 Assistant Professor, D. R. G. Solapur, Department of Electronics Engineering. Today in this video, we are going to see discussion over L-section filter. In this video, we are going to understand the working of LC filter, then we will analyze the performance of LC filter, then the concept of bleeder resistance. Now, we will start here, we already know about this the filter that uses capacitance and we know here the ripple factor of that filter is given by 1 upon 4 root 3 fc into RL, which confirm that the value of ripple is inversely proportional to the value of load. And that is why we say that this filter is recommended for those circuit in which the value of RL is very high that is it is nothing but called as a light load. Secondly, we also can use inductor to provide the better ripple that is we are going to define the ripple factor for inductor filter that is the ratio of RL upon 3 root 2 omega L. That is in this case ripple is directly proportional to the value of load. So, we know here the value of ripple is very low when the value of RL is very small. So, we say that this inductor filter is suitable for a heavy load. Now, here we are going to combine the features of L filter and the C filter hoping that the ripple value will be independent of the load variable. Now, in this circuit the inductor is connected in the series with the load as we know inductor offers a very high impedance to the flow of AC current that is why it is connected in the series of the load. And whereas the capacitance is connected across the load that is in the parallel with the load because we know the capacitance is providing me a very low reactance path to the flow of AC current. And that is why when we combine the features of inductor and the capacitor we are going to get a better ripple reduction. Now, this is a circuit that uses a full rectifier circuit along with this L and C component. Friends just see closely the inductor is in the series with the load whereas C is across the load. So, the output voltage of the rectifier circuit without any load is defined by the Fourier series. So, in that we are having number of components there you see that the 2 Vm by pi component it has got no frequency component. So, it is normally called the DC component whereas the terms in the bracket carries frequency terms that is in terms of omega. So, these are basic terms. Now, we know here normally in the output the harmonics above second harmonics are not going to contribute much in the output and they are also seen that they are going to decay very rapidly. So, usually the output of this rectifier circuit is defined by the equation we limited only up to second harmonic. Now, we can modify this DC term here that is suppose if I put a load in the circuit because I say that VDC is 2 Vm by pi under no load condition. So, when I put a load in the circuit the DC output voltage available across load will be equal to the 2 Vm minus IDC into r where r is nothing but the sum of rf then the r choke and the rs. But usually the values of these all components that is they define this value of r they are quite small and practically we normally negate them. Whereas, we also can define the AC component from the equation 1 that defines the AC voltage across the load here also can be called as a ripple. So, V ripple is defined by this value 4 Vm by pi 3 pi is magnitude and cos 2 omega d is the frequency component. Now, we want to define the ripple current in the circuit. So, I will define that is 4 Vm by 3 pi divided by the value of impedance. The impedance is not defined by this Xl and Xc and Rl or we can define this z equal to Xl plus Xc in parallel with this Rl. Normally the value of Xl is very high as complete with the parallel combination of this Xc and Rl and therefore, the ripple current available in the load is given as 4 Vm upon 3 pi into Xl. Here, Xl is the reactance offered by the inductance is given by this. Now, after that we are going to define the ripple factor. Now, we will express this ripple value of a current in the form of Rms here and we know here the ripple value its Rms form is given by I ripple upon root 2. Hence, we can redefine the above equation as root 2 by 3 into Xl into Vdc. Now, we know here this ripple current is now flowing through the parallel combination of Xc and Rl. Now, we know here for the given frequency the value of Xc is very small as compared with the value of Rl and therefore, the AC current will prefer the path to this Xc. Therefore, this ripple current is going to provide AC voltage only across the capacitor. That is we rearrange all the equations that we written above. But, we know here the value of ripple is defined as the ratio of Vr Rms and Vdc. So, we can rearrange all these terms here. So, we can define this ripple as Vr Rms upon Vdc which is same as root 2 into Xc upon 3 into Xl. Now, we can provide the values of Xc that is 1 upon 2 j 2 omega c and the value of this Xl is nothing but 2 omega l. When we replace the value of these two factors in the equation, the value of ripple factor is coming equal to 1 over 6 root 2 omega square l into c. Now, when we look this formula of this ripple factor closely, you will see that there is no any load value is available. That is the equation is independent of the value of load and that is why we can say that this type of filter is very good in the case of a load which is not the constant that is a variable. So, it is a solution for that power supply in which we are going to get variable loads there. Now, friends we got one more difficult is experienced here in the circuit. That is we know here whenever a current is flowing in the inductance it is suppose interrupted here we found that we generate some emf across the inductance here. Now, we know here the emf we generate across inductor is given as Ldi by dt which confirm that in the case of direct current means a constant current here there is no change in the current which means there is no di by dt that becomes 0 here. So, the value of emf across the inductor is always 0. But we have to confirm that this value always remains 0 otherwise suppose current is get interrupted so that we can create some voltage across inductor and that is get applied as a black emf across the diode and the capacitor. Now, the value of this emf may be so large that it can exceed the value of PIV of a diode that can burn the diode here. Even this voltage will be above the voltage rating of the capacitor that can damage the capacitor. So, we have to carefully design the value of resistance or some value of the component here that confirm that they will be always a flow of current in the inductor and there is no interruption in the current here. So, we recommend one resistance that is called bleeder resistor here which is connected across the load and normally we choose the value of RB very high as compared with the value of load. Friends, this RB is connected across the load here that is shown in the given diagram. The value of RB is very high as compared with the load. Now, how we can design the value of RB? Let us concentrate on this waveform here. Now, dotted line shows me the DC current flowing in the choke and whereas this sine wave shows me the AC component coming across the choke here. Now, actually in the current example here, the value of R is such that the negative peak of this AC current which is flowing in the choke here is exactly touching to the 0 here. So, still I can say that the current is flowing here continuously in the given load. But suppose if I decide if I change the value of this IDC below this given value here, then we can see that the variation in the current will be stopped here because that goes below 0 and that provides me some back MF here and that can damage the choke or the capacitor. Let us try to analyze this happening here. Now, to ensure that this IDC is always at the given level at least here and we can confirm that the current will not be interrupted at all here. So, this IDC should be always more than IP to peak upon 2. So, when I replace the values of this IDC and this IP peak by 2 and by solving this equation, we will see that the value of this RB is always coming more than 943 times L or practically we can use the value of RB is equal to 900 times L. Now, these are my references friend. I hope that you understand the working of this LC filter. You understand the working of this reader. I hope that this video will help you to understand the L section filter. Thank you friends. Thank you.