This is the third chapter in the Rectifier series. In the last two chapters, we have covered the basics of rectifiers, and Half wave rectifier. Though we have covered various properties in our last article but kept aside a mathematical derivation of the half-wave rectifier efficiency equation. Because we think it needs to be explained separately.
In this article, we are going to learn to calculate the equation of efficiency of the half-wave rectifier and about its various applications in detail.
Before learning ahead, let’s have a little recap about the working of a half-wave rectifier.
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. Read our article on a half-wave rectifier circuit for full information.
Now let’s go ahead and get to the mathematical part.
What is Rectifier Efficiency
The ratio of DC power obtained at the 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
Where η is Rectifier Efficiency.
Half-wave Rectifier Efficiency Mathematical Calculation
For the determination of the efficiency of the half-wave rectifier consider the circuit diagram shown in the Figure below.
Let rf and RL be the forward resistance & load resistance of the diode. v = Vm sin θ be the voltage appearing across the secondary of the power transformer.
During the positive half cycle, the diode is forward biased making the current flow through the load resistor. While during the Negative half cycle the diode is reverse biased so it stops the current flow through the load resistor.
The waveform diagram at the right side of the above figure shows only a positive waveform at the output and a suppressed negative waveform. During the conduction period its instantaneous value is given by the equation:
i = v / (rf + RL)
v = i (rf + RL)
As we know,
v = Vm sin θ
i = Vm sin θ / (rf + RL)
When sin θ = 1, current = maximum. Therefore,
Im = Vm / (rf + RL)
i = Im sin θ
Since the output is obtained across RL, therefore
D.C power output = Idc2 RL
= *Iav2 RL
Iav = ʃ (i dθ) / 2π ….. (i)
Integrate equation (i) from 0 to π,
Iav = (1 / 2π) * ʃ Im sin θ dθ
= (Im / 2π) * ʃ sin θ dθ
= (Im / 2π) [ – cos θ ]
= (Im / 2π) [ -(-1-1)]
= 2 (Im / 2π)
= (Im / π)
Therefore, DC power output is given as,
Pdc = Idc2 RL = (Im / π)2 RL
And AC power input is given as,
Pac =Irms2 (rf + RL)
** Irms = ʃ (i2 dθ) / 2π ….. (ii)
Integrate equation (ii) from 0 to π,
= √ (1 / 2π) * ʃ Im2 sin2 θ dθ
= √ (Im2 / 2π) * ʃ ( 1- cos 2θ)/ 2 dθ
= √ (Im2 / 4π) * [ ʃ dθ – ʃ cos 2θ dθ ]
= √ (Im2 / 4π) * [[θ] – [sin 2θ / 2]]
= √ (Im2 / 4π) * [π – 0]
= Im / 2
Therefore, AC power input is given as,
Pac =Irms2 (rf + RL)
=( Im / 2)2 (rf + RL)
Rectifier Efficiency (η) = Pdc / Pac
Put the values of Pdc and Pac from above equations, therefore,
η =[ (Im / π)2 * RL ] / ( Im / 2)2 * (rf + RL)
= 0.406 RL / (rf + RL)
= 0.406 / (1+ rf RL)
If rf is neglected as compare to RL 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. Hence the half wave rectifier efficiency is 40.6%
In fact, 50% power in the negative half cycle is not converted and the remaining 9.4% is lost in the circuit.
You can watch this video by Neso Academy for more info.
Half wave rectifier Applications
Half wave rectifier is not so good as compared to Full-wave or Bridge rectifier, but sometimes we require this rectifier depending on the requirements. Some of the applications of half-wave rectifier are
- It is used for the detection of amplitude modulated radio signals.
- For the welding purpose, it supplies polarized voltage.
- It is used in many signal demodulation processes.
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