or equal to pi by 3 or 6 degree and this mode is the continuous conduction mode. Vdc is equal to 3 upon 2 pi pi by 6 plus alpha 2 pi by 2 Vab d omega t and pi by 2 2 pi pi by 6 plus alpha Vac d omega t. After solving this we obtain Vdc is equal to 3 root 3 Vm upon 2 pi 1 plus cos alpha. Vdm is equal to maximum average DC output voltage. When alpha is equal to 0 the Vdc is equal to Vdm. So, normalized output is equal to Vdc upon Vdm is equal to 0.5 into bracket 1 plus cos alpha. Rms value of output voltage is calculated by using the equation V0 Rms is equal to 3 upon 2 pi and the limits of integration pi by 6 plus alpha 2 pi by 2 Vab square d omega t plus pi by 2 to pi pi by 6 plus alpha Vac square d omega t raise to half. Now, after solving this we obtain V0 Rms is equal to root 3 Vm into bracket 3 upon 4 pi in sub bracket 2 pi by 3 plus root 3 cos square alpha raise to half. Why? 3 phase semi-converter acts as a 6 pulse converter for alpha is equal to 6 less than or equal to 60 degree. If for alpha is equal to less than or equal to 60 degree for one complete cycle 6 pulses are appears across the load that is why the converter acts as a 6 pulse converter. These are the waveforms for alpha is equal to 90 degree and the mode is the discontinuous conduction mode. These are the voltage and current waveforms. Why? 3 phase semi-converter acts as a 3 pulse converter for alpha greater than 60 degree. If alpha is greater than 60 degree for one complete cycle 3 pulses are appears across the load that is why the converter acts as a 3 pulse converter. Expression for the average output voltage of 3 phase semi-converter for alpha greater than 60 degree or pi by 3 and the mode is the discontinuous conduction mode. Vdc is equal to 3 upon 2 pi limits of integration pi by 6 plus alpha to sin pi by 6 now here substitute the value of Vac Vac is equal to root 3 Vm sin omega t minus pi by 6 after solving this we obtain Vdc is equal to 3 root 3 Vm upon 2 pi 1 plus cos alpha Vdc is equal to 3 Vm upon 2 pi 1 plus cos alpha where Vm is equal to root 3 Vm is equal to maximum value of line to line supply voltage. The maximum average output voltage that occurs at del angle alpha is equal to 0 degree so Vdc max is equal to Vdm is equal to 3 root 3 Vm upon pi the normalized average output voltage is Vn is equal to Vdc upon Vdm is equal to 0.5 into bracket 1 plus cos alpha the RMS output voltage is found from V0 RMS is equal to into bracket 3 upon 2 pi limits of integration pi by 6 plus alpha to sin pi by 6 Vac square d omega t rest to half and after solving this we obtain V0 RMS is equal to root 3 Vm into bracket 3 upon 4 pi into sub bracket pi minus alpha plus sin 2 alpha upon 2 rest to half. Why this converter acts as a single quadrant converter? This converter provides the positive direction of voltage and positive direction of current that is why the converter acts as a single quadrant converter these are references. Thank you.