 myself, Giridhar Jain, Assistant Professor in Electronics and Telecommunication Engineering, Walchen Institute of Technology, Sholapur. Today, I am going to explain single-phase full-wave control rectifier with RL load. Learning outcomes of this session are, at the end of this session, students will be able to explain waveforms for full-wave control rectifier with RL load and second outcome is, students will be able to derive expression for load voltage for the above converter. Contents of the session are, circuit diagram and waveforms for the converter, circuit explanation and derivation of average load voltage. Figure shows, figure A shows the circuit diagram and the figure B is the equivalent circuit. Now, if we look at figure A, that is circuit diagram, the input is applied through a center cap transformer, primary is fed with 230 volt AC, this is secondary is center tap, point A, N is center tap and this is another is B. Now, T1 and T2 are the two SCRs connected as shown in circuit, cathode of two SCRs connected together and from that point, L load is connected, RL load is connected between this point and the center tap of the secondary of center tap transformer as shown in circuit. And in the equivalent circuit, VAN and VBN are shown as two sources, this is thyristor SCR T1 and SCR T2 and this is load. Now, circuit operation, when VAN is positive, SCR T1 is forward bias and T2 is reverse bias, because this is the center tap secondary, VAN and VBN are 180 degree out of phase, that is specialty of center tap transformer. Therefore, in positive half cycle of input, SCR T1 can conduct and SCR T2 is reverse bias. Similarly, in negative half cycle of input, VAN is negative. Therefore, SCR T1 will be reverse bias and T2 will be forward bias, which can conduct if firing pulse is applied. Now, let us understand the circuit operation in detail with the help of waveforms. Now, this figure shows the waveforms for input, this is the input, sign of VAN, this is VBN. So, VAN and VBN are 180 degree out of phase, as shown in figure. This is the waveform for load voltage for continuous current mode. So, this is waveform for load current, this is waveform for load current for discontinuous current mode and this is load voltage waveform for discontinuous current mode. Now, here question arises in mind is that, what is a continuous and discontinuous current mode? Now, for RL load, when inductance is sufficiently large, so that load current is continuous, means load current is always greater than holding current of the SCR. Therefore, SCR pair, SCR will continue to conduct till another SCR is fired. Now, this is continuous current mode and for discontinuous current mode, the load inductance is not sufficiently large. So, as to maintain the continuous load current, that is for some part of the output, the load current is zero. So, in positive half cycle of input, SCR T1 is triggered at firing angle alpha. So, it will conduct from this point and the output is part of input voltage neglecting potential drop across SCR. At zero cross-off line, we think the load current will become zero, but due to highly inductive load, it will not become zero and SCR 1 will continue to conduct in the negative half cycle of input. So, input voltage reverses polarity after this zero cross-off, but load current is still positive. Therefore, load voltage will be the negative voltage and SCR 1 will continue to conduct till SCR 2 is triggered by application of positive gate pulse at omega t is equal to pi plus alpha. So, at omega t is equal to pi plus alpha, SCR T2 will conduct, which will apply reverse voltage across conducting SCR T1 and turn it off immediately by natural commutation. Then, SCR 2 will continue to conduct till SCR 1 is triggered again at omega t is equal to 2 pi plus alpha. Load current is continuous. So, this is the input voltage VAN VBN and this is the waveform for the load voltage for continuous current mode. Now, circuit operation is this is the circuit operation for continuous and discontinuous current mode, which we have explained earlier. Now, pause this video and think on how to obtain average value of sinusoidal waveform given by Pheom sine of omega t. Now, answer to this question. Let us go back to the waveform of the input voltage. Now, let us consider Pheom, which is given by Pheom sine of omega t sinusoidal waveform and now we are interested in obtaining the average whole average of this waveform. So, for the sinusoidal waveform, the period is 2 pi. So, obtain the area under the curve for the period. So, for positive half cycle, this is the area and for negative half cycle, this is the area. Now, area under the curve is obtained by integration means we obtain integration of Pheom sine of omega t d omega t from 0 to 2 pi. Now, if sinusoidal waveform is ideal, then area under the curve for positive half cycle and negative half cycle are equal. Therefore, average that the sum of these two areas will come 0. Now, this is divided by the time period of the waveform, which is 2 pi. So, in this way, the average of the sinusoidal waveform can be obtained. Now, same method is applied for obtaining the average voltage of any periodic waveform. Now, this is load voltage waveform for continuous current mode. Now, if we observe this load voltage waveform, the period this is alpha 2 pi plus alpha, then pi plus alpha to 2 pi plus alpha means this part is repeated here. Therefore, the period of output is pi minus alpha minus alpha that is pi. Now, let us make the derivation for average load voltage for continuous current mode. Now, average load voltage for continuous current mode is given by Pheo dc is equal to integration Pheom sine of omega t d omega t and the limits of integration are from alpha to pi plus alpha and this is to be divided by the period which is equal to pi that is 1 upon pi. This is equal to Pheom is constant taken outside Pheom by pi integral alpha to pi plus alpha sine of omega t d omega t. This is equal to Pheom by pi integral of sine omega t is minus cos of omega t limits alpha pi plus alpha. Substituting these limits, we get minus cos of put upper limit pi plus alpha minus minus become plus cos of alpha that is equal to Pheom by pi in the bracket minus minus cos alpha plus alpha cos alpha. This will become equal to 2 Pheom by pi cos of alpha means average load voltage for continuous current mode is given by 2 Pheom by pi into cos of alpha. Now, these are the references power electronics by M. D. Singh and K. B. Khan Chandani McGrahill publication and second is industrial and power electronics by Dev Dutta Shingare Electrotech publications. So, these are references. Thank you for watching this video.