 Hello everyone, myself, Sanjay Utke, Assistant Professor, Department of Electronics Engineering, Valchand Institute of Technology, Solapur. Today, we are going to discuss Full Wave Rectifier. Learning outcome, at the end of this session, students will be able to analyze full wave rectifier circuit, efficiency, ripple factor, and form factor. Guidelines, full wave rectifier circuit, derivation for efficiency, derivation of ripple factor, derivation of form factor, assignment, references. In full wave rectifier circuit, four diodes are now used, two for each half of the cycle. A transformer is used, whose secondary winding is connected to bridge network. This configuration results in two diodes conducting in turn, when their anode terminals are positive, producing an output during both half cycles. Twice that for the half wave rectifier, so it is 100 percent efficient. Full wave rectifier using center type transformer. Here in bridge rectifier, we used four diodes D1, D2, D3, D4 in the Vistone's bridge manner, but using a center type, it requires only two diodes D1 and D2. The only modification over here is, we are using a center type transformer. That means, we are having two windings at the secondary, rather two equal windings at the secondary, the total secondary voltage is equally divided across two secondary winding. In the positive half cycle, primary top primary is positive, so point A will be at positive potential, then C negative, again lower part of C will be positive and B will be negative potential. So, in the positive half cycle, A being at positive potential, diode D1 will conduct, so the current will flow through diode D1, then lower resistance RL and back to the point C. So, in this way we got this positive half cycle. In the negative half cycle of the AC input, the polarities at ACB will get changed, so here B will be at positive potential, this lower C will be at negative, again C positive and A will be negative. So, in summary B at positive potential, A at negative potential, so diode D1 whose anode is at negative potential will be in reverse bias condition, whereas anode of D2 which is connected to point B, now which is at positive potential, so now it will be in the forward bias condition, so in negative half cycle D2 forward bias current will flow from diode D2 and then low resistance RL. Look at the direction for the positive and the negative half cycle of the AC input, the direction at the secondary that is through D1 and D2 or direction of the current in the load resistance is same for both half cycle of the AC input, so in this way we got the train of positive half cycles at the load resistance RL. This circuit consists of two power diodes connected to a single load resistance RL with each diode taking it in turn to supply current to the load, when point A of the transformer is positive with respect to point C diode D1 conducts in the forward direction as indicated by the arrow, when point B is positive in the negative half cycle of the cycle with respect to point C diode D2 conduct in the forward direction and the current flowing through the resistor R is in the same direction for both half cycles. As the output voltage across the resistor R is the phasor sum of the two waveforms obtained. Derivation of efficiency, the ratio of DC power obtained at the output to the applied input AC power is known as the rectifier efficiency, let RF and RL be the forward resistance and load resistance of the diode respectively, V is equal to Vm sin theta with the voltage appearing across the secondary of the power transformer where small v is the instantaneous value capital Vm is the maximum value. Efficiency derivation, during the conduction period it is instantaneous value is given by the equation I is equal to V upon RF plus RL or V is equal to I into RF plus RL. As we know V is equal to Vm sin theta I is equal to Vm sin theta upon RF plus RL, when sin theta is equal to 1 current is maximum, so I m is equal to Vm upon RF plus RL where I is equal to I m sin theta, since the output is obtained across load resistance RL, therefore DC power output is equal to I square DC RL is equal to I square average into RL. Let us derive the equation for the average value for full wear rectifier, I average it is again the integration of the half wave, so I average mathematical can be expressed as integration of I into d theta upon pi equation number 1, integrating equation 1 from 0 to pi, I average is equal to 1 upon pi integration of I m sin theta d theta is equal to I m upon pi integration of sin theta d theta is equal to I m upon pi into bracket minus cos of theta that is the integration of sin theta, which equals to I m upon pi integrate minus of minus 1 minus 1 that is 2 I m upon pi. So, this is what we can find out DC power is equals to I m upon pi bracket square into RL, now to find out the AC power we must find out the RMS value, RMS value I RMS can be given as square root of integration of I square d theta upon 2 pi, integrating above equation from 0 to 2 pi is equals to I RMS is equals to square root of 1 upon 2 pi integration of I square m sin square theta d theta. So, instead of sin square theta now we will use that 1 minus cos 2 theta upon 2, by solving this integration we got finally, it is the I RMS is equals to square root of I square m upon 4 pi into 2 pi is equals to I m upon root 2 that is 0.707 I m. Now, let us come to the derivation of efficiency, which is P dc upon P ac can be given as P dc in terms of average value that is 2 I m upon pi bracket square into RL divided of P ac in terms of RMS value I m upon root 2 bracket square into RL plus RL comes out to be 0.812 RL upon RL plus RL. Since RL is very greater than RL, so RL upon RL plus RL is equals to 1, we got the finally, efficiency as 81.2. So, that indicates the full way rectifier can convert maximum 81.2 percent of the ac part into dc, which is the choice of the half way rectifier. Question, what is the difference between the bridge full way rectifier and the center tap? In bridge rectifier full diodes are used, two diodes are conducting simultaneously in one half cycle remaining two conducting in another half cycle. In center tap, two diodes are used, each will conduct in one half cycle locating a center tap on transformers is difficult. Advantages, efficiency is double, there is reduction in the ac ripples at the output. Applications, it is used for the construction of constant DC voltage power supplies especially in general power supplies. It is ideal for any type of general power supply applications like charging a battery, powering a DC device like motor, LED, higher for an audio application or general power supply may not be enough. This is because of the residual ripple factor in a bridge rectifier. Ripple factor is the ratio of residual AC input component to the DC components and it is 0.482. Form factor, again it is ratio of IRMS upon I average comes out to be 1.11. This is the acknowledgement.