1. PREPRAIRED BY :Solanki Kishan(130590111008) SUB NAME :EDC TOPIC : Half And Full Wave Rectifier GUIDED BY : Nikhil Sir

2. * introduction  A rectifier is an electrical device that converts alternating current (AC), which periodically reverses direction, to direct current (DC), which is in only one direction, a process known as rectification.

3. * Types of Rectifiers  Half wave Rectifier Full wave Rectifier Bridge Rectifier

4. Half wave rectifier  In half wave rectification, either the positive or negative half of the AC wave is passed, while the other half is blocked.  Because only one half of the input waveform reaches the output, it is very inefficient if used for power transfer.

5. RECTIFIERS  Half-wave Rectifiers  In a half-wave rectifier, the output waveform occurs after each alternate half-cycle of the input sinusoidal signal.  The half-wave rectifier will generate an output waveform vo. Between the time interval t = 0 to T/2, the polarity of the applied voltage vi is such that it makes the diode forward-biased.  As a result the diode is turned on, i.e., the forward voltage is more than the cut-in voltage of the diode.

6. Conduction region (0 to T/2) RECTIFIERS Non-conducting region (T/2 to T )

7. RECTIFIERS Half-wave rectified signal

10. Example: Half Wave Rectifier Given a half wave rectifier with input primary voltage, V p = 80 sin  t and the transformer turns ratio, N 1 /N 2 = 6. If the diode is ideal diode, (V  = 0V), determine the value of the peak inverse voltage. 1.Get the input of the secondary voltage: 80 / 6 = 13.33 V 1.PIV for half-wave = Peak value of the input voltage = 13.33 V

11. full wave rectifier  A full-wave rectifier converts the whole of the input waveform to one of constant polarity (positive or negative) at its output.  Full-wave rectification converts both polarities of the input waveform to DC (direct current), and is more efficient.0

12. RECTIFIERS  Full-wave Rectifier  The full-wave rectifier can be classified into two distinct types. (i) Centre-tapped transformer full-wave rectifier:- It comprises of two half-wave circuits, connected in such a manner that conduction takes place through one diode during one half of the power cycle and through the other diode during the second half of the cycle. Full-wave rectifier

13. RECTIFIERS (ii) Bridge type full-wave rectifier:- The most important disadvantage of the centre-tapped rectifier is that it brings in the use of a heavy transformer with three terminals at its output, i.e., a centre-tapped transformer. The centre tapping may not be perfect in most cases. This problem can be solved by designing another circuit with four diodes and a simple transformer. This is called a bridge rectifier. Waveform for full-wave rectifier

14. RECTIFIERS Bridge rectifier Advantages of a bridge rectifier (i) In the bridge circuit a transformer without a centre tap is used. (ii) The bridge circuit requires a smaller transformer as compared to a full- wave rectifier giving the identical rectified dc output voltage. (iii) For the same dc output voltage, the PIV rating of a diode in a bridge rectifier is half of that for a full -wave circuit. (iv) The bridge circuit is more appropriate for high-voltage applications, thus, making the circuit compact.

15. RECTIFIERS Disadvantages of a bridge rectifier (i) Two or more diodes are required in case of a bridge rectifier, as a full- wave rectifier uses two diodes whereas a bridge rectifier uses four diodes. (ii) The amount of power dissipated in a bridge circuit is higher as compared to a full-wave rectifier. Hence, the bridge rectifier is not efficient as far as low voltages are concerned. A full-wave capacitor-filtered rectifier

16. A full-wave center-tapped rectifier circuit is shown in Fig. 3.1. Assume that for each diode, the cut-in voltage, V  = 0.6V and the diode forward resistance, r f is 15 . The load resistor, R = 95 . Determine: – peak output voltage, V o across the load, R – Sketch the output voltage, V o and label its peak value. ( sine wave )

17. SOLUTION peak output voltage, V o V s (peak) = 125 / 25 = 5V V  +I D (15) + I D (95) - V s(peak) = 0I D = (5 – 0.6) / 110 = 0.04 AV o (peak) = 95 x 0.04 = 3.8V 3.8V Vo t

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