A simple Half Wave Rectifier is nothing more than a single pn junction diode connected in series to the load resistor. As you know a diode is to electric current like a one-way valve is to water, it allows electric current to flow in only one direction. This property of the diode is very useful in creating simple rectifiers which are used to convert AC to DC.

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 the 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).

### Half Â Wave Rectifier Operation

Simply put, a half wave rectifier removes the negative half cycle of an AC input and allows only the positive cycles to pass creating a DC flow.

To understand the operation of a half wave rectifier perfectly, you must know the **theory part** really well. If you are new to the concepts of a PN junction and its characteristics, I recommend you to read the half wave rectifier theory part first.

The operation of a half wave rectifier is pretty simple. From the theory part, you should know that a pn junction diode conducts current only in 1 direction. In other words, a pn junction diode conducts current only when it is forward biased. The same principle is made use of in a half wave rectifier to convert AC to DC. The input we give here is an alternating current. This input voltage is stepped down using a transformer. The reduced voltage is fed to the diode ‘D’ and load resistance RL. During the positive half cycles of the input wave, the diode ‘D’ will be forward biased and during the negative half cycles of input wave, the diode ‘D’ will be reverse biased. We take the output across load resistor RL. Since the diode passes current only during one-half cycle of the input wave, we get an output as shown in the diagram. The output is positive and significant during the positive half cycles of the input wave. At the same time output is zero or insignificant during negative half cycles of the input wave. This is called **half wave rectification.**

#### Explaining Half Wave Rectification in academic words!

When a single rectifier diode unit is placed in series with the load across an ac supply, it converts alternating voltage into a uni-directional pulsating voltage, using one-half cycle of the applied voltage, the other half cycle being suppressed because it conducts only in one direction. Unless there is an inductance or battery in the circuit, the current will be zero, therefore, for half the time. This is called * half-wave rectification. *As already discussed, a diode is an electronic device consisting of two elements known as cathode and anode. Since in a diode electrons can flow in one direction only

*i.e.*from the cathode to anode, the diode provides the unilateral conduction necessary for rectification. This is true for diodes of all types-vacuum, gas-filled, crystal or semiconductor, metallic (copper oxide and selenium types) diodes.

**Semiconductor diodes,**because of their inherent advantages are usually used as a rectifying device. However, for very high voltages, vacuum diodes may be employed.

### Working of a Half wave rectifier

The half-wave rectifier circuit using a semiconductor diode (D) with a load resistance R_{L}Â but no smoothing filter is given in the figure. The diode is connected in series with the secondary of the transformer and the load resistance R_{L.}Â The primary of the transformer is being connected to the ac supply mains.

The ac voltage across the secondary winding changes polarities after every half cycle of the input wave. During the positive half-cycles of the input ac voltage *i.e. *when the upper end of the secondary winding is positive w.r.t. its lower end, the diode is forward biased and therefore conducts current. If the forward resistance of the diode is assumed to be zero (in practice, however, a small resistance exists) the input voltage during the positive half-cycles is directly applied to the load resistance R_{L}, making its upper-end positive w.r.t. its lower end. The waveforms of the output current and output voltage are of the same shape as that of the input ac voltage.

During the negative half cycles of the input ac voltage *i.e. *when the lower end of the secondary winding is positive w.r.t. its upper end, the diode is reverse biased and so does not conduct. Thus during the negative half cycles of the input ac voltage, the current through and the voltage across the load remains zero. The reverse current, being very small in magnitude, is neglected. Thus for the negative half cycles, no power is delivered to the load.

Thus the output voltage (VL) developed across load resistance R_{L} Â is a series of positive half cycles of alternating voltage, with intervening very small constant negative voltage levels, It is obvious from the figure that the output is not a steady dc, but only a pulsating dc wave. To make the output wave smooth and useful in a DC power supply, we have to use a **filter** across the load. Since only half-cycles of the input wave are used, it is called a ** half wave rectifier**. Â

### Half Wave Rectifier Theory

Rectification is an application of the pn junction diode. A half wave rectifier is a device which makes use of key properties of a pn junction diode. So to understand the underlying theory behind a half wave rectifier, you need to understand the pn junction and the characteristics of the pn junction diode. We have developed two articles to help you understand both of them.

**1)****Understanding the PN Junction** Â – This article will help you to understand the pn junction and the underlying theory behind using PN junction as a rectifier.

**2)** **Characteristics of PN junction diode** – This article will help you to understand the characteristics of a pn junction diode with help of graphs. You can understand the behavior of a diode across various voltage levels and how it conducts.

**Note:-** There is an interesting **story behind the invention of PN junction diode**. Â The story revolves around the perseverance of a young scientist at Bell Laboratories in USA, Mr. Russel Ohl. In the story, you will learn how great inventions happen and how some bright minds of the 1930’s like Walter Brattain (one among the 3 who invented transistor) worked together to bring great inventions t our life

**Power Supply Specifications of a rectifier**

The most important characteristics which areÂ requiredÂ to be specified for a power supply are the required output dc voltage, the average and peak currents in the diode, the peak inverse voltage (PIV) of Â diode, the regulation and the ripple factor.

### Advantages and Disadvantages of Half wave rectifier:

A half wave rectifier is rarely used in practice. It is never preferred as the power supply of an audio circuit because of the very high ripple factor. High ripple factor will result in noises in the input audio signal, which in turn will affect audio quality.

The advantage of a half wave rectifier is only that its cheap, simple and easy to construct. It is cheap because of the low number of components involved. Simple because of the straight forwardness in circuit design. Apart from this, a half wave rectifier has more number of disadvantages than advantages!

**Disadvantages of Half wave rectifier**

**1.** The output current in the load contains, in addition to dc component, ac components of basic frequency equal to that of the input voltage frequency. Ripple factor is high and an elaborate filtering is, therefore, required to give steady dc output.

2. The power output and, therefore, rectification efficiency is quite low. This is due to the fact that power is delivered only during one-half cycle of the input alternating voltage.

3. Transformer utilization factor is low.

4. DC saturation of the transformer core resulting in magnetizing current and hysteresis losses and generation of harmonics.

The Â DC output available from a half-wave rectifier is not satisfactory to make a Â general power supply. However, it can be used for some applications like battery charging.Â

### Half Wave Rectifier with Capacitor Filter Â

The output of half wave rectifier is not a constant DC voltage. You can observe from the output diagram that its a pulsating dc voltage with ac ripples. In real life applications, we need a power supply with smooth waveforms. In other words, we desire a DC power supply with the constant output voltage. A constant output voltage from the DC power supply is very important as it directly impacts the reliability of the electronic device we connect to the power supply.

We can make the output of half wave rectifier smooth by using a filter (a capacitor filter or an inductor filter) across the diode. Â In some cases, a resistor-capacitor coupled filter (RC) is also used. The circuit diagram below shows a half wave rectifier with capacitor filter.

### Half Wave Rectifier Analysis

The following parameters will be explained for the analysis of Half Wave Rectifier:-

**1.Â Â Â Â Â Â ****Peak Inverse Voltage (PIV)**

Peak Inverse Voltage (PIV) rating of a diode is important in its design stages. It is the maximum voltage that the rectifying diode has to withstand, during the reversely biased period.

When the diode is reverse biased, during the negative half cycle, there will be no current flow through the load resistor RL. Hence, there will be no voltage drop through the load resistance RL which causes the entire input voltage to appear across the diode. Thus V** _{SMAX}**, the peak secondary voltage, appears across the diode. Therefore,

Peak Inverse Voltage (PIV) of half wave rectifier = V_{SMAX}

**2.Â Â Â Â Â ****Average and Peak Currents in the diode**

By assuming that the voltage across the transformer secondary be sinusoidal of peak values V** _{SMAX}**, the instantaneous value of the voltage given to the rectifier can be written as

Assuming that the diode has a forward resistance of RF ohms and infinite reverse resistance value, the current flowing through the output load resistance RL is

I** _{MAX }**= V

**/(R**

_{SMAX}**R**

_{F}+**)**

_{L}**3.Â Â Â Â Â ****DC Output Current**

The dc output current is given as

**Substituting the value of **I** _{MAX}Â for the equation **I

**= V**

_{MAX }**/(R**

_{SMAX}**R**

_{F}+**), we have**

_{L}I** _{dc }= **V

**Â = V**

_{SMAX}/**Â R**

_{SMAX}/**R**

_{L }if**R**

_{L }>>

_{F}**4.Â Â Â Â Â ****DC Output Voltage**

Dc value of voltage across the load is given by

V** _{dc} **= I

**R**

_{dc}**V**

_{L}=**X R**

_{SMAX}/pi(R**R**_{F}Â +Â_{L)}**V**

_{L }=**{1+R**

_{SMAX}/

_{F/}_{R}

**}**

_{L}If**Â **R** _{L }>> **R

**V**

_{F},**= V**

_{dc}

_{SMAX}/pi**5.Â Â Â Â Â ****Root Mean Square (RMS) Value of Current**

RMS value of the current flowing through the diode is given as

**6.Â Â Â Â Â ****Root Mean Square (RMS) Value of Output Voltage**

RMS value of voltage across the load is given as

V** _{Lrms}** = I

**R**

_{rms}**= V**

_{L}**R**

_{SMAX}**/2(R**

_{L}**R**

_{F}+**) = V**

_{L}**{1+R**

_{SMAX}/2

_{F/}_{R}

**}**

_{L}If**Â **R** _{L }>> **R

**V**

_{F},**= V**

_{Lrms}

_{SMAX}/2**7.Â Â Â Â Â Â ****Rectification Efficiency**

Rectification efficiency is defined as the ratio between the output power to the ac input power.

Efficiency, È = DC power delivered to the load/AC input power from the transformer = P** _{dc}**/P

_{ac}DC power delivered to the load, P_{dc}** = **I^{2}** _{dc}** R

**(I**

_{L }=_{max/pi})

^{2 }R

_{L}AC power input to the transformer, P** _{ac }**= Power dissipated in diode junction + Power dissipated in load resistance R

_{L}**= **I** ^{2}_{rms}** R

**I**

_{F }+**R**

^{2}_{rms}**{I**

_{L }=**/4}[ R**

^{2}_{MAX}**R**

_{F}+**]**

_{L}So, Rectification Efficiency, È = P** _{dc}**/P

_{ac }= {4/

^{2}**}[**R

**(R**

_{L}/**R**

_{F}+**)**

_{L}**] = 0.406/**{1+ R

_{F/}_{R}

**Â }**

_{L}The maximum efficiency that can be obtained by the half wave rectifier is 40.6%. This is obtained if R** _{F}** is neglected.

**8. Â Â Â Ripple FactorÂ **

Ripple factor is in fact a measure of the remaining alternating components in a filtered rectifier output. It is the ratio of the effective value of the ac components of voltage (or current) present in the output from the rectifier to the dc component in output voltage (or current).

The effective value of the load current is given as

I** ^{2 }=**I

**I**

^{2}_{dc}+**I**

^{2}_{1}+**I**

^{2}_{2}+**I**

^{2}_{4 }Â =^{ Â }**I**

^{2}_{dc}+

^{2}_{ac}Where, I** _{1},**I

**I**

_{2},^{ Â }**and so onare the rms values of fundamental, second, fourth and so on harmonics and I**

_{4 }**is the sum of the squares if the rms values of the ac components.**

^{2}_{ac}So, ripple factor, Î³ = I** _{ac}/** I

**I**

_{dc}=**I**

^{2 }–**I**

^{2}_{dc})/**{( I**

_{dc }=**I**

_{rms}/**)-1}**

_{dc}^{2}**=**K

**â€“ 1)**

_{f}^{2}Where K** _{f }**is the form factor of the input voltage. For half wave rectifier, form factor is given as

K** _{f }= **I

**/I**

_{rms}**(I**

_{avg }=_{max}/

_{2})/ (I

_{max}/pi)Â = pi/2 = 1.57

So, ripple factor, Î³ = Â (1.57** ^{2} **â€“ 1) = 1.21

**9. Â Â Â Â Regulation**

The variation of the output voltage as a function of dc load current is called regulation. Percentage regulation is given as

% Regulation = {(Vno-load â€“ Vfull-load)/ Vfull-load}* 100

Fror an ideal power supply, the output voltage should be independent of load current and the percentage regulation should be equal to zero.

**Applications of Half wave rectifier**

Any rectifier is used to construct DC power supplies. The practical application of any rectifier (be it half wave or full wave) is to be used as a component in building DC power supplies. Â A half wave rectifier is not special than a full wave rectifier in any terms. In order to build an efficient & smooth DC power supply, a full wave rectifier is always preferred. Â However, for applications in which a constant DC voltage is not very essential, you can use power supplies with half wave rectifier.

## Comments