Alternating Current — Question 6

Question 6

Half-wave rectification changes both mean and rms in different proportions

An ideal rectifier passes $V_p\sin\omega t$ during the positive half-cycle and outputs zero during the negative half-cycle. Ignore diode drop.

1. Find the cycle mean and rms output voltage.

2. Find the AC-only rms ripple after subtracting the mean.

Question 6 - Solution

Half-wave rectification changes both mean and rms in different proportions

(a) Method and calculation

$$\bar v=\frac{V_p}{2\pi}\int_0^\pi\sin\theta d\theta=V_p/\pi.$$$$\boxed{V_{\mathrm{rms}}^2=\frac{V_p^2}{2\pi}\int_0^\pi\sin^2\theta d\theta=V_p^2/4}.$$

(b) Interpretation and further calculation

$$V_{{\mathrm{ripple}},\mathrm{rms}}^2=\langle(v-\bar v)^2\rangle=V_{\mathrm{rms}}^2-\bar v^2.$$$$\boxed{V_{{\mathrm{ripple}},\mathrm{rms}}=V_p\sqrt{1/4-1/\pi^2}}.$$

Physical check

Removing the DC mean reduces rms ripple but does not alter the definition of total rms.