Electronic – What’s the difference between Clamper and Rectifier

Personally I wouldn't consider your final circuit to be a full wave rectifier at all. It's two half wave rectifiers the first charges C1 during the positive half cycle and the second charges C2 during the negative half cycle.

This circuit does indeed double the voltage and a slight variant of it used to be common in consumer electronics. As shown below.

^{simulate this circuit – Schematic created using CircuitLab}

With SW1 open the output is close to the peak of the AC input voltage and all 4 diodes provide full-wave rectification. R1 and R2 are needed to ensure that C1 and C2 share the voltage equally.

If Sw1 is closed then D3 and D4 do not conduct and R1 and R2 are not doing necessary but the output voltage roughly twice the input AC.

Bearing in mind that European mains voltages are typically 220V or 230Vrms while in the US its 120Vrms. This circuit provided an easy way to make devices work anywhere in the world just by configuring SW1 to be open or closed as appropriate for the country it was going to be used in.

It's less common now as for low power devices it's relatively easy to design a wide range input fly-back converter and for higher power units devices typically need some form of power factor circuitry. I'm sure it still does get used occasionally though.

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}