 Hello. Myself Sunil Kalshetti, Assistant Professor, Department of Electronics Engineering, Walshan Institute of Technology, Swelapur. Today, I am going to explain the three-phase full-wave controlled rectifier, learning.com. At the end of this session, students can analyze three-phase full-wave controlled rectifier, three-phase full-wave controlled rectifier. In DC terminals of two three-pulse converters are connected in series, the converter formed is called as a six-pulse bridge converter. It is used in industrial application up to 120 kilowatt output power level. Two quadrant operation is possible. This is the circuit diagram of three-phase full-wave controlled rectifier with highly inductive load. It consists of six thyristor. The circuit consists of two groups T1, T3, T5 forms positive group thyristor and T2, T4, T6 forms negative group thyristor. The positive group thyristor turns on when supply is positive and the negative group thyristor turns on when supply is negative. The thyristors are triggered at an interval of 60 degree or pi by 3 radian. Each thyristor can conduct for 120 degree or 2 pi by 3 radian. At any instant of the time, two thyristor can conduct that is one from upper half and another form lower half. Thyristors are numbered in order in which they are triggered. The thyristor triggering sequence is 1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 1, 2. When two thyristors are conducting that is one from upper group and another from lower group, the corresponding line voltage is appears across the load. In T1 conducts the current flows through the source T1 load and T6. The phase VAB appears across the load. T1 is triggered at omega t is equal to pi by 6 plus alpha, T6 is already conducting when T1 is turned on. During the interval pi by 6 plus alpha to pi by 2 plus alpha, T1 and T6 conduct together and the output load voltage is equal to VAB is equal to VAN minus VBL. T2 is triggered at omega t is equal to pi by 2 plus alpha, T6 turns off naturally as it is reverse bias as soon as T2 is triggered. During the interval pi by 2 plus alpha to pi by 6 plus alpha, T1 and T2 conduct together and the output load voltage V0 is equal to VAC is equal to VAN minus VCN. We define three line neutral voltages, three phase voltages as follows. VRN is equal to VAN is equal to VM sin omega t, where VM is the maximum phase voltage. VBN is equal to VM sin omega t minus 2 pi by 3 and VBN is equal to VCN is equal to VM sin omega t plus 2 pi by 3, VM is the peak phase voltage of star connected source. 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, VIB is equal to VBC is equal to VBN minus VCN is equal to 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, these are the waveforms for highly inductive load. At omega t is equal to pi by 6 plus alpha, the pulse is applied to the T1 and before this T6 is in conducting state. At omega t is equal to pi by 6 plus alpha, the pulse is applied to the T1 and before omega t is equal to pi by 6 plus alpha, T6 is in conducting state, therefore the current flows through the source T1 load T6 back to the source. So effect of this the phase VAB appears across the load and the T6 T1s are continuously conduct up to pi by 2 plus alpha. At pi by 2 plus alpha the pulse is applied to the T2, so effect of this the T6 turns off because of the line commutation and once the T6 turns off T1, T2 conducts and when the T1, T2 conducts the VAC appears across the load, in this way the operation repeats and the thyristor T1, T2 conducts and after that T2, T3, T3, T4, T4, T5, T5, T6 conducts, these are the current waveforms. What is the ripple frequency of 3 phase full wave controlled rectifier? For one complete half cycle 6 pulses are appears across the load, effect of this the output ripple frequency is the 6FS, why this converter is called as 6 pulse converter? For one complete cycle 6 pulses are appears across the load that is why name is the 6 pulse converter. The output load voltage consists of 6 voltage pulses over the period of 2 pi radian, hence the average output voltage is calculated as V0 dc is equal to Vdc is equal to 6 upon 2 pi, the limits of integration pi by 6 plus alpha 2 pi by 2 plus alpha V0 du omega t, V0 is equal to line voltage Vab is equal to root 3 Vm sin omega t plus pi by 6. Now substitute the value of V0 in our main equation, after solving this we obtain Vdc is equal to 3 root 3 Vm upon pi cos alpha is equal to 3 Vml upon pi cos alpha, where the Vml is equal to root 3 Vm is equal to Pi V is equal to maximum line to line supply voltage. The maximum average dc output voltage is obtained for delay angle alpha is equal to 0. To substitute alpha is equal to 0 in this equation therefore, Vdc max is equal to Vdn is equal to 3 root 3 Vm upon pi is equal to 3 Vml upon pi. The normalized average output voltage is Vdcn is equal to Vn is equal to Vdc upon Vdm is equal to cos alpha, the RMS value of output voltage is found from V0 RMS is equal to 6 upon 2 pi limits of integration pi by 6 plus alpha to pi by t plus alpha V0 square d omega t square. So, V0 RMS is equal to 6 upon 2 pi pi by 6 plus alpha to pi by 2 plus alpha Vsb square d omega t raise to half. After solving this we obtain V0 RMS is equal to root 3 Vm into bracket half plus 3 root 3 upon 4 pi cos 2 alpha raise to half. These are references. Thank you.