# Efficiency of Half wave Rectifier and input RMS value

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.

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: $$P_{rms} = \sqrt{\frac{1}{T}\int_0^{T}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt} + \frac{1}{T}\int_{T/2}^{T}\left[V\sin(\omega t)\cdot0\right]^2\mathrm{dt}} = \sqrt{\frac{1}{T}\int_0^{T/2}\left[VI\sin^2(\omega t)\right]^2\mathrm{dt}+0}$$

• Thanks for the answer, Yes about the DC output power, answer is absolutely correct. When we consider the RMS value of the input why do we not consider one full cycle.Is the absence of current flow in the negative half cycle the reason? Commented Oct 29, 2016 at 12:49
• Because the RMS value of the input would just be the same whether calculated on one half or full cycle, as the formula in the answer shows. I think you should not neglect the portion in which no current flows because you have to consider the power totally available, and that "effect" is due to the rectifier in use. Commented Oct 29, 2016 at 13:21
• Yes, that should be the case, but The RMS of the Ac input in fullwave rectifier is Vm/sqrt(2), and for half wave rectifier it is Vm/2, where Vm is the peak input voltage. This is because only the conducting part(only one half cycle) is considered for half wave rectifier, which is what I am confused about! Commented Oct 29, 2016 at 13:55
• You are correct, sorry for the nonsense I wrote. The efficiency is computed considering the power actually given to the circuit, not the available one. In this way, efficiency measures how much of the absorbed power is transferred to the output. Thus, yes, only the half cycle during which the rectifier conducts is to be considered. Commented Oct 29, 2016 at 14:19
• No issues, But yes how much of the input applied appears across the load makes sense. And thanks. Commented Oct 29, 2016 at 14:34

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.

• Thanks for the answer, true there is no current flowing for negative half cycle. It is the case in the secondary part of the transformer. So the input we consider is at the secondary of the transformer or at the primary part. Commented Oct 29, 2016 at 12:46
• You cannot have a DC output behind a transformer. If you place a single diode that way, the transformer will go into saturation and does not work as expected. So you either have to use two diodes and a transformer with a tap in the middle or use a rectifier bridge. Your half-wave power calculation would be true for either half of the circuit. Commented Oct 31, 2016 at 23:14

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.

• Thanks for the answer. But about the first part of the answer, the RMS voltage in the positive cycle is exactly same as negative cycle. If we calculate for only half cycle then we make the calculations as 2*one half cycle times the RMS value of the half cycle. Which is what the case in full wave rectifier. Commented Oct 29, 2016 at 12:43
• It's true, the RMS of the AC waveform is the same as the RMS of the full wave rectified waveform (assuming ideal diodes). The heating effect is the same and it would illuminate an incandescent lamp identically. Commented Oct 29, 2016 at 13:01
• The RMS of the Ac input in fullwave rectifier is Vm/sqrt(2), and for half wave rectifier it is Vm/2, where Vm is the peak input voltage. This is because only the conducting part(only one half cycle) is considered for half wave rectifier, which is what I am confused about! Commented Oct 29, 2016 at 13:54
• What are you possibly talking about! The AC input to a full or half wave rectifier is unchanged by whether it is either - it is the AC input and always is. Please do try to get the words right because this is very important to understanding what you are trying to say. Commented Oct 29, 2016 at 14:01
• Sorry if I am not being clear, for instance please take a look at these links. visionics.a.se/html/curriculum/Experiments/FW%20Rectifier/… and visionics.a.se/html/curriculum/Experiments/HW%20Rectifier/…. The efficiency is the ratio of DC output power by AC input power. But they calculate the RMS value across the load, not at the input as input for both HW and FW is same while across the load it is not same. Why consider the power at load, while defination is RMS value at the input and should be same for both. Commented Oct 29, 2016 at 14:23