 Hello everyone, myself Sanjay Utke, assistant professor, department of electronics engineering, Valchand Institute of Technology, Solapur. Today, we are going to discuss halfway rectifier, learning outcome. At the end of this session, students will be able to analyze halfway rectifier circuit, efficiency, ripple factor and form factor guidelines, introduction, halfway rectifier circuit, derivation for efficiency, derivation of ripple factor, derivation of form factor, assignment, references, introduction. What is rectification? The most popular application of the diode is rectification, conversion of alternating current to direct current. This involves a device that only allows one way flow of electric charge. This is exactly what a semiconductor diode does. The simplest kind of rectifier circuit is the halfway rectifier. It only allows one half of an AC waveform to pass through the load. This is the circuit diagram for halfway rectifier having on the left hand side AC input, then a symbol of a transformer. AC input is connected to the primary whereas secondary is given to a series network of diode and a load resistance RL. Below is the waveforms. In the blue color it is the AC input waveform consisting of train of positive and negative half cycles. When the secondary, top secondary of the transformer is at positive potential, diode will be forward biased since it acknowledges positive with respect to cathode and the current will flow through the diode and through the load resistance RL and we are getting the output that is V output at the output side. So, we are getting a positive half cycle corresponding to the AC input positive half cycle. In negative half cycle of the AC input, the diode the upper part of this secondary will be at negative potential, the diode will be reverse biased. So, it will be in reverse biased condition, it will act as an open switch, no current will flow. Hence, no output voltage is available across the load resistance RL. So, output will be 0. Hence, at the secondary of the halfway rectifier, the negative half cycle is completely suppressed. So, we are getting a straight line 0 volts. So, this is how we are getting a train of positive half cycles by suppressing negative half cycles. A type of rectifier that derivation of efficiency, the ratio of DC power obtained at the output of 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. V is equal to Vm sin theta be the voltage appearing across the secondary of the power transformer. V is equal to Vm sin theta is the equation of a sinusoidal quantity alternating quantity, where capital Vm is the maximum amplitude and small v is the instantaneous value. This is let us consider a waveform same circuit diagram with a waveform, which is shown at the bottom waveform at the output side by considering an elemental strip of width dq with height i. Let us start for deriving the average and the RMS values. Efficiency during the conduction period, it is instantaneous value is given by the equation i is equals to V upon RF plus RL. V is equals to i into RF plus RL. As we know V is equal to Vm sin theta. So, i will be equals to Vm sin theta divided by RF plus RL. When sin theta is equals to 1, the current will be maximum. Therefore, i m will be equals to Vm divided by RF plus RL, where i is equal to i m sin theta. Since the output is obtained across RL therefore, dc power output is equals to i square dc RL or it is equals to i square average into RL. Now, let us determine the value of average of the instantaneous value for halfway rectifier. Average value can be given by integration of i into d theta divided by 2 pi. See average is what it is the area under the curve. That is why we are going for the integration. Integrate equation 1 from 0 to pi by putting the limits that is from 0 to pi because in this the full cycle is of base from 0 to 2 pi, but we do not have area between pi to 2 pi. So, we are integrating this equation number 1 by 0 to pi. So, i average will be equals to 1 upon 2 pi into integration of i m sin theta d theta. i average is equals to i m upon 2 pi into integration of sin theta d theta is equals to i m upon 2 pi into bracket minus cos theta. So, integration of sin theta d theta will be minus cos theta. By putting the limits from 0 to pi we got the equation i average equals to i m upon 2 pi into bracket minus of minus 1 minus 1 is equal to 2 i m upon 2 pi comes out to be i m upon pi. So, this is what we got the average value I can find out the dc power. Now, let us coming to the rms value. What is rms? It is a square root of mean of squares. So, rms value can be found out by the equation square root of integration of i square d theta upon 2 pi equation number 2 and integrating this equation from the limits 0 to pi. rms value i rms is equal to square root of 1 upon 2 pi integration of i square m sin square theta d theta is equals to square root of i square m upon 2 pi integration of 1 minus cos 2 theta upon 2 into d theta. So, by putting the limits and solving this integration we got the average rms value equals to i m upon 2. Now, the efficiency can be found out as average rms power is equal to i square rms into r plus rl rectifier efficiency is p dc upon p ac. So, efficiency is equals to what is p dc? It is i m upon pi bracket square into rl divided by i m upon 2 bracket square into rf plus rl comes out to be 0.406 rl upon rf plus rl. Since, load resistance is very very greater than the forward resistance of the diode. So, rl upon rf plus rl will be equals to 1 hence the efficiency comes out to be 40.6 percent conclusion for efficiency this indicates that the half way rectifier can convert maximum 40.6 percent of ac power into dc power and the remaining power 59.4 percent is lost in the rectifier circuit hence the efficiency 40.6. In fact, 50 percent power is in the negative half cycle is not connoted and the remaining 9.40 percent loss in the circuit. What is conducting state of the diode in negative half cycle? In negative half cycle the id is reverse bias. So, it will act as open switch it will not conduct. Applications half way rectifier is not so good as compared to full wave or bridge, but sometimes we request these rectifier depends on the requirements. It is used for detection of amplitude modulated radio signals. For the welding purpose it supplies polarized voltage it is used in many signal demodulation process. Ripple factor equals to 1.21 is the ratio of rms value fluctuating ac components to the average or dc value. The ripple this is the formula for a ripple factor comes out to be 1.21. Form factor is the ratio of rms value upon average value comes out to be 1.57. Acknowledgement