**Half wave rectifier** is the type of **rectifier** that rectifies only one half cycle of the waveform i.e. either positive or negative cycle. It is nothing more than a single **p-n junction** diode connected in series to the load resistor. To rectify the signal these rectifiers used in the AM radios. In our last tutorial, we explain **rectifier**, **diode**, **diode types **in detail. In this article, we are going to explain *half wave rectifier circuit, operation, characteristics and efficiency* in detail.

**Half-wave Rectifier Theory:**

To understand the basic theory of half wave rectifier exactly, you must know rectifier operation, circuit etc. Here we discuss one by one in brief.

**(i) Half wave rectifier circuit:**

For **half-wave rectification**, only one crystal diode is used. As shown in Figure above, we supply alternating current as an input which generally get rectified through a transformer.The transformer is used to step-up or step-down the mains supply voltage as per requirement. At the anode side of the diode, the **transformer** is connected while at the cathode side load resistance is connected. The main purpose of the transformer is to isolate the rectifier circuit from power lines, obtain the desired level of dc voltage and reduces the risk of electric shock.

**(ii) Half wave rectifier working principle:**

The **working principle of Half wave rectifier** is quite simple i.e convert alternating current to direct current. The reduced voltage is fed to the diode and load resistance and the input voltage is stepped down using a transformer. The diode will be forward biased during the positive half cycle of the input wave while it is reversed biased during the negative half cycle of the input wave. Across the load resistor, the output is taken out indicates that the diode passes current only during one-half cycle of the input wave. During the positive half cycle, the output is positive & significant while during negative half cycle output is zero & insignificant.

**(iii) Half wave rectifier Operation:**

As shown in Figure above, the **half wave rectifier circuit **comprises of **semiconductor** diode with a load resistance RL. The diode is connected in series with the secondary of the transformer and the load resistance while AC supply main connected to the primary of the transformer. It basically works in two cycles; positive and negative.

**(i) Positive Half cycle:**

When AC supply is switched on, the alternating voltage V_{in }starts appearing across the terminals AB at secondary winding. During the positive half cycle, the **crystal diode** is forward biased makes the terminal A positive w.r.t B terminal. Therefore, it conducts and current flows through the load resistor. As shown in waveform diagram, the current varies in magnitude and shows a positive half cycle and suppressed negative half cycle.

**(ii) Negative Half cycle:**

During the negative half cycle, the crystal diode is reversed biased makes the terminal A negative w.r.t B terminal. Under this condition, the diode does not conduct and no current flows through the circuit. Therefore, in the negative half cycle of the input no voltage appears across the load resistor.

**Half wave rectifier Characteristics:**

For the analysis of half wave rectifier following parameters or properties are considered. Following of its characteristics are:

**(i) Ripple Factor:**

It is defined as the ratio of the effective value of the AC components of current or voltage present in the output from the rectifier to the dc component in output voltage. Simply, it is the measure of remaining alternating components in a filtered rectifier output.

**Calculation of Ripple factor:**

The effective value of ripple factor can be calculated as given below. The effective of load current is given as,

**I ^{2} =I^{2}_{dc} +I^{2}_{1}+I^{2}_{2}+I^{2}_{4} **

= I^{2}_{dc} +I^{2}_{ac}

Where I_{1, }+I_{2, }+I_{4} are the RMS values of the second, fourth and so on harmonics.

Ripple factor (γ)** **is given as follows;

**γ = I _{ac} / I_{dc}**

= (I^{2} – I^{2}_{dc})/ I_{dc}

= {( I_{rms}/ I_{dc}^{2})-1}

= K_{f}^{2} – 1

where K_{f }is the form factor of the input voltage.For half wave rectifier form factor is calculated as:

**K _{f }= I_{rms} /I_{avg }**

= (I_{max}/2)/ (I_{max}/π)

= π/2 = 1.57 ( π = 3.14)

As we know,

Ripple factor = K_{f}^{2} – 1

= (1.57 * 1.57) – 1

= 1.21

**(ii) Peak inverse voltage:**

Peak inverse voltage is defined as the maximum value of the voltage coming out of the diode when it is reverse biased during the negative half cycle. The diode used must have higher **PIV** rating than the voltage which is coming across it.

**(iii) Transformer Utilization Factor: **

It is defined as the ratio of the ratio of power delivered to load and VA rating of the transformer. Half wave rectifier has TUF of around 0.287.

Note: 1 / TUF signifies that the transformer must be 1.23 times higher then that when it is used to deliver power from a pure AC voltage.

**(iv) Regulation:**

Regulation is defined as the ratio of percentage of the difference between the no-load voltage & full load voltage to full load voltage. It is given as:

**% Regulation = {(V _{no-load}**

**–**V

_{full-load})/ V_{full-load}}* 100Note: For an ideal power supply, the output voltage should be independent of load current and the percent regulation is equal to zero.

**(v) Efficiency:**

It is defined as the ratio of the DC power output to the AC power input. Mathematically, it is given as:

**η = DC Power Output / AC power input**

where η = Rectifier Efficiency.

**Advantages of Half Wave rectifier:**

Following of its advantages are:

- It is cheap because it requires lesser components.
- It is simple and easy to construct.

Note: Due to high ripple factor, half wave rectifier is rarely used in practice because high ripple factor will result in noises in input audio signal.

**Disadvantages of Half wave rectifier:**

Following of its disadvantages are:

- The output is low because AC supply delivers power only half the time.
- The output contains more alternating components, therefore, it needs heavy filter circuit to smooth out the output.
- It has low transformer utilization factor.
- Hysteresis losses and harmonics occur due to DC saturation of transformer core.

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