I'm trying to derive the efficiency of a half-wave rectifier using the definition for efficiency, $$\eta = \frac{\text{output power}}{\text{input power}}$$

And I also know that $$P = V_\text{rms} \cdot I_\text{rms}$$

So when I computed these for the output and input, I got V0 / 2 and I0 / 2 for the output, and V0 / sqrt(2) and I0 / sqrt(2) for the input.

Plugging in everything, the efficiency should be 0.5, but every source I look at (like this one) tells me that it's 40.6%. Where am I going wrong?

EDIT: Here's what I did to get the RMS values. I assumed that the rectifier is connected to an external resistance R.

I_0 is the maximum current of the input, V_0 is I_0 * R

For the input, $$I_{rms}^2 = \frac{\int_0^{T} I_0^2 \sin^2\left(\frac{2\pi}{T} t\right) \mathrm{d} t}{T} = \frac{I_0^2 \int_0^{T} 1 - \cos\left(\frac{4\pi}{T} t\right) \mathrm{d} t}{2T} = \frac{I_0^2}{2}$$ $$\implies I_{rms} = \frac{I_0}{\sqrt{2}}$$

$$V_{rms} = \frac{I_0 R}{\sqrt{2}} = \frac{V_0}{\sqrt{2}}$$

For the output, $$I_{rms}^2 = \frac{\int_0^{\frac{T}{2}} I_0^2 \sin^2\left(\frac{2\pi}{T} t\right) \mathrm{d} t + \int_\frac{T}{2}^T 0 \mathrm{d} t}{T} = \frac{I_0^2 \int_0^{\frac{T}{2}} 1 - \cos\left(\frac{4\pi}{T} t\right) \mathrm{d} t}{2T} = \frac{I_0^2}{4}$$ $$\implies I_{rms} = \frac{I_0}{2}$$

$$V_{rms} = \frac{I_0 R}{2} = \frac{V_0}{2}$$

This gives the efficiency as $$\frac{\frac{V_0 I_0}{4}}{\frac{V_0 I_0}{2}} = 0.5$$

  • 1
    \$\begingroup\$ The linked webpage doesn't contain the word "efficiency". Did you link to the correct page ? \$\endgroup\$
    – AJN
    Commented Nov 13, 2020 at 14:00
  • \$\begingroup\$ \$I_0/\sqrt 2\$ for the input is incorrect. The current is same for input and output side (if there is no capacitor). If the diode is ideal and load is pure resistor, there is no energy absorbing element other than the load. So efficiency should be 100% ??? Give more detailed calculations for voltage and current on input and output side. \$\endgroup\$
    – AJN
    Commented Nov 13, 2020 at 14:06
  • \$\begingroup\$ Current, whether it is input or output is flowing only in one half cycle. So the integral for the input current should also be up to T/2; not T. Also please put a circuit diagram. \$\endgroup\$
    – AJN
    Commented Nov 13, 2020 at 16:02
  • 4
    \$\begingroup\$ The new link given doesn't look like a good learning resource. e.g. "Why half-wave rectifiers are not used in dc power supply? Answer: Half-wave rectifiers are not used in dc power supply because the supply provided by the half-wave rectifier is not satisfactory." !!! It just provides a list of claims without explanation or proof. \$\endgroup\$
    – AJN
    Commented Nov 13, 2020 at 16:12
  • 2
    \$\begingroup\$ @AJN So true. But sad to say that this particular learning resource is now the most popular paid learning resource in my country. With millions of students enrolling in per year. \$\endgroup\$
    – Mitu Raj
    Commented Nov 13, 2020 at 19:42

2 Answers 2


The 'efficiency' they are referring to is Conversion Ratio as I found in the wikipedia article about Rectifiers -

Conversion ratio (also called "rectification ratio", and confusingly, "efficiency") η is defined as the ratio of DC output power to the input power from the AC supply.

If we go by this convention, assuming transformer and diodes are ideal, and if \$R_L\$ is the load, then "efficiency" would be -


where \$I_{dc}\$ is the DC component of the current thru \$R_L\$, and \$I_{rms}\$ is the rms component. $$\implies e =\frac{(\frac{1}{T}\int_0^TI_m\text{sin}\omega t)^2}{\frac{1}{T}\int_0^TI_m^2\text{sin}^2\omega t}$$

$$\implies e =\frac{\frac{1}{T^2}(\int_0^{T}I_m\text{sin}\omega t)^2}{\frac{1}{T}(\int_0^{T/2}I_m^2\text{sin}^2\omega t+0)}$$

$$\implies e =\frac{\frac{1}{T^2}(\int_0^{T/2}I_m\text{sin}\omega t+0)^2}{I_m^2/4}$$

$$\implies e =\frac{\frac{1}{\omega ^2T^2}([-I_m\text{cos}\omega t]_0^{T/2})^2}{I_m^2/4}=\frac{\frac{1}{\omega ^2T^2}.4I_m^2}{I_m^2/4}$$ putting \$\omega=2\pi/T\$ $$\implies e =\frac{4}{\pi^2}\approx0.405=40.5\%$$


For anything else other than resistive loads driven with linear devices the power equation you used is correct. For bridge rectifiers (that's a full wave rectifier) then it's nearly correct for real diodes, but for half wave rectifiers it's out by a good margin.

If the diode were ideal then during its conducting half cycle the power into the load transfers 100%. But for the non conducting half cycle there is no power taken from the supply so, calculate power if it's a full wave ideal bridge rectifier then divide it by 2 to get the half wave power transferred.

And, what you will find is that the power efficiency is nearly 100% in either the full bridge or the half bridge. If the diodes were ideal then it's 100% efficiency in both cases.

  • 3
    \$\begingroup\$ Exactly. Conservation of energy. EnergyOut = EnergyIn - EnergyLost. A perfect diode won't lose any energy (no heat). \$\endgroup\$
    – Mattman944
    Commented Nov 13, 2020 at 14:20

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