Efficiency of Half wave Rectifier and input RMS value

Question

Efficiency of the half wave rectifier is given by \begin{align} \eta &= \frac{dc\ output\ power}{ ac\ input\ power} \\\\ \end{align} with ideal diodes for the given Vin, we get the Vout as in the figure.

So for calculating the dc output power we consider only one half cycle, since diode is conducting during one half cycle.

My doubt is why do we consider one half cycle of the AC input to calculate Vrms and why not one full cycle. Peak inverse voltage of the diode is given by negative peak value of the input.

Answer 1

The "DC output power" is not really calculated on only one cycle: it is computed on the whole \$2\pi\$ cycle, but only one half cycle contributes to it (indeed, because the diode is conducting). If it were actually computed on only one half cycle, it would have the same power of a full-wave rectified sinusoid, which it doesn't.

Consider the RMS formula: \$V_{rms}=\sqrt{\frac{1}{T}\int_0^{T}v^2(t)\mathrm{dt}}\$ where \$v(t) = V\sin(\omega t)\$ is \$T\$-periodic (\$V\$ is the maximum value of the voltage). Obviously, \$v^2(t) = V^2\sin^2(\omega t)\$ is only \$T/2\$-periodic; thus \$V_{rms}=\sqrt{\frac{1}{T}\int_0^{T}\sin^2(\omega t)\mathrm{dt}} = \sqrt{\frac{1}{T/2}\int_0^{T/2}\sin^2(\omega t)\mathrm{dt}}\$ doesn't depend on its calculation being made in a half or full cycle.

However, what is of interest in evaluating the efficiency is not \$V_{rms}\$, but \$P_{rms}\$. The latter is obtained by calculating the root mean square of the absorbed power \$p(t) = v(t)i(t)\$, which is \$0\$ during the half cycle in which the half wave rectifier doesn't conduct; it is still evaluated considering a whole cycle, but only half a cycle contributes to it: \begin{equation} P_{rms} = \sqrt{\frac{1}{T}\int_0^{T}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt} + \frac{1}{T}\int_{T/2}^{T}\left[V\sin(\omega t)\cdot0\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}+0} \end{equation}

Answer 2

Power isn't proportional to voltage but to voltage times current. When there's no current, there's no power.

Vrms is a measure for the area below the U(t) curve. You place a rectangle with the length of one full cycle and the height Vrms into the graph. Then you choose Vrms so the rectangle and the original graph have the same area below.

Answer 3

My doubt is why do we consider one half cycle of the Ac input to calculate Vrms and why not one full cycle

The RMS voltage in the positive cycle is exactly the same in the negative cycle so there's no need to do a full cycle calculation for AC.

So for calculating the dc output power we consider only one half cycle, since diode is conducting during one half cycle.

No, that would be incorrect for a half wave rectifier because the average power over one complete cycle is half the value of the power in the conducting half cycle.