The Junction Rectifier — Question 4

Question 4

An ideal half-wave rectifier has distinct mean and rms outputs

A sinusoidal source $v_s=V_m\sin\theta$ drives an ideal diode and resistive load R. The output is $v_o=V_m\sin\theta$ for $0<\theta<\pi$ and zero for $\pi<\theta<2\pi$.

1. Derive the mean and rms output voltages by integrating over a complete period.

2. Find the ratio of dc-component power to total load power and evaluate both voltages for $V_m=12.0\,\mathrm V$.

Question 4 - Solution

An ideal half-wave rectifier has distinct mean and rms outputs

(a) Method and calculation

$$\boxed{V_{\mathrm{dc}}=\frac1{2\pi}\int_0^\pi V_m\sin\theta\,d\theta=V_m/\pi}.$$$$\boxed{V_{\mathrm{rms}}=\sqrt{\frac1{2\pi}\int_0^\pi V_m^2\sin^2\theta\,d\theta}=V_m/2}.$$

(b) Interpretation and further calculation

$$\boxed{\frac{P_{\mathrm{dc}}}{P_{\mathrm{load}}}=\frac{V_{\mathrm{dc}}^2/R}{V_{\mathrm{rms}}^2/R}=4/\pi^2=.4053}.$$$$\boxed{V_{\mathrm{dc}}=3.820\,\mathrm V,\quad V_{\mathrm{rms}}=6.00\,\mathrm V}.$$

Physical check

The remaining load power is carried by the ac ripple component. This ratio is not the energy efficiency of an ideal lossless diode.