What is Half wave rectifier and how does diode act as a half wave rectifier?

What is a Half Wave Rectifier?

 A half wave rectifier is defined as a type of rectifier that only allows one half-cycle of an AC voltage waveform to pass, blocking the other half-cycle. Half-wave rectifiers are used to convert AC voltage to DC voltage, and only require a single diode to construct.

Circuit Diagram of a Half Wave Rectifier :

A half wave rectifier circuit diagram looks like this:

The above figure shows a half wave rectifier circuit in series with a resistance RL . If the applied voltage,  e = VmSinθt  then

i=V_msinθt/(Rf + RL) = ImSinθt (for 0≤θt≤π)

                                  =0 ( for π≤ θt≤2π)

Where Rf = resistance of diode in forward direction

Also Im=Em/(Rf+RL)

                              (Current waveform in Half wave rectifier circuit)

Let Rf and RL be the forward resistance & load resistance of the diode. v = Vm sin θ be the voltage appearing across the secondary of the power transformer. During the positive half cycle, the diode is forward biased making the current flow through the load resistor. While during the Negative half cycle the diode is reverse biased so it stops the current flow through the load resistor. The waveform diagram shows only a positive waveform at the output and a suppressed negative waveform. During the conduction period its instantaneous value is given by the equation: 

i = v / (Rf+RL )                                                                                                                      v = i(Rf+RL ) As we know, v = Vm sin θ Therefore, i = Vm sin θ /(Rf+RL ) When sin θ = 1, current = maximum. Therefore, Im = Vm /(Rf+RL )

Where, i = Im sinθ                                                                                                          As the output is obtained across RL, therefore 

                                              D.C power output = I^²_dcRL

                                                                           = *Iav2 RL 

 Where, Iav = ʃ (i dθ) / 2π   ….. (i) 

 Integrate equation (i) from 0 to π, 

                                                     Iav = (1 / 2π) * ʃ Im sin θ dθ 

                                                           = (Im / 2π) * ʃ sin θ dθ 

                                                           = (Im / 2π) [ – cos θ ]

                                                           = (Im / 2π) [ -(-1-1)]

                                                           = 2 (Im / 2π) 

                                                           = (Im / π)

 

Therefore, DC power output is given as, 

                                                            Pdc = I^²_dcRL  

                                                                  = (Im / π)²RL

 AC power input is given as,

                                                         Pac =I²_rms(Rf+ RL) 

Where,  Irms = ʃ (i²dθ) / 2π ….. (ii) 

Integrate equation (ii) from 0 to π,

                                                            = √ (1 / 2π) * ʃ I²_msin² θ dθ 

                                                            = √ (Im2 / 2π) * ʃ ( 1- cos 2θ)/ 2  dθ 

                                                            = √ (Im2 / 4π) * [ ʃ dθ – ʃ cos 2θ dθ ] 

                                                            = √ (Im2 / 4π) * [[θ] – [sin 2θ / 2]]

                                                            = √ (Im2 / 4π) * [π – 0]

                                                            = Im / 2 

Therefore, AC power input is given as, 

                                                               Pac =I²_rms(Rf+ RL)

                                                                     = ( I_m / 2)² (Rf+ RL)

Rectifier Efficiency (η) = Pdc / Pac 

                                 η =[ (I_m / π)² * RL ] /( I_m / 2)² * (Rf+ RL) 

                                    = 0.406 RL / (Rf+ RL) 

                                    = 0.406 / [1+(Rf /RL )] 

If Rf is neglected as compare to RL then the efficiency of the rectifier is maximum. Therefore   ηmax =0.406 = 40.6%

This indicates that the half wave rectifier can convert maximum 40.6% of AC power into DC power, and the remaining power of 59.4% is lost in the rectifier circuit. Hence the half wave rectifier efficiency is 40.6% In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit.