Why is the efficiency of a half wave rectifier equal to 40.6% and not 50%?

Question

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 $$\frac{\frac{V_0 I_0}{4}}{\frac{V_0 I_0}{2}} = 0.5$$

Answer 1

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 -

$$e=\frac{P_{dc}}{P_{ac}}=\frac{I_{dc}^2.R_L}{I_{rms}^2.R_L}=\frac{I_{dc}^2}{I_{rms^2}}$$

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\$ $$\implies e =\frac{4}{\pi^2}\approx0.405=40.5\%$$

Answer 2

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.