Chapter 3 – Diode Circuits – Part 2 Rectifier Circuits 1. Half Wave Rectifier 2. Full Wave Rectifier 3. Bridge Rectifier

Power Supply Many electronic applications such as radio sets, toys, walkmans, MP3, etc. require a d.c. power supply (usually 6V or 3V). One way to achieve it is through the use of dry cells in series. But an economically more convenient solution would be the use of the a.c. supply to generate the desired d.c. output. Power supplies are used to achieve precisely this result.

Given: A 50 Hz (time period = 20 ms), 230 V r.m.s (i.e. 230 x 2 Vpeak) a.c. supply Our Objective : To design a circuit which would take this as input and give as output a constant d.c. voltage, say 6 V.

Steps Involved Reduce the amplitude of the input to say 6 V through the use of a transformer. Coverts ac 6 V to dc 6 V (?)

Half-Wave Rectifier The voltage equation around the loop acdba is Nonlinear device Linear circuit element The voltage equation around the loop acdba is v - vd - i RL= 0 ------- (1) Or the current i in the circuit is i = v / RL – 1 / RL vd ----- (2) Equation (2) is a straight line with slope -1/RL and intercept at i = 0, vd = v.

Load line Diode current, i = 0 Then from eq. (1), we get v = vd (ii) Diode voltage, vd = 0. We have i = v / RL (iii) Join these two points. This line is called the load line which intersects the diode device characteristic at point Q. (iv) The slope of the load line is – 1/RL. (v) At point Q, the parameters of the nonlinear device (diode) and the parameters of the linear circuit (RL) match.

By plotting the load line and current-voltage characteristic together, the voltage-time input signal is transferred to corresponding current-time out put signal. The output-current waveform is half sinusoids. The frequency of output waveform is equal to the frequency of the input voltage. (iii) For most practical applications, the applied voltage, v is much larger than the forward voltage drop across the diode so that the second term in equation (2) is neglected. (iv) In this situation, the diode acts as an ideal rectifier.

Disadvantages of Half-wave rectifier: Full-Wave Rectifier Disadvantages of Half-wave rectifier: (a) In the Half wave rectification, we got an output only in on of the half cycles. (b) The rectifier is inactive during one-half of the input cycle. (c) Therefore, it is less efficient. To Overcome this situation: The half-wave circuit is modified to full-wave rectifier. In the full wave, we get it for both the half cycles. This is achieved by the use of two diodes instead of one as now, one of the two diodes remains in conduction in both of the half cycles.

 2   

Requirements & Operation: we require a center tapped transformer to give us two  shifted sinusoids so that exactly one of the waveforms is positive at one time. Diode D1 conducts on one half-cycle while diode D2 is reverse biased. On the alternate half-cycle, the conditions are interchanged. Out-put current waveform both diodes has only momentary zero values. In the full wave rectification, the negative part of the input waveform changes to the positive. So, you have 'two positive humps for each input cycle. Each hump is +ive and in the time taken to give one complete waveform you now have two (doubled). The frequency of output waveform is twice the supply frequency. However, the output is not the desired pure d.c., but better than the half-wave rectifier.

Bridge Rectifier Bridge rectifier is a full-wave rectifier without a ceter-tapped transformer.

Principle of operation During the half-cycle, suppose upper terminal of the transformer is positive. - + The diodes D2 and D3 conduct and current is present in the load resistor RL.

2. On the alternate half-cycle, the lower terminal of the transformer is positive. + - The diodes D4 and D1 conduct and current is present in the load resistor RL. The current direction in the load resistance is the same as before.

3. The current direction in the load resistance is the same as before. 4. The voltage across RL corresponds to full-wave rectification. 5. Peak output voltage is equal to the peak transformer voltage minus the potential drops across the diodes. 6. Since two diodes are in series with the load, the output voltage is reduced by twice the drop. 7. The un-rectified waveform is a 60 Hz sine wave. In the full wave rectification provided by the bridge, the negative part of the input waveform changes to the positive. So, you have 'two positive humps for each input cycle. Each hump is +ive and in the time taken to give one complete waveform you now have two (doubled).