**Full wave rectifier** basically uses both half cycles of the applied AC voltage and converts an AC voltage into a pulsating DC voltage. In **full wave rectification**, one **diode** conducts during one half-cycle while other conducts during the other half cycle of the applied AC voltage. It has some advantages over **half wave rectifier** like it has high average output DC voltage and has much fewer ripples resulting in the smoother output waveform. In our previous article, we explain **Rectifier**, **half wave rectifier efficiency**, **diodes**, **p-n junction** etc in detail. In this article, we are going to explain the *mathematical expression of an efficiency of Full wave rectifier equation and its applications* etc.

**Full wave Rectifier Equations:**

The efficiency of a full wave rectifier is defined as the ratio of DC power output to the applied input AC power. Therefore, it is given as;

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

**(i) Mathematical derivation of Full wave rectifier efficiency:**

In this Article You will learn:

Let the v = V_{m} sin θ be the AC voltage to be rectified, r_{f} be the forward resistance of crystal diode and R_{L} be the load resistance. The diagram given below shows the input voltage wave and rectified output wave.

The instantaneous value of current is given by the equation:

**i = v / (r _{f }+ R_{L})**

As we know,

v = V_{m} sin θ

Therefore,

i = V_{m} sin θ / (r_{f }+ R_{L})

When sin θ = 1, then current = maximum. Therefore,

I_{m} = V_{m} / (r_{f }+ R_{L})

Where,

i = I_{m} sin θ

Since output is obtained across R_{L} , therefore

**D.C power output = I _{dc}^{2} R_{L}**

= *I_{av}^{2} R_{L}

Where,

I_{av} = ʃ (i dθ) / π ….. (i)

Integrate equation (i) from 0 to π,

I_{av }= (1 / π) * ʃ I_{m} sin θ dθ

= (I_{m }/ π) * ʃ sin θ dθ

= (I_{m }/ π) [ – cos θ ]

= (I_{m }/ π) [ -(-1-1)]

= 2 (I_{m }/ π)

= (2I_{m }/ π)

Therefore, DC power output is given as,

P_{dc} = I_{dc}^{2} R_{L} = (2I_{m }/ π)^{2} R_{L}

And AC power input is given as,

P_{ac} =I_{rms}^{2} (r_{f }+ R_{L})

Where,

** I_{rms }= **√** ʃ (i^{2} dθ) / π ….. (ii)

= √ (1 / π) * ʃ I_{m}^{2} sin^{2} θ dθ

= √ (I_{m}^{2} / π) * ʃ ( 1- cos 2θ)/ 2 dθ

= √ (I_{m}^{2} / 2π) * [ ʃ dθ – ʃ cos 2θ dθ ]

= √ (I_{m}^{2} / 2π) * [[θ] – [sin 2θ / 2]]

= √ (I_{m}^{2} / 2π) * [π – 0]

= I_{m }/ √ 2

Therefore, AC power input is given as,

**P _{ac} =I_{rms}^{2} (r_{f }+ R_{L})**

=( I_{m }/ √2)^{2} (r_{f }+ R_{L})

As we know,

**Rectifier Efficiency (η) = P _{dc} / P_{ac}**

Put the values of P_{dc} and P_{ac} from above equations, therefore,

η =[ (2I_{m }/ π)^{2} * R_{L} ] / [( I_{m }/ √2)^{2} * (r_{f }+ R_{L})]

= 0.812 R_{L }/ (r_{f }+ R_{L})

= 0.812 / (1+ r_{f }/R_{L})

If r_{f }is neglected as compared to R_{L }then the efficiency of the rectifier is maximum. Therefore,

**η _{max }=0.812 = 81.2%**

**(ii) Calculation of Form Factor and Peak factor:**

**Form factor:**

It is defined as the ratio of RMS value of the output voltage to the average value of the output voltage. Its equation can be given as:

Form Factor = RMS value of the output voltage / average value of the output voltage

= (V_{m} / √2) / (2 V_{m }/ π) = π / 2 √2

**Form Factor = 1.11**

**Peak Factor:**

It is defined as the ratio of the peak value of the output voltage to the RMS value of the output voltage. Its equation can be given as:

Peak factor = Peak value of the output voltage / RMS value of the output voltage

= V_{m} / (V_{m} / √2)

**Peak Factor = √2**

**Applications of Full wave Rectifier:**

Full wave rectifier is of two types; centre tap and **bridge rectifier**. Both these rectifiers are used for following purposes depends upon the requirement. Following of **full wave rectifier applications** are:

- It can be used to detect the amplitude of modulated radio signal.
- It can be used to supply polarised voltage in welding.
- It is mainly used in power supply circuits.

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Thanks very much for the lesson