Half Wave Rectifier Circuit Working, Analysis and Characteristics.

A simple Half Wave Rectifier is nothing more than a single pn junction diode connected in series to the load resistor. If you look at the above diagram, we are giving an alternating current as input. Input voltage is given to a step down transformer and the resulting reduced output of transformer is given to the diode ‘D’ and load resistor RL. The output voltage is measured across load resistor RL.

As part of our “Basic Electronics Tutorial” series, we have seen that rectification is the most important application of a PN junction diode. The process of rectification is converting alternating current (AC) to direct current (DC).

pr1nctple:
It is a circuit which convert alternating voltage or current into pulsating
voltage or current for half the period of input cycle hence it is named as
"halfwave rectifier".

Construction

• It consists of step down transformer, semiconductor diode and the load resistance RL.

• The need of step down transformer is that, reducing the available ac voltage (230V, 1ϕ, AC) in to required level of smaller ac voltage.

• The diode can be used to convert the ac into pulsating dc.

• A halfwave rectifier and its input, output waveforms are shown in figure

(I) When diode is forward biased (ii) Diode reverse biased Half Wave Rectifier Circuit and INPUT and OUTPUT Waveforms

Operation

• During the positive half cycle of input, the diode D is forward biased, it offers very small resistance and it act as closed switch and hence conducts the current through the load resistor RL.

• During the negative halfcycle of the input diode D is heavily reverse biased, it offers very high resistance and it acts as open switch hence it does not conduct any current. The rectified output voltage will be in phase [i.e., starting and ending points are same for both wave forms] with AC input voltage for completely resistive loads and the waveforms is shown in figure

Analysis of halfwave rectifier

The input sinusoid.al voltage applied at the input of the transformer is given by

 Vi  =  Vmsinωt

The diode current or load current is given by

it  =   Imsinωt          for    0 < ωt < π

it  = 0        for    0 < ωt < π

Maximum or peak current through the circuit

IM=VmRf+RL

Where    Rf    =    forwardresistanceofthediode

RL   =   loadresistance

(a) The average or DC current

Idc    =     Iav    =     1T∫T0i(t)dt

             =    12π∫2π0i(t)d(t)    =     12π[∫π0Imsinωt.d(ωt)+∫2ππ0d(ωt)]

Where                i(t)=0 for the interval  π ≤ ωt ≤ 2π

Hnece   ,Idc=12π∫π0Imsinωtd(ωt)=Im2π[−cosωt]π0

                      =Im2π[−cosπ+cos0]

Finally, 

 Iav  =  Idc  =  Imπ  =  Vmπ(RF+RL)

Similarly the DC output voltage or Average voltage is given by

Vdc  =  IdcRL  =  IMRLπ  =  Vmπ(RLRL+Rf)

Vdc=Vmπ(1+RfRL)

(b) RMS current is calculated as follows

Irms=[1T∫T0i2(t)d(ωt)]−−−−−−−−−−−−−−−√=12π∫2π0i2td(ωt)−−−−−−−−−−−−−√

using equations  2 and 3

Irms = [12π∫b0I2msin2ωtd(ωt) + ∫2ππ0d(ωt)]−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

=I2m2π∫π0(1−cos2ωt)2d(ωt)−−−−−−−−−−−−−−−−−−−−−√

We Know, sin2ωt=1−cos2ωt2

=I2m4π[ωt−12sin(2ωt)]π0−−−−−−−−−−−−−−−−−−√=[Ibm4]−−−−−√

Irms=Im2=Vm2(RF+RL)

(c) The DC out put power developed across the load RL is given by

Pac=I2dc.RL=I2m.RLπ2=V2mRLπ2(Rf+RL)2

(d) Rectifier Efficiency

Rectifier efficiency is defined as the ratio of the DC output power [Pde] to the AC input power [Pac] Hence the rectifier efficiency is given by

%η=0.4061+RfRL×100=40.61+RfRL

From the equation it is observed that rectifier efficiency increases if the ratio Rf/RL is negligible. The maximum theoretical value of rectifier efficiency of a half-wave rectifier is 40.6% when RL is very high or for a fixed RL, RF should be low.

(e) Ripple Factor

The ripple factor is defined as the ratio of the effective. value or rms value of the ac component of voltage or current to the average value of vo ltage or current.

Mathematically it can be written as

γ =  VrmsVdc=IrmsIdc

γ =  IrmsIdc  =  IacIdc  =  I2rms−I2dc−−−−−−−−√Idc  =  (IrmsIdc)2−1−−−−−−−−−−−√


Hence ripple factor for half-wave rectifier is,

γ=(Im2)2(Imπ)2−1−−−−−−−−−⎷=(π2)2−1−−−−−−−−√=1.21

The "ripple frequency" in a half wave rectification circuit is same as the fundamental AC frequency and hence half-wave method of rectification with single phase AC supply is not used for power charging. Also the rms ripple voltage exceeds the dc voltage and hence it is a poor conductor to dc.

(f) Peak Inverse Voltage (PIV)

This is the maximum voltage with which the rectifier has to withstand during nonconduction period or reverse biasing. In a half wave rectifier, the peak inverse voltage equals the peak value of the applied voltage. i.e., PIV = Vm un d er reverse biased condition.