In **Half wave rectifier**, the diode conducts only for positive input voltages i.e. one-half of the AC wave is removed because it cannot pass through the diode. As the name indicates it rectify only half of the input wave through the phenomenon name as rectification. In our previous tutorial, we gave information about **half wave rectifier circuit**, **Rectifier**,** Diodes**, **Diode types **etc. In this article, we explain the *mathematical expression of an efficiency of Half wave rectifier and its uses*.

**Efficiency of Half wave rectifier:**

The ratio of DC power output to the applied input AC power is known as **rectifier efficiency**. Mathematically it can be given as:

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

**Calculation of Half wave rectifier Efficiency:**

For the determination of the **efficiency** of half-wave rectifier consider the circuit diagram shown in Figure below. Let r_{f} and R_{L} be the forward resistance of** diode** & load resistance, v = V_{m} sin θ be the voltage appearing across the secondary of the power transformer. During the positive half cycle, the diode is forward biased makes the current to flow through the load resistor while during the Negative half cycle the diode is reverse biased stops the current flow through the load resistor.

The waveform diagram above shows only positive waveform at the output and suppressed or no negative waveform. During conduction period its instantaneous value 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) / 2π ….. (i)

Integrate equation (i) from 0 to π,

I_{av }= (1 / 2π) * ʃ I_{m} sin θ dθ

= (I_{m }/ 2π) * ʃ sin θ dθ

= (I_{m }/ 2π) [ – cos θ ]

= (I_{m }/ 2π) [ -(-1-1)]

= 2 (I_{m }/ 2π)

= (I_{m }/ π)

Therefore, DC power output is given as,

**P _{dc} = I_{dc}^{2} R_{L} = (I_{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) / 2π ….. (ii)

Integrate equation (ii) from 0 to π,

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

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

= √ (I_{m}^{2} / 4π) * [ ʃ dθ – ʃ cos 2θ dθ ]

= √ (I_{m}^{2} / 4π) * [[θ] – [sin 2θ / 2]]

= √ (I_{m}^{2} / 4π) * [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 =[ (I_{m }/ π)^{2} * R_{L} ] / ( I_{m }/ 2)^{2} * (r_{f }+ R_{L})

= 0.406 R_{L }/ (r_{f }+ R_{L})

= 0.406 / (1+ r_{f} R_{L})

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

**η _{max }=0.406 = 40.6%**

This indicates that the half wave rectifier can convert maximum 40.6% of AC power into DC power, and the remaining power of 59.4% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit.

**Applications of Half wave rectifier:**

Half wave rectifier is not so good as compare to full wave rectifier, but sometimes we required this rectifier depend on requirement. Following of **rectifier applications** are:

- It is used for detection of amplitude modulated radio signals.
- For the welding purpose, it supplies polarized voltage.
- It is used in many signal demodulation processes.

*Hope you all like this article. For any suggestions please comment below. We always appreciate your suggestions.*

## 7 Comments

Efficiency is DC power to the load / ac power obtained by the secondary transformer …but the ac power obtained is not DC…it is full wave …then y is the rms value of rectified wave taken into consideration??

See for the negative half cycle there is voltage and current in the primary of the transformer but there is no current in the secondary due to high resistance of reverse biasing. So the waveform of the current is having only positive cycle before entering the diode and coming out of the diode, but no negative cycle. So coming out of it, there is pulsating dc whose output power we calculate. Now before entering the diode the waveform is same, thus if there would be an ac that had the rms value of that current waveform, then what power had it produced is calculated. And thus we solve for the efficiency that what would have been the dc’s effect to that of the ac’s.

Thanks, Shivam for the comment. I don’t know how I skipped this question from Preeti. We are glad that people like you are here to help others.

Thank You for this

Welcome Saptarshi,

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Sir,

In the input ac power expression why we take RMS value of I. Input current is a full sine wave, but in Irms expression we take only 0 to pi value(half wave o/p).

Please guide me, sir.

Hi Ritesh,

Sorry for late reply… here is the solution.

Irms expression is related from the AC power Type. For full-wave, it is Irms = Ipk/sqrt(2) but for half wave, it is Irms = Ipk/2