 Myself Giridhar Jain, Assistant Professor in Electronics and Telecommunication Engineering, Valchin Institute of Technology, Sholapur. Today, I am going to explain single-phase fully controlled bridge rectifier. Learning outcomes of the session are, at the end of the session, students will be able to explain circuit and waveforms for fully controlled bridge rectifier with R and RL load, and second outcome is, at the end of the session, students will be able to derive expression for load voltage for above converter. Contents of the session are circuit diagram and waveforms for above converter, second is circuit explanation and derivation for average load voltage. This is circuit diagram for fully controlled bridge rectifier. Input is fed as shown in circuit. T1, T2, T3 and T4 are the four SCRs. Cathode of the upper two SCRs will act as a positive of the load voltage and the anode of the lower two SCRs, negative. And this is the load. RL load is connected as shown in circuit and this is output voltage view. This is input voltage. This is IG1 and IG2 means SCR1 and 2 are forward bias in positive half cycle of input. Therefore, they are gated together. These are the gate pulses for SCR1 and 2 and this is gate pulses for SCR3 and 4. 3 and 4 are forward bias in negative half cycle of input. This is load voltage waveform for the continuous current mode. Means if load is highly inductive, then at omega t is equal to alpha, positive gate pulses applied to the gate of SCR T1 and T2. T3 and T4 are reverse bias. Therefore, they will not conduct. SCR1 and 2 will conduct. Therefore, load voltage is equal to the part of input voltage neglecting voltage drop across SCR. After zero cross-off line, input voltage reverses the polarity. But due to inductive nature of load, the load current is still positive and greater than holding current. Therefore, SCR1 and 2 will continue to conduct till SCR3 and 4 are fire at omega t is equal to pi plus alpha. At omega t is equal to pi plus alpha, positive pulse is applied at the gate of SCR3 and 4. Therefore, SCR3 and 4 will conduct. As soon as SCR3 and 4 will conduct, they will apply reverse voltage across conducting SCR1 and 2. Therefore, SCR1 and 2 will turn off by natural commutation. Then SCR3 and 4 will continue to conduct till SCR1 and 2 are triggered again at 2 pi plus alpha as shown in figure. So, this is waveform for the load voltage for continuous current mode and this is waveform for load current. Pause this video and think on the following question. What is the frequency of ripple? And the second is what is the time period of output? Now to answer this question, let us come back to the waveform. This is the waveform for load voltage. Now frequency of ripple means the frequency of output. Now this is one cycle of input. So, time period of input is 2 pi and here this cycle, this part alpha 2 pi plus alpha is repeated again from pi plus alpha to 2 pi plus alpha. That means period of the output is pi. Means within one cycle of input, 2 cycles of the output are completed. Therefore, frequency of output that is frequency of ripple is twice the frequency of input. If input is 230 whole 50 hertz means frequency of input is 50 hertz, then frequency of ripple that is frequency of output is 100 hertz. That is the answer to the first question. And second is what is the time period of output? So, time period of output is pi. Now let us make derivation for the average load voltage for a continuous current mode. So, for a continuous current mode, if we go back to the waveform, so in the waveform output is present from alpha to pi plus alpha. And this is the part of input that is Vm sin of omega t. So, these limits of integration are taken in the derivation. So, Vodc is equal to integration Vm sin of omega t d omega t and the limits of integration are taken from alpha to pi plus alpha and the period is pi. So, divide this by period pi, so 1 upon pi. This is equal to Vm is constant taken outside Vm by pi integral alpha to pi plus alpha sin of omega t d omega t. This is equal to Vm upon pi integral of sin is minus cos of omega t alpha to pi plus alpha. This is equal to Vm by pi putting the limits minus cos of pi plus alpha plus cos of alpha. And after simplifying, we get Vodc is equal to 2 Vm by pi into cos alpha. If we compare this result with the earlier result that is the output of full wave control rectifier using center tap transformer, the average load voltage for continuous current mode is same that is 2 Vm by pi into cos alpha. So, what is the difference between this converter and earlier converter that is full wave converter using center tap transformer? So, basic difference is this control rectifier, bridge control rectifier does not require center tap transformer while as earlier will require center tap transformer. And second is PIV that is peak inverse voltage. So, PIV of the earlier circuit using center tap transformer is 2 Vm and PIV peak inverse voltage for the circuit is Vm. So, this is the difference between these two. Now, we have made the derivation for average load voltage for continuous current mode. And now we are going to make the derivation for average load voltage for resistive load. Now, for resistive load if we go back to the waveforms. So, these are the waveforms for continuous current mode. For resistive load L is equal to 0. So, inductance is 0 only resistance is there. So, as far as these waveforms are concerned in positive half cycle of input, SCR 1 and 2 are triggered at omega t is equal to alpha. They will continue to conduct till 0 cross-off line. At 0 cross-off line load current will become 0 due to resistive nature of load. Therefore, SCR 1 and 2 will turn off by natural commutation at omega t is equal to pi. And from pi to alpha plus pi load voltage will be 0. Then SCR t3 and t4 are triggered. They will continue to conduct from this point up to omega t is equal to 2 pi. And then 2 pi to 2 pi plus alpha load voltage will be 0. So, this is the change in the output voltage waveform. Therefore, limits of integration are from alpha to pi. Previously it is alpha to pi plus alpha. And the time period of the output is again pi. So, this is the derivation for average load voltage for a resistive load. So, for a resistive load VODC is given by integration alpha to pi Vm sin of omega t d omega t. And divided by the time period of the output that is pi. So, it is 1 upon pi. Vm is taken outside Vm by pi. Then integration of sin of omega t is minus cos of omega t. So, Vm by pi minus cos of omega t alpha to pi. Substituting the limits we get this is equal to Vm by pi cos of pi minus cos of pi minus minus plus cos of alpha. Cos of pi is equal to minus 1. So, minus 1 and this minus will become plus 1. And this is cos alpha as it is. Therefore, average load voltage is given by Vm upon pi in the bracket 1 plus cos alpha. So, this is the expression for average load voltage for a resistive load. These are references. Power Electronics by M.D. Singh and K.B. Khan-Chandani, McGrile publication. And this is one web link. Thank you for watching this video.