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• The need of step down transformer is that, reducing the available ac voltage (230V, 1

• 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• 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 byThe diode current or load current is given by

Maximum or peak current through the circuit

Where Rf = forwardresistanceofthediode

### (a) The average or DC current

**Iav = Idc = Imπ = Vmπ(RF+RL)**

Similarly the DC output voltage or Average voltage is given by

**Vdc=**

**Vm**

**π(1+RfRL)**

### (b) RMS current is calculated as follows

using equations 2 and 3

Irms=Im2=Vm2(RF+RL)

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

**Pac=I2dc.RL=I2m.RLπ2=**

**V2mRL**

**π2(Rf+RL**

**)**

**2**

### (d) Rectifier Efficiency

Rectifier efficiency is defined as the ratio of the DC output power [%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

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.