 Assistant Professor, Department of Electronics Engineering, Walsh and Institute of Technology Solapur. Today, I am going to explain the filters from electronic circuit analysis and design 1. Learning outcome. At the end of this session, students can analyze and design filter circuits. Filters. A circuit which is used to remove the AC components from the output. A filter is a circuit which transmits or blocks some desired frequency range. A filter is an electronic circuit composed of C, L or both and connected in between rectifier and load so as to convert the pulsating DC to the pure DC. The circuit is used to obtain the ripple free DC voltage. Filters of filter circuit, capacitor filter, inductor filter, LC filter and CLC filter. CLC filter is also called as a pi filter. These are all are the passive filters because here the passive components are LC are used that is why these filters are called as a passive filters. Capacitor filter. This is the circuit diagram of capacitor filter. Here the main filtering component is the capacitor. It is connected in between the rectifier and load. Why? In filter circuits, capacitors are connected in C in parallel and inductors are in series. As we know, capacitor reactance is equal to 1 upon 2 pi fc. For AC signal, the frequency is large. So capacity reactance is small means it indicates that the capacitor allows the AC for DC, f is minimum, capacity reactance is maximum means it indicates that the capacitors blocks the DC. Here in this filter circuit, capacitor is connected in parallel means it is connected in between the rectifier and load. The output of rectifier is the pulsating DC and here capacitor is connected across the rectifier. The input of capacitor is the DC. That DC is effectively blocked due to this capacitor and if there are any ripples that ripples are passed towards the ground. Suppose if we connect the capacitor in series, here the input of the capacitor is pulsating DC that totally that totally DC is blocked because of this capacitor. So the circuitry becomes open circuit and output current is 0 that is why always capacitor is connected in parallel and why inductors are connected in series. As we know inductive reactance is equal to excel is equal to 2 pi fl. For AC f is large, inductive reactance is large means it indicates that the inductor opposes strongly for the AC. For DC the frequency is minimum, inductive reactance is minimum means the inductors allows the DC. Generally in the filter circuit inductors are connected in series because the input of inductor is DC that DC is passed towards the output. Suppose if we connect the inductor in parallel what happens? The output of rectifier is the pulsating DC and if we connect the inductor in parallel, this inductor actually short for the DC. So effect of this the load gets shorted. So it is not possible to connect the inductor in parallel. So always in the filter circuit inductors are connected in series these are the waveforms. The first waveform represents the input signal, the second one is the output of rectifier and output of the filter circuit and third one represents the ripples are in the form of triangular wave. This is the ripple voltage waveforms. Let us see how it works. When we apply the AC for this circuit, so during positive half cycle the current flows through the upper half winding secondary diode D1 capacitor. During positive half cycle diode D1 is forward biased so capacitor C short charges and here the capacitor C charges up to its peak input value. So immediately capacitor C gets charged up to its peak input value. Once the capacitor C gets charged up to its peak input value further the diode D becomes reverse biased while the cathode potential becomes greater than anode potential and capacitor C starts discharges through the load resistance and it discharges until in the next half cycle when the diode D2 becomes forward biased. In the next half cycle when the anode potential of D2 crosses the capacitor voltage diode D2 conducts and the current flows through the lower half winding and the current path is the secondary D2 capacitor and back to center tap and again capacitor C charges and immediately capacitor C gets charged completely up to its peak input value. So effect of this the ripples are in the form of triangular wave. Here in this waveform the D by 2 represents the half time period. D1 represents the charging time of the capacitor and in this duration diode D conducts and D2 represents the discharging time of the capacitor and Vr peak to peak is the peak to peak value of ripple voltage and Vm is the peak value of the output voltage expression for ripple factor. Consider output waveform here T is the time period of AC input voltage. D by 2 is the half time period T1 conduction period of diode and charging time of capacitor. T2 non conduction period of diode and discharging time for capacitor. Here discharging time constant is large therefore T2 is greater than greater than T1. Expression for Vr. Let Vr be the peak to peak value of ripple voltage which is assumed to be triangular as shown. During time interval T2 capacitor C discharges through the RL. Therefore charge lost is equal by Q is equal to C into Vr. Vr is the peak to peak value of ripple voltage but I is equal to DQ by DT rate of change of charge. Therefore Q is equal to integration of I into DT and here the limits of integrations are 0 to T2 because capacitor C discharges in this dilation T2 therefore Q is equal to I DC into T2. Here in this time period T2 capacitor C discharges through the load resistance therefore the current is I is equal to I DC into I DC therefore Q is equal to I DC into T2 for I DC into T2 is equal to C into Vr but T1 plus T2 is equal to T by 2 is equal to 1 upon 2f but as we know the discharging time is large therefore T2 is greater than greater than T1 therefore simply neglect the T1 therefore Vr is equal to I DC upon 2fc or Vr is equal to Vdc upon 2fcr just substitute the value of I dc is equal to Vdc upon r. For this filter the ripples are in the form of triangular wave. Let us use the standard formula which states the rms value of ripple voltage Vrms is equal to Vr upon 2 root 3 and according to our previous equation Vr is equal to Vdc upon 2fcr just substitute that value here therefore Vrms becomes Vdc upon 4 root 3fcr expression for ripple factor. So, what is the ripple factor? The amount of ripples present in the output is expressed mathematically by the factor that factor is called as a ripple factor. So, ripple factor is equal to Vrms upon Vdc. According to our previous equation as we know just substitute the value of Vrms therefore ripple factor is equal to 1 upon 4 root 3fcr. Here the ripple factor is depends upon the c and r. The ripple factor is reduces by increasing the value of c and increasing the value of r expression for DC output voltage. Just consider the waveforms here. In this waveform Vm is the peak value and Vr to peak is the peak value of ripple voltage therefore Vdc is equal to Vm minus Vr peak to peak divided by 2. Therefore Vdc is equal to Vm minus Vr by 2. Just substitute the value of Vr as we know Vr is equal to idc upon 2fc therefore equation becomes Vdc is equal to Vm minus idc upon 4fc. Davo equation indicates that for the no load condition Vdc is equal to Vm because for no load condition idc is equal to 0 therefore Vdc is equal to Vm and for the full load condition the finite value of current flows through the circuitry means idc is maximum for the full load therefore Vdc is less than Vm means for no load condition Vdc is equal to Vm and for full load condition the Vdc is less than Vm means as load increases Vdc reduces. Why capacitor filter is applicable for the light load condition? As we know the ripple factor of capacitor filter is equal to 1 upon 4 root 3fcr. The ripple factor is inversely proportional to load resistance. For the light load the current is small the resistance is large so effect of this the ripple factor of capacitor filter is minimum and the ripple factor is minimum means the performance is the best. Suppose if load is heavy for the heavy load the current is large the resistance is minimum and the ripple factor is encourages means the amount of ripples present in the outputs are more means it affects the performance. So capacitor filter is applicable only for the light load condition these are the references. Thank you.